WBCHSE Chemistry Class 11 Equilibrium Notes

Equilibrium Introduction

Dynamic Equilibrium Concept: Under a given set of experimental conditions, a system is said to be at equilibrium if the macroscopic properties of the system, such as temperature, pressure, concentration, etc. do not show any change with time.

There are two types of equilibria: Static equilibrium and Dynamic equilibrium. Equilibrium involving physical and chemical changes is dynamic.

A dynamic equilibrium is established when two or more opposing processes occur in a system at the same rate. For example, if the decomposition of hydrogen iodide [2HI(g)⇌ H₂(g) +I₂(g)] is carried out in a closed vessel,  it is found that the reaction is never complete.

At the onset of the reaction, the system contains only hydrogen iodide (reactant) molecules. With time, the concentration of molecules gradually decreases.

In contrast, the concentrations of H₂ and I₂ (product) molecules gradually increase till a stage is reached at which no further change in concentrations of either the reactants or the products takes place.

Dynamic Equilibrium Concept

At this stage, the reaction appears to have stopped. This state of the system at which no further change occurs is called a state of equilibrium. This state of equilibrium is not static, but it is dynamic because the forward and backward reactions are still going on at the same rate.

Due to this dynamic nature of equilibrium, no change in concentration and other properties of the system occurs at the equilibrium state.

Equilibrium involving chemical reaction (i.e., chemical equilibrium) is represented as Reactants; F=± Products The double half arrows indicate that the reactions in both directions are going on simultaneously.

The mixture consisting of reactants and products in the equilibrium state is called an equilibrium mixture. Dynamic equilibrium is also observed in case of physical changes, particularly during the transition of state example melting of solids, vaporization of liquids, etc.

Physical Equilibrium

Equilibrium involving physical processes is called physical equilibrium. Thus, the equilibria attained during the dissolution of a salt, the evaporation of a liquid, etc., are examples of physical equilibria. Different types of physical equilibria are briefly discussed in the following section.

Solid-liquid Equilibrium

At the melting point (or freezing point) of a pure substance, both its solid and liquid phases co-exist and a dynamic equilibrium develops between the two phases: solid-liquid When the system with the above equilibrium mixture is heated, the temperature of the system remains constant until the whole solid transforms into liquid. Similarly, if heat is withdrawn from this system, the temperature of the system remains constant until the whole liquid transforms into a solid.

Dynamic Equilibrium Concept

Melting point Or freezing point

At Normal atmospheric pressure, the temperature at which the solid and the liquid states of a pure substance remain in equilibrium is called the normal melting point (or normal freezing point) of the substance.

When the solid and liquid phases of a pure substance are kept in contact with each other at its melting point in a closed insulated container, no exchange of heat takes place between the system and its surroundings.

However, a state of dynamic equilibrium is established between the solid and the liquid phases inside the container. It is also observed that the masses of solid and liquid phases do not change with time and the temperature of the system remains constant.

Example: Let us take some ice cubes together with some water inside a thermos flask at 0°C and 1 atm pressure and leave the mixture undisturbed. After some time it will be seen that the masses of ice and water are not changing with time and also the temperature remains unchanged. This represents an equilibrium between ice and water

Dynamic Equilibrium Concept

Equilibrium: H₂O (s)⇌H₂O(l)

• Although we observed apparent change inside the thermos flask, a careful examination shows that some activity is still going on between the two phases of water.
• Some molecules of ice convert into water, while at the same time, the same number of molecules of water convert into ice.
• However, as the masses of ice and water remain unchanged, it can be concluded that the two opposite processes (i.e., melting of ice and freezing of water) occur at the same rate.

The rate of melting of ice = The rate of freezing of water

Thus, the equilibrium that is established in the solid-liquid system is dynamic in nature.

Liquid-vapour equilibrium

The equilibrium between a liquid and its vapor can be better understood if we consider the vaporization of water in a closed vessel. Let us take a closed vessel connected to a manometer and a vacuum pump as shown in

The closed vessel is first evacuated. The levels of mercury are the same in both the limbs of the manometer.

Dynamic Equilibrium Concept

Then some pure water is introduced into the vessel and the whole apparatus is kept at room temperature (or any desired temperature by placing it in a thermostat).

After some time it is seen that the level of mercury in the left limb begins to fall and the right begins to rise and eventually the levels of mercury in both limbs become fixed at two different levels. Under this condition, the system is said to have attained equilibrium

Dynamic Equilibrium Concept

Equilibrium: H₂O(Z) ⇌H₂O(g)

Molecular interpretation: At the initial stage of the experiment, as more and more water changes into vapor (by evaporation), the pressure inside the vessel gradually increases. This is indicated by the fall in mercury level in the left limb of the manometer.

The molecules of water vapor, so produced, collide among themselves, with the walls of the vessel and also with the surface of the water.

Water vapor molecules with lower kinetic energy get converted into liquid states when they come in contact with the surface of water. This is called condensation.

At the beginning of the experiment, the rate of evaporation of water is greater than the rate of condensation of its vapor. However, with time, the rate of condensation increases, and that of evaporation decreases.

After some time, the rates of evaporation and condensation become equal. It is said that a dynamic equilibrium is established between water and its vapor

Dynamic Equilibrium Concept

At equilibrium: The rate of evaporation = The rate of condensation

• The difference in the levels of mercury in the two limbs gives a measure of the equilibrium vapor pressure or saturated vapor pressure of water at the experimental temperature.
• At a fixed temperature, if a liquid remains in equilibrium with its vapor, then the pressure exerted by the vapor is called the vapor pressure of the liquid at that temperature.
• An equilibrium between a liquid and its vapor is established only in a closed vessel. If the liquid is placed in an open vessel, then its vapor diffuses into the air. Consequently, no equilibrium is established between a liquid and its vapor in an open container.

Solid Vapour equilibrium

In general, solid substances have verylow vapor pressure compared with liquid substances at the same temperature.

However, some substances, e.g., iodine, camphor, solid CO₂, naphthalene, etc. have high vapor pressure even at ordinary temperatures. Such substances can convert directly from the solid to the vapor state without passing through the liquid state.

In this process of transformation of a solid directly to the When sublimation of volatile solids done in a closed vessel, equilibrium is established between the solid and its vapor.

Example: If we take some solid iodine [I₂(s)] in a closed vessel and heat it below its melting point (113.6°C), it is found that the vessel gets filled with violet vapor of iodine.

Initially the intensity of color increases and eventually it becomes constant. Under this condition, the rate of sublimation of solid iodine is equal to the rate of condensation of iodine vapor. This results in a state of dynamic equilibrium as below,

Dynamic Equilibrium Concept

Equilibrium: I₂(s)⇌ I₂(g)

Other substances showing this kind of equilibrium are:

• NH₄Cl(s)⇌NH₄Cl(g)
• Camphor (s)⇌ Camphor (g)

Equilibrium involving dissolution of solid in liquid

• Suppose, at a fixed temperature, an excess amount of a solid substance, say sugar (solute), is added to a definite volume of a suitable solvent (say water) taken in a beaker and then the mixture is stirred well with a glass rod.
• The particles (i.e., molecules in case of non-electrolytes and ions in case of electrolytes) of solute gradually pass into the solvent, thereby increasing the concentration of the solute in the solution. This process is called dissolution of solute.
• Then a stage comes when no more solute dissolves in the solvent. Instead, the solute settles down at the bottom of the beaker i.e., a saturated solution is obtained.
• During the process of dissolution of solute, the reverse process also occurs simultaneously, i.e., the solute particles from the solution get deposited on the surface of the undissolved solute (a process called crystallization).
• Initially, the rate of dissolution of the solid solute is higher than the rate of crystallization of the dissolved solute.
• However, with time, as the solution becomes more and more concentrated, the rate of dissolution decreases, and that of crystallization increases. Finally, the rate of dissolution becomes equal to that of crystallization.
• Under this condition, a state of equilibrium is established between the dissolved solute particles and the undissolved solid solute. The equilibrium can be represented as,
• Dynamic Equilibrium Concept

Solute (solid)⇌Solute (in solution)

This state of equilibrium is said to be dynamic because the process of dissolution and crystallization continues as long as the temperature and other external conditions remain unchanged.

The solution obtained at equilibrium is called a saturated solution. The concentration of the saturated solution depends on the temperature.

Equilibrium involving the dissolution of gas in liquid

Dynamic Equilibrium Concept

1. The solubility of a gas in a given liquid depends on the experimental temperature and pressure and also on the nature of the liquid and the gas under consideration.
2. When a gas (say CO₂) comes in contact with a liquid, the molecules of the gas begin to collide with the surface of the liquid. Consequently, some of the gas molecules get attracted by the molecules of the liquid and ultimately pass into the liquid phase.
3. % At a fixed temperature and pressure, if a gas is passed continuously through a fixed amount of a liquid kept in a closed vessel, gas molecules get dissolved in the liquid, and eventually, a saturated solution of the gas in the liquid is obtained.
4. In this solution, a dynamic equilibrium is established between the dissolved gas and the gas over the liquid surface.
5. Under this condition, the rate of dissolution of the gas molecules in the liquid is equal to the rate at which the dissolved gas molecules escape from the solution.
6. Thus, it is a dynamic equilibrium; Liquid + Gas Dissolved gas.

Taking CO₂ as the gaseous substance it can be represented as:

Dynamic Equilibrium Concept

At a fixed temperature, the amount of gas dissolved in a liquid depends upon the pressure of the gas over the liquid. The concentration of the dissolved gas increases with the increase in the pressure of the gas. Henry’s law demonstrates how the solubility of a gas in a liquid varies with pressure at a given temperature.

Henry’s law: The mass (or mole fraction) of a gas dissolved in a given mass of a solvent at a given temperature is directly proportional to the pressure of the gas over the solvent.

In a fixed amount of a liquid, if a gas with a pressure of p dissolves by an amount of w, then according to Henry’s law, wp or, w = kp [k is the proportionality constant]

The reason for fizzing out of CO₂ gas when a soda water uc is exposed: In a sealed soda water bottle, CO₂ remains dissolved in liquid under high pressure, and there exists an equilibrium between the dissolved CO₂ and CO₂ gas present over the liquid. As soon as the bottle is opened, the pressure of CO₂ gas over the liquid decreases and becomes equal to the atmospheric pressure.

Since the solubility of a gas in a liquid is proportional to the pressure of the gas, the solubility decreases considerably because of the lowering of pressure. As a result, a large amount of dissolved CO₂ escapes from the solution until a new equilibrium is established.

Dynamic Equilibrium Concept

This phenomenon of escaping CO₂ gas is associated with a fizzing sound. This is also the reason why a soda water bottle turns flat when left open in the air for some time.

Dynamic Equilibrium Concept

Irrversible And Reversible Reactions

Irreversible reactions

A reaction in which the products formed do not react together to revert to the reactants despite the changes in reaction conditions is called an irreversible reaction. Examples: When potassium chlorate (KC1O₃) is heated in an open vessel, potassium chloride (KC1) and oxygen (O₂) are produced.

However, KC1 and O₂ do not react with each other to regenerate KC1O₃. So, the thermal decomposition of KC1O₂ is an example of an irreversible reaction.

Most of the ionic reactions are irreversible. For example, when an aqueous solution of KC1 is treated with an aqueous AgNO₃ solution, a curdy white precipitate of AgCl and KNO₃ are produced, but the precipitated AgCl and KN03 do not react back to AgNO₃ and KC1.

Dynamic Equilibrium Concept

⇒ \(\mathrm{AgNO}_3(a q)+\mathrm{KCl}(a q) \rightarrow \mathrm{AgCl}(s) \downarrow+\mathrm{KNO}_3(a q)\)

Characteristics of an irreversible reaction:

1. The products in an irreversible reaction do not show any tendency to react together. So, the reaction in the opposite direction can never happen. For this reason, an irreversible reaction attains completion in course of time.
2. Since an irreversible reaction undergoes completion, the reactants participating the reaction in equivalent amounts are completely exhausted.
3. Irreversible reactions are accompanied by a decrease in Gibbs free energy (i.e., AG<0).

Reversible reactions

A Reaction in which the products formed react together to regenerate the reactants, and an equilibrium is established between the reactants and products under the condition of the reaction is called a reversible reaction.

Dynamic Equilibrium Concept

Example: The thermal decomposition of NH₄C1 vapour in a closed vessel is a reversible reaction.

⇒ \(\mathrm{NH}_4 \mathrm{Cl}(\text { vapour }) \rightleftharpoons \mathrm{NH}_3(g)+\mathrm{HCl}(g)\)

Explanation: When NH₄C1( vapour) is heated in a closed vessel at 350°C,it undergoes thermal decomposition, producing NH3 and HC₁ gases. However, even after a long time, it is observed that the reaction mixture contains not only NH₃ and HC₁ but also NH₄C₁ vapour.

This proves that the decomposition of NH₄C₁ vapour in a closed container never gets completed. In another closed vessel, if an equimolar mixture of NH₃ and HC₁ gases are heated at 350°C for a long time, the vessel is found to contain NH₄C₁ vapour along with NH₃ and HC₁ gases.

This means that the reaction between NH₃ and HC₁ in a closed vessel never gets completed. Thus, it can be concluded that on heating NH₄C₁ vapour, it decomposes to produce NH₃ and HC₁ gases which again react partially to form NH₄C₁ vapour.

Hence, the thermal decomposition of NHC₁ vapour is a reversible reaction. The following reactions show reversibility when carried outin a closed vessel.

⇒ \(\begin{aligned}
& \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) ; \mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \\
& \mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) ; \mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)
\end{aligned}\)

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Characteristics of reversible reaction:

In a reversible reaction, both the forward and the backward reactions occur simultaneously. In the forward reaction, the reactants react together to yield the products, while the products react together to produce the reactants in the backward reaction.

For example, when an equimolecular mixture of H₂ gas and I₂ vapour is heated in a closed container, the following reaction takes place—

⇒ \(\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) .\)

Here, the forward reaction is: H₂(g) + I₂(g) →2HI(g) and the backward reaction is: 2HI(g)→H₂(g) + I₂(g)

Since a reversible reaction does not complete, the reactants are not completely consumed in such reactions. Instead, a mixture containing both the reactants and the products is obtained.

Dynamic Equilibrium Concept

Such reactions achieve equilibrium state when the rate of the forward reaction becomes equal to that of the backwaed reaction.

At a given temperature and prfessure, when a reversble reaction reaches equilibrum, the gibbs free energy change becomes zero i.e., ΔGp,T=0

Reversibility and irreversibility of chemical reactions when carried out in open and closed containers

Many chemical reactions which are found to be irreversible when carried out in open containers become reversible when they are carried outin closed containers.

Different results are obtained when solid calcium carbonate (CaC03) is heated separately in a closed container andin an open container.

Explanation: On strong heating solid CaCO₂ decomposes to solid CaO and CO₂ gas. If CaCO₃ is decomposed in an open container, CO₂ gas escapes from the container into the air, and only solid CaO remains as residue. As the reactant (solid CaCO₃) in this reaction gets converted into products completely, the reaction is considered as an irreversible reaction.

⇒ \(\mathrm{CaCO}_3(s) \xrightarrow{\Delta} \mathrm{CaO}(s)+\mathrm{CO}_2(\mathrm{~g})\)

If the same quantity of CaCO₃ is decomposedin a closed container, some quantity of CaC03 is still found to remain in undecomposed state. This is because CO₂ produced in the reaction cannot escape from the container.

As a result, a portion of CO₂ gas reacts with an equivalent amount of CaO to form CaCO₃ again. Hence, in the closed vessel, the reaction occurs reversibly. Consequently, a mixture of CaCO₃, CaO and CO₂ are found to be presentin the reaction vessel.

Dynamic Equilibrium Concept

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{\Delta}{\rightleftharpoons} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

When steam is passed over the red hot iron, ferrosoferric oxide (Fe₃O₄) and H₂ gas are produced.

Explanation: If the reaction is carried out in an open vessel, H₂ gas produced diffuses into air. So, Fe₃O₄ resulted from the reaction cannot have hydrogen gas to react with.

Consequently, the reverse reaction cannot take place. For this reason, at the end of the reaction, only Fe₃O₄ is left behind as residue in the reaction vessel.

⇒ \(3 \mathrm{Fe}(s)+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Fe}_3 \mathrm{O}_4(s)+4 \mathrm{H}_2(\mathrm{~g}) \uparrow\)

However, if the reaction is carried out in a closed vessel, H₂ gas produced cannot escape from the container. As a result, a certain amounts of H₂ gas and Fe₃O₄ together and regenerate Fe and H₂O.

Hence, in a closed vessel, the reaction occurs bothin the forward and reserve. directions, giving a mixture of Fe(s), H₂O(g), Fe₃O₄(s) and H₂(g).

⇒ \(3 \mathrm{Fe}(s)+4 \mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_3 \mathrm{O}_4(s)+4 \mathrm{H}_2(g)\)

Dynamic Equilibrium Concept Chemical Equilibrium

Let us consider a hypothetical reversible reaction A +B⇌C+D, Which is started with 1 mol of A and lmol of B in a closed container at a given temperature.

1. At the beginning, the reaction system does not contain C and D (products). It contains only A and B (reactants). So, the reaction occurs only in the forward direction (A + B→C+ D).
2. At the outset of the reaction, since the concentrations of the reactants are maximum, the rate of the forward reaction is also maximum. This is because the rate of a reaction is directly proportional to the concentrations of the reactants.
3. As there are no C and D molecules at the start, the backward reaction (C+D→A + B) does not occur. However, the backward reaction starts occurring with the formation of A and B in the forward reaction. As C and D accumulate is the reaction system, they begin to react together to form A and B.
4. With time, the concentrations of C and D increase, while the concentrations of A and B decrease. As a result, the rate of the backward reaction increases, while that of the forward reaction decreases.
5. Eventually a moment comes when the rate of the forward reaction becomes equal to the rate ofthe backward reaction.
6. When the rate of the forward reaction is equal to that of the reverse reaction, the reaction is said to have reached the state of equilibrium. However, at equilibrium, the reaction does not stop; instead both the forward and backward reactions occur simultaneously at the same rate
7. The mixture of reactants and products at equilibrium of a reaction is called the equilibrium mixture. The concentrations of the reactants and products in the equilibrium mixture are called their equilibrium concentrations. If the conditions {i.e., temperature, pressure etc.) of the reaction remain undisturbed the relative concentrations of the reactants and products in the equilibrium mixture do not change with time.

Dynamic Equilibrium Concept

Chemical equilibrium The state of a reversible chemical reaction at a given temperature and pressure when the rates of the forward and reverse reactions become the same, and the concentrations of the reactants and products remain constant with time then the particular state is called the state of chemical equilibrium.

Chemical equilibrium is a dynamic

After the attainment of equilibrium of a reversible chemical reaction, if the reaction system is left undisturbed for an indefinite period at constant temperature and pressure, then the relative amounts of the reactants and the products are found to remain unaltered.

This observation leads to the impression that a reaction stops completely at equilibrium. However, it has been proved experimentally that the reaction does not cease rather both the forward and the backward reactions continue at the same rate. This is the reason why chemical equilibrium is designated as a dynamic equilibrium.

Experimental proof of the dynamic nature of chemical equilibrium:

When some quantity of pure CaCO₃ is heated strongly above 827°Cin a closed vessel (A), it decomposes into CaO and CO_2, and an equilibrium is established

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

At equilibrium, the temperature and pressure of the reaction vessel remain unaltered with time.

Now this vessel is connected with another closed vessel 14 containing CO₂ at the equilibrium pressure and temperature in such a way that there will be no effect on the equilibrium of the reaction occurring in vessel (A).

Dynamic Equilibrium Concept

After some time, a small quantity of solid is collected from the vessel (A) and analyzed. The analytical data indicates the presence of 14C in CaCO₃.

This is possible only if some amount of 14CO₂ and CaO combine to form Ca14CO₃ at equilibrium. At the same time, some quantity of CaCO₃ decomposes to produce CaO and CO₂.

So, the pressure on the reaction vessel remains constant. Thus, this experiment proves that even after the attainment of equilibrium, the reactions do not cease, both the forward and the reverse reactions proceed simultaneously at the same rate.

Characteristics of chemical equilibrium

Permanency of chemical equilibrium: As long as the conditions under which a reaction attains equilibrium remain unaltered, no further change in equilibrium takes place, that is to say, the composition of the equilibrium mixture and other properties of it remains the same with time.

Dynamic Equilibrium Concept

Dynamic nature of equilibrium: Even after the attainment of equilibrium, a chemical reaction does not cease; both the forward and the reverse reactions continue at equal rates.

Incompleteness of the reaction at equilibrium: At the equilibrium of a reaction, both the forward and reverse reactions take place simultaneously at the same rate. If any one of these reactions goes to completion, then the term equilibrium becomes irrelevant. Hence, for the equilibrium to exist, the reactions of both directions will have to be incomplete.

Approachability of equilibrium from either direction: Under a given set of conditions, a reversible reaction attains the same equilibrium state irrespective of whether the reaction is started with its reactants or products.

Example: H₂ gas and I₂ vapor are allowed to react with each other in a closed vessel at 445°C. Eventually, the following equilibrium is established,

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

The equilibrium mixture is found to contain H₁, H_2, and I₂ with mole percents of 80%, 10%, and 10%, respectively. In a separate container with the same volume, if 2 moles of H₁ gas are heated at 445°C, then H₁ gas decomposes to H₂ and I₂ gases, and eventually, the following equilibrium is established, 2HI(g) H₂(g) + I₂(g).

Dynamic Equilibrium Concept

Here also, the equilibrium consists of H₁, H_2, and I₂ with mole percents of 80%, 10%, and 10%, respectively. Thus, the same equilibrium mixture is obtained, no matter whether we start the reaction with HI(g) or H₂(g) and l₂(g).

A catalyst cannot alter the state of equilibrium: A catalyst is a substance that enhances the rate of a reaction without being used up in the reaction. A catalyst does not affect the position of equilibrium in a reaction. Its only function is to reduce the time that a reaction takes to reach an equilibrium state.

In the presence of a catalyst, the forward and reverse reactions of a reversible reaction are speeded up to the same extent. Under a given set of conditions, if a reaction is carried out in the presence or absence of a catalyst, then the same equilibrium mixture is obtained. That is, in both cases, the concentrations of the reactants and products are found to be the same.

Homogeneous and heterogeneous equilibria

Homogeneous equilibrium: An equilibrium in which all the substances, Z.e., reactants, and products, are in the same phase is known as homogeneous equilibrium.

Examples 1. N₂(g) + 3H₂(g)⇌ 2NH₃(g)

2SO₂(g) + O₂(g) ⇌2SO₃(g)

CH₃COOH(Z) + C₂H₅OH(l) ⇌CH₃COOC₂H₅(Z) + H₂0(l)

Dynamic Equilibrium Concept

Heterogeneous equilibrium: An equilibrium in which the reactants and products are in different phases is known as heterogeneous equilibrium.

Examples CaCO₃(s) ⇌CaO(s) + CO₂(g)

2HgO(s)⇌ 2Hg(l) + O₂(g)

Dynamic Equilibrium Concept The Law Of Mass Action

In 1864, C.W. Guldberg and P. Waage formulated a law regarding the dependence of the reaction rate on the concentration of the reactant. This law is known as the law of mass action.

At a constant temperature, the rate of a chemical reaction at any instant during the reaction is directly proportional to the active mass of each of the reactants at that instant.

So, the rate of a reaction increases with the increase in active masses of the reactants, while it decreases with the decrease in active masses of the reactants.

Active mass: The active mass of a substance is generally considered as the same as its molar concentration. The active mass is expressed in different ways.

• In case of a dissolved substance is a solution, the active mass of the substance is taken to be the same as its molar concentration.
• If a VL solution contains n mol of a substance, then the active mass (or molar concentration) of the substance is n/v .
• In the case of a component gas in a gas mixture, the active mass of the component can be expressed either in terms of its molar concentration or partial pressure in the mixture.
• This is because the partial pressure of a component gas in a gas mixture is directly proportional to its molar concentration.

For a pure solid or liquid, the active mass is always taken as unity (1).

Dynamic Equilibrium Concept

The molar concentration of a pure solid or liquid is directly proportion to its density:

⇒ \(\begin{aligned}
& =\frac{\text { number of moles of the substance }}{\text { volume of the substance (in } \mathrm{L})} \\
& =\frac{\text { mass of the substance }}{\text { molar mass of the substance }} \times \frac{1}{\text { volume of the substance (in L) }} \\
& =\frac{\text { mass of the substance }}{\text { volume of the substance (in } \mathrm{L})} \times \frac{1}{\text { molar mass of the substance }} \\
& =\frac{\text { density of the substance }}{\text { molar mass of the substance }}
\end{aligned}\)

As the molar mass of a pure substance is a fixed quantity, the molar concentration of a pure solid or liquid is directly proportional to its density. The density of a pure solid or liquid is constant at a given temperature, so its molar concentration.

Dynamic Equilibrium Concept

Mathematical expression of the law of mass action: Let us consider the following simple chemical reaction in which one mole of A reacts with one mole of B, forming one: mole of C: A + B→C According to the law of mass action, at a particular moment during die reaction, the rate of the reaction, or, r = k[A] [B] Where [A] and [B] are the active masses or molar concentration + of A and B, respectively at that moment, and k is proportionality constant, known as the rate constant of the reaction. Equation [1] represents the rate equation of the said reaction. Here is a table is which some reactions and their rate equations are given.

General statement of the law of mass action: At constant temperature, the rate of a chemical reaction at any instant is directly proportional to the product of molar concentrations (active masses) of the reactants at that instant, each concentration (active mass) term being raised to a power which appears as a stoichiometric coefficient of the species in the balanced chemical equation of the reaction.

Mathematical Form Of The Law Of Mass Action For A Reversible Reaction

Suppose, a reaction is started with ‘ a’ mol of A and ‘b 1 mol of B, and the reaction of A with B leads to the formation of C and D. Let the reaction occur according to the following equation and form an equilibrium: aA + bB cC + dD According to the law of mass action, at equilibrium, the rate of forward reaction (rf)∝[A]a x [B]b or rf=Kf [and that of backward reaction (rb)oc[D]d x [E]e or> rb = kb [D]dx[E]e where k and kb are the rate constants of the forward and the backward reactions respectively. [A], [B], [D], and [E] are the respective molar concentrations or active masses (mol.L-1) of A, B, D, and E at equilibrium.

At equilibrium, the rate of the forward reaction (ry) = the rate of the backward reaction (rb).

⇒ \(\text { So, } k_f[A]^a \times[B]^b=k_b[D]^d \times[E]^e\)

⇒ \(\text { or, } \frac{k_f}{k_b}=\frac{[D]^d \times[E]^e}{[A]^a \times[B]^b} \text { or, } \boldsymbol{K}=\frac{[\boldsymbol{D}]^d \times[E]^e}{[A]^a \times[B]^b}\)

The ratio of the two rate constants (fcy and kb) is a constant quantity at a given temperature.

Dynamic Equilibrium Concept

So, K is a constant. The constant ‘K’ is called the equilibrium constant of the said reversible reaction. Equation [1] expresses the mathematical form of the law of mass action of the given reversible reaction.

In the expression of the equilibrium constant, the reaction. concentration terms of the reactants and the products represent their respective molar concentrations at equilibrium. They do not denote their initial concentrations

At constant temperature, the equilibrium constant (K) of a chemical reaction has a definite value. The value of K changes with temperature. This is because the changes of values of kJ- and kJ with the temperature change do not occur to the same extent.

Equilibrium constant

The equilibrium constant of a reaction is the ratio of the product of the active masses of products at equilibrium to the product of the active masses of reactants at equilibrium, with each active mass term raised to a power equal to its stoichiometric coefficient in the balanced equation of the reaction.

The equilibrium constant is also represented as Kc, Kp, or Kx depending on whether the active mass is expressed in terms of molar concentrations partial pressure, or mole fraction.

The law of chemical equilibrium: This law states that when a reversible chemical reaction reaches equilibrium at a particular temperature, the ratio of the product of active masses of the products to that of the reactants, with each active mass term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation is constant.

The mathematical expression for the law of chemical equilibrium can be obtained by applying the law of mass action to a reversible reaction at equilibrium. For a general reversible reaction, aA + bB dD + cE; the mathematical expression can be written as, (constant at a particular temperature) [A]n[B];’

⇒ \(\frac{[D]^d[E]^e}{[\mathrm{~A}]^a[B]^b}=K\) = K (constant at a particular temperature

Expression of the equilibrium constant in case of a heterogeneous equilibrium

At a given temperature, the active mass or molar concentration of a pure solid or liquid is always taken as unity (1). For this reason, in the case of a heterogeneous equilibrium, the active mass or molar concentration term of a pure solid or liquid does not appear in the expression of the equilibrium constant.

Dynamic Equilibrium Concept

Examples: The thermal decomposition of solid CaCO., in a closed container leads to the following equilibrium:

CaCO₃(s) CaO(s) + CO₂(g)

⇒ \(\text { Equilibrium constant, } K_c=\frac{[\mathrm{CaO}(s)]\left[\mathrm{CO}_2(\mathrm{~g})\right]}{\left[\mathrm{CaCO}_3(s)\right]}=\left[\mathrm{CO}_2(\mathrm{~g})\right]\)

[CaCO₃(s)] = 1, [CaO(s)] = 1 ] and Kp = PCO₂{gy As the value of Kc or Kp is constant of a given temperature, the molar concentration or the partial pressure of CO₂(g) at the equilibrium formed on the decomposition of solid CaC03 at a given temperature is always constant.

The following equilibrium is established during the vapo¬ risation ofwaterin a closed vessel: H₂O(l) H₂O(g)

Here, equilibrium constant \(K_c=\frac{\left[\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right]}{\left[\mathrm{H}_2 \mathrm{O}(l)\right]}=\left[\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right]\)

Therefore, the molar concentration or the partial pressure of water vapor remaining in equilibrium with pure water at a particular temperature is always constant.

Dynamic Equilibrium Concept Relation Between Different Equilibrium Constants

Relation between Kp and Kc

Let the following reversible gaseous reaction is at equiUbriumin a closed container at a certain temperature: aA(g) + bB(g);=± dD(g) + eE(g) If the molar concentrations of A(g), B(g), D(g) and E(g) at equilibrium be [A], [B], [D] and [£] respectively, and the partial pressures of A(g), B(g), D(g) and E(g) at equilibrium be pA, PiB pD and pp respectively, then

⇒ \(K_c=\frac{[D]^d \times[E]^e}{[A]^a \times[B]^b} \quad \cdots[1] \quad \text { and } \quad K_p=\frac{p_D^d \times p_E^e}{p_A^a \times p_B^b} \quad \cdots[2]\)

If the reaction mixture behaves as an ideal gas, then the ideal gas equation can be applied to each of the species present in the mixture.

Dynamic Equilibrium Concept

if the pressure, temperature, and volume of n mole of an ideal gas are P, T, and V respectively, then \(P V=n R T \quad \text { or, } P=\left(\frac{n}{V}\right) R T=C R T\)

Now by applying this equation to each species of the reaction mixture, we get \(p_A=[A] R T, p_B=[B] R T, p_D=[D] R T \text { and } p_E=[E] R T \text {. }\)

Putting the values of pA, pB, PD, and pE into equation (2), we have

⇒ \(\begin{aligned}
K_p & =\frac{\{[D] R T\}^d \times\{[E] R T\}^e}{\{[A] R T\}^a \times\{[B] R T\}^b} \\
& =\frac{[\mathrm{D}]^d \times[E]^e}{[A]^a \times[B]^b}(R T)^{(d+e)-(a+b)} \\
\text { or, } K_p & =K_c(R T)^{\Delta n}
\end{aligned}\)

where, An = (d+ e)-(a + b) = the total number of moles of gaseous products – the total number of moles of gaseous reactants.

Dynamic Equilibrium Concept

Inequation [3], the value of An may +ve, -ve, or zero. When the number of moles of the gaseous products is greater than, less than, or equal to the number of moles of the gaseous reactants, then the values of An become positive, negative, or zero, respectively.

If An is positive, Kp is greater than Kc. If An is negative, then the value of Kp is smaller than Kc. If An = 0, then Kp and Kc have the same value.

Dynamic Equilibrium Concept

Relation between K_P and K_X

Let, at a constant temperature, the following reversible gaseous reaction is at equilibrium in a closed vessel:

⇒ \(a A(g)+b B(g) \rightleftharpoons d D(g)+e E(g)\)

⇒ \(\text { For this reaction, } K_p=\frac{p_D^d \times p_E^e}{p_A^a \times p_B^b} \cdots[1] ; K_x=\frac{x_D^d \times x_E^e}{x_A^a \times x_B^b} \cdots[2]\)

Where, P_A, P_B, P_D, and P_E are the partial pressures of A, B, D, and E respectively, at equilibrium, and xA, xB, xD and xE are their respective mole fractions at equilibrium. The partial pressure of a component gas in a gas mixture is its mole fraction times the total pressure of the mixture.

Dynamic Equilibrium Concept

Hence, the relation between the partial pressures and mole fractions of the different components in the said gas mixture is P_A = x_A xP, P_B = x_B xP, P_D = x_D xP, and P_E = x_E x P where P is the total pressure of the gas mixture at equilibrium.

Putting the values of pA, pB, PD, and pE into equation (1), we have \(\begin{aligned}
K_p & =\frac{\left(x_D \times P\right)^d \times\left(x_E \times P\right)^e}{\left(x_A \times P\right)^a \times\left(x_B \times P\right)^b}=\frac{x_D^d \times x_E^e}{x_A^a \times x_B^b} \times(P)^{(d+e)-(a+b)} \\
& =K_x \times(P)^{(d+e)-(a+b)}
\end{aligned}\)

∴ \(K_p=K_x \times P^{\Delta n}\)

Dynamic Equilibrium Concept

Here An = the total number of moles of gaseous products – the total number of moles of gaseous reactants. Equation (3) denotes the relation between Kp and Kx.

Relation between K_C and K_X

Let the following reversible gaseous reaction be at equilibrium at a temperature T and pressure P in a closed vessel: aA(g) + bB{g) dD(g) + eE(g) For this reaction, the relation between Kp and Kc is, Kp = Kc(RT)Δn, and the relation between Kp and Kx is, Kp = Kx(P)Δn where An = total number of moles of the gaseous products- total number of moles of the gaseous reactants. From equations (1) and (2), we have, Kc(RT)ÿn = Kx(P)ÿn

∴ \(K_c=K_x\left(\frac{P}{R T}\right)^{\Delta n}\)

Equation (3) gives the relation between K_P and K_k.

Relation Between  K_P, K_C & K_X:

K_P = K_c(RT)Δn = K_X(P)Δn When An = 0 for a reaction [e.g.,H₂(g) +I₂(g)⇌2HI(g)

or, N₂(g) + O₂(g) 2NO(g)], ⇌then, K_P= K_C = Kx

Dynamic Equilibrium Concept

Characteristics of the equilibrium constant

At constant temperature, the value of the equilibrium constant for each chemical reaction has a definite value. Temperature change brings about an increase or decrease in the value equilibrium constant.

In the case of an endothermic reaction, the value of the equilibrium constant increases with the rise in temperature, while in the case of an exothermic reaction, the value of the equilibrium constant decreases with the rise in temperature

The values of K_P and K_C are not influenced by pressure. If the temperature remains constant, the increase or decrease in pressure does not alter the values of K_P and K_P. However, except for a reaction for which An = 0, the value of depends upon pressure.

Dynamic Equilibrium Concept

The value of the equilibrium constant for any reaction neither increases nor decreases in the presence of a catalyst because the rate of both the forward and the backward reactions increase equally.

The value of the equilibrium constant of a chemical reaction at a given temperature does not depend on the initial concentration of the reactants.

Example: PCl₅(g)PCl₃(g) + Cl₂(g) ⇌In this reaction, Kp at 450°C is 0.19. At 450°C, if the reaction is started with PCl5(g) at any concentration, the value of Kp for the reaction is always 0.19.

The value of the equilibrium constant of any reaction depends on how the balanced equation for the reaction is written.

Consider the reaction in which NH₃(g) is synthesized from N₂ and H₂ gases. The balanced equation for this reaction is: N₅(g) + 3H₂(g) 2NH₃(g)

One can also write the equation as \(\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})\)

In case of(1), equilibrium constant \(K_{c_1}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3},\)

Dynamic Equilibrium Concept

while in case of (2), equilibrium constant, \(K_{c_2}=\frac{\left[\mathrm{NH}_3\right]}{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{H}_2\right]^{3 / 2}}\)

Thus, the values of Kc1 and Kc2 are not the same. Comparingÿ andÿ gives = \(K_{c_1}=K_{c_2}^2 \text {, i.e., } K_{c_2}=\sqrt{K_{c_1}}\)

Suppose, the equilibrium constant of the reaction,

aA + bB dD+eE is K.

If the coefficients of the reactants and products are multiplied by then the equation becomes: maA + mbB mdD + meE For this equation, the equilibrium constant, K = Km. Again, if the coefficients of the reactants and products are divided by m then the equation becomes:

⇒ \(\frac{1}{m} a A+\frac{1}{m} b B \rightleftharpoons \frac{1}{m} d D+\frac{1}{m} e E\)

Dynamic Equilibrium Concept

For this equation, the equilibrium constant, K” = K1/m

For any reversible chemical reaction, the values of AT equilibrium constants for the forward and the reverse reactions are reciprocal to each other.
Example: \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\) for this reaction equilibrium constant \(K_c=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right] \times\left[\mathrm{H}_2\right]^3} \quad \cdots[1]\)

If the reaction is started with NH3 gas (product), then the reaction is: 2NH3(g) N2(g) + 3H2(g) \(\text { Here, equilibrium constant, } K_c{ }^{\prime}=\frac{\left[\mathrm{N}_2\right] \times\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2} \quad \cdots[2]\)

Dynamic Equilibrium Concept

From equations [1] and [2], we get \(K_c=\frac{1}{K_c{ }^{\prime}}\)

if a given reaction is expressed as the sum of two or more individual reactions, then the equilibrium constant of the given reaction equals the product of the equilibrium constants of the individual reactions.

If reaction [3] =reaction [2] + reaction [1], then equilibrium constant of reaction [3] = equilibrium constant of reaction [2] x equilibrium constant of reaction [1].

Example:

• Reaction 1: \(A+B \rightleftharpoons C ; K_1=\frac{[C]}{[A][B]}\)
• Reaction 2: \(C \rightleftharpoons D ; K_2=\frac{[D]}{[C]}\)

Dynamic Equilibrium Concept

• Reaction 3. \(A+B \rightleftharpoons D ; K_3=\frac{|D|}{[A][B]} .\)
• Reaction 1+ reaction 2: \(A+B \rightleftharpoons D\)

∴ \(K_1 \times K_2=\frac{[C]}{[A][B]} \times \frac{[D]}{[C]}=\frac{[D]}{[A][B]}=K_3\)

Unit of the equilibrium constant

The unit of equilibrium constant depends on the difference between the sum of the exponents of the concentration (or partial pressure) terms in the numerator and that is the denominator of the equilibrium constant expression. Suppose, this difference is Ax. If Cl Ax = 0, then neither Kc nor Kp has a unit

Ax=0, then both Kp and Kc have units.

Dynamic Equilibrium Concept

Dynamic Equilibrium Concept

The equilibrium constant is unitless: The term ‘active mass’ mentioned in the law of mass action is unitless. Consequently, Kp or Kc is also unitless. With the help of thermodynamics, it can be shown that the partial pressure of any species present in the expression of Kp is the ratio of measured pressure (P⁰) of that species at equilibrium to its standard pressure (P⁰). Since (P/P⁰) is unitless, Kp is also unitless.

Dynamic Equilibrium Concept

In the case of a pure gas, standard pressure (P⁰) is taken as 1 atm. Similarly, the concentration of any species present in the expression of Kc is the ratio of the measured concentration (C) of that constituent at equilibrium to its standard concentration (C⁰). Since (C/C⁰) is unitless, Kc Is also unitless. The standard concentration (C⁰) of a solute dissolved in a solution is taken as 1(M) or 1 mol-L-1.

Graphical representations of some reversible reactions

Dynamic Equilibrium Concept

Dynamic Equilibrium Concept

Dynamic Equilibrium Concept

Question 1. At a particular temperature, the values of rate constants of forward and backward reactions are 1.5 X 10-2 L -mol-1 -s-1 and 1.8 x 10-3 L-mol-1 -s-1 respectively for the reaction A + B — C + D. Determine the equilibrium constant of the reaction at that temperature
Answer: Equilibrium constant,

⇒ \(K=\frac{\text { Rate constant of forward reaction }}{\text { Rate constant of backward reaction }}\)

⇒ \(=\frac{1.5 \times 10^{-2} \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}}{1.8 \times 10^{-3} \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}}=8.33\)

Question 2. At a particular temperature, the equilibrium constant of the reaction 2A + B 2C is 8.0 x 104 L-mol-1. If the rate constant of the reverse reaction is 1.24 L-mol-1 -s-1, then find the value of the rate constant of the forward reaction at that temperature.
Answer: Equilibrium constant

⇒ \(K=\frac{\text { Rate constant of forward reaction }}{\text { Rate constant of reverse reaction }}\)

∴ The rate constant of the forward reaction = K x rate constant of the reverse reaction = 8 x 104 x 1.24 L2-mol-2-s-1

Dynamic Equilibrium Concept

= 9.92 X 104 L2-mol-2-s-1

Question 3. For the reaction 2SO₂(g) +O₂(g)=±2SO₃(g), Kp = 3 X 1024 at 25°C. Find the value of ICc.
Answer: For the given reaction, An = 2-(2 + 1) = -1 As given, Kp = 3 X 1024, T = (273 + 25) = 298K and R = 0.0821 L-atm-moH-K-1 We knowKp = Kc(RT)n or, 3 X 1024 = Kc(0.0821 X 298)-1 /. Kc = 3 X 1024 X 0.0821 X 298 = 7.34 X 1025

Question 4. At 1500 K, Kc = 2.6 X 10-9 for the reaction 2BrFg(g) Br2(g) + 5F2(g). Determine the Kp of the reaction at that temperature.
Answer: For the given reaction, An = (1+5)-2 = +4.

As per given data, Kc = 2.6 x 10-9 and T = 1500K Using R = 0.0821 L-atm-mol-1-K-1 in the equation Kp = Kc{RT)Δn, Kp = 2.6 X 10″9 X (0.0821 X 1500)4= 0.598

Dynamic Equilibrium Concept

Question 5. Find the temperature at which the numerical values of Kp and Kc will be equal to each other for the reaction \(\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})\)
Answer: In the case of the given reaction, \(\Delta n=1-\left(\frac{1}{2}+\frac{3}{2}\right)=-1\). If the numerical values of Kp and Kc be x, then for the above reaction, Kp = x atm-1 and = x(mol-L-1)-1=x L-mol-1

Therefore, Kp = Kc(RT)-l

Or, \(x \mathrm{~atm}^{-1}=x \mathrm{~L} \cdot \mathrm{mol}^{-1} \times \frac{1}{0.0821 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times T}\)

∴ T=12.18K

∴ The numerical values of Kp and Kc for the given reaction will be equal to 12.18K.

Question 6. At 400°C, H₂(g) and I₂(g) are allowed to react in a closed vessel of 5 L capacity to produce 111(g). At equilibrium, the mixture in the flask Is found to consist of 0.6 mol H₂(g), 0.6 mol I₂(g), and 3.5 mol I₂(g). Determine the value of Kc of the reaction.
Answer: Equilibrium of the reaction is: H₂(g) + 1₂(g)⇒2HI(g)

⇒ \(\text { Therefore, } K_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}\)

The capacity of the vessel = 5 L. Hence, molar concentrations of H2, 12 and HI are: \(\left[\mathrm{H}_2\right]=\frac{0.6}{5}=0.12 \mathrm{~mol} \cdot \mathrm{L}^{-1} ;\) \(\left[\mathrm{I}_2\right]=\frac{0.6}{5}=0.12 \mathrm{~mol} \cdot \mathrm{L}^{-1} \text { and }[\mathrm{HI}]=\frac{3.5}{5}=0.7 \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

∴ \(K_c=\frac{(0.7)^2}{(0.12) \times(0.12)}=34.03\)

Dynamic Equilibrium Concept

Question 7. At a particular temperature, CO(g) reacts with Cl₂(g) in a closed container to produce COCI₂(g). In the equilibrium mixture, partial pressures of CO(g), Cl₂(g), and COCl₂(g) are found to be 0.12, 1.2, and 0.58 atm respectively. Find the value of Kp of the reaction, \(\mathrm{CO}(g)+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}_2(\mathrm{~g}).\)
Answer: For the given equilibrium \(K_p=\frac{p_{\mathrm{COCl}_2}}{p_{\mathrm{CO}} \times p_{\mathrm{Cl}_2}}\)

As given, pCQ = 0.12 atm, pc = 1.2 atm, and Pcoc2 = 0.58 atm at equilibrium.

∴ \(K_p=\frac{0.58}{0.12 \times 1.2}=4.03\)

Question 8. In a closed vessel of 1 dm₃ capacity, 1 mol N₂(g) and 2 mol H₂(g) interact to produce 0.8 mol NH₃(g) in the equilibrium mixture. What is the concentration of H₂(g) in the equilibrium mixture?
Answer: Equation of the equilibrium reaction:

⇒ \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\)

It is observed from the reaction that 1 mol N2(g) and 3 mol H2(g) are necessary for the production of 2 mol NH3(g). Therefore, for the production of 0.8 mol NH3(g), the
number of moles of H2(g) required \(=\frac{3}{2} \times 0.8=1.2 \mathrm{~mol}\)

Hence, the number of moles of H2(g) remaining in the equilibrium mixture = 2-1.2 = 0.8 and its molar concentration =0.8 mol.L-1 [since 1 dm3=1L]

Dynamic Equilibrium Concept

Question 9. At 20°C, 0.258mol A(g) and 0.592 mol 5(g) are mixed in a closed vessel of capacity to conduct the following reaction: A(g) + 2B(g) C(g). If 0.035 mol C(g) remains in the equilibrium mixture, then determine the partial pressure of each constituent at equilibrium.
Answer: According to the equation, 1 mol A(g) reacts with 2 mol 5(g) to produce 1 mol C(g). Hence, 0.035 mol A(g) and 2 x 0.035 = 0.07 mol 5(g) are required to produce 0.035 mol C(g). Therefore, equilibrium molar concentrations of different constituents will be as follows:

⇒ \(\begin
& \multicolumn{1}{c}{A(g)+} & \multicolumn{1}{c}{2 B(g) \rightleftharpoons C(g)} \\
No. of moles & 0.258-0.035 & 0.592-0.07 & 0.035 \\
at equilibrium: & =0.223 & =0.522 &
\end{tabular}\)

⇒ \(\begin{aligned}
& \begin{array}{llll}
\text { Equilibrium } & \left(\frac{0.223}{5}\right) & \left(\frac{0.522}{5}\right) & \left(\frac{0.035}{5}\right)
\end{array} \\
& \text { conc. }\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right): \quad=0.0446 \quad=0.1044=7 \times 10^{-3} \\
&
\end{aligned}\)

As given, T = (273 + 20) K = 293 K

∴ At equilibrium,

PA = [A]BT =0.0446 x 0.0821 x 293 = 1.072 atm

PB = [B]5T = 0.1044 x 0.0821 x 293 = 2.511 atm

pc = [C]5T = 7 X 10-3 x 0.0821 X 293 = 0.168 atm

Dynamic Equilibrium Concept

Question 10. 2 mol of were heated in a sealed tube at 440°C until the equilibrium was reached. HI was found to be 22% dissociated. Calculate the equilibrium constant for the reaction \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \cdot\)
Answer: For the given reaction \(K_c=\frac{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}\)

As given in the question, HI(g) undergoes 22% dissociation. Hence, out of 2 mol HI(g), 2 x 0.22 = 0.44 mol HI(g) dissociates.

As obtained from the equation, 2 mol HI(g) dissociates to produce 1 mol H₂(g) and 1 mol I₂(g). Therefore, 0.44 mol HI(g) dissociates to form 0.22mol of each of H₂(g) and I₂(g).

If the volume of the container is V L, then the equilibrium molar concentrations of different constituents will be as follows:

⇒ \(\begin{gathered}
2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_2(g)+\mathrm{I}_2(g) \\
\text { Equilibrium concern. }\left(\mathrm{mol} \cdot \mathbf{L}^{-1}\right):(2-0.44) / V(0.22) / V(0.22) / V \\
=1.56 / V
\end{gathered}\)

∴ \(K_c=\frac{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}=\frac{\frac{0.22}{V} \times \frac{0.22}{V}}{\left(\frac{1.56}{V}\right)^2}=0.0198\)

Dynamic Equilibrium Concept

Question 11. 1 mol PCl₅(g) is heated in a closed container of 2-litre capacity. If at equilibrium, the quantity of PCl₅(g) is 0.2 mol then calculate the value of equilibrium constant for the given reaction \(\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)\)
Answer: For the above reaction \(K_c=\frac{\left[\mathrm{PCl}_3\right] \times\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}\)

As per the given data, the number of moles of PCl₅(g) at equilibrium =0.2. Hence, number of moles of PCl₅(g) dissociated =(1- 0.2) = 0.8.

According to the equation, 1 mol PCl₅(g) dissociates to produce 1 mol of each of PCl₃(g) and Cl₂(g). Therefore, 0.8 mol PCl₅(g) will dissociate to give 0.8 mol of each ofPCl₃(g) and Cl₂(g).

As given, the volume of the container = 2 L. So, the equilibrium concentrations of different constituents are as follows:

⇒\(\begin{array}{lll}
\text { Equilibrium } & \mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) \\
\text { conc. }\left(\mathrm{mol} \cdot \mathrm{L}^{-1}\right): & \frac{0.2}{2}=0.1 & \frac{0.8}{2}=0.4 \quad \frac{0.8}{2}=0.4
\end{array}\)

∴ \(K_c=\frac{\left[\mathrm{PCl}_3\right] \times\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{0.4 \times 0.4}{0.1}=1.6\)

Dynamic Equilibrium Concept

Question 12. The following reaction is carried out at a particular temperature in a closed vessel of definite volume: CO₂(g) +H2(g)s=± CO(g) + H₂O(g). Initially, partial pressures of CO₂(gj and H₂(g) are 2 atm and 1 atm respectively and that of CO₂(g) at equilibrium is 1.4 atm. Calculate the equilibrium constant of the reaction.
Answer: For the given reaction \(K_p=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_2 \mathrm{O}}}{p_{\mathrm{CO}_2} \times p_{\mathrm{H}_2}}\)

As given, partial pressure of CO₂(g) at equilibrium (Pco₂) = 1-4 atm. Hence, decrease in pressure of CO₂(g) until the equilibrium is reached =(2- 1.4)atm = 0.6atm According to the equation, 1 mol CO₂(g) reacts with 1 mol H₂(g) to form 1 mol CO(g) and 1 mol H₂O(g).

Therefore, if the pressure of CO₂(g) is reduced by 0.6 atm, then the pressure of H₂(g) will also be reduced by 0.6 atm and the pressure of each of CO(g) and H₂O(g) will be 0.6 atm Hence, partial pressures of different constituents at equilibrium will be as follows:

⇒ \(\begin{array}{cccc}
\mathrm{CO}_2(g)+\mathrm{H}_2(g) \rightleftharpoons & \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) \\
\text { Pressure (atm) }: 1.4 & (1-0.6)=0.4 & 0.6 & 0.6
\end{array}\)

∴ \(K_p=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_2 \mathrm{O}}}{p_{\mathrm{CO}_2} \times p_{\mathrm{H}_2}}=\frac{0.6 \times 0.6}{1.4 \times 0.4}=0.64\)

Dynamic Equilibrium Concept

Question 13. B(g) + C(g)= A(g). At constant temperature, a mixture of 1 mol A(g), 2 mol E(g), and 3 mol C(g) is left to stand in a closed vessel of 1 L capacity. The equilibrium mixture is found to contain B(g) of 0.175 molar concentration (mol-L-1). Find the value of the equilibrium constant at that temperature.
Answer: For the above reaction \(K_c=\frac{[A]}{[B] \times[C]}\) As given, at equili-brium, [B] =0.175 mol-L-1 . So, 2-0.175=1.825 mol-L 5 of B participated in the reaction.

Now according to the equation, 1 mol B and 1 mol C combine to form 1 mol A. So, 1.825 mol of B and 1.825 mol of C combine to produce 1.825 mol of A. Thus at equilibrium, the concentration of A, B, and C will be

⇒ \(\begin{array}{lcccc}
& B(g)+C(g) & \rightleftharpoons A(g) \\
\text { Initial conc. }\left(\mathrm{mol} \cdot \mathbf{L}^{-1}\right): & 2 & 3 & 1 \\
\text { Equilibrium conc. }\left(\mathrm{mol} \cdot \mathbf{L}^{-1}\right): & 0.175 & 3-1.825 & 1+1.825 \\
& & =1.175 & =2.825
\end{array}\)

[since volume of the container = 1l]

∴ \(K_c=\frac{[A]}{[B] \times[C]}=\frac{2.825}{0.175 \times 1.175}=13.74\)

Dynamic Equilibrium Concept

Reaction Quotient, Q

The reaction quotient of a reaction has the same form as the equilibrium constant expression. However, the values of molar concentrations (or partial pressures) in the expression of the reaction quotient are the values at any instant of the reaction, whereas these values in the equilibrium constant expression represent the equilibrium values.

Hence, at a particular temperature, the value of the equilibrium constant of a reaction is fixed but that of the reaction quotient is not.

Reaction Quotient: At constant temperature, the reaction quotient of a reaction at any instant may be defined as the ratio of the product of molar concentrations (or partial pressures) of the products to that of the reactants, with each concentration (or partial pressure) term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.

The reaction quotient is denoted by Q. When the reaction quotient is expressed in terms of the molar concentration of the reactants and the products, then it is represented by Qc. The reaction quotient expressed by the partial pressures of the reactants and the products is represented by Q_P.

At the start of the reaction, only reactants are present in the reaction system. So, the value of the numerator in the expression of Q (reaction quotient) is zero, and consequently Q = 0.

If a reaction goes to completion, then only the products are present in the reaction system. So, the value of the denominator in the expression of Q is zero, and hence Q→∞.

If the reaction system contains both the reactants and products, then Q assumes a value in between zero and ∞.

1. For the reaction, \(a A+b B \rightleftharpoons d D+e E\) , the reaction quotient at any moment \(Q_c=\frac{[D]^d \times[E]^e}{[A]^a \times[B]^b},\). [B], [D], and [E] are molar concentrations of A, B, D, and E respectively, at that moment.
2. For the same reaction, the expression of Kc: \(K_c=\frac{[D]_{e q}^d \times[E]_{e q}^e}{[A]_{e q}^a \times[B]_{e q}^b}\) Where [A]eq, [B]eq, [D]eq, and [E]eq are the molar concentrations of A, B, D, and E respectively, at equilibrium.
3. For the reaction,\( A(g)+b B(g) \rightleftharpoons d D(g)+e E(g),\) the reaction quotient at any instant, \(Q_p=\frac{p_D^d \times p_E^e}{p_A^a \times p_B^b}\) PA, pB, pD and pE are the partial pressures of A, B, D
   and E respectively, at that instant.
4. For the same reaction, the expression of Kp: \(K_p=\frac{\left(p_D\right)_{e q}^d \times\left(p_E\right)_{e q}^e}{\left(p_A\right)_{e q}^a \times\left(p_B\right)_{e q}^b}\) where (pA)eq, (PB)eq, {pc)vq and (Pp)eq are the partial pressures of, B, D, and E, respectively, at equilibrium.

Dynamic Equilibrium Concept

Significance of reaction quotient: At a particular temperature, by comparing the values of the reaction quotient (Q) of a chemical reaction at any instant and the equilibrium constant (K) of the reaction at that temperature, one can predict the extent to which the reaction has proceeded.

From the values of Q and AT, it is possible to predict whether the reaction has already reached or will reach the state of equilibrium. If equilibrium is not attained, then it can also be predicted whether the reaction will proceed in the forward or backward direction to achieve equilibrium.

Dynamic Equilibrium Concept

Example: At 700 K, for the following reaction which is carried out in a closed vessel: H2(g) + I2(g) 2HI(g). Kc = 55.0 at 700K. Analysis of the reaction mixture at a given moment during the reaction shows that the molar concentrations of II2(g), I2(g), and HI(g) are 1.8, 2.8, and 0.4 mol-L-1, respectively. Is the reaction at equilibrium at that moment? If not, in which direction will we proceed to attain equilibrium?
Answer: For the given reaction, \(Q_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}\)

At the moment of analysis \(Q_c=\frac{(0.4)^2}{1.8 \times 2.8}=0.0317\)

Hence, Qc< Kc. Therefore, the reaction is not at equilibrium. To attain equilibrium, the value of Qc will increase until it becomes equal to Kc.

Again, the value of Qc will increase if, in its expression, the value of the Numerator increases and that of the denominator decreases. This is possible if the forward reaction occurs to a greater extent Therefore, the reaction will proceed more in the forward direction to attain equilibrium.

Applications Of Equilibrium Constant

Application-1: At a given temperature, the value of the equilibrium constant indicates the extent to which a reaction has proceeded before it attains equilibrium.

Explanation: In the expression equilibrium constant (K) of a reaction, the concentrations of the products appear in the numerator, while that of the reactants appear in the denominator.

Hence, a larger value of equilibrium constant (K) for any reversible reaction signifies higher concentrations of the products than reactants in the equilibrium mixture. This means reactants convert into products to a large extent before attaining equilibrium. Hence, the position of equilibrium lies far to the right.

Dynamic Equilibrium Concept

Application 2: If the value of the equilibrium constant of a reaction at a given temperature is known, then the concentrations of the reactants and products at equilibrium can be calculated from the known initial concentrations of the reactants

Question 1. At 550 K, the value of equilibrium constant (K£) is 0.08 for the given \(\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)\) occurring in a closed container. If the equilibrium concentration of PClg(g) and Cl2(g) are 0.75 and 0.32 mol-L-1 respectively, then find the concentration of PCl3(g).
Answer: for the given reaction \(K_c=\frac{\left[\mathrm{PCl}_3\right] \times\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}\)

As given, [PC15] = 0.75mol-L-1, [Cl2] = 0.32 mol-L-1 and Kc = 0.08.

∴ \(\left[\mathrm{PCl}_3\right]=K_c \times \frac{\left[\mathrm{PCl}_5\right]}{\left[\mathrm{Cl}_2\right]}=0.08 \times \frac{0.75}{0.32}=0.187 \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

∴ The equilibrium concentration of PC13 = 0.187mol-L-1.

Dynamic Equilibrium Concept

Question 2. At a given temperature, Kp is 0.36 for the reaction, 2SO₂(g) + O₂(g) 2SO₃(g) occurring in a closed vessel. If at equilibrium, the partial pressures of SO₂(g) & O₂(g) are 0.15 atm & 0.8 atm respectively, then calculate the partial pressure of SO₃(g).
Answer: For the given reaction, Kp \(K_p=\frac{\left(p_{\mathrm{SO}_3}\right)^2}{\left(p_{\mathrm{SO}_2}\right)^2 \times p_{\mathrm{O}_2}}\)

As Given Kp=0.36,pSO₂= 0.15 atm and pO₂= 0.8 atm

∴ (pso3)2 = KP x (Pso2>2 x Po₂ = 036 x (0.15)2 x (0.8)

= 6.48 x10-3 atm or PSO₃= 0.08 atm

Question 1. For the reaction, \(\mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})\) the value ofequilibrium constant is 50 at 100°C.If a flask of 1 L capacity containing 1 mol A2 is connected with another flask of 2 L capacity containing 2 mol B2, then calculate the number of moles of AB at equilibrium.
Answer: Equilibrium constant, \(K_c=\frac{[\mathrm{AB}]^2}{\left[\mathrm{~A}_2\right] \times\left[\mathrm{B}_2\right]}\)

The equation of the reaction shows that the reduction of x mol of A2 leads to the reduction of x mol of B2 and the formation of 2x mol of AB. Hence, the number of moles of A2, B2, and AB at equilibrium will be as follows.

Dynamic Equilibrium Concept

⇒ \(\begin{array}{lccc}
& \mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g}) \\
\text { Initial no. of moles: } & 1 & 2 & 0 \\
\text { No. of moles at equilibrium: } & 1-x & 2-x & 2 x
\end{array}\)

Now if the two flasks are connected, then the total volume of the reaction system becomes (1 + 2)L = 3L.

Hence, at equilibrium \(\left[\mathrm{A}_2\right]=\left(\frac{1-x}{3}\right) \mathrm{mol} \cdot \mathrm{L}^{-1}\)

⇒ \([B]=\left(\frac{2-x}{3}\right) \mathrm{mol}^{-1} \mathrm{~L}^{-1} \text { and }[\mathrm{AB}]=\left(\frac{2 x}{3}\right) \mathrm{mol} \cdot \mathrm{L}^{-1} \text {. }\)

Therefore, \(K_c=\frac{\left(\frac{2 x}{3}\right)^2}{\left(\frac{1-x}{3}\right) \times\left(\frac{2-x}{3}\right)}=50 \text { or, } \frac{4 x^2}{(1-x)(2-x)}=50\)

or, Ax2 s 50×2- 1 50x + J 00 or, 46×2- 150x+ 100 = 0

Or, \(x=\frac{150 \pm \sqrt{(150)^2-4 \times 46 \times 100}}{2 \times 46}=2.32 \text { or, } 0.934\)

∴ 2x = 4.64 or, 1.868

By the Initial number of moles of A2 and B2, the number of moles of AB cannot be 4.64. Therefore, the number of moles of AB at equilibrium = 1 .868.

Dynamic Equilibrium Concept

Question 2. At a particular temperature, the value of Kp is 100 for the reaction, \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_2(g)+\mathrm{O}_2(g)\)occurringin a closed container. If the initial pressure of NO(g) is 25 atm then calculate the partial! pressures of NO, N₂, and O₂ at equilibrium.
Answer: For the given reaction \(K_p=\frac{p_{\mathrm{N}_2} \times p_{\mathrm{O}_2}}{\left(p_{\mathrm{NO}}\right)^2}\)

After the attainment of equilibrium, if the reduction in pressure of (g) is p atm, then partial pressures of different constituents at equilibrium will be as follows:

⇒ \(\begin{array}{lccl}
& 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \\
\text { Initial pressure }(\mathrm{atm}): & 25 & 0 & 0 \\
\text { Pressure at equilibrium }(\mathrm{atm}): & 25-p & p & p \\
& 2 & 2
\end{array}\)

[The equation shows that at constant temperature and pressure if the pressure reduction of NO(g) is p atm then the increase in pressure of each of N₂(g) and O₂(g) will be P/2 atm.

⇒ \(\text { Hence, } K_p=\frac{\left(\frac{p}{2}\right) \times\left(\frac{p}{2}\right)}{(25-p)^2}=100 \text { or, }\left(\frac{p}{25-p}\right)^2=400\) \(\text { or, } \frac{p}{25-p}=20 \text { or, } 21 p=25 \times 20\) ∴ P= 23.8 atm

Therefore, at equilibrium, partial pressure of NO(g) = (25- 23.8) atm = 1.2 atm andpartial pressure of N2(g) = partial pressure Of \(\mathrm{O}_2(\mathrm{~g})=\frac{23.8}{2}=11.9 \mathrm{~atm}\)

Dynamic Equilibrium Concept

Application-3: If the value of the equilibrium constant of any reaction at a constant temperature is known, then it is possible to predict whether the mixture of reactants and products is in equilibrium and if not, in which direction the reaction will proceed Formore discussion, to attains equilibrium article number at that 7.6. temperature.

Free Energy Change (ΔG) Of A Reaction and Equilibrium Constant

Suppose, at a temperature of TK, A and B react together to produce D and E according to the following equation: \(a A+b B \rightleftharpoons d D+e E\)

⇒ \(K=\frac{[D]_{e q}^d \times[E]_{e q}^e}{[A]_{e q}^a \times[B]_{e q}^b}[e q=\text { equilibrium }]\)

where [A)eq, [B]eq [D]eq, and [FJÿare the molar concentrations of A, B, D, and E respectively, at equilibrium.

If AG is the from the energy of the system, then with the help of can show that \(\Delta G=\Delta G^0+N T \ln \frac{[D]^a \times[B]^b}{[A]^a \times[B]^b}\)

or, ΔG = ΔG0+RT in Q

Dynamic Equilibrium Concept

⇒ \(\text { where, } \left.\left.Q=\frac{[D]^d \times[B]^e}{[A]^a \times[B]^b} \text { and } \mid A\right], \mid B\right],[D] \text { and }[B]\)

Represent the active masses or molar concentrations of A, B, I) and respectively at a given moment during the reaction. Is the reaction quotient.

AG° Is the standard free energy change of the reaction. If In a reaction, (lie molar concentration of each of the reactants and products. If unity, then the free energy change of the reaction is called (IK; standard free energy change (AG°).

For a reaction at constant temperature and pressure, if—

1. ΔG is negative, the reaction is spontaneous as written.
2. ΔG is positive, the reaction Is non-spontaneOus as written.
3. ΔG is zero, the reaction Is at equilibrium.

Now, at the equilibrium of a reaction, AG = 0 and

⇒ \(Q=\frac{[D]_{e q}^a \times[B]_{e q}^e}{[A]_{e q}^a \times[B]_{e q}^b}=K \text { [equilibrium constant] }\)

Hence, from equation (2) we get, 0 = AG° + RTlnK or, AG° = -RTlnK —[3] or,AG° = -2.303RTlogK

Dynamic Equilibrium Concept

or K= e^-ΔG0/Rt

Equations (3), (4), and (5) represent the relation between equilibrium constant {K) and the standard free energy change (AG°) of a reaction at a given temperature. From equations (2) and (3), we get, AG = -RTlnK +RTlnQ

or ΔG = RT In \(\frac{Q}{K}\)

Equation (6) is called reaction isotherm.

If the equilibrium constant expression of a reaction is written in terms of the molar concentrations of reactants and products then K=Ke. This gives \(\Delta G^0=-R T \ln K_c \text { or, } K_c=e^{-\Delta G^0 / R T}\)

Dynamic Equilibrium Concept

For gaseous reactions, equilibrium constant expression is written in terms of the partial pressures of reactants and products. So, for such reactions K=K and AG° = -RTlnK p or, Kp = e-ΔG°/RT.

In the equation, reactant products taken the standards I molstatel, of 1, the equation the; -RT in K p, the standard state of each reactant and products Is considered to be 1 atm.

Significance Of the reaction ΔG0=-RT in K

1. At a particular temperature If the value of AG° is known, then the value of the equilibrium constant (K) can be calculated by using this equation. Similarly, If the equilibrium constant of a reaction at a given temperature is known, then the value of A <7° can be determined by using this reaction.
2. If ΔG° < 0, i.e., ΔG° = – ve, then K will be greater than [K> 1]. Under such conditions, the amount of products will be relatively more than that of the reactants present in the equilibrium mixture, i.e., the forward reaction predominates.
3. If ΔG° >0, i.e., ΔG = + ve, then K will be less than 1 [K < 1]. In such cases, the concentration of the reactants is greater than that of the products, i.e., the backward reaction predominates.

Dynamic Equilibrium Concept

Dynamic Equilibrium Concept Numerical Examples

Question 1. For the reaction A(g) + 2B(g) ⇌2D(g), ΔG°- 2kJ.mol-1 at 500 K. What is the value of Kp for the reaction iyl(g) + B(g)?=±D(g) at that temperature?
Answer: From the equation AG° = -RTlnKp, we get

⇒ \(\ln K_p=-\frac{\Delta G^0}{R T}=-\frac{2000 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{8.314 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1} \times 500 \mathrm{~K}}=-0.4811\)

∴ Kp= 0.6181

∴ For the given reaction \(K_p=\frac{\left(p_D\right)^2}{p_A \times\left(p_B\right)^2}\)

If the equilibrium constant for the reaction

⇒ \(\frac{1}{2} A(g)+B(g) \rightleftharpoons D(g) \text { be } K_p^{\prime} \text {, then } K_p^{\prime}=\frac{p_D}{p_A^{1 / 2} \times p_B}\)

∴ \(K_p^{\prime}=\sqrt{K_p} \quad K_p^{\prime}=\sqrt{0.6181}=0.7862\).

Question 2. Find the value of AG° and ICc for the following reaction at 298K. \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)\) Given: standard free energy of formation (ΔG) of NO₂ and NO are 52.0 and 87.0 kj.mol-1 respectively.
Answer: AG° of a reaction = EAG° of products -EAG° of reactants. For the given reaction,

⇒ \(\Delta G^0=\Delta G_f^0\left(\mathrm{NO}_2\right)-\Delta G_f^0(\mathrm{NO})-\frac{1}{2} \Delta G_f^0\left(\mathrm{O}_2\right)\)

We know that, AG° =-RT In Kc

or, -35 x 103 J-mol-1 = – S.SMJ-K-ÿmol-1 x 298K ln Kc

Dynamic Equilibrium Concept

or; lnKc = 14.126

∴ Kc = 1.365 x 10 6

Question 3. At 298K, for the attainment of equilibrium of the reaction N₂O₄(g) ⇌2NO₂5mol of each of the constituents is taken. Due to this, the total pressure of the mixture turns 20 atm. If AGOF(N₂O4)= 100 kJ-mol-1 and 1G(NO₂) = 50 k(J-moI_1) then— 0 Give the value of AG of the reaction. In which direction will the reaction proceed to equilibrium?
Answer: For the reaction, \(\begin{aligned}
\Delta G^0 & =2 \Delta G_f^0\left(\mathrm{NO}_2\right)-\Delta G_f^0\left(\mathrm{~N}_2 \mathrm{O}_4\right) \\
& =(2 \times 50-100)=0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Total number of moles in the reaction mixture =5 + 5 =10

\(\text { So, } p_{\mathrm{N}_2 \mathrm{O}_4}=\frac{5}{10} \times 20=10 \mathrm{~atm} \& p_{\mathrm{NO}_2}=\frac{5}{10} \times 20=10 \mathrm{~atm}\)

Dynamic Equilibrium Concept

Therefore, Qp f the reaction \(=\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{(10)^2}{10}=10\)

We know, AG = AG° + RT]nQp

∴ ΔG =0 + 8.314 x 298 In 10 = 5.706 kJ

Since ΔG0 =0

Again, AG° =-RT In Kp;

So, 0 = -RT In Kp [since ΔG0=0]

or, Kp = 1

Since, Qp > Kp, the reaction wall proceeds more toward the left to attain equilibrium.

Determination Of Equilibrium Constants Of Some Reactions

Esterification of alcohol:

⇒ \(\mathrm{CH}_3 \mathrm{COOH}(l)+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l) \rightleftharpoons \mathrm{CH}_3 \operatorname{CoOC}_2 \mathrm{H}_5(l)+\mathrm{H}_2 \mathrm{O}(l)\)

Let at a particular temperature, a mol of acetic acid (CH₃COOH) reacts with b mol of ethyl alcohol (C₂H₅OH) to produce x mol of ethyl acetate (CH₃COOC₂H₅) and x mol of water (H₂O) at equilibrium.

According to the balanced equation, for the formation of x mol of ester and x mol of H₂0, x mol of CH₃COOH and x mol of C₂H₅OH are required. If the volume of the reaction mixture is by V L, the die concentration of the different species at equilibrium as well as the expression of equilibrium constant will be as given in the following table:

Dynamic Equilibrium Concept

Formation of NO(g) from N₂(g) and O₂(g): For the reaction: \(\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\); let us assume, at a constant temperature a mol of N₂ reacts with b mol of O₂ in a closed container of VL to produce 2xmol of NO at equilibrium. According to the equation,ifx mol of N₂ and O₂ reacts with each other, then 2x mol of NO are formed.

Let, the total pressure of the reaction mixture at equilibrium = P. The molar concentrations and partial pressures of the reactants and products at equilibrium and the expressions of Kp and Kc are given in the table.

Dissociation of PCl₅ gas: For the reaction: PCl₅(g)⇌PCl₃(g) + Cl₂(g); let us assume, at a particular temperature, 1 mol of PC1₅ gas is heated in a closed vessel of V L capacity so that x mol of PCI5 gets dissociated at equilibrium.

Dynamic Equilibrium Concept

According to the given equation, x mol of PC1₅ on dissociation produces x mol of PC1₃ and x mol of Cl₂.

1. Let us assume that die total pressure of the reaction mixture at equilibrium is P. Hence, the molar concentrations and partial pressures of the reactant and products at equilibrium and the expression equilibrium constants are given in the following table.

Dynamic Equilibrium Concept

Formation of NH₃ from N₂ and H₂: For the reaction: N₂(g) + 3H₂(g) v=± 2NH₃(g); let us assume, at a particular temperature, 1 mol of N₂(g) is allowed to react with 3 mol of H₂(g) in a closed vessel of V L capacity. After the attainment of equilibrium, if 2x mol of NH₃(g) is produced, then according to the equation x mol of N₂(g) and 3x mol of H₂(g) would be consumed.

Let, the total pressure of the reaction mixture be at equilibrium. Therefore, the molar concentrations and partial pressures of the reactants and product at equilibrium and the expression of equilibrium constants are given in the following table

Dynamic Equilibrium Concept

Thermal decomposition of ammonium carbamate: Reaction: \(\mathrm{NH}_2 \mathrm{CO}_2 \mathrm{NH}_4(\mathrm{~s}) \Rightarrow 2 \mathrm{NH}_3(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})\). Let us assume, the l mol of NH₂CO₂NH₄(s) is heated at a particular temperature in a closed vessel of volume VL.

At equilibrium, if x mol of NH₂CO₂NH₄(g) undergoes decomposition, then, according to the equation, 2x mol of NH₃ and x mol of CO₂, will be produced.

The total pressure of the reaction mixture at equilibrium = P, then the molar concentrations and partial pressures of products and the expression of equilibrium constants are given in the table:

Dynamic Equilibrium Concept

 

Relation Between Degree Of Dissociation And Degree Of Association With Vapour Density

When a compound undergoes incomplete dissociation, an equilibrium is established between the undissociated molecules of the die compound with the species formed on dissociation.

The extent of dissociation of a compound is usually quantized in terms of the degree of dissociation. It is defined as the ratio of the number of moles of the compound that dissociates to the initial number of moles of the compound.

Similarly, if a compound undergoes association, its extent of association is quantized in terms of degree of association. It is defined as the ratio of the number of moles of die compound associates to the initial number of moles of the compound.

Relation between the vapor density and degree of dissociation

Suppose, each molecule of gas A₂ on dissociation forms n molecules of gas B. If the reaction is initiated with 1 mol of gas A in a closed vessel of volume V at constant temperature and pressure, then the die number of moles of A₂ and B at equilibrium are as follows-

Dynamic Equilibrium Concept

Reaction: Initial No.of moles No. of moles at equilibrium

Where = the degree of dissociation of A₂

Total number of moles of the mixture at equilibrium =1- a + no =1 + o(n- 1) mol At constant temperature and pressure, the total volume of [1 + o(n- 1)] mol of the gas mixture is [1 + <r(n- 1) x V].

Let the actual density and vapor density of gas A2 before dissociation be d and D, respectively. Since the volume of the system increases on the dissociation of gas A2 the density and vapour density of the gas mixture at equilibrium will be different from the actual values of these two quantities for gas A2.

Suppose, the observed density and vapor density of the gas mixture at equilibrium be d’ and D’, respectively. Since the mass of the system remains the same, we can write

⇒ \(\begin{array}{rlr}
d \times V=d^{\prime} \times V[1+\alpha(n-1)] & \\
\text { or, } \frac{d}{d^{\prime}} & =1+\alpha(n-1) & \cdots[1] \\
\frac{D}{D^{\prime}} & =1+\alpha(n-1) & \cdots[2] \quad\left[\frac{d}{d^{\prime}}=\frac{D}{D^{\prime}}\right]
\end{array}\)

From equation [1] we get \(\alpha=\frac{d-d^{\prime}}{d^{\prime}(n-1)}\)

From equation [2] we get \(\alpha=\frac{d-d^{\prime}}{d^{\prime}(n-1)}\)

Relation between the vapor density and degree of association

Dynamic Equilibrium Concept

Let us consider n mol of gas A associated to form 1 mol of gas B. If the reaction is initiated with 1 mol of gas A in a closed vessel of volume V at constant temperature and pressure, then the number of moles of A and B at equilibrium are as follows—

⇒ \(\begin{array}{lcc}
\text { Reaction: } & n A \rightleftharpoons B \\
\text { Initial no. of moles: } & 0 & 0 \\
\text { No. of moles at equilibrium: } & 1-\alpha & \underline{\alpha} \\
& & n
\end{array}\)

where a = the degree of association of A

∴ Total number of moles of the mixture at equilibrium

⇒ \(=1-\alpha+\frac{\alpha}{n}=1-\alpha\left(1-\frac{1}{n}\right) \mathrm{mol}\)

At constant temperature and pressure, the total volume of,\(\left[1-\alpha\left(1-\frac{1}{n}\right)\right]\) mol gas \(\left[1-\alpha\left(1-\frac{1}{n}\right) \times v\right]\) Now, let actual density and vapor density of gas A before association be d and D respectively.

Dynamic Equilibrium Concept

If the observed density and vapor density of the gas mixture at equilibrium are d’ and D’, respectively, then—

⇒ \(\begin{aligned}
& \quad d \times V=d^{\prime} \times V\left[1-\alpha\left(1-\frac{1}{n}\right)\right] \\
& \text { or, } \quad \frac{d}{d^{\prime}}=1-\alpha\left(1-\frac{1}{n}\right)
\end{aligned}\)

∴ \(\frac{D}{D^{\prime}}=1-\alpha\left(1-\frac{1}{n}\right)\) …..[2] [since \(\frac{d}{d^{\prime}}=\frac{D}{D^{\prime}}\)

From equation [1] we get ,\(\alpha=\frac{d^{\prime}-d}{d^{\prime}\left(1-\frac{1}{n}\right)}\)

From equation [2] we get \(\alpha=\frac{D^{\prime}-D}{D^{\prime}\left(1-\frac{1}{n}\right)}\)

Le Chateuer’s Principle

Equilibrium of a chemical reaction is established under some conditions such as pressure, temperature, Concentration, etc. Le Chatelier, a celebrated French chemist studied the effect of such conditions on a large number of chemical equilibria.

He summed up his observations regarding the effect of these factors on equilibrium in the form of generalization which is commonly known as Le Chatelier’s principle.

Le Chateuer’s Principle If a system under equilibrium Is subjected to a change In pressure, temperature, or concentration then the equilibrium will shift Itself in such a way as to reduce the effect of that change.

Dynamic Equilibrium Concept

The effect of the change in the various conditions of the chemical reactions at equilibrium is discussed below based on Le Chatelier’s principle.

Effect of change in concentration of reactant or product at equilibrium of a reaction

According to Le Chatelier’s principle, at a constant temperature, keeping the volume fixed, if the concentration of reactant or product at equilibrium is changed, the equilibrium will shift in the direction in which the effect of change in concentration is reduced as far as possible.

With the help of the following general reaction, let us discuss how the change in concentration of reactant or product affects the equilibrium of a chemical reaction: A + B⇌ C+D

Dynamic Equilibrium Concept

Effect of addition of reactant to the reaction system at equilibrium at constant volume and temperature

Reaction: A+B⇌C+D

1. At constant temperature, keeping the volume un¬ changed, if a certain amount of reactant (A or J3) is added to the system at equilibrium, the concentration of that reactant will increase.
2. As a result, the reaction will no longer remain in the state of equilibrium.
3. According to Le Chatelier’s principle, the system will arrange itself in such a manner so that the effect of increased concentration of that reactant is reduced as far as possible. Naturally, the equilibrium will tend to shift in a direction that causes a decrease in the concentration of the added reactant.
4. Since the concentrations of the reactants reduce in the forward direction, the net reaction will occur in this direction until a new equilibrium is established when the rates of both the forward and reverse reactions become the same.
5. Therefore, the addition of reactant to the reaction system at equilibrium causes the equilibrium to shift to the right. As a result, the yields of the products (C andD) increase.

Conclusion: At constant volume and temperature, when some quantity of reactant is added to a reaction system at equilibrium, the equilibrium shifts to the right, and the yield of product (s) increases.

Dynamic Equilibrium Concept

Example: Reaction: N₂(g) + 3H₂(g) 2NH₃(g)

At constant temperature, keeping the volume fixed, if some quantity of H₂(g) is added to the above system at equilibrium, the net reaction will occur In the forward direction until a new equilibrium is established.

This means that the addition of some H₂(g) to the reaction system at equilibrium causes the equilibrium to shift to the tire right. As a result, the yield of NH₃(g) will increase.

Dynamic Equilibrium Concept

Effect of addition of the product to the reaction system at equilibrium at constant volume and temperature:

Reaction: A+B ⇌C+D

1. At constant temperature, keeping the volume fixed, when some quantity of the product (C or D) is added to the system at equilibrium, the concentration of that product increases.
2. As a result, the reaction will no longer remain in the state of equilibrium.
3. According to Le Chatelier’s principle, the system will adjust itself in such a way that the effect of an increase in concentration of that product is reduced as far as possible. Naturally, the equilibrium of the system will try to shift in a direction that reduces the concentration of the added product.
4. Since the concentrations of the product decrease in the reverse direction, the net reaction will occur in this direction until a new equilibrium is established when the rates of both the forward and reverse reactions become identical.
5. Therefore, the addition of aproduct to the reaction system at equilibrium makes the equilibrium shift to the left. As a result, the concentrations of the reactants (4 and B) increase.
6. Conclusion: At constant volume and temperature, if some quantity of product is added to a reaction system remaining at equilibrium, then the equilibrium will shift to the left. As a result, the concentration of product(s) decreases, whereas that of the reactant(s) increases.

Dynamic Equilibrium Concept

Effect of removal of reactant from the reaction system at equilibrium at constant volume and temperature:

Reaction A+B⇌C+D

1. At constant temperature without changing the volume If some amount of reactant (4 orB) is removed from the system at equilibrium, the concentration of that reactant decreases.
2. As a result, the reaction will no longer exist in the state of equilibrium.
3. According to Le Chatelier’s principle, the system will adjust itself in such a manner so that the effect of a decrease in concentration of that reactant is reduced as far as possible. Naturally, the equilibrium of the system will shift in a direction that increases the concentration of that reactant.
4. Since the concentration of the reactants increases in the reverse direction, the net reaction will occur in this direction until a new equilibrium is established when both the forward and reverse reactions take place at equal rates.
5. Therefore, the removal of a reactant from the reaction system results in shifting the equilibrium position to the left. As a result, the yield of the products (C and D) will decrease and that of the reactants (A and B) will increase.
6. Conclusion: At constant volume and temperature, if some quantity of reactant(s) is removed from the reaction system at equilibrium, the equilibrium will shift to the left. As a result, the yield of product(s) decreases, whereas that of the reactant(s) increases

Dynamic Equilibrium Concept

Effect of removal of the product from the reaction system at equilibrium at constant volume and temperature:

Reaction A+B⇌C+D

1. At fixed temperature, keeping the volume unaltered, when some quantity of product (C or D) is removed from the system at equilibrium, the concentration of that product will decrease.
2. As a result, the reaction will no longer remain in the equilibrium state.
3. According to Le Chatelier’s principle, the system will adjust itself in such a manner, so that the effect of a decrease in the concentration ofthatproduct is reduced as far as possible. Hence, the equilibrium will shift in a direction that increases the concentrations of that removed product.
4. Since the concentration of the products increases in the forward direction, the net reaction will occur in this direction until a new equilibrium is established.
5. Therefore, the removal of a product from the reaction system results in shifting the equilibrium position to the right. As a result, the yields of products( C or D) increase.
6. Conclusion: At constant volume and temperature, if some quantity of a product is removed from the reaction system at equilibrium, the equilibrium will shift to the right. As a result, the yield of product(s) increases and that of the reactant(s)
7. Explanation of the effect of addition or removal of reactant or product on equilibrium in terms of reaction quotient: Let us suppose, at a given temperature, the following reaction Is at equilibrium: A+B⇌C+D

Dynamic Equilibrium Concept

Equilibrium constant, \(K_c=\frac{[C]_{e q} \times[D]_{e q}}{[A]_{e q} \times[B]_{e q}}\) …[1]

Where [A]eq, [B]gq, [C]eq and [D]eq are equilibrium molar concentrations of A, B, C, and D respectively.

Effect of addition of the reactant: Suppose, keeping the temperature and volume fixed, some amount of A is added to the reaction system at equilibrium. Consequently, the concentration of A in the mixture will increase.

Let the concentration of A increase from [A]eq to [A]. At this condition, the reaction quotient will be—

⇒ \(Q_c=\frac{[C]_{e q} \times[D]_{e q}}{[A] \times[B]_{e q}}\)

Since,[A] > [A]eq, QC<KC. Thus, the reaction is not in equilibrium now (because at equilibrium, Qc = Kc). Due to the increase in concentration of A, the forward reaction will take place to a greater extent compared to the reverse reaction. As a result, the value of the numerator in equation (2) increases and that of the denominator decreases, leading to a net increase in the value of Qc.

Dynamic Equilibrium Concept

A time comes when Qc = K£ and the equilibrium is re-established. Therefore, the addition of reactant (A) to the above reaction system at equilibrium will result in a shift of the equilibrium to the right. As a result, the yields of the products (C and D) increase.

Effect of addition of the product: At constant temperature and volume, if some amount of C is added to the reaction at equilibrium, then the concentration of C in the reaction mixture will increase. Suppose, the concentration of C increases from [C]. at this condition, the reaction quotient Will Be-

⇒ \(Q_c=\frac{[C] \times[D]_{e q}}{[A]_{e q} \times[B]_{e q}}\)

Since, [C]>[C]eq, Qc>Kc. As Qd=Kc, the reaction is no longer at equilibrium. Re-establishing of the equilibrium occurs when Qc = Kc.

This is possible if the shifting of equilibrium occurs to the left because this will cause the numerator to decrease and the denominator to increase in the equation [3]. As a result of the shifting of equilibrium to the left, the yields of the products decrease.

Dynamic Equilibrium Concept

Effect Of Removal Of The Reactant: Keeping the temperature and volume fixed, let some amount of A be removed from the reaction system at equilibrium.

Consequently, the concentration of A in the reaction mixture will be reduced and eventually, the equilibrium will be disturbed. At this condition, Qc will be greater than Kc i.e., Qc > Kc. This will cause the equilibrium to shift to (lie left. As u result, the yields of the products (C and I) decrease.

Effect of removal of the product: At constant temperature and volume, If some amount of product C is removed from the reaction system, the equilibrium will be disturbed because of a decrease in the concentration of C in the reaction mixture.

At this condition, Qc < Kc. As a result, equilibrium will shift to the right. Consequently, the yields of the products ( C and D) increase.

Dynamic Equilibrium Concept

Some examples from everyday life:

Drying of clothes: Clothes dry quicker when there is a breeze or we keep on shaking them. This is because water vapor of the nearby air is removed and cloth loses more water vapor to re-establish equilibrium with the surrounding air.

Transport of O₂ by hemoglobin in blood: Oxygen breathed in combines with the hemoglobin in the lungs according to equilibrium, Hb(s) + O₂(g) HbO₂(s). In the tissue the pressure of oxygen is low. To re-establish equilibrium oxyhemoglobin gives up oxygen. But in the lungs, more oxyhemoglobin is formed due to the high pressure of oxygen.

Removal of CO₂ from tissues by blood: This equilibrium

⇒ \(\begin{aligned}
\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l) & \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3(a q) \\
& \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HCO}_3^{-}(a q)
\end{aligned}\)

In tissue, partial pressure of CO₂ is high thus, CO₂ dissolves in the blood. In the lungs, the partial pressure of CO₂ is low, it is released from the blood.

Dynamic Equilibrium Concept

Tooth decay by sweets: Our teeth are coated with an enamel of insoluble substance known as hydroxylapatite,

⇒ \(\begin{gathered}
\mathrm{Ca}_5\left(\mathrm{PO}_4\right)_3(\mathrm{OH})(s) \\
\text { Demineralisation } \rightleftharpoons \text { Remineralisation } \\
5 \mathrm{Ca}^{2+}+3 \mathrm{PO}_4^{3-}+\mathrm{OH}^{-}
\end{gathered}\)

If we do not brush our teeth after eating sweets, the sugar gets fermented on the teeth and produces H ions which combine with the OH ions shifting the above equilibrium in the forward direction thereby causing tooth decay.

Dynamic Equilibrium Concept

Effect of pressure on equilibrium at constant temperature

The effect of change in pressure at equilibrium is observed only for those chemical reactions whose reactants and products are in a gaseous state and have different numbers of moles.

The effect of change in pressure is not significant for the chemical reactions occurring in a solid or liquid state because the volume of a liquid or a solid does not undergo any appreciable change with the variation of pressure.

Dynamic Equilibrium Concept

Effect of increase in pressure:

1. At constant temperature, a reaction exists at equilibrium under a definite pressure. Keeping the temperature constant, if the pressure on the system at equilibrium is increased, then the reaction will no longer exist at equilibrium.
2. According to Le Chatelier’s principle, the system will tend to adjust itself in such a way as to minimize the effect of the increased pressure as far as possible.
3. At constant temperature, the only way to counteract the effect of the increase in pressure is to decrease the volume or to reduce the number of moles (or molecules).
4. Hence, at a constant temperature, if the pressure on the system at equilibrium is increased, then the net reaction will take place in a direction that is accompanied by a decrease in volume or several moles (or molecules).

Dynamic Equilibrium Concept

Effect Of decrease in pressure:

1. At constant temperature, if the pressure of a reaction system at equilibrium is decreased, then the reaction will no longer remain at equilibrium.
2. According to Le Chatelier’s principle, the net reaction will tend to occur in a direction that is associated with an increase in the volume or number of molecules.
3. So, at a constant temperature, the decrease in pressure at the equilibrium of a reaction will result in a shifting of equilibrium in a direction that is accompanied by an increase in volume or an increase in the number of molecules (or moles).

Example: Let, at a constant temperature, the following reaction is at equilibrium: N2(g) + 3H2(g) 2NH3(g). In the reaction, the number of moles of the product is fewer than that of the reactants. So, the forward reaction is accompanied by a decrease in volume.

Dynamic Equilibrium Concept

Effect of increase in pressure at equilibrium: At constant temperature, if the pressure of the reaction system at equilibrium is increased, then according to Le Chatelier’s principle, the equilibrium of the reaction will shift to the right i.e., the forward reaction will occur to a greater extent compared to the reverse reaction.

So the yield of NH₃(g) will increase. Effect of decrease in pressure at equilibrium: Keeping the temperature constant, if the pressure of the reaction system at equilibrium is decreased, then according to Le Chatelier’s principle, the equilibrium will shift to the left i.e., the backward reaction will occur to a greater extent, leading to a reduction in the yield of NH3.

For gaseous reactions in which the total number of mole of reactants is equal to that of the products (i.e., An = 0), equilibrium is unaffected by the change in pressure.

Dynamic Equilibrium Concept

This is because these types of reactions are not accompanied by any volume change. Some examples are:

⇒ \(\begin{aligned}
& \mathrm{H}_2(g)+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(g) ; \quad \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \\
& \text { and } \mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HCl}(\mathrm{g}) \text {. }
\end{aligned}\)

Dynamic Equilibrium Concept

Effect of temperature on equilibrium

Chemical reactions are usually associated with the evolution or absorption of heat. A reaction in which heat is evolved is called an exothermic reaction, while a reaction in which heat is absorbed is called an endothermic reaction.

In a reversible reaction, if the reaction in any one direction is endothermic, then the reaction in the reverse direction will be exothermic. The temperature of a system at equilibrium can be increased by supplying heat from an external source while the temperature of the system can be lowered by cooling.

Dynamic Equilibrium Concept

Effect of increase in temperature: At equilibrium, if the temperature of a system is increased, then according to Le Chatelier’s principle, the system will try to offset the effect of the increase in temperature as far as possible. As a result, when the temperature of a reaction system at equilibrium is raised, equilibrium will shift in a direction in which heat is absorbed because it is possible to neutralize the effect of an increase in temperature through the absorption of heat.

Effect of increase In temperature in case of endothermic reactions: If the temperature is increased at the equilibrium of an endothermic reaction, the forward reaction will take place to a greater extent compared to the reverse reaction until a new equilibrium is established when both the forward and backward reaction occur at equal rates. As a result, equilibrium shills to the right, and the yields of products Increase.

Effect of Increase In Temperature In the case of exothermic reactions: The temperature Is Increased at the equilibrium of an exothermic reaction, and then the reverse reaction will occur to a greater extent than the forward reaction until a new equilibrium is established when both the forward and backward reactions take place at equal rates. This makes the equilibrium of the reaction shift to the left, resulting In decreased yields of the products.

Dynamic Equilibrium Concept

Effect of decrease in temperature: According to I.C. Chntolicr’s principle, when the temperature of any system at equilibrium is decreased, the system will try to offset the effect of a decrease in temperature as far as possible.

Therefore, when the temperature is decreased at equilibrium, the equilibrium will shift in a direction in which heat is evolved because it is possible to neutralize the effect of a decrease in temperature through the evolution of heat.

Effect of decrease in temperature In case of endothermic reactions: As heat is absorbed in an endothermic reaction, a decrease in temperature at equilibrium of such a reaction causes the backward reaction to take place to a greater extent compared to the forward reaction until a new equilibrium is established. As a result, the equilibrium shifts to the left, and the yields of products decrease.

Dynamic Equilibrium Concept

Effect of decrease in temperature in case of exothermic reactions: If the temperature is decreased at the equilibrium of an exothermic reaction, the forward reaction will take place to a greater extent compared to the reverse reaction until a new equilibrium is established. As a result, shifts to the right, and the yields of products increase.

Example: Manufacturing of ammonia (NH₃) by Haber’s process is an example of an exothermic reaction: N₂(g) + 3H₂(g) ?=± 2NH₃(g); AH = -22kcl. In this case, the forward reaction is exothermic.

Hence, the backward reaction is endothermic. Effect of increase in temperature at equilibrium: If the temperature is increased at the equilibrium of the reaction, then according to Le Chatelier’s principle, the backward reaction will take place to a greater extent compared to the forward reaction’ until a new equilibrium is formed. Hence, the equilibrium will shift to the left, causing a decrease in the yield of NH₃.

Dynamic Equilibrium Concept

Effect of decrease In temperature at equilibrium: If the temperature  Is decreased at the equilibrium of the reaction, then according to i.e., Cliuteller’sprinciple, the equilibrium will be In (lie direction In which heat Is generated ie., in ibis case, the forward reaction will be favored, and consequently the yield of N 1 L, will be higher.

Formation of NO(g) from N₂(g) and O₂(g) Is an ondothormlc reaction: N₂(g) + O₂(g) ?=2NO(g); All = -idd kcnl. Mere the forward reaction is endothermic. Hence, the backward reaction Is exothermic.

Effect of increase In temperature on equilibrium: if the temperature is Increased at the equilibrium of the reaction, then according to Lc Cliatelier’s principle, the forward reaction will take place to a greater extent compared to the backward reaction until a new equilibrium is formed. Hence, the equilibrium will shift to the right, leading to a higher NO.

Effect of decrease in temperature on equilibrium: If the temperature is decreased at the equilibrium of the reaction, then according to Le Cliatelier’s principle, the equilibrium will shift in the direction in which heat is generated i.e., in this case, the backward reaction will be favored over the forward reaction. Consequently, the yield of NO will be reduced.

Effect Of Catalyst On Equilibrium

A catalyst has no role in the equilibrium of a reaction. At a given temperature, when a reaction is conducted separately in the presence and the absence of a catalyst, the composition of the equilibrium mixture formed in either case remains the same. This is because the catalyst increases the rates of the forward and backward reactions equally.

Dynamic Equilibrium Concept

The catalyst functions to make the attainment of equilibrium faster by accelerating the rates of both the forward and reverse reactions to the same extent. The yield of product in a reaction cannot be increased with the use of a catalyst.

Effect of addition of inert gas on equilibrium

At constant temperature, adding an inert gas (He, Ne, Ar, etc.) to an equilibrium reaction system can be done at constant volume or pressure.

Effect of addition of inert gas at constant volume: At constant temperature, keeping the volume fixed, when an inert gas is added to a reaction system at equilibrium, the total number of molecules (or moles) in the system increases. So, the total pressure of the system increases, but the partial pressure of the components does not change. Hence, the equilibrium of the system remains undisturbed.

Dynamic Equilibrium Concept

Effect of addition of inert gas at constant pressure: Keeping both temperature & pressure fixed, the addition of inert gas to a reaction system at equilibrium causes an increase in the volume of the system (because the total number of moles in the system increases) with a consequent decrease in partial pressures of the components.

So, the sum of the partial pressures of the reactants and the products also decreases. In this situation, equilibrium will shift in a direction that increases the volume of the reaction system i.e., the number of molecules in the reaction system.

Depending on An, three situations may arise—

Numerical Examples

Question 1. At 986°C, 3 mol of H2O(g) and 1 mol of CO(g) react with each other according to the reaction, CO(g) + H₂O(g)⇌CO₂(g) + H₂(g). At equilibrium, the total pressure of the reaction mixture is found to be 2.0 atm. If Kc = 0.63 (at 986°C), then at equilibrium find O the number of moles of H₂(g), 0 the partial pressure of each of the gases.
Answer: Let, a decrease in several moles of H₂O(g) be x after the reaction attains equilibrium. Consequently, number of moles of CO also decreases by x. According to the reaction, each of CO₂(g) and H₂(g) increases by x number of moles.

Therefore, number of moles of different substances will be as follows:

So, the total number of moles of different substances at equilibrium =1-x+3-x+x+x = 4

Partial pressures of different substances at equilibrium:

⇒ \(\begin{aligned}
& p_{\mathrm{CO}}=\frac{(1-x)}{4} \times 2=\frac{1}{2}(1-x) ; \\
& p_{\mathrm{H}_2 \mathrm{O}}=\left(\frac{3-x}{4}\right) \times 2=\frac{1}{2}(3-x)
\end{aligned}\)

Dynamic Equilibrium Concept

⇒ \(p_{\mathrm{CO}_2}=\left(\frac{x}{4}\right) \times 2=\frac{x}{2} ; p_{\mathrm{H}_2}=\left(\frac{x}{4}\right) \times 2=\frac{x}{2}\)

In the given reaction, Δn = 0.

Therefore, Kp – Kc = 0.63.

In the given, ,K_p=\(\frac{p_{\mathrm{CO}_2} \times p_{\mathrm{H}_2}}{p_{\mathrm{CO}} \times p_{\mathrm{H}_2 \mathrm{O}}}\)

⇒ \(\text { So, } 0.63=\frac{\left(\frac{x}{2}\right) \times\left(\frac{x}{2}\right)}{\left(\frac{1-x}{2}\right) \times\left(\frac{3-x}{2}\right)}=\frac{x^2}{(1-x) \times(3-x)}\)

or, x2= 0.63×2- 2.52x+ 1.89

or, x2 + 6.81x- 5.108 = 0 x = 0.681

∴ At equilibrium, the number of moles of H2(g) = 0.68

∴ At equilibrium \(p_{\mathrm{CO}}=\frac{1}{2}(1-0.681)=0.1595 \mathrm{~atm}\)

Dynamic Equilibrium Concept

⇒ \(\begin{aligned}
& p_{\mathrm{H}_2 \mathrm{O}}=\frac{1}{2}(3-0.681)=1.1595 \mathrm{~atm} \\
& p_{\mathrm{CO}_2}=p_{\mathrm{H}_2}=\frac{0.681}{2}=0.3405 \mathrm{~atm}
\end{aligned}\)

Question 2. For the reaction, N₂O₄(g), and – 2NO₄(g) occurring in a closed vessel at 300K, the partial pressures of N₂O₄(g) and NO₂(g) at equilibrium are 0.28 atm and 1.1 atm respectively. What will be the partial pressures of these gases if the volume of the reaction system is doubled keeping the temperature constant?
Answer: Equilibrium constant, \(K_p=\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{(1.1)^2}{0.28}=4.32\)

If the volume of the reaction system is doubled at constant temperature, then partial pressures of N₂O₄ and NO₂ will decrease to half of their initial values. Therefore, partial pressures of N₂O₄ and NO₂ will be 0.14 and 0.55 atm respectively. So, the equilibrium of the reaction will be disturbed. Now, according to Le Chaterlier’s principle, a reaction will attain a new equilibrium by shifting to the right because in such a case the number of moles as well as the volume will increase.

Dynamic Equilibrium Concept

Let, at the new equilibrium, the partial pressure of N₂O₄(g) decreases to p atm. According to the equation, the partial pressure of NO₂(g) will increase to 2p atm. Therefore, at a new equilibrium, the partial pressure of each of the component gases will be:

⇒ \(\begin{array}{rrr}
& \mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \\
\text { Partial pressure at new } & & \\
\text { equilibrium }(\mathrm{atm}): & 0.14-p & 0.55+2 p
\end{array}\)

⇒ \(K_p=\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{(0.55+2 p)^2}{(0.14-p)}=4.32\)

Or, 4p2 + 2.2p + 0.3025 = 4.32(0.14-p) = 0.6048-4.32p

or, p2+ 1.63p- 0.0755 = 0

∴ p = 0.045 atm

Dynamic Equilibrium Concept

So, at new equilibrium, partial pressure of N₂O₄(g) = (0.14-0.045) atm = 0.095 atm and partial pressure of NO₂ = (0.55 + 2 X 0.045) atm = 0.64 atm

Question 3. PCl₆(g)⇌PCl₃(g) + Cl₂(g); Kp=1.8 At 250°C if 50% of PC1₅ dissociates at equilibrium then what should be the pressure of the reaction system?
Answer: Let the initial number of moles of PCl₅₂g) be a. At equilibrium, 50% dissociation of PCl₅(g) will occur if the pressure of the reaction system = P atm. After 50% dissociation of PCl₅(g), the number of moles of PCl₅(g) decreases by an amount of 0.5a, and for PCl₃(g) and Cl₂(g) it increases by 0.5a. So at equilibrium:

⇒ \(\begin{array}{ccc}
\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) \\
\text { Number of moles: } \quad a-0.5 a=0.5 a & 0.5 a \quad 0.5 a
\end{array}\)

At equilibrium, total no. moles = 0.5a + 0.5a + 0.5a = 1.5a

∴ At equilibrium, partial pressure of different components \(\text { are, } p_{\mathrm{PCl}_5}=\frac{(0.5 a)}{(1.5 a)} P=\frac{P}{3} ; p_{\mathrm{PCl}_3}=\frac{P}{3} \text { and } p_{\mathrm{Cl}_2}=\frac{P}{3}\)

Equilibrium constant of the reaction \(K_p=\frac{p_{\mathrm{PCl}_3} \times p_{\mathrm{Cl}_2}}{p_{\mathrm{PCl}_5}}\)

Dynamic Equilibrium Concept

So at 5.4 atm pressure, 50% of PCl5(g) will be dissociated at 250°C.

Question 4. At a particular temperature and 0.50 aim pressure, NH) Ami some amount of .solid NH4Hs arc present In a rinsed container. Solid NH₃(g) dissociates to give NH₄(g) and H₂S(g). At equilibrium, the total pressure of (ho reaction-mixture Is found to be 0.8 1 atm. Hud the value of the equilibrium constant of this reaction at that temperature.
Answer: Reaction: NH₄HS(s)⇌NH₃(g) + H₂S(g)

Prom the reaction, it is dear that, ) mol NM₄HS(s) dissociation produces 1 mol NH(g) and 1 mol HS(g). So, at a particular temperature and volume, the partial pressure of NHa(g) and I H₂S(g) will be the same and independent of the amount of NH₄HS(s).

Let, partial pressure of H2S(g) at equilibrium =p atm. Therefore, partial pressure of NH3(g) and H2S(g) at Initial stage and at equilibrium arc as follows:

⇒ \(\begin{array}{ccc}
& \mathrm{NH}_4 \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_3(g)+\mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \\
\text { Initial pressure }(\mathrm{atm}): & 0.5 & 0 \\
\text { Equilibrium pressure }(\mathrm{atm}): & 0.5+p \quad p \\
\text { At equilibrium, total pressure of reaction mixture }
\end{array}\)

∴ 0.5 + 2p = 0.04 or, p = 0.17 atm

Dynamic Equilibrium Concept

∴ At equilibrium, partial pressure of NH₃,

pNH₃ = (0.5 + 0.17) atm =0.67 atm and partial pressure of

H₂S,PH₂S = 0.17 atm

So, the equilibrium constant of the reaction, Kp = PNH₃ X PH₂0 = 0.67 X 0.17 = 0.1139 atm2

Dynamic Equilibrium Concept Ionic Equilibrium Introduction

The compounds that conduct electricity in a molten state or solution and dissociate chemically into new substances are called electrolytes. Various acids (e.g., HC1, HNO₃, H₂SO₄), bases (e.g., NaOH, KOH), and salts (e.g., NaCl, KC1, CuSO₄, AgNOg) dissolve in water to conduct electricity. Hence, these are electrolytes.

On the other hand, substances that are unable to conduct electricity either in a molten state or in solution are known as non-electrolytes. Glucose, sugar, alcohol, benzene, etc., are examples of non-electrolytes. In an aqueous solution (or in a molten state), electrolytes undergo spontaneous dissociation or ionization to produce positively and negatively charged particles or ions.

Dynamic Equilibrium Concept

This is known as electrolytic dissociation or ionization. The ions can conduct electricity in the solution. In a particular solvent and at a certain temperature and concentration, the degree of dissociation or ionization of any electrolytic substance depends on the nature of that substance. The fraction of the total amount of dissolved electrolyte that exists in a dissociated or ionized state is called the degree of dissociation or ionization.

Depending on the degree of ionization in an aqueous solution, electrolytes can be classified into two categories Strong electrolytes and Weak electrolytes. The electrolytes which dissociate almost completely in aqueous solution at all concentrations are called strong electrolytes.

Dynamic Equilibrium Concept

Strong acids (e.g., HC1, HNO₃, H₂SO₄ ), strong bases (e.g., NaOH, KOH), and most of the salts (e.g., NaCl, KC1, NH₄C1, CuSO₄ ) are strong electrolytes. Electrolytes that dissociate partially aqueous solution are called weak electrolytes.

Some salts (Example- BaSO₄, HgCl₂), most of the organic acids (example HCOOH, CHgCOOH ), a few in organic bases (example Fe(OH)₃, NH₃ ), and some inorganic acids (e.g., HCN, H₂CO₃ ) are weak electrolytes. In aqueous solutions, strong electrolytes undergo almost complete ionization. Hence, such ionisations are represented by a single arrow(→). For example— NaCl(ag)- Na+(ag) + Cl-(ag); HCl(ag)→H+(ng) + Cl-{aq).

Dynamic Equilibrium Concept

On the other hand, weak electrolytes undergo partial ionization in aqueous solution. Hence, such ionisations are reversible. Consequently, in an aqueous solution of weak electrolytes, a dynamic equilibrium exists between the dissociated ions and unionized molecules. This is known as ionic equilibrium.

Due to its irreversible nature, such ionisations are represented by double arrows(→). For example—HCN(ag)⇒H+(ag) + CN-(ag); CH₃COOH(ag) CH₃COO-(ag) + H+(ag)

Acids And Bases

Acids and bases according to Arrhenius’s theory

Acids Hydrogen-containing compounds that ionize in an aqueous solution to produce H+ ions are called acids.

Example: The hydrogen-containing compounds such as HC1, HNO₃, H₂SO₄, CH₃COOH, etc., ionize in aqueous solutions to form H+ ions. Thus, these compounds are acids according to Arrhenius’s theory.

Dynamic Equilibrium Concept

Bases A compound that ionizes in an aqueous solution to produce hydroxyl ions (OH-) is called a base.

Example: The compounds such as NaOH, KOH, Ca(OH)₂ NH4OH, etc., ionize in aqueous solution to produce OH’ ions and hence are termed as bases according to Arrhenius theory.

1. NaOH(ag) Na+(ag) + OH-(aq)
2. KOH(ag)→ K+(aq) + OH-(aq)
3. Ca(OH)2(aq)→ Ca2+{aq) + 20H-(aq)
4. NH4OH(aq)→ NH+(aq) + OH-(aq)

Limitations of Arrhenius theory: Arrhenius theory is useful for defining acids and bases. It explains the acid-base neutralization reaction by the simple equation: H₃O+(a<7) + OH-(aq)→2H₂O(Z). However, there are certain limitations of this theory.

1. According to this theory, the presence of water is essential for a compound to exhibit its acidic or basic properties. However, the fact that acidic or basic property of a compound is its characteristic property, which is independent of the presence of water.
2. The acidic or basic properties of a substance that is insoluble in water cannot be explained by this theory.
3. The acidity or basicity of any compound in non-aqueous solvents cannot be explained by Arrhenius’s theory. For example, the acidity of NH₄C1 orbasicity of NaNH₂ in liquid
   ammonia cannot be explained by this theory.
4. According to Arrhenius’s theory, compounds containing only hydroxyl ions are considered as bases. Consequently, the basicity of ammonia (NH₃), methylamine (CH₃NH₂), aniline (C₆H₅NH₂), etc., cannot be explained by this theory.
5. According to Arrhenius’s theory, only the hydrogen-containing compounds that ionize in aqueous solution to produce H+ ions are considered acids. Consequently, the acidity of compounds such as PC1₅, BF₃, and A1C1₃ cannot be explained by this theory.

Acids And Bases According To Bronsted Lowry Concept (Protonic Theory)

Definitions of acids and- bases according to the theory proposed by J.N. Bronsted and T.M. Lowry are given below:

1. Acid: An acid is a substance that can donate a proton (or H+ ion)
2. Base: A base is a substance that can accept a proton (or J+ ions)

So, according to this theory, an acid is a proton donor and a base is a proton acceptor.

Example: HCl(aq) + H₂O(aq)→H₃O+(aq) + Cl-(aq)

In this reaction, HC1 donates one proton, behaving as acid, while H20 accepts a proton, behaving as a base.

Dynamic Equilibrium Concept

According to this theory, apart from the mill compounds (HCI, HNO₃, CH₃COOH, etc cations example; NH+4, C₆H₅NH+3,[Fe(H2O)₆]3, [A](H2O)₆]3+, etc.) and anions (example HSO-4, HCO-3, HC₂O-4, etc.] and The acidic properties of these three types of substances are shown by the following reactions ;

1. CH₃COOOH(aq)+ H2O(l) ⇌ H3O+(aq)+CH₃COO-(aq)
2. H₂SO₄(aq) + 2H2O(l)⇌ 2H3O(aq)+SO₂-4(aq)
3. NH+4{aq) H- H₂O(/)⇌ H₃O+(aq) +NH₃(aq)
4. [Fe(H₂O)₆]3+(H₂O) + H₂O(l)⇌
5. [Fe(H₂O)₅OH]₂(aq)+ H₃O+(aq)
6. H₂PO-4(aq) + H₂O(l)⇌ H₃O+(aq)+HPO₂-4(aq)

Similarly, apart from the neutral compounds (e.g., NH₃, C₆H5NH₂, H₂O, etc.), a large number of unions (e.g., OH-, CH₃COO-, CO₂-3, etc.) can act as a base,

The following reactions indicate the basic properties of these two lands of substances:

1. CH₃COO-(aq) + H₂O(l)⇌ CH₃COOH(aq) + OH-(aq)
2. NH₃(aq) +H2O(l)⇌NH+4(aq) + OH-(aq)
3. HPO₂-4-(aq) + H2O(l)⇌ (aq) + OH-(aq)

Concept of conjugate acid-base pair:

Concept of conjugate acid-base pair Definition: A pair of species (a neutral compound and the Ion produced from it or, an ion and a neutral compound formed from it or, an ion and the other ion produced from it) having a difference of one proton is called a conjugate acid-base pair.

Examples: (H₂O, H₃O+), (H2POJ, HPO₂-4), (CH₃COO-, CH₃COOH ), etc., are some examples of conjugate acid-base pairs.

Explanation: To get an idea about conjugate acid-base pair, let us consider the ionization of CH3COOH in an aqueous solution:

CH₃CO₂H(aq) + H2O(l) ⇌CH₃CO₂ (aq) + H₃O+(aq)

Dynamic Equilibrium Concept

Since CH3COOH is a weak acid, it undergoes partial ionization in the solution, and the above equilibrium is tints established. In the forward reaction, the CH₃COOH molecule donates a proton (H+ion) which is accepted by the H₂O molecule.

Therefore, according to the Bronsted-Lowry concept, CH₂COOH is an acid and H2O is a base. In the reverse reaction, the H₃O+ ion donates a proton which is accepted by a CH₃COO- ion. Therefore, in the reverse reaction, H₃O+ ion acts as an acid and CH₃COO- as a base.

⇒ \(\underset{\text { acid }}{\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}}(a q)+\underset{\text { base }}{\mathrm{H}_2 \mathrm{O}}(l) \underset{\text { base }}{\mathrm{CH}_3 \mathrm{CO}_2^{-}(a q)}+\underset{3}{\mathrm{H}_3 \mathrm{O}^{+}}(a q) \cdots[1]\)

In equation [1], CH₃COO- ion is the conjugate base of CH₆COOH and CH₃COOH is the conjugate acid of CH₃COO- ion. Hence, (CH₃COOH, and CH₆COCT) constitute a conjugate acid-base pair.

Dynamic Equilibrium Concept

Similarly, in equation (1), H₂O is the conjugate base of the H₃O+ ion and the H₃O+ ion is the conjugate acid of H₂O. Therefore, (H₃O+, H₂O ) constitutes a conjugate acid-base pair.

An acid donates a proton to produce a conjugate base and a base accepts a proton to produce a conjugate acid. The conjugate base of an acid has one fewer proton than the acid. On the other hand, the conjugate acid of the base has one more proton than the base

Dynamic Equilibrium Concept

Dynamic Equilibrium Concept

Dynamic Equilibrium Concept

Strength of conjugate acid-base pair or Bronsted acid-base pair in aqueous solution:

The stronger an acid, the greater its ability to donate a proton. Similarly, a base with greater proton-accepting ability exhibits stronger basicity.

1. The acids HC1, HNO₃, H₂SO₄, etc., undergo complete ionization in aqueous solution to form H₃O+ ions and the corresponding conjugate bases. Hence, these are considered strong acids in an aqueous solution. In an aqueous solution, the conjugate base produced from a strong acid has less tendency than H2O to accept a proton. Therefore in an aqueous solution, the conjugate base of a strong acid is found to be very weak.
2. The acids HF, HCN, CH₆COOH, HCOOH, etc., undergo slight ionization in an aqueous solution to produce H₂O+ions and the corresponding conjugate bases. As these acids have little tendency to donate protons in aqueous solution, they are called weak acids.
3. In an aqueous solution, the conjugate base produced from a weak acid has more tendency’ than H₂O to accept a proton. Therefore, in aqueous solution, the conjugate base of a weak add is found to be stronger than H₂O.
4. In aqueous solution, strong bases like NH₂, O₂-, H” etc., react completely with water to form their corresponding conjugate adds and OH- ions. These conjugate acids are later than H₂0. Hence, in aqueous solution, the conjugate acid of a strong base is very weak. On the other hand, in an aqueous solution, weak bases like NH₃, CH₃NH₂, etc., react partially with water to produce the corresponding conjugate acids and OH- ions. These conjugate acids are stronger than H₂O. Therefore, in an aqueous solution, the conjugate acid of a weak base is strong.

Dynamic Equilibrium Concept

The conjugate acid of a strong base has little tendency to accept protons. On the other hand, the conjugate acid of a weak base has a high tendency to accept protons.

Acid-base neutralization reactions according to Bronstedlowry concept: According to Bronsted-Lowry concept, in an acid-base neutralization reaction, a proton from an acid molecule gets transferred to a molecule of a base.

As a result, the acid converts to its conjugate base, and the base changes to its conjugate acid by accepting a proton.

Example:

Dynamic Equilibrium Concept

Limitations of Bronsted-Lowry concept: With the help of this theory, the reaction of an acid with a base is explained in terms of the gain or loss of proton(s). However, there are many acid-base reactions in which the exchange of proton(s) does not take place. Such types of acid-base reactions cannot be explained by this theory.

The acidic properties of many non-metallic oxides (for example; CO₂, SO₂ ) and basic properties of many metallic oxides (for example; CaO, BaO) cannot be explained with the help of the Bronsted-Lowry concept. Also, the acidic properties of BF₃, A1C1₂, SnCl₂, etc., cannot be explained with the help of this theory.

Lewis’s Concept Of Acids And Bases

On the electronic theory of valency, scientist Gilbert N. Lewis proposed the following definitions of acids and bases.

Acid: An acid is a substance which can accept a pair of electrons

Dynamic Equilibrium Concept

Examples: The compounds that have a central atom with incomplete octets can act as Lewis acids, such as BF₃ BC1₃, A1C1₃, etc.

In some compounds due to the presence of vacant orbitals in the central atom, the octet can be expanded. These compounds can also behave as Lewis acids, such as PCl₃, SnCl₃, SiF₄, etc.

SiF₄ (Lewis acid) + 2F’ (Lewis base) → [SiF₆]2-

Cations like H+, Ag+, Cu2+, Fe3+, Al3+, etc., behave as Lewis acids.

Dynamic Equilibrium Concept

Molecules containing multiple bonds between two atoms of different electronegativities behave as Lewis acids, such as CO₂, SO₂, etc.

Dynamic Equilibrium Concept

Base: A base Is a Substance that can donate a pair of electrons

Example: Compounds that contain an atom having one or more lone pairs of electrons behave as Lewis bases, such as NH₃, H₂O: CH₃OH, etc.

Anions like NH-2, Cl-, I-, OH-, CN- etc., are considered as Lewis bases.

A Lewis add is an acceptor of a pair of electrons and forms a coordinate bond with a Lewis base.

A Lewis base is a donor of a pair of electrons and forms a coordinate bond with Lewis acid.

Dynamic Equilibrium Concept

Limitations of Lewis’s concept:

1. This concept provides no idea regarding the relative strengths of acids and bases.
2. This theory contradicts the general concept of acids by considering BF₃ A1C1₃, and simple cations as acids.
3. The behaviour ofprotonic adds such as HC₁, H₂SO₄ etc., cannot be explained by this concept. These acids do not form coordinate bonds with bases which is the primary requirement Lewis concept.
4. Normally, the formation of coordination compounds is slow, therefore acid-base reactions should also be slow, but acid-base reactions are extremely fast, this cannot be explained by the Lewis concept.

Degree Of Ionisation And Ionisation Constant Of Weak Electrolyte

Weak electrolytes partially dissociate into ions in solutions and there always exists a dynamic equilibrium involving the dissociated ions and the undissociated molecules.

This state of equilibrium is called an ionic equilibrium. The equilibrium constant associated with an ionic equilibrium is known as the ionization or dissociation constant of the weak electrolyte.

Dynamic Equilibrium Concept

Degree Of Ionisation

During the ionization of a solution of a weak electrolyte, the fraction of its total number of molecules that get dissociated at equilibrium is called the degree of ionization or dissociation of the electrolyte.

Degree of ionization of an electrolyte (α)= Number of dissociated molecules of the electrolyte at equilibrium/ Total number of molecules of the electrolyte.

Suppose, a fixed volume of solution contains 0.5 mol of a dissolved electrolyte. If 0.2 mol of this electrolyte gets dissociated at equilibrium, then the degree of dissociation (a) O₂ of the electrolyte \(=\frac{0.2}{0.5}=0.4\), i.e., 40 % of the electrolyte exists in an ionized state in the solution.

As strong electrolytes dissociate completely in solutions, their degree of dissociation (a)= 1 but, the degree of dissociation of weak electrolytes is always less than 1.

Dynamic Equilibrium Concept

Ionization or dissociation constant of weak electrolytes: Let us consider the following equilibrium which is established by a weak AB because of its partial ionization in water

⇒ \(\mathrm{AB}(a q) \rightleftharpoons \mathrm{A}^{+}(a q)+\mathrm{B}^{-}(a q)\)

Applying the law of mass action to the equilibrium we have equilibrium constant, K,\(K=\frac{\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right]}{[\mathrm{AB}]}\)

where [A+], [B-], and [AB] are the molar concentrations (mol L-1) of A+, B-, and AB, respectively at equilibrium.

Dynamic Equilibrium Concept

K represents the ionization or dissociation constant of the weak electrolyte AB. The value of equilibrium constant (K) changes well with a variation of temperature,  at a constant uimporniure, It has a fixed value.

Oslwald’s dilution law

Let the initial concentration (before dissociation) of an aqueous solution of a weak electrolyte ΔH he c mold- 1,

Dynamic Equilibrium Concept

The following equilibrium Is established due to partial dissociation of ΔH in an aqueous solution: ΔH(aq) Δ+(aq)+ B-(aq) Suppose, at equilibrium, the degree of ionization or dissociation of ΔH is a.

So, a mol of ΔH on Its ionization will result In a mol of Δ1 Ion and a mol of H- Ions. The number of moles of AB that remain unlonloniscd is (l — α).

Hence at equilibrium, the molar concentrations of different species will be as follows:

Dynamic Equilibrium Concept

By applying the law of mass action to this equilibrium, we get equilibrium constant \((K)=\frac{\left[\mathrm{A}^{+}\right] \times\left[\mathrm{B}^{-}\right]}{[\mathrm{AB}]}=\frac{\alpha c \times \alpha c}{c(1-\alpha)}=\frac{\alpha^2 c}{1-\alpha} \cdots[1]\)

Equation [1] represents the mathematical expression of Ostwald’s dilution law. The equilibrium constant (K) is called the Ionisation constant of the weak electrolyte, AB. At ordinary concentration, the value of the degree of dissociation (a) of a weak electrolyte is generally very small. So, (1 – a) s: 1 . With this approximation equation [1] can be written as, \(K=\alpha^2 c \quad \text { or, } \quad \alpha=\sqrt{\frac{K}{\boldsymbol{c}}}\)

Equation [2] is the simplified mathematical expression of Ostwald’s dilution law Conclusion,’ From equation [2), it can be concluded that—

Dynamic Equilibrium Concept

The degree of dissociation (or) of a weak electrolyte in a solution is inversely proportional to the square root of the concentration of the solution (since at constant temperature, K has a definite value).

So, at a fixed temperature, the degree of dissociation of a weak electrolyte in its solution increases with the decrease in the concentration of the solution and decreases with the caraway in the concentration of the solution, It roof of weak electrolyte remains dissolved In 6, of the solution, then the molar concentration of the solution, \(c=\frac{1}{V}.\) Substituting \(c=\frac{1}{V} .\) In equation [2], we have a Thin equation showing that when v Increases (as happens when the solution Is diluted), the degree of dissociation of the electrolyte also Increases.

Dynamic Equilibrium Concept

Ostwald’s dilution law: At a certain temperature, the degree of Ionisation of a weak electrolyte in a solution Is Inversely proportional to the square root of the molar concentration of the solution.

Or, At a certain temperature, the degree of ionization of a weak electrolyte In a solution is directly proportional to the square root of the volume of the solution containing mol of the electrolyte.

Dynamic Equilibrium Concept

Limitation of Ostwald’s dilution law: Ostwald’s dilution law applies to weak electrolytes only. As strong electrolytes ionize almost completely at all concentrations, this law does not apply to them.

Ionization or dissociation constant of a weak acid and concentration of H30+ ions in its aqueous solution

Weak acids partially dissociate into ions in aqueous solutions, and there always exists a dynamic equilibrium involving the dissociated ions and the undissociated molecules. Like any other equilibrium, such type of equilibrium also has an equilibrium constant, known as the ionization or dissociation constant of the corresponding weak acid.

Dynamic Equilibrium Concept

The ionization constant of a weak acid is designated by the Ionisation constant of a weak monobasic acid: Let HA be a weak monobasic acid which on partial ionization in an aqueous solution forms the following equilibrium

HA(aq) + H₂O(l) ⇌H₃O+(aq) + A-(aq)

Using the law of mass action to the above equilibrium, we get equilibrium contract \(K=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]\left[\mathrm{H}_2 \mathrm{O}\right]}\)

Dynamic Equilibrium Concept

where [H₃O+], [A-], [HA], and [H₂0] represent the molar concentrations (mol-L-1) of H3O+, A-, HA, and H₂O, respectively, at equilibrium in solution.

In the solution, the concentration of H₂O is much higher than that of HA and its concentration does not change significantly due to partial ionization of HA. The concentration of H₂O, therefore, remains constant.

⇒ \(\text { So, } K \times\left[\mathrm{H}_2 \mathrm{O}\right]=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)

Since [H2O] = constant, the Kx(H₂O] constant is known as the ionization or dissociation constant of the weak acid and Is designated by ‘Ka’.

Dynamic Equilibrium Concept

Therefore \(K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+} \| \mathrm{A}^{-}\right]}{[\mathrm{II} \mathrm{A}]}\)

Significance of ionization constant of a weak acid:

Like any other equilibrium constant, the value of the ionization constant (Ka) of a weak acid is constant at a fixed temperature.

The higher the tendency of a weak acid HA to donate proton in water, the higher the concentration of H₃O+ and A- ions at equilibrium in the solution.

Hence from equation (1), it can be said that the stronger the acid, the larger the value of its ionization constant.

Dynamic Equilibrium Concept

If the solutions of two monobasic acids have the same molar concentration, then the solution containing the acid with a larger value of Ka will have a higher concentration of H30+ ions than the other solution.

Ionization constants (Ka) of some weak monobasic acids at 25°C [in water]

Dynamic Equilibrium Concept

The concentration of H₃O+ ions in an aqueous solution of a weak monobasic acid: Let a weak acid HA on its partial ionization water form the following equilibrium:

HA(aq) + H₂O(l)⇌ H3O+(aq) + A-(aq)

If the initial concentration of HA in its aqueous solution is Cmol-L-1 and the degree of ionization of HA at equilibrium is then the concentrations of different species nl equilibrium will be as follows:

(During ionization of a weak acid, the concentration of H20 remains unchanged.) Therefore, the Ionisation constant of the weak add HA,

⇒ \(K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{c \alpha \times c \alpha}{c(1-\alpha)}=\frac{\alpha^2 c}{1-\alpha}\)

At ordinary concentration, the degree of dissociation (a) of a weak acid is negligible. So, (1 – or)1

Dynamic Equilibrium Concept

⇒ \(\text { Hence, } K_a=\alpha^2 c \text { or, } \alpha=\sqrt{\frac{K_a}{c}}\)

Therefore, if the concentration (c) of a weak monobasic acid and its ionization constant (Ka) are known, then the degree of ionization (a) of the acid can be calculated by using the equation [1].

Concentration of H₃O+ ion [H₃O+αc]

⇒ \(\text { or, }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\alpha c=\sqrt{\frac{K_a}{c}} \times c=\sqrt{c \times K_a}\)

In an aqueous solution of a weak acid, if the concentration of the acid and its degree of ionization (or) are known, then the concentration of H₃O+ ions in the solution can be calculated by using equation [2].

Dynamic Equilibrium Concept

Alternatively, if the concentration (c) of the acid and its ionization constant (Ka) is known, then the concentration of H3O+ ions can be calculated by using the equation [3].

Relative strengths of two weak monobasic acids: Let us consider the aqueous solutions of two weak acids HA and HA-, each with the same molar concentration of cool-L-1.

At a given temperature, if the ionization constants of HA and HA’ are Ka and K’a respectively and a and a are their degrees of ionization in their respective solutions, then

⇒ \(\alpha=\sqrt{\frac{K_a}{c}} \quad \text { and } \quad \alpha^{\prime}=\sqrt{\frac{K_a^{\prime}}{c}} \quad
\)

∴ \(\frac{\alpha}{\alpha^{\prime}}=\sqrt{\frac{K_a}{K_a^{\prime}}}\)

Dynamic Equilibrium Concept

Therefore, at a particular temperature, if the molar concentrations of the solutions of two weak monobasic acids are the same, then the acid having a larger ionization constant will have a higher degree of ionization than the other.

The degree of dissociation (a) of a weak acid in its solution of a given concentration is a measure of its strength (or proton donating tendency). The higher the degree of dissociation of an acid in its solution, the stronger the acid.

It means that the strength of an acid in its solution is proportional to its degree of dissociation in the solution.

Hence, at a particular temperature, for a definite molar concentration \(\frac{\text { Strength of acid } \mathrm{HA}}{\text { Strength of acid } \mathrm{HA}^{\prime}}=\sqrt{\frac{K_a}{K_a^{\prime}}}\)

Dynamic Equilibrium Concept

Ionization constant of weak polybasic acids: An acid with more than one replaceable H-atom is called a polybasic acid, e.g., H2C03, H3PO4, H2S. H2SO4 and H2S are dibasic acids as they have two replaceable H-atoms, whereas H3P04 is a tribasic acid as it has three replaceable hydrogen atoms. These acids dissociate in a series of steps, each of which attains an equilibrium and has its characteristic equilibrium constant.

Example: Ionisation of H₃PO₄ in water: In water, H₃PO₄ ionizes in the following three steps as it contains three replaceable hydrogen atoms:

⇒ \(\begin{aligned}
& \mathrm{H}_3 \mathrm{PO}_4(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{H}_2 \mathrm{PO}_4^{-}(a q) \\
& \text { Ionisation constant, } K_{a_1}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right]}{\left[\mathrm{H}_3 \mathrm{PO}_4\right]}
\end{aligned}\)

Dynamic Equilibrium Concept

⇒ \(\begin{aligned}
& \mathrm{H}_2 \mathrm{PO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{HPO}_4^{2-}(a q) \\
& \text { Ionisation constant, } \boldsymbol{K}_{a_2}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{HPO}_4^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right]}
\end{aligned}\)

Dynamic Equilibrium Concept

⇒ \(\begin{aligned}
& \mathrm{HPO}_4^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{PO}_4^{3-}(a q) \\
& \text { Ionisation constant, } \boldsymbol{K}_{a_3}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{PO}_4^{3-}\right]}{\left[\mathrm{HPO}_4^{2-}\right]}
\end{aligned}\)

Overall ionization constant, Ka = Ka1 x Ka2 x Ka3

Dynamic Equilibrium Concept

At constant temperature, Ka1 > Ka2 > Ka3 for a weak tribasic acid. Due to the electrostatic force of attraction, a singly charged anion (examples- or H₂PO4) has less tendency to lose a proton than a neutral molecule (for example H₂S ). Similarly, it is more difficult for a doubly charged anion (e.g., HPO₂-4 ) to lose a proton than a singly charged anion.

Ionization or dissociation constant of a weak base and concentration of 0H- ions in its aqueous solution

When a weak base is dissolved in water, it reacts with water to form its conjugate acid and OH- ions. Eventually, an equilibrium involving the conjugate acid and unreacted base is established. Such an equilibrium has its characteristic equilibrium constant known as the ionization constant of the weak base. The ionization constant of weak basis is denoted by.

Ionisation Constant of a weak monoacidlc base: Let us consider an aqueous solution of a weak monoacidic base B. Since B is a weak base, a small nili fiber of molecules reacts with an equal number of H20 molecules to form the conjugate acid, BH+, and OH- ions.

Dynamic Equilibrium Concept

A dynamic equilibrium is thus established between BH+, OH- and unionized molecules as follows:

⇒ \(\mathrm{B}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{BH}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Applying the law of mass action to the above equilibrium, we have, equilibrium constant \(K=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]\left[\mathrm{H}_2 \mathrm{O}\right]}\)

Where [BH+], [OH-], [B], and [H20] represent the molar concentrations of BH+, OH-, B, and H20 respectively, at equilibrium. In an aqueous solution, the concentration of H₂O is much higher than that of B. Thus, any change in concentration that occurs because of the reaction of H₂O with B can be neglected. So, [H₂O] remains essentially constant.

So, \(K \times\left[\mathrm{H}_2 \mathrm{O}\right]=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}\)

Since [H₂O] = constant, K x [H₂O] = constant This constant is known as the ionization constant of the weak base B and is denotedby’Kb ‘

Dynamic Equilibrium Concept

Therefore \(K_b=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}\)

Equation [1] expresses the ionization constant for the weak monoacidic base, B.

Significance of ionization constant of a weak base:

As in the case of other equilibrium constants, the ionization constant of a weak base (Kb) has a definite value at a particular temperature.

If the weak base, B, has a high tendency to accept protons in water, it reacts with water to a greater extent. This results in high concentrations of BH+ and OH- in the solution and gives rise to a large value of Kb for the base. Therefore, the larger the value of Kb for a weak base, the stronger the base.

Dynamic Equilibrium Concept

At a certain temperature, if the aqueous solution of two weak bases has some molar concentration, then the solution containing the base with a larger value of Kb will have a higher concentration of OH” ions at equilibrium in the solution.

Dynamic Equilibrium Concept

The concentration of OH- in an aqueous solution of a weak monoacidic base: Let a weak base, B on partial ionization in an aqueous solution form the following equilibrium:

B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)

Let the initial concentration of B in the aqueous solution be c mol-L-1. According to the above equation, at equilibrium, if the concentration of OH- is x mol-L-1, then the concentrations of BH+ & B will be x mol-L-1 and (c-x) mol-L-1 respectively because according to the given equation, 1 molecule of B and 1 molecule of H20 react to form one BH+ and one OH- ion.

Since the base is weak, a very small amount of it reacts with water. So, x is negligible in comparison to c, and hence c-xx.

⇒ \(K_b=\frac{x^2}{c} \quad \text { or, } x=\sqrt{K_b \times c}\)

Dynamic Equilibrium Concept

Therefore, in an aqueous solution of the weak base, B \(\left[\mathrm{OH}^{-}\right]=\sqrt{K_b \times c}\)

If the ionization constant (Kb) of a monoacidic weak base and the concentration (c) of its aqueous solution are known, then the concentration of OH- ions in the solution can be determined with the help of equation [1].

Determination of [H₃O+] in a solution of strong acid and [OH-] in a solution of a strong base

Strong acids (e.g., HCl, HBr, HC1O₄, HNOs, H₂SO₂) and strong bases [e.g., NaOH, KOH, Ca(OH)₂] completely ionize in their aqueous solutions. Hence, the concentration of H30+ ions (or, OH- ions) in the aqueous solution of a strong acid (or a strong base) can be calculated from the initial concentration of the acid (or base) in the solution.

Dynamic Equilibrium Concept

When the concentration of the solution is in ‘molar’ unit: Suppose, the molarity of an aqueous solution of a strong acid (or base) is M. If each acid (or base) molecule in the solution produces x H₃0+ (or OH-) ions, then in the case of solution of an acid, [H₃O+] = x M and in the case of solution of a base [OH-] = x x M.

Examples: 1 molecule of strong monobasic acid on its ionization (HCl, HBr, HC10₄, HNO₃, etc.) gives an H₃O+ ion. Hence, in such a solution, the molar concentration of H30+ ions = the molar concentration of the acid solution.

For example, the molar concentration of H₃O+ ions in 0.1(M) HC1 or HNO₃ solution = 0.1(M). Each molecule of a strong dibasic acid (for example H₂SO₄) on its ionization produces two H₃O+ ions.

Therefore, in such a solution, [HsO+] =2x molar concentration of the solution. For example, in 0.1(M) H₂SO₄ solution, [H₃O+] =2×0.1= 0.2(M).

Dynamic Equilibrium Concept

Similarly, in 0.1(M) NaOH solution, [OH-] =0.1(M) and in O.l(M) Ca(OH)2 solution, [OH-] =2x 0.1 = 0.2(M).

When the concentration of the solution is in ‘normal unit: If the concentration of a solution of strong acid (or base) is given in ‘normal’ unit, then the concentration of H₃O+ (or OH- ) ion in that solution will be equal to the normal concentration of the solution.

Example 1. The concentration of H₃O+ ionizing.l(N) H₂SO₄ solution= 0.1(N). Since the H₃O+ ion is monovalent, the molar concentration of the H₂O+ ion in 0.1(N) H₂SO₄ solution is 0.1(M).

Dynamic Equilibrium Concept

The concentration of OH- ion in 0.1(N) Ca(OH)₂ solution= 0.1(N). Since OH- is monovalent, the molar concentration of OH- ion in 0.1(N) Ca(OH)₂ solution= 0.1(M) Normality of a solution =nx Molarity; where n= basicity in case of an acid; acidity in case of a base; total valency location (or anion) per formula unit in case of a salt.

Examples: 1 (M) HC1 = 0.1(N) HC1 solution.

Since the basicity of Hcl =1

0.1(M) H₂SO₄ = 2 x 0.1 = 0.2 (N)H₂SO₄ solution.

Since Basicity Of H₂SO₄ =2

Dynamic Equilibrium Concept

0.1(M) H₂SO₄ = 2 x 0.1 = 0.2 (N)H₂SO₄ solution.

Since Acidity Of CO(OH₂)=2

0.1(M) Ca₂+ = 2 X 0.1= 0.2 (N) Ca₂+

Since Valency Of Ca2+ in its salt =2

0.1(M)Al₂(SO₄)₃ solution = 6 x 0.1 = 0.6(N) A1₂(SO₄)₃ solution [Y in each molecule of AL₂(SO₄)3, total valency of cation (Al3+) or anion (SO-) = Number of ions of Al₃+(or SOl-) x Valency of Al₃+ (or SOl-) =6]

Dynamic Equilibrium Concept

Dynamic Equilibrium Concept Numerical Examples

Question 1. At 25°C temperature, the molar concentrations of NH₃, NH+4 and OH- at equlibrium are 9.6 X 10-3(M), 4.0 x 10-4(M) and 4.0 X 10-4(M) respectively. Determine the ionization constant of NH₃ at that temperature.
Answer: In aqueous solution of NH₃, the following equilibrium is established
\(\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q)\)

∴ Ionisation Constant Of NH₃, Kb = \(=\frac{\left[\mathrm{NH}_4^{+}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_3\right]}\)

As [NH₃] = 9.6×10-3+(m),

[NH+4] = 4.0 X 10-4(M) and [OH-] = 4.0 x 10-4(M)

Dynamic Equilibrium Concept

⇒ \(K_b=\frac{\left(4 \times 10^{-4}\right) \times\left(4 \times 10^{-4}\right)}{9.6 \times 10^{-3}}=1.67 \times 10^{-5}\)

Question 2. A 0.1(M) solution of acetic acid is 1.34% ionized at 25°C Calculate the ionization constant of the acid.
Answer: We know, the ionization constant of a weak monobasic acid example CH₃COOH is \(K_a=\frac{\alpha^2 c}{1-\alpha}\) where- a = degree of ionization and c = initial concentration of the acid solution.

As \(\alpha=\frac{1.34}{100}=1.34 \times 10^{-2} \text { and } c=0.1(\mathrm{M})\)

⇒ \(K_a=\frac{\alpha^2 c}{1-\alpha}=\frac{\left(1.34 \times 10^{-2}\right)^2 \times 0.1}{\left(1-1.34 \times 10^{-2}\right)}=1.82 \times 10^{-5}\)

∴ At 25°C, ionisation constant of CH₃COOH = 1.82 x 10-5

Dynamic Equilibrium Concept

Question 3. In a 0.01(M) acetic acid solution, the degree of ionization of acetic acid is 4.2%. Determine the concentration of HgO+ ions in that solution
Answer: Acetic acid is a weak monobasic acid. In such a solution, [H30+] = ac; where, c and a are the initial concentration of the acid and its degree of ionization respectively.

⇒ \(\text { As } \alpha=\frac{4.2}{100}=4.2 \times 10^{-2} \text { and } c=0.01(\mathrm{M}) \text {, }\) in 0.01(M) acetic acid solution.

[H₃O+] =ac = 4.2 X IQ-2 X 0.01 = 4.2 x 10-4 (M)

Dynamic Equilibrium Concept

Question 4. The value of the ionization constant of pyridine (C₆H₆N) at 25°C is 1.6 X 10-9. what is the concentration of OH- ions in a 0.1(M) aqueous solution of pyridine at that temperature
Answer: Pyridine is a monoacidic base. In aqueous solutions of such bases \(\left[\mathrm{OH}^{-}\right]=\sqrt{c \times K_b}\); where c = initial concentration of the base and Kb = ionization constant of the weak base.

As c = 0.1(M) and Kb = 1.6 x 10-9 in 0.1(M) aqueous pyridine solution,

Dynamic Equilibrium Concept

⇒ \(\left[\mathrm{OH}^{-}\right]=\sqrt{0.1 \times 1.6 \times 10^{-9}}=1.26 \times 10^{-5}(\mathrm{M}) .\)

Question 5. At 25°C, the value of the ionization constant of a weak monobasic acid, HAis 1.6 X 10-4. What is the degree of ionization of HA in its 0.1(M) aqueous solution?
Answer: Degree of ionization of weak monobasic acid (a) \(=\sqrt{\frac{K_a}{c}}.\)

Given, c = 0.1(M) and Ka = 1.6 x 10-4

The degree of ionisation of HA in its 0.1(M) aqueous solution \(=\sqrt{\frac{K_a}{c}}=\sqrt{\frac{1.6 \times 10^{-4}}{0.1}}=0.04\)

Dynamic Equilibrium Concept

∴ Degree of ionisation of HA in its 0.1(M) aqueous solution = 0.04 x 100%= 4%

Question 6. Ionisation constant of ammonia is 1.8 x 10-5 at 25°C. Calculate the degree of ionization of ammonia in its 0.1(M) aqueous solution at that temperature.
Answer: NH₃ is a weak monoacidic base. In aqueous solutions degree of ionization (a) of such base \(=\sqrt{\frac{\kappa_b}{c}}.\) As c = 0.1(M) and k’b = 1.8 x 10-5 tire degree of ionisation of NH3 in its 0.1(M) aqueous solution \(=\sqrt{\frac{1.8 \times 10^{-5}}{0.1}}=0.0134=0.0134 \times 100 \%=1.34 \%\)

Ionic Product Of Water

Pure water is a very poor conductor of electricity, indicating its verylow ionization. Due to the die self-ionization of pure water, H+ and OH- ions are formed and the following dynamic equilibrium involving H+, OH- ions, and unionized water molecules is established:

Dynamic Equilibrium Concept

⇒ \(\mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Applying the mass action to this equilibrium, we get

⇒ \(K_d=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{O}^2\right.}\)

where [H₃O+], [OH ], and [H₂O] are the molar concentrations of H₃O+ OH (aq) and H₂O(l) at equilibrium, respectively, and Kd is the ionization or dissociation constant of water.

Dynamic Equilibrium Concept

As the degree of ionization of water Is very small, its equilibrium concentration is almost the same as Its concentration before Ionisation. Thus, at equilibrium, [H₂O]2 = constant.

From equation [1 ] we have, kd[H₂O)2 = [H₃O+] x [OH-] Since, [H₂O]2 = constant, Kd X [H₂O]2 = constant. This constant is called the Ionic product of water & is denoted by K_w.

therefore K_w=[H₃O+] x[OH-]

Dynamic Equilibrium Concept

Concentrations of H₃O+and OH- in aqueous solution

Any aqueous solution, whether it is acidic or basic, always contains both H₃O+ and OH ions. A concentrated acid solution also contains OH- ions although its concentration is much lower compared to H₃O+ ions. Likewise, a concentrated alkali solution also contains H₃O+ ions but with a much lower concentration than OH- ions.

On the other hand, in a neutral solution, the concentrations of H₃O+ and OH ions are always the same. At a particular temperature, if the ionic product of water Kw, then for

\(\text { a neutral solution: }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\sqrt{K_w}\) \(\text { an acidic solution: }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]>\sqrt{K_w} \text { and }\left[\mathrm{OH}^{-}\right]<\sqrt{K_w}\) \(\text { a basic solution: }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]<\sqrt{K_w} \text { and }\left[\mathrm{OH}^{-}\right]>\sqrt{K_w}\)

At 25°C, Kw = 10’14. So, at this temperature, in case of —

an acidic solution: [H₃O+] > 10-7 mol.L-1 and [OH-] < 10-7 mol.L-1

a basic solution: [OH-] > 10-7 mol.L-1 and [H₃O+] < 10-7 mol.L-1

Dynamic Equilibrium Concept

Determination of [H₂O₄] and [OH-] in aqueous solution:

At a particular temperature, if the value of and any one of [H₃O+] or [OH-] are known, then the other can be determined using either of the following equations:

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_w}{\left[\mathrm{OH}^{-}\right]} \text {or, }\left[\mathrm{OH}^{-}\right]=\frac{K_w}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}\)

For example, in an aqueous solution at 25°C if [H₃O+] = 10-4(M),

Dynamic Equilibrium Concept

⇒ \(\text { then }\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{10^{-4}}=10^{-10}(\mathrm{M})\)

[since K_w(25C)=10-14]

PK_w = pK_w = -log10K_w At 25°C,K_w=10-14

Therefore, at 25°C, pK_w = -log 10( 10-14) = 14.

Concept Of PH And PH Scale

In the case dilute solution of an acid or a base, the concentration of H30+ or OH’ ions Is generally expressed in terms of the negative power of 10.

For instance, the concentration of H₃O+ negative power of 10. For instance, the concentration of H₃O+ concentration of OH- ions In 0.0002(M) NaOH solution is 2 x 10-4 mol. L-1 . However, it is very inconvenient to express the concentration of H₃O+ or OH- ions in terms of such negative power.

Dynamic Equilibrium Concept

To overcome such difficulty encountered in the case of dilute solutions, Sorensen introduced the system of expressing the concentration of H₃O+ ions by pH (pH stands for Potein of hydrogen ion; the German word ‘Potenz’ means ‘power’).

Concept Of PH And PH Scale Definition: The ph of a solution is defined as the negative i logarithm to the 10 of Its H30′ Ion concentration In mol- 1,

Therefore, pH=-log 10[H₃O+]

Example: If the concentration of If. O- Ions In a solution is 10-3(M), then all of the solution a-log10(10-3)=3

Important points to remember about all of the solutions:

pH – log10[H₃O+1 . According to this equation, If (the concentration of Ions in a solution Increases, then the of the solution decreases, and vice-versa. Thus, the higher the value for a solution, [H₃O+] lower the pH of the solution. Conversely, the lower the value for a solution, the higher the pH of the solution.

Dynamic Equilibrium Concept

Example: If [H₃O+ ] = 10-3 (M) In an aqueous solution, then all of the solution = 3. Now, If the solution Is diluted such that [H₃O+ = 1()-G (M), then the pH of that solution increases and becomes 5.

If the pH of an aqueous solution is increased or decreased by one unit, then the concentration of H₈O+ Ions In the solution undergoes a ten-fold decrease or increase in its value.

Example: Let the pH of an aqueous solution be 3. So, [H3O4] in the solution = 10-PW= 1()-:,(M). Now, by diluting the solution, If the pH of (lie solution Is made 4, then, the concentration of H₃O+ ions i.e., H₃O+ will be =10-P-= 10-I (M).

Dynamic Equilibrium Concept

Hence, when the pH of (lie solution is increased by one unit, the concentration of H₃O4 Ions In the solution decreases by a factor of ten. Similarly, the decrease In pH by one unit corresponds to a ten-fold Increase In [H₃O4].] The acidity of a solution Increases with a decrease In pH and decreases with an Increase In pH.

The POH of a solution Is defined as the negative logarithm to the base 10 of Its OH- Ion concentration In mol- L-1.

Therefore POH = – log 10[OH-]

Important points about the pH of a solution:

1. With a decrease or Increase In (lie concentration of OH- ions In the solution, the pOH of the solution Increases or decreases respectively
2. The basicity of a solution increases with a decrease in pOH and decreases with an increase in pOH.

Relation holen pH, pOH, and pKw

At a fixed temperature, for pure water or an aqueous solution,| H3O+|x|OH-| as Kw

Dynamic Equilibrium Concept

Taking negative logarithms on both sides, we get

-log₁₀[H3O+]-log₁₀[OH-]= – log₁₀ Kw Or, Ph+POH = pkw

Therefore, at a fixed temperature, for pure water or an aqueous solution, pH+POH= pk

At 25C, pKw = J 4. Hence, at 25°C, for pure water or any aqueous solution pH+POH =14

Dynamic Equilibrium Concept

Values of pH and pOH for pure water: Pure water is neutral. Hence, in pure water [H3O+1 = [OH-] \(\text { or, }-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10}\left[\mathrm{OH}^{-}\right] \text {or, } p \mathrm{HOH}\)

From the relation, pH + pOH = pKw, we have 2pH= 2pOH = pKw

or \(p H=p O H=\frac{1}{2} p K_w\)

At 25 °C, pKw = 14 . Hence, in the case of pure water at 25 °C,

⇒ \(p H=p O H=\frac{1}{2} \times 14=7\)

At 100°C, pKw = 1 2.26. Hence, in the case of pure water at 100 °C,

Dynamic Equilibrium Concept

⇒ \(p H=p O H=\frac{1}{2} \times 12.26=6.13\)

pH of pure water at 100 °C is lower than that at 25 °C. Hence, at 100 °C, the molar concentration of H₂O+ ions is higher than that at 25°C. However, this does not mean that pure water is acidic at 100 C. Because the concentration of H₂O+ and OH- ions are always the same in pure water, it Is always neutral irrespective of temperature

pH of a basic solution

To calculate the pH of a basic solution, the equation pH + pOH = pK w is used. If the temperature is 25°C, then pKw = 14. Therefore, at 25°C, if the pOH of an aqueous solution lies at 3, then, pH = 14 -pOH =14-3 = 11.

Dynamic Equilibrium Concept

PH -scale

PH -scale Definition: The scale in terms of which the acidity or basicity of any aqueous solution is expressed by its pH value is called the pH scale.

Range of pH-scale at 25°C: Generally in dilute solution, the concentration of H₃O+ or OH- ions is not more than 1mol-L-1. If in the solution, [H30+] = 1 mol-L-1, then pH = —log 10- = 0.

If in the solution, [OH-] = 1 mol-L-1, then [H₃0+] = 10-14 mol-L- [since At 25°C, Kw = 10-14]. Hence, for a dilute solution.

pH=-log 10-14=14

Dynamic Equilibrium Concept

Therefore, at 25 °C, the pH of a dilute aqueous solution ranges from 0 to 14.

Range of pW-scale vs. temperature: The value of pKw of water determines the range of pH -scale. Since the value of pKw varies with the temperature change, the range of pH -scale also changes with the temperature change.

Dynamic Equilibrium Concept

The range of the pH -scale can generally be expressed in the following way:

The value of pKw decreases with an increase in temperature. As a result, the range of pH -scale also decreases. For instance, at 25°C, pK_w = 14. Hence, at this temperature, the pH scale extends from 0 to 14. On the other hand, at 100°C, pK_w = 12.26. Thus, at this temperature pH scale extends from 0 to 12.26. For a neutral aqueous solution at 25°C, pH = 7, and at 100°C, pH =6.13.

Dynamic Equilibrium Concept

a pH of neutral, acidic, and basic solutions at 25’C:

pH of neutral aqueous solution: In case of a neutral aqueous solution at 25°C, [H3O+]=[OH-] =, O-14

= 10-7 mol-L-1. Hence, for a neutral solution at 25°C, pH=-Iog₁₀ [H₃O+] =-log₁₀10-7 = 7.

pH of aqueous acidic solution: In the case of an aqueous acidic solution at 25 °C, [H₃O+] > 10-7 mol.L-1 or, -log₁₀ [H₃O+]<7 or, pH < 7.

pH of aqueous basic solution: For an aqueous basic solution at 25 °C, [H₂O+] < 10-7 mol.L-1 or, -log₁₀[H₃O+] >7 or, pH> 7.

At 25 °C, Forneutralaqueoussolution: pH = 7

Dynamic Equilibrium Concept

For acidic aqueous solution: pH< 7

Dynamic Equilibrium Concept

The pH of a solution of a weak acid and that of a weak base

Weak acids or weak bases ionize partially in their aqueous solutions. So, the concentration of H₃O+ or OH- ions in an aqueous solution of a weak acid or a weak base cannot be determined directly from their initial concentrations.

To determine the pH (or pOH) of a weak acid (or weak base), the initial concentration of the acid (or base) as well as the degree. of ionization of the add (or base) or ionization constant of the acid (or base) should be known.

Dynamic Equilibrium Concept

Determination of a solution of a weak monobasic acid: Let a weak monobasic acid be HA. In an aqueous solution,

Hapartiallyionises to establish the given equilibrium: HA(a<7) + H₂O(1) =H₃O+(a<7) + A-(aq) If the initial concentration of HA = c (M), its degree of ionisation at equilibrium = a and its ionization constant

⇒ \(=K_a \text {, then }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\alpha c=\sqrt{\frac{K_a}{c}} \times c=\sqrt{c \times K_a}\)

∴ For HA, pH = -log₁₀[H₂O+] = -log₁₀(orc)

⇒ \(\text { or, } p H=-\log _{10}\left(c \times K_a\right)^{1 / 2}=-\frac{1}{2} \log _{10} K_a-\frac{1}{2} \log c\)

∴ \(p H=\frac{1}{2} p K_a-\frac{1}{2} \log c\)

Dynamic Equilibrium Concept

Thus, if the initial concentration (c) of the solution and the degree of ionization (or) of the acid are known, the pH of the solution can be determined by applying equation (1) Or, from the knowledge of the initial concentration (c) of the solution and the ionization constant (Ka) of the acid at the experimental temperature, it is possible to determine the of that solution with the help of equation [2]

Since the ionization constant of a weak acid (or base) is very small similar to the concentration of H30+ (or OH-) ions in very dilute solutions, the ionization constant can also be expressed in terms of ‘p’ pKa = -log₁₀Ka and pKb = -log₁₀Kb.

So, a smaller value of Ka (or Kb) corresponds to a large value of pKa (or pKb) and vice-versa. The stronger an acid, the larger its Ka, and hence the smaller its pKa.

For this reason, between two weak acids, the one with a smaller value of pKa is stronger than the other in water. Similarly, between two weak bases, the one with a small value of pKb is stronger than the other in water.

Dynamic Equilibrium Concept

Determination of pH of a solution of a weak monoacidic base: Let B be a weak monoacidic base. In an aqueous solution, B reacts with water and forms the following equilibrium.

⇒ \(\mathrm{B}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{BH}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

In aqueous solution \(\left[\mathrm{OH}^{-}\right]=\sqrt{c \times K_b}\)

∴ \(-\log _{10}\left[\mathrm{OH}^{-}\right]=-\log _{10}\left(c \times K_b\right)^{1 / 2}\)

⇒ \(\text { or, } p O H=-\frac{1}{2} \log _{10} K_b-\frac{1}{2} \log c\)

∴ \(p O H=\frac{1}{2} p K_b-\frac{1}{2} \log c\)

Dynamic Equilibrium Concept

Hence, if we know the initial concentration (c) of the weak monoacidic base in the solution and its ionization constant (Kb) at the experimental temperature, then we can calculate the pOH of the solution by applying equation (1).

Now, pH+ pOH = 14 [at 25°C]

∴ For a solution of weak monoacidic base, \(p H=14-p O H=14-\left(\frac{1}{2} p K_b-\frac{1}{2} \log c\right)\)

∴ \(p H=14-\frac{1}{2} p K_b+\frac{1}{2} \log c\)

Therefore, by putting the values of pKb and concentration (c) of the solution in equation (2), the pH of the solution can be determined.

Dynamic Equilibrium Concept

pH (or pOH) of an aqueous solution of acid (or base) having concentration <10-7(m)

It is apparent that for an aqueous solution of 10-7(M)HC1, pH = 7, and an aqueous solution of 0-8(M) NaOH, pOH = and i.e., pH = 14 -pOH = 14-8 = 6. However, these values are not acceptable because the pH of an acidic solution and the pOH of a basic solution are always less than 7. Similarly, the pOH of an acidic solution and the pH of a basic solution are always greater than 7.

Generally, in the calculation of the pH or pOH of an aqueous acidic or basic solution, the concentration of H₃O+ or OH- ions produced by the ionization of water is considered to be negligible. However, we cannot neglect them when the concentration of the acidic or basic solutions is very small [<10-7 (M)].

Dynamic Equilibrium Concept

The total concentration of HgO+ ions in a very dilute aqueous acid solution = the concentration of H₆O+ ions produced by the ionization of acid + the concentration of H₆O+ ions produced by ionization of water. If the acid solution is calculated by using this total concentration of H₃O+, then the value of pH is always found to be less than 7.

The total concentration of OH- ions in a very dilute aqueous base solution is the concentration of OH- ions produced by the ionization of base + the concentration of OH- ions produced by the ionization of water.

Dynamic Equilibrium Concept

If pOH of the baste solution is calculated by using this total concentration of OH-, then the value of pOH is always found to be less than 7.

Numerical Examples

Determine the pH of the following solutions:

1. 0.01(N)
2. HC1
3. 0 0.05(M) H₂SO₄
4. 0 0.001(N)
5. H₂SO₄.

Answer: 0.01(N) HC1 = 0.01(M)I-IC1 solution HC1 is a monobasic acid] In 0.01(M) [since HCl is a monobasic acid]

[since 1 molecule of HC1 ionizes to give a single H₃O+ ion] In case of an aqueous 0.01(M) HC1 solution, pH = -log₁₀[H₃O+] = -log₁₀(0.01) = 2.0

In 0.05(M) H₂SO₄ solution, [H30+]= 2 X 0.05= 0.1(M)  [since Each H₂SO₄ molecule ionises to give two H₃0+ ions] For an aqueous 0.05(M)H2S04 solution,

pH = -log10[H₃O+] = -log10(0.1) = 1.0

Dynamic Equilibrium Concept

In 0.001(N) H₂SO4 solution, [H₃O+] = 0.001(M) In case of an aqueous 0.001(N) H₂SO₄ solution,
pH = -log₁₀[H₃O+] = -log₁₀(0.001) = 3.0

Question 2. Determine the pH of the following solutions: 0.1(N)NaOH 0 0.005(M) Ca(OH)
Answer: 0.1(N)NaOH = 0.1(M)NaOH [NaOH is amino acid base]

In 0.1(M)NaOH solution, [OH ] = 0.1(M)

∴ For an aqueous 0.1(M)NaOH solution pOH = -log10[OH-] = -log10(0.1) = 1.0

pH = 14-pOH = 14-1 = 13

Dynamic Equilibrium Concept

In case of 0.005(M) Ca(OH)2 solution, [OH-] = 2X0.005 = 0.01(M)

[since Each Ca(OH)₂ molecule ionizes to give 2 OH- ions]

∴ In case of 0.005(M) Ca(OH)₂ solution,

pOH = -log₁₀[OH-] – -log10(0.01) = 2

∴ pH = 14- pOH = 14- 2 = 12

Dynamic Equilibrium Concept

Question 3. Calculate the concentrations of HaO+ and OH- ions in the solutions with the following pH values at 25 °C. 0 pH = 5.0 pH = 12
Answer: In the case of a solution with pH = 5,

[H₃O+1 = 10-PH(M) = 10-5(M)

∴ In this solution,

⇒ \(\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}=\frac{10^{-14}}{10^{-5}}(\mathrm{M})=10^{-9}(\mathrm{M})\)

In case of a solution with pH – 12 [H₃O+] = 10-(M) = 10-12(M)

Dynamic Equilibrium Concept

∴ In the solution \(\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}=\frac{10^{-14}}{10^{-12}}(\mathrm{M})=10^{-2}(\mathrm{M})\)

Question 4. Calculate the amount of Ca(OH)₂ required to be dissolved to prepare 250mL aqueous solution of pH- 12.
Answer: As given in the question, the pH of the solution =12.

∴ pOH = 14- 12 = 2

∴ In the solution, [OH-] = 10-POH = 10-2(M)

Common Ion Effect On The Ionisation Of Weak Acids And Weak Bases

In a solution of two electrolytes, the ion which is common to both electrolytes, is called the common ion.

Example: Acetic acid on its partial ionization forms CH₃COO-(aq) and H₃O+(aq), and sodium acetate on its complete ionization forms CH₃COO-(aq) and Na+(aq). As the CH₃COO-(aq) ion is common to both CH₃COOH and CH₃COONa, it is a common ion in this system.

Dynamic Equilibrium Concept

Common Ion Effect When a strong electrolyte is added to a solution of a weak electrolyte having an ion common with a strong electrolyte, the extent of ionization of the weak electrolyte decreases. This phenomenon is called the common ion effect.

Common ion effect on ionization of a weak acid:

Effect of common anion: Acetic acid is a weak acid. It partially ionizes in its aqueous solution, forming the following equilibrium:

CH₃COOH(aq) + H₂O(Z)⇌HgO+(aq) + CH₃COO-(aq)

If CH₃COONa is added to this solution, then CH₃COONa, being a strong electrolyte, ionizes almost completely into Na+ and CH₃COO- ions (common ion) in the solution. Consequently, the equilibrium involved in the ionization of acetic acid gets disturbed.

So, according to Le Chatelier’s principle, some of the CH₃COO- ions combine with an equal number of H₃O+ ions to form the unionized CH₃COOH and H₂O molecules, thereby causing the equilibrium to shift to the left. As a result, the degree of ionization of CH₃COOH and the concentration of H₃O+ ions in the solution decreases. This leads to increased pH of the solution.

Dynamic Equilibrium Concept

⇒ \(\begin{aligned}
& \text { Common ion } \\
& \mathrm{CH}_3 \mathrm{COONa}(a q) \longrightarrow \begin{array}{l}
\mathrm{CH}_3 \mathrm{COO}^{-}(a q) \\
\mathrm{CH}_3 \mathrm{COO}^{-}(a q)
\end{array}+\mathrm{Na}^{+}(a q) \\
&+\mathrm{H}_3 \mathrm{O}^{+}(a q)
\end{aligned}\)

Effect of common cation: When a strong acid such as HC1 is added to a solution of acetic acid, it almost completely ionizes into H₃O+ and Cl- ions. The complete dissociation of HC1 increases the concentration of H₃O+(ag) ions (common ion) in the solution. As a result, the equilibrium formed by the ionization of CH₆COOH gets disturbed.

According to Le principle, some of the H₃O+ ions combine with an equal number of CH₆COO- ions to form unionized CH₆COOH and H₂O molecules. As a result, the equilibrium shifts to the left, causing a decrease in the degree of ionization of CH₃COOH.

Dynamic Equilibrium Concept

⇒ \(\begin{gathered}
\mathrm{HCl}(a q)+\mathrm{H}_2 \mathrm{O}(l) \\
\mathrm{CH}_3 \mathrm{COOH}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathbf{H}_3 \mathbf{O}^{+}(a q)+\mathrm{Cl}^{-}(a q) \\
\mathbf{H}_3 \mathbf{O}^{+}(a q)+\mathrm{CH}_3 \mathrm{COO}^{-}(a q)
\end{gathered}\)

Common ion effect on ionization of a weak base: Effect of common cation: Ammonia (NH₃) is a weak base. In aqueous solution, NH₃ reacts with water to establish the following equilibrium:

⇒ \(\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q)\)

If a strong electrolyte such as, NH₄C1 is added to this solution, it almost completely ionizes to form NH+4(a<7) and Cl-(aq). The complete ionization of NH₄C1 gives rise to a high concentration of NH₄(a₃) ions (common ions) in the solution. As a result, the equilibrium involved in the ionization of NH₃(ag) gets disturbed.

According to Le Chatelier’s principle, some NH₃(aq) ions combine with an equal number of OH- ions to form unionized NH₃ and H₂O molecules and thereby cause the equilibrium to shift to the left. Consequently, the degree of ionization of NH3 as well as the concentration of OH- ions in the solution decreases. This results in a decrease in the pH of the solution.

⇒ \(\mathrm{NH}_4 \mathrm{Cl}(a q) \longrightarrow \mathrm{NH}_4^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

Dynamic Equilibrium Concept

⇒ \(\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Effect of common anion: When NaOH (a strong base) is added to an aqueous solution of NH₃, it almost completely ionizes into Na+ and OH- ions. This increases the concentration of OH- ions (common ions) in the solution.

As a result, the equilibrium involved in the ionization of NH₃ gets disturbed. According to Le Chatelier’s principle, some OH- ions combine with an equal number of NH- ions to form unionized NH₃ and H₂O molecules, thereby causing the equilibrium to shift
to the left. This results in a decrease in the degree of ionization of NH3.

Dynamic Equilibrium Concept

⇒ \(\begin{gathered}
\text { Common ion } \\
\mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathbf{O H}^{-}(a q)
\end{gathered}\)

⇒ \(\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathbf{O H}^{-}(a q)\)

Hydrolysis Of Salts

A normal salt (Example; NaCl, KC1, Na₂SO4 CH₃COONa, NH₄C1 ) does not contain any ionizable H-atom or OH- ion. The solutions of normal salts are therefore expected to be neutral as they are formed by the complete neutralization of an acid. and a base. However, the aqueous solutions of many normal salts are found to be acidic or basic.

Dynamic Equilibrium Concept

Example: Aqueous solutions of NH₄C1, FeCl₃, A1C1₃, etc., acidic, and those of CH₆COONa, NaF, and KCN are basic. This is because the cations or the anions produced by the dissociation of these salts react partially with water to produce an H₃O+ or OH- ions solution. This increases the concentration of H₃O+ or OH- ions in the solution. As a result, the solutions become acidic or basic. Such a phenomenon is known as hydrolysis.

Definition: The process in which the cations or the anions or both of a normal salt in its aqueous solution react with water to furnish H₃O+ or OH- ions, thus making the solution acidic or alkaline is known as hydrolysis of the salt.

Types of normal salts that undergo hydrolysis

A salt forms due to the reaction of an acid with a base. An acid or a base may be strong or weak. Four different normal salts are possible depending upon the nature of acids and bases involved in their formations.

Dynamic Equilibrium Concept

Among these salts, those produced by reactions of strong acids and strong bases do not undergo hydrolysis. So, if the acid and the base that react to form a salt are weak or one of them is weak, then the salt formed will undergo hydrolysis.

Consequently, aqueous solutions of these salts are either acidic (pH < 7) or basic (pH > 7).On the other hand, an aqueous solution of a salt strong acid, and strong base is always neutral (pH = 7).

Hydrolysis of salts obtained from strong acids and strong bases

A salt of this type does not undergo hydrolysis in its aqueous solution because neither its cation nor its anion reacts with water.

As a result, the concentration of H+ ions or OH ions in the solution is not affected; that is, the equality of their concentrations is not disturbed. This is why an aqueous solution of a salt derived from a strong acid and a strong base is neutral {pH = 7).

Dynamic Equilibrium Concept

Explanation: NaCl is a salt of strong acid (HC1) and strong base (NaOH). NaCl dissociates almost completely In aqueous solution to produce Na+(ag) and Cl {aq) ions [NaCl(a<7)-Na+(ag) + cr(ag)].

Water also ionizes slightly to produce an equal number of H30+(aq) and OH-(aq) ions [2H₂O(Z) s=± H₃O+{aq) + OH-{aq)].

Na+(aq) is a very weak Bronsted acid and it is unable to react with H20 to produce proton: Na+{aq) + 2H₂O(Z) NaOH(aq) + H₃O+{aq).

On the other hand, Cl- ion is the conjugate base of strong acid HC1. Hence, it is a very weak Bronsted base. For this reason, Cl- is unable to react with water to produce OH- ions [Cl-{aq) + H₂O(Z) HC1{aq) + OH-{aq)].

Dynamic Equilibrium Concept

Thus, an aqueous solution of NaCl contains H₃O+ and OH- ions in equal concentration. As a result, the aqueous solution of NaCl is neutral. For the same reason, other salts obtained from strong acids and strong bases form neutral aqueous solutions.

Hydrolysis of salts obtained from weak acids and strong bases

A salt of this type undergoes hydrolysis in its aqueous solution as its anion reacts with water to form OH- ions. As a result, the concentration of OH- in the solution becomes higher than that of H3O+, thereby making the solution basic (H > 7). Since anion of such a salt takes part in hydrolysis, this type of hydrolysis is called anionic hydrolysis.

Explanation: KCN is a salt-weak acid (HCN) and strong base (KOH)

Dynamic Equilibrium Concept

It dissociates almost completely in its aqueous solution and produces K+{aq) and CN-{aq) ions [KCN(a<7)⇌K+(ag) + CN-(ag)]. Water also ionises slightly to produce equal number of H₃O+(aq) and OH-{aq) ions \(\mathrm{K}^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{KOH}(a q)+\mathrm{H}^{+}(a q)\)  K+(aq) is a very weak Bronsted acid.

Dynamic Equilibrium Concept

So, it cannot react with to produce proton: K+(aq) +H₂O KOH(aq) + H+(aq).

On the other hand, CN- ion is the conjugate base of a weak acid (HCN). Hence, it has sufficient basic character to abstract a proton from an H₂O molecule to form OH-ions: [CN-(aq) + H₂O(l) ⇌ HCN(ag) + OH-{aq)]

The formation of OH- ions increases the concentration of OH- ions in the solution and makes the solution basic (pH > 7). For the same reason, other salts obtained from weak acids and strong bases form basic aqueous solutions.

Dynamic Equilibrium Concept

Hydrolysis of salts obtained from strong acids and weak bases

A salt of this type undergoes hydrolysis in its aqueous solution because its cation reacts with water to form H₃O+ ions. As a result, the concentration of H₃O+ ions in the solution becomes higher than that of OH- ions. This makes the solution acidic (pH < 7). Since the cation of such a salt takes part in hydrolysis, this type of hydrolysis is called cationic hydrolysis.

Hydrolysis of salts obtained from strong acids and weak bases Explanation: Ammonium chloride (NH₄C1) is formed by the reaction of HC1 (a strong acid) with NH3 (a weak base).  NH4C1 almost completely dissociates in its aqueous solution to produce,NH+(nq) and Cl-{aq) ions [NH4Cl(aq)→NH+(aq) + cr(aq)]. H₂O also ionislightly to produce equal number of  \(\left[2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\right] .\)

Dynamic Equilibrium Concept

Cl- ion is the conjugate. the base of strong acid:  and hence, is a very weak Bronsted base. So, Cl- ion fails to react with H2O to produce OH- ions \(\left[\mathrm{Cl}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HCl}(a q)+\mathrm{OH}^{-}(a q)\right]\)

NH+ ion is the conjugate acid of a weak base, NH₃, and has sufficient acidic character to donate a proton to H₂O molecules. Thus, NH+4 ion reacts with water to form unionised molecules of NH₃ and H₃O+ ions \(\left[\mathrm{NH}_4^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_3(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)\right]\)

The formation of HaO+ ions increases the concentration of HgO+ ions in the solution and makes the solution acidic (pH < 7). For the same reason, other salts obtained from
strong acids & weak bases form acidic aqueous solutions.

Dynamic Equilibrium Concept

Aqueous solution of FeCI₃ [or Fe(NO₂)g] is acidic: Being a strong electrolyte, FeCl₃ undergoes complete dissociation in the solution to form [Fe(H₂O)g]3+(oÿ) & Cr(aq): eCl₃(aq) + 6H₂O(Z) [Fe(H₂O)g]3+(at₂) + 3Cl- Water also ionises slightly to produce equal number of H₃O+ ions and OU-(aq) ions [2H₃O(Z) H₃O+(aq) + OH-{aq)] . Cl- ion is the conjugate base of strong acid (HC₁).

Hence, it is a very weak Bronsted base and fails { to react with water in aqueous solution. Due to the small size and higher charge of Fe3+ ion, its charge density is very high. As a result, the H₂O molecules bonded to the Fe3+ ion are polarised and their O—H bonds become very weak.

Dynamic Equilibrium Concept

These O—H bonds are easily dissociated to produce H+ ions, which are accepted by H₂0 molecules to form H₃O+ ions. [Fe(H₂O)6]3+(a<7) + H₂0(Z) [Fe(H₂0)50H]+ H₃O+(aq) Among the H₂O molecules attached to the Fe3+ ion, the one which releases proton finally gets attached to the Fe3+ ion as OH- ion. For the same reason, aqueous solutions of AIGb₃, CuSO₃, etc., are acidic. A1C1₃→ + 6H₂O(Z) [Al(H₂O)6]₃+→ + 3Cl- [A1(H₂O)6]₃+H₂O(Z) [A1(H₂O)5OH]₃+H3O+(aq)

Hydrolysis of salts obtained from weak acids and weak bases

In the case of a salt of this type, both the cation and anion react with water. In reaction with water, the cation forms H₃O+ ions, and the anion forms OH- ions. Aqueous solutions of such a salPipay be acidic, basic, or neutral, depending upon the relative acid strength of the cation and base strength of anion. If the strengths of the cation and anions are the same, the solution will be neutral.

On the other hand, if the strength of the cation is greater or less than that of the anion, then the solution will be acidic or basic. When the aqueous solution of the salt is neutral: CH₃COONH₄ is a salt of weak acid, CH₃COOH, and a weak base, NH₃.

Dynamic Equilibrium Concept

Explanation: Being a strong electrolyte, CH₃COONH₄ dissociates almost completely in aqueous solution to produce NH₄ and CH₃COO-(aq) ions:

CH₃COONH₄(aq)→NH+(a<7)]. H₂O, a weak electrolyte, also ionizes slightly in solution to produce an equal number of U₃0+(aq) and OH-(aq) ions [2H₂0(l) H₃0+(a<7) + OH-(aq)].

NH+ ion is the conjugate acid of a weak base (NH₃) and CH₃COO- is the conjugate base of a weak acid, (CH₃COOH). In an aqueous solution, both NH₄ and CH₃COO- are stronger than H₂O (which can act both as a weak Bronsted acid and base). As a result, CH₃COO- and NH₃ ions react with water to establish the following equilibria:

⇒ \(\begin{aligned}
\mathrm{NH}_4^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) & \rightleftharpoons \mathrm{NH}_3(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q) \\
\mathrm{CH}_3 \mathrm{COO}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) & \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

Dynamic Equilibrium Concept

At ordinary temperature, both CH₃COO- and NH₃ ions have the same value of dissociation constants. As a result, they get hydrolyzed to the same extent in aqueous solution.

Therefore, the aqueous solution of CH₃COONH₄ contains equal concentrations of H₃O+ (produced by the hydrolysis of NH₄ ions) and OH- (produced by the hydrolysis of CH₃COO- ions), and hence, the solution is neutral.

Dynamic Equilibrium Concept

When the aqueous solution of the salt is acidic: Ammonium formate (HCOONH₄ strong electrolyte) is a salt of weak acid, HCOOH, and weakbase, NH3. Explanation:

HCOONH₄, dissociates almost completely in its aqueous solution to form HC₃CT(ag) and NH₃[HCOONH₄(a<7H>HCOO-(a<7) + NH+(a<7)].

H₂O, a weak electrolyte, also ionizes slightly to form an equal number of H₃O+{aq) and OH-(aq) ions. NH₄ and HCOO- react with water to form the given equilibria:

⇒ \(\begin{aligned}
\mathrm{NH}_4^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) & \rightleftharpoons \mathrm{NH}_3(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q) \\
\mathrm{HCOO}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) & \rightleftharpoons \mathrm{HCOOH}(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

At ordinary temperature, Ka for NH+4 is larger than Kb for HCOO- i.e., in aqueous solution NH- hydrolyses to a greater extent than HCOO-. The concentration of H₃O+ is more than that of OH-. So, the aqueous solution of HCOONH₄ is acidic.

Dynamic Equilibrium Concept

When the aqueous solution of the salt is basic: Ammonium bicarbonate, NH4HCO₃ is a salt of a weak acid, H₂CO₃, and weak base, NH₃.

Explanation: strong electrolyte, NH₄HCO₃ dissociates almost completely solution to form NH+ and HCO₃– ions:,\(\left[\mathrm{NH}_4 \mathrm{HCO}_3(a q) \rightarrow \mathrm{NH}_4^{+}(a q)+\mathrm{HCO}_3^{-}(a q)\right] \cdot \mathrm{H}_2 \mathrm{O}\) a weak electrolyte, also Ionises slightly to form equal number of HgO + and OH→ 7) ions [2H2O(l)x=i H30+) + OH~(aq) ]. NH3 and HCO3 ions react with water to establish die following equilibria.

⇒ \(\begin{aligned}
\mathrm{NH}_4^{+}(a q)+\mathrm{H}_2 \mathrm{O}(l) & \rightleftharpoons \mathrm{NH}_3(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q) \\
\mathrm{HCO}_3^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) & \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3(a q)+\mathrm{OH}^{-}(a q)
\end{aligned}\)

Dynamic Equilibrium Concept

At ordinary temperature, Kb for HCO₄ Is much greater than Ka for NH+. This means that in an aqueous solution, HCO₃ ions hydrolyze to a greater extent than NH- ions. So, the concentration of OH- ions is more than that of H₃O 1 ions 111 an aqueous solution of NHhCO-. Thus, the aqueous solution of NH₂HCO, Is music in nature.

Hydrolysis constant, degree of hydrolysis, and pH of an aqueous solution of a salt

Hydrolysis constant: In the hydrolysis of a salt, u dynamic equilibrium is established Involving the unhydrolyzed salt and the species formed by hydrolysis. The equilibrium constant corresponding to this equilibrium Is called the hydrolysis constant.

Dynamic Equilibrium Concept

Degree of hydrolysis: The degree of hydrolysis (ft) of a salt may be defined as the fraction of the total number of moles of that salt hydrolyzed in its aqueous solution at equilibrium. In an aqueous solution, the degree of hydrolysis of salt is ft— it means that out of 1 mol of the salt dissolved in the solution, ft mol undergoes hydrolysis.

Relation Between Ionisation Constants Of Conjugate Acid-Base

Let’s consider a weak acid, HA. The conjugate base of this acid is A-. Hence, the conjugate acid-base pair is (HA, A-).

In an aqueous solution, HA and A- form the following equilibrium: \(\mathrm{HA}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{A}^{-}(a q)\)

The ionisation constant of HA, Ka \(=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \quad \cdots[1]\)

In aqueous solution, the base A- reacts with water to form the following equilibrium:

A-(aq) + H₂O(l)⇌ HA(aq) + OH-(aq)

Dynamic Equilibrium Concept

The ionisation constant of \(\mathrm{A}^{-}, K_b=\frac{[\mathrm{HA}] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{A}^{-}\right]}\)

Multiplying equations (1) and (2), we have,

KaxKb= [H₃O+]x[OH-]

Again, Kw = [H₃O+] X [OH-]

Dynamic Equilibrium Concept

∴ Ka x Kb = Kw

Equation (3) represents the relation between ionization constants of a conjugate acid-base. This equation applies to any conjugate acid-base pair in aqueous solutions

at 25°C, Kw = 10-14

Therefore, at this temperature, Ka X Kb = 10-14]

Therefore, at a given temperature, we know the value of Kw and the ionization constant of any one member of the conjugate acid-base pair, then the ionization constant of the other can be determined by applying equation (4)

Dynamic Equilibrium Concept

Examples: 1 At 25°C, Ka (CH₃COOH) = 1.8 x 10-5

So, at 25°C, Kb (CH₃COO- ) \(=\frac{10^{-14}}{K_a}=\frac{10^{-14}}{1.8 \times 10^{-5}}\)

At 25°C, Kb (NH₃) = 1.8 x lO-5

Dynamic Equilibrium Concept

\(\text { So, at } 25^{\circ} \mathrm{C}, K_a\left(\mathrm{NH}_4^{+}\right)=\frac{10^{-14}}{K_a}=\frac{10^{-14}}{1.8 \times 10^{-5}}=5.5 \times 10^{-10}\)

Derivation of the relation, pKa + pKb = pKw: We know that, Ka x Kb = Kw. Taking negative logarithm on both sides

we have, -log10 (Ka x Kb) = -log10 Kw

or, -log₁₀ Ka- log₁₀  Kb = -log₁₀ Kw

∴ PK_a+PK_b = PK_w

Dynamic Equilibrium Concept

At 25°C, pK_w = 14. So, at 25°C, pK_a + pK_b = 14

Hence, if the pKa of the acid (or pKb of the base) in a conjugate acid-base pair is known, then the pKb of the base (or pKa of the acid) can be determined by using equation (5)

Dynamic Equilibrium Concept Buffer Solutions

At ordinary temperature, the pH of pure water is 7. Now, when 1 mL of (M) HC1 solution is added to 100 mL of pure water, it is observed that the value of pH decreases from 7 to 2. Again, if 1 mL of 1 (M) NaOH solution is added to 100 mL of pure water the value of the pH of the solution increases from 7 to 2. However, many solutions resist the change in pH even when a small quantity of acid or base is added to them. Such solutions are called buffer solutions.

Buffer Solutions Definition: A buffer solution may be defined as a solution that resists the change in its pH when a small amount of acid or base is added to it.

Various types of buffer solutions depending on the nature of the acid/base and the salt:

Solution of a weak acid and its salt: An aqueous solution of a weak acid and its salt acts as a buffer solution. Example: Aqueous solutions of CH₃COOH and CH₃COONa, H₂CO₃ and NaHCO₃, citric acid and sodium citrate, boric acid, and sodium borate, etc.

Dynamic Equilibrium Concept

Solution of a weak base and its salt: An aqueous solution of a weak base and its salt acts as a buffer solution. Example: Aqueous solutions of NH₃ and NH₄C1, aniline and anilinium hydrochloride, etc.

Solution of two salts of a polybasic acid: An aqueous solution containing two salts of polybasic acid can act as a buffer solution.

Dynamic Equilibrium Concept

Example: Aqueous solutions of Na₂CO₃ and NaHCO₃ (NaHCO₃ is a weak acid and Na₂CO₃ is its salt), NaIT₂PO₄ and Na₂HPO₄ etc.

Solution of a salt derived from a weak acid and a weak base: An aqueous solution of a salt formed by a weak acid and a weak base can function as a buffer solution.

Example: CH₃COONH₄ is a salt of weak acid (CH₃COOH) and weak base (NH₄OH). A solution of this salt acts as a buffer solution.

Dynamic Equilibrium Concept

Types of buffer solutions depending on their pH values: Depending on pH values, buffer solutions are of two types—

Acidic buffer: Buffer solutions with a pH lower than 7 are called acidic buffers. Aqueous solutions of CH₃COOH and CH₃COONa, aqueous solutions of lactic acid and sodium lactate, etc., are some common examples of acidic buffer solutions.

Basic buffer: Buffer solutions with a pH higher than 7 are called basic buffers. An aqueous solution of NIT₄O11 and NH₄C1 is an example of a basic buffer solution.

Mechanism of buffer action

Mechanism of buffer action Definition: The ability of a buffer solution to resist the change of its pH value on the addition of a small amount of an acid or a base to it is called buffer action.

Mechanism of action of an acidic buffer: To understand the mechanism of buffer action of an acidic buffer, let us consider an acidic buffer solution that consists of CH₃COOH and CH₃COONa.

Dynamic Equilibrium Concept

In the solution, CH₃COONa almost completely dissociates into CH₃COO_(at/) and Na+ ions, whereas acetic acid, being a weak electrolyte, partially dissociates into CH₃COO-(£7<7) and H30+(flt/). The partial dissociation of acetic acid leads to the formation of the following equilibrium.

⇒ \(\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH}(a q)+ \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \\
& \mathrm{CH}_3 \mathrm{COO}^{-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)
\end{aligned}\)

As CH₃COOH ionizes partially and CH₃COONa ionizes almost completely, the solution consists of high concentrations of CH₃COOH and CH₃COO- ions

Addition of a small amount of strong acid: If a small amount of strong acid (e.g., HC1 ) is added to this buffer solution, H₃O+ ions produced from the strong acid combine with an equal number of CH₃COO- ions present in the solution to form unionized CH₃COOH molecules: CH₃COO→+ H₃O+(l)→CH₃COOH(l) + H₂O(l)

This causes the equilibrium involved in the ionization of CH₃COOH (equation 1) to shift to the left. As a result, the concentration of H₃O+ Ionsin in the solution virtually remains the same, i.e., the pH of the solution remains almost unchanged.

Addition of a small amount of strong base: When a small amount of a strong base like NaOH, KOIT, etc., is added to this buffer solution, the OH- ions obtained from the strong base combine with an equal number of H₄,f ions (which results mainly from the dissociation of CH₃COOH ) present in the solution, producing unionized water molecules.

Consequently, the equilibrium associated with the ionization of CH₃COOH [eqn. (1)] gets disturbed. So, according to Le Chatcller’s principle, some molecules of CH₃COOH ionize to produce H₃0- ions, and the equilibrium, shifts towards the right. Therefore, the addition of a small quantity of a strong base with almost no change in are to the solution causes the concentration of either OH- ions or H₂O+ ions, and consequently pH of the solution remains almost unaltered.

Mechanism of action of basic buffer: Let us consider a basic buffer solution consisting of a weak base NH₃ and its salt, NH₄C1. Being a strong electrolyte, ammonium chloride (NH4C1) dissociates almost completely in solution: NH4Cl(a<7)→ NH₂(ar7) + Cl-(aq). However, NH₃, being a weak base, reacts partially with water and forms the following equilibrium solution.

⇒ \(\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q)\)

As NH3 reacts partially with water and NH4C1 gets completely dissociated, the concentration of NH3 and NH3 in the solution is sufficient.

Addition of a small amount of strong acid: If a small quantity of strong acid (for example HC1, H₂SO₄) is added to this buffer solution, H₃O+ ions obtained from the strong acid react almost completely with an equal number of OH- ions (which results mainly from the reactions of NH₃ with water) present in the solution and produce unionized water molecules.

As a result, the equilibrium [eqn. (1)] gets disturbed. According to Le Chatelier’s principle, some NH3 molecules react with water to form OH- ions. As a result, the equilibrium [eqn. (1)] shifts to the right. So, the addition of a small amount of a strong acid to the buffer solution causes almost no change in concentration of either OH- ions or HsO+ ions, i.e., the pH of the solution remains unchanged.

Addition of a small amount of strong base: When a small quantity of a strong base (e.g., NaOH, KOH is added to the buffer solution, OH- ions produced by the strong base react almost completely with an equal number of NHJ ions present in the solution and form unionized NH3 molecules[NH|(ag) + OU~(aq) – NH3(l) + H2O(Z)].

As a result the equilibrium [eqn. (1)] shifts to the left. Consequently, the concentration of OH- ions in the solution virtually remains the same, and hence pH of dissolution remains unchanged.

Buffer action of two salts of polybasic acid (in solution): Let us consider the buffer solution comprising of two salts NaH₂PO₄ and Na₂HPO₄ (which are the salts of polybasic acid, H₃PO₄ ). In this solution, NaH₂PO₄ acts as an acid while Na₂HPO₄ as a salt, and both of them undergo complete dissociation as given below:

⇒ \(\begin{aligned}
& \mathrm{NaH}_2 \mathrm{PO}_4(a q) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{H}_2 \mathrm{PO}_4^{-}(a q) \\
& \mathrm{Na}_2 \mathrm{HPO}_4(a q) \rightarrow 2 \mathrm{Na}^{+}(a q)+\mathrm{HPO}_4^{2-}(a q)
\end{aligned}\)

H2PO4 itself is a weak acid and due to a common ion (HPO-), it undergoes slight ionization in the solution forming the following equilibrium:

⇒ \(\mathrm{H}_2 \mathrm{PO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HPO}_4^{2-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q) \cdots[1]\)

As H₂PO₄ undergoes slight ionization and Na2HP04 gets completely ionized, the concentration of H₂PO₄ and HPO4- in the solution is sufficient.

Addition of a small amount of strong acid: When a small quantity of a strong acid is added to this buffer solution, H₃O+ ions produced by the strong acid react almost completely with an equal number of HPO- ions and form unionized H₂PO₄ ions

⇒ \(\mathrm{HPO}_4^{2-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q) \rightarrow \mathrm{H}_2 \mathrm{PO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

As a result, the equilibrium [eqn. (1)] shifts towards the left. Hence, the effect of the addition of a small amount of strong acid is neutralized. Thus, the pH of the solution remains unchanged.

Addition of a small amount of strong base: When a small quantity of a strong base is added to the buffer solution, OH- ions from the added base almost completely react with an equal number of H₃O+ ions (which results mainly from the ionization of H₂PO₄ ) present in the solution and form unionized water molecules, This disturbs the ionization equilibrium of NaH₂PO₄.

As a result, according to Le Chatelier’s principle, some molecules of NaH₂PO₄ ionize to form H3O+ ions and cause the equilibrium to shift towards the right. Therefore, the addition of a small quantity of a strong base makes no change in the concentration of either OH- ions or H₃O+ ions, and consequently, the pH of the solution remains almost unchanged.

Determination of pH of a buffer solution: Henderson’s equation

Let us consider a buffer solution consisting of a weak acid (HA) and its salt (MA). In the solution, MA dissociates completely, forming M+(aq) and A-(aq) ions, while the weak acid, HA, because of its partial ionization, forms the following equilibrium \(\mathrm{HA}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{A}+(a q)\) Applying the law of mass action to the abbveÿequilibrium, we have ionization constant of HA \(K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]},\)

Where, [H3O+], [A-], and [HA] are the molar concentrations of H₃O+, A-, and HA respectively at equilibrium.

∴ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=K_a \times \frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}\)

Taking negative logarithms on both sides we get,

⇒ \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10} K_a-\log _{10} \frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}\)

⇒ \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10} K_a-\log _{10} \frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}\)

Or, ,\(p H=p K_a+\log _{10} \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)

As HA is a weak acid, its ionization in the solution is very small, which gets further decreased in the presence ofthe common ion, A-. Therefore, the molar concentration of unionized HA can be considered to be the same as the initial molar concentration of HA.

Again, the molar concentration of A- ions produced by complete dissociation of MA is much higher than that of A- ions produced by partial ionization of HA. Therefore, the total molar concentration of A- ions in the solution is almost the same as the initial molar concentration of MA in the solution.

∴ From equation [1] \(p H=p K_a+\log \frac{[\text { salt }]}{[\text { acid }]}\)

where Ka is the ionization constant of the weak acid; [acid] and [salt] are the initial molar concentrations of acid and salt respectively the buffer solution. Equation [2] is called the Henderson’s equation.

It is used to determine the pH of a buffer solution consisting of a weak acid and its salt. Similarly, the equation for determining the pOH of a buffer solution consisting of a weak base and its salt is-

⇒  \(p O H=p K_b+\log \frac{[\text { salt] }}{\text { [base] }}\)

where, K, b is the ionization constant ofthe weak base; [salt] and [base] represent the initial molar concentrations of salt and base respectively. The above equation can be rewritten as,

⇒ \(p H=14-p O H=14-p K_b-\log \frac{[\text { salt] }}{\text { [base] }}\)

At a fixed temperature, the value of pATa of a weak acid is fixed. Hence, at a fixed temperature, the pH of a buffer solution consisting of a weak add and its salt depends upon the die value of the pKa of the weak acid as well as the ratio of the molar concentrations of the salt to the molar concentration of the acid.

Similarly, at a particular temperature, the pH of a buffer solution consisting of a weak base and its salt depends on the value of pKb of the weak base and the ratio of molar concentrations of the salt to the base.

Applications of Henderson’s equation:

If the molar concentrations of a weak acid (or base) and its salt present in a buffer solution as well as the dissociation constant of that acid (or base) are known, then that solution can be determined by using Henderson’s equation.

If in an acidic or basic buffer solution, the molar concentrations of the weak acid (or base) and its salt and pH of that solution are known, then the dissociation constant of the weak acid (or base) can be calculated with the help ofHenderson’s equation.

For the preparation of a buffer solution with a desired value of pH, the ratio of the weak acid and its salt (or weak base and its salt) in the solution can be determined by Henderson’s equation.

Buffer Capacity

Buffer Capacity Definition: The buffer capacity of a buffer solution is defined as the number of moles of a strong base or an acid required to change the pH of 1L of that buffer solution by unity.

When an acid is added to a buffer solution, its pH value decreases, whereas when a base is added to a buffer solution, its pH value increases.

Mathematical explanation: When the pH of 1 L of solution increases by d(pH) due to the addition of db mol of any strong base, then the buffer capacity of that buffer
solution \(\beta=\frac{d b}{d(p H)}\)

Dynamic Equilibrium Concept

Similarly, when the pH of 1L of buffer solution decreases by d(pH) due to the addition of da mol of any acid, then the buffer capacity of that buffer solution \(\beta=\frac{d a}{d(p H)}\)

Some important facts regarding buffer capacity:

1. If the buffer capacity of any buffer solution is high, then a greater amount of a strong add or base will be required to bring about any change in the pH of that solution.
2. Between two buffer solutions having the same components, the buffer capacity of that solution will be higher when the concentrations of the components are
   higher.
3. In a buffer solution, if the difference in molar concentration of die components is small, then the addition of a small amount of strong acid or base causes a small change in molar concentrations of die components.

Consequently, the change in pH of the die solution also becomes small. For this reason, the die buffer capacity of a buffer solution becomes maximum when the molar concentrations ofthe components are the same.

Dynamic Equilibrium Concept

The buffer capacity of a buffer solution consisting of a weak acid and its salt becomes maximum when die molar concentrations of the weak acid and its salt become equal. Under this condition, the pH of the buffer solution = pKa of the weak acid, e.g., the buffer capacity of CH3COOH/CH3COONa solution will be maximum when its pH = 4.74 (since, pKa of CH₃COOH =4.74).

For the same reason, the buffer capacity of a buffer solution consisting of a weak base and its salt becomes maximum when the pOH of the buffer becomes equal to the pKb of the weak base. For example, the buffer capacity of NH₃/NH₄C1 solution will be maximum when its pOH = 4.74 (since, pKb ofNH3 = 4.74)

pH range of buffer capacity: A buffer solution consisting of a weak acid and its salt will work properly only when the ratio of the molar concentration of the salt to the acid lies between 0.1 to 10. Therefore, according to Henderson’s equation, the pH -range for the buffer capacity of such a buffer will vary from (pKa – 1) to (pKa + 1). 0 Similarly the range ofpOH for buffer capacity of a buffer consisting of a weak base and its salt will vary from (pKb – 1) to (pKb+ 1).

Importance of butter solutions

Dynamic Equilibrium Concept

The pH of human blood lies between 7.35 to 7.45, i.e., human blood is slightly alkaline. In spite, ofthe presence of acidic or basic substances produced due to various metabolic processes, the pH of human blood remains almost constant because ofthe presence of different buffer systems like bicarbonate-carbonic acid buffer (HCO₂-4 / H2PO-4), phosphate buffer (HPO2-4 /H2PO-4), etc.

Excess acid in the blood is neutralized by the reaction: HCO-3 + H₃O+⇌ H₂CO3 +H₂O. H₂CO₃ formed decomposes into CO₂ and water. The produced CO₂ is exhaled out through the lungs. Again, if any base (OH-) from an external source enters the blood, it gets neutralized by the reaction: H₂CO₃ + OH- HCO3 +H₂O.

Buffers find extensive use in analytical works as well as in chemical industries. In these cases, a buffer solution is used to maintain pH at a certain value. For instance, in qualitative analysis, in electroplating, tanning of hides and skins, fermentation, manufacture of paper, ink, paints, and dyes, etc., pH is strictly maintained. Buffer solution is also used for the preservation of fruits & products derived from fruits.

Solubility And Solubility Product Dynamic Equilibrium Concept

The terms ‘solubility’ and ‘solubility product’ are often used in the context of the dissolution of a solute in a liquid. It is important to note that these two terms do not have the same meaning and therefore cannot be used interchangeably.

Comparison between solubility and solubility product:

1. The term solubility applies to all kinds of solutes (ionic, neutral, sparingly soluble, or highly soluble) whereas the term solubility product is mainly used for sparingly soluble compounds.
2. The solubility of a solute in a solution may change due to common ion effects or formation of complex salt but its solubility product remains constant at a fixed temperature.
3. Both solubility and solubility products of a solute in a liquid change with the temperature variation.

Solubility products of sparingly soluble salts

Salts having solubilities less than 0.01 mol-L-1 at ordinary temperature are commonly known as sparingly soluble salts. Some examples of such salts are AgCl, BaSO₄, CaSO₄, etc. A salt of this type in its saturated aqueous solution remains virtually insoluble and only a small part of it gets dissolved but the dissolved portion of the salt, however small it may be, gets completely dissociated.

Solubility Product: The Solubility Product of A sparingly soluble salt at a given temperature is defined as the product of the molar concentrations of the constituent ions in its saturated solution, each concentration term raised to a power representing the number of ions of that type produced by dissociation of one formula unit of the salt.

Dynamic Equilibrium Concept

Characteristics of solubility product:

1. At a particular temperature, the solubility product of a sparingly soluble salt has a fixed value. Its value changes with temperature variation.
2. At a particular temperature, the solubility product of a sparingly soluble salt in its saturated solution remains unaltered if the concentrations of the constituent ions are changed. It remains fixed even in the presence of other ions in the solution.

Equation of solubility product: Let’s consider the equilibrium that exists in a saturated solution of the sparingly soluble salt, AxBy. In this solution, undissolved AxBy remains in equilibrium with the ions (A+ and B- ) produced from the dissociation of dissolved AxBy.

⇒ \(\mathrm{A}_x \mathrm{~B}_y(s \text {, undissolved }) \rightleftharpoons x \mathrm{~A}^{y+}(a q)+y \mathrm{~B}^{x-}(a q)\)

Applying the law of mass action we have, equilibrium constant, \(K=\frac{\left[\mathrm{A}^{y+}\right]^x \times\left[\mathrm{B}^{x-}\right]^y}{\left[\mathrm{~A}_x \mathrm{~B}_y(s)\right]}\)

where, [Ay+], [Bx-] and [AxBy(s)] are the molar concentrations of AJ,+, Bx- and AxBy(s) respectively, at equilibrium.

Now, the molar concentration of pure solid at a particular temperature is constant. Again, at a given temperature, the equilibrium constant, K is also constant.

So, at a given temperature, Kx [AxBy(s)] = constant.

Dynamic Equilibrium Concept

∴ Kx[AxBy(s)]=[Ay+]x [Bx-]y

or Ksp = [Ay+]x x [Bx-]y [ sp = solubility product ]

where, Ksp = Kx [AxByO)] = constant.

Ksp is called the solubility product or solubility product constant of the sparingly soluble salt, AxBy.

Dynamic Equilibrium Concept

Relation between solubility and solubility product

In case of a sparingly soluble salt of the type AB: Equilibrium formed by a sparingly soluble salt of the type AB in its saturated aqueous solution:

AB(s)⇌ A+(aq) + B-(aq)

∴ The solubility product of, Ksp = [A+] x [B-]

Let the solubility of the salt, AB, in its saturated aqueous solution at a given temperature be S mol.L-1.

Hence, in the solution, the molar concentration of A+ ions [A+] = S mol.L-1 and that of B- ions, [B-] = S mol-L-1 [v one formula unit of AB on dissociation gives one A+ and one B- ion].

∴ Ksp=[A+][B-]=S x S = S₂

Dynamic Equilibrium Concept

∴ Ksp = s2 ……[1] and S = \(\sqrt{K_{s p}}\)….[2]

Equation [1] is the relation between the solubility the and solubility product of a sparingly soluble salt of the type AB.

Equation [2] indicates that the solubility of a sparingly soluble salt, AB is equal to the square root of its solubilityproduct.

Example: AgCl is a sparingly soluble salt. In a saturated solution of AgCl, the following equilibrium is formed:

⇒ \(\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

So, the solubility product of AgCl, Ksp = [Ag+] x [Cl-] Suppose, at a certain temperature, the solubility of AgCl in its saturated aqueous solution = S mol.L-1. Hence, in the solution, [Ag+] tS mol-L-1 and [Cl-] =S mol-L-1.

∴ Ksp = [Ag+] x [Cl-] = S x S = S2

Dynamic Equilibrium Concept

Similarly, in the case of other sparingly soluble salts ofthe type AB (CaSO4, BaSO4, AgBr, etc.), Ksp = S2

In the case of a sparingly soluble salt of type AB2: The following equilibrium exists in a saturated aqueous solution of a sparingly soluble salt of the type AB2:

⇒ \(\mathrm{AB}_2(s) \rightleftharpoons \mathrm{A}^{2+}(a q)+2 \mathrm{~B}^{-}(a q)\)

∴ The solubility product of AB2, Kgp = [A2+] X [B-]2 Let the solubility of AB2 in its saturated solution at a given temperature be S mol.L-1. Therefore, in the solution, [A2+] = S mol-L-1, [B-] = 2S mol-L-1 formula unit of AB2 dissociates into one A2+ ion and two B_ ions].

∴ Ksp = [A₂+] X [B-]₂ = S X (2S)₂ = 4S3

∴ Ksp=4S₃

Dynamic Equilibrium Concept

Equation [1] represents the relation between the solubility and the solubility product of the sparingly soluble salt AB2.

Example: In a saturated aqueous solution of CaF₂, the following equilibrium exist: CaF₂(s) Ca.2+(aq) + 2F-(aq)

The solubility product of CaF₂, Ksp = [Ca₂+] [F-]2 Let the solubility of CaF2 in its saturated aqueous solution at a certain temperature = S mol.L-1. So, in the saturated solution,

[Ca2+] = S mol-L-1 and [F-] =2S mol-L-1 Ksp = [Ca₂+] x [F-]2 = S X (2S)₂ = 4S3

Similarly, in the case of other sparingly soluble salts ofthe type AB₂ (BaF2, PbCl₂, etc.), Ksp = 4S3 sparingly soluble salt of the type A_xB_y: in a saturated aqueous solution of a salt of the type, AxBy, the following equilibrium is established: AxBy(s) xkVÿaq) +yBx-(aq).

∴ The solubility product of AxBy, Ksp = [Ay+]x x [Bx-y]y

Suppose, the solubility of its saturated solution is = S mol.L-1.

Hence, in the solution, [Ay+] = xS mol-L-1 and [Bx-] = yS mol.L-1

Dynamic Equilibrium Concept

[since one formula unit of AxBy dissociates into x number of A-V+ and y number of Bx- ions]

∴ Ksp=[Ay+]xx[Bx-]y=(xs)xx(ys)y=(xxyy) sx+y

∴ Ksp = xx yy(S)x+y

Equation [1] represents the relation between the solubility and the solubility product of sparingly soluble salt AxBy

Dynamic Equilibrium Concept

Dynamic Equilibrium Concept

Dynamic Equilibrium Concept

Common Ion Effect On The Solubility Of A Sparingly Soluble Salt 

Ag+ and Cl are always constant at a particular temperature.

Therefore, if the molar concentration of one of the ions between Ag+ and Cl- is increased, then the molar concentration of the other will be decreased to maintain a constant value of Ksp.

Let strong electrolyte KC1 (having common ion Cl- ) be added to the saturated solution AgCl. This will increase the molar concentration of the common ion (Cl-) in the solution.

According to Le Chatelier’s principle, to maintain the constancy of Ksp, some of the Cl- ions combine with an equal number of Ag+ ions to form solid AgCl. This causes the equilibrium to shift to the left. As a result, the solubility ofAgCl in the solution decreases.

Dynamic Equilibrium Concept

At the new equilibrium, [Cl-] <sc[Ag+], and the solubility of AgCl becomes equal to [Ag+]. If an aqueous AgNO3 solution instead of KC1 is added to the saturated solution of AgCl, then the solubility of AgCl will also decrease because of the communion (Ag+) effect.

Increase in solubility due to the presence of common ion: AgCN is a sparingly soluble salt and its solubility increases when KCN is added to its aqueous solution. However, this increase in solubility does not contradict the principle of solubility product. KCN reacts with sparingly soluble AgCN, and forms a water-soluble complex, K [Ag(CN)2]. Thus, the solubility of AgCN increases. KCN(a<7) + AgCN(aq) K [Ag(CN)2] (aq) A similar phenomenon is observed in the case of mercuric iodide (Hgl2). It shows greater solubility in an aqueous KI solution due to the formation of a water-soluble complex, K2[HgI4]. 2KI(aq) + Hgl2(a<?) K2[HgI4](a<7)

Effect of pH on solubility

The solubility of salts of weak acids like phosphate increases at lower pH. This is because, at a lower pH, the concentration of the anion decreases due to its protonation. This in mm increases the solubility of that salt. Example: Consider the solubility equilibrium: CaF2(s) Ca2+(ai7) + 2F-{aq) If the solution is made more acidic, it results in the protonation of some ofthe fluoride ions.

⇒ \(\mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}(a q)+\mathrm{H}_2 \mathrm{O}(l)\) As acid is added, the concentration of F- ion decreases due to the reaction [2]. This causes the equilibrium [1] to shift to the right, thereby increasing the concentration of Ca2+ ions. Thus, more CaF₂ dissolves in a more acidic solution.

Dynamic Equilibrium Concept

Principle of solubility product and precipitation of sparingly soluble salts

Principle of solubility product: At a particular temperature, the precipitation of salt from its solution or its dissolution in the solution depends on the value ofthe solubility product of the salt at that temperature.

If the product of the molar concentrations of the constituent ions, with suitable powers, of the salt in the solution exceeds the value of the solubility product of the salt at that temperature, then the salt will be precipitated from the solution.

If this value does not exceed the solubility product of that salt, then the salt will not be precipitated. This principle is known as the principle of solubility product. This principle is extensively used in different chemical analysis and industrial processes.

Explanation: At 25°C, the solubility product of AgCl is 1.8 x 10-10. Hence, in the saturated solution of AgCl at 25°C, the highest value of the product of molar concentrations of Ag+ and Cl- ions is 1.8 xlO-10.

If the product of molar concentrations of Ag+ and Cl- ions is less than, then the solution will be unsaturated. Under this condition, AgCl will be dissolved in the solution to increase the concentration of Ag+ and Cl- ions and this will continue until the product of ofmolar concentrations of Ag+ and Cl- becomes equal to. If the product of molar concentrations of Ag+ and Cl- is greater than kTip(AgCl), then AgCl will be precipitated and this will continue unless the product ofmolar concen¬ trations of Ag+ and Cl- becomes equal to KspfAgCl).

Dynamic Equilibrium Concept

Applications of solubility product

Application-1: At a certain temperature, if the solubility product of a sparingly soluble salt is known, then the solubility of the salt, as well as the molar concentrations of the constituent ions of the salt in its saturated solution, can be determined.

Conditions for precipitation: In the solution, if the product of the molar concentrations (with appropriate powers) of the constituent cation and anion of the salt at a particular temperature.

Is greater than the solubility product of the salt at that temperature, then the salt will be precipitated, Is less than the solubility product of the salt at that temperature, then the salt will remain dissolved in the solution, Is equal to the solubility product of the salt at that temperature, then the solution will just be saturated. In this case, the dissolved portion of the salt remains in equilibrium with the undissolved part.

Applications of solubility product principle

Preparation of pure NaCl from impure NaCl: pure sodium chloride can be prepared from impure sodium chloride by applying this principle. When dry HC1 gas is introduced into the saturated aqueous solution of impure sodium chloride, the concentration of Cl- ions increases to a great extent and the ionic product of Na+ and Cl-, i.e., [Na+] x [Cl-] exceeds the solubility product of NaCl. Consequently, sodium chloride gets separated as pure crystals.

During this precipitation, impurities like MgCl₂, CaCl₂, etc., do not precipitate out as the values of the ionic product of the constituent ions of these impurities do not exceed their respective solubility products.

Separation of NaHCO3 by Solvay process: in this process, a saturated solution of sodium chloride is added to a solution of NH4HC03. The reaction concerned is: \(\mathrm{NH}_4 \mathrm{HCO}_3(a q)+\mathrm{NaCl}(a q) \rightarrow \mathrm{NaHCO}_3(a q)+\mathrm{NH}_4 \mathrm{Cl}(a q)\) Among the compounds participating in the reaction, NaHCO3 has the least value of solubility product. The product of the molar concentrations of Na+ and HCO-3 ions present in the solution can easily exceed the solubility product of NaHCOg, and hence NaHCO. gets precipitated.

[KHCO₃ cannot be prepared by the Solvay process because the solubility product of KHCO₃ Is very high.]

Manufacture of soaps: Soap consists of sodium or potassium salts of organic fatty acids of high molecular mass, such as stearic acid, oleic acid, etc. During the manufacturing of soaps, some amount of soap remains in a colloidal state in the mother liquor. If some NaCl is added to this solution, then the concentration of Na+ ions increases, thereby decreasing the solubility of soap. As a result, soap gets precipitated.

Dynamic Equilibrium Concept

Application in analytical chemistry: The principle of solubility product is extremely helpful in the qualitative analysis of basic radicals. Based on the values of the solubility product, the basic radicals of inorganic salts are divided into different groups.

Group 1: Basic radicals \(\mathbf{A g}^{+}, \mathbf{P b}^{2+}, \mathbf{H g}_2^{2+}(\text { ous })\)

Group reagent: 6(N) HC1 As the values of the solubility product of chlorides of group-I metals are sufficiently low [At 25°C, Ksp(AgCl)= 1.8 X 10-10, Ksp(Hg₂Cl₂) = 1.2 X lO-5, Ksp(PbCl2) = 1.7 X 10-18 ], on addition of6(N)HCI, the solubility products of these chlorides are exceeded by the products of molar concentration of the corresponding ions. So, these chlorides get precipitated.

On the other hand, the chlorides of other groups have higher values of solubility product and thus are not precipitated. [The solubility product of PbCl2 is much higher than that of both AgCl and Hg2Cl2. So, Pb2+ ions are not completely precipitated as PbCl2 in group 1. Pb2+ ions left in the solution are separated as precipitate in group 2.

Group 2: Basic radicals: Cu2+, Pb2+, Cd2+, Hg2+ (ic), Bi3+, As3+, Sb3+, Sn2+.

Dynamic Equilibrium Concept

Group reagent: H₂S gas in the presence of dilute HC1 In an aqueous solution, H₂S ionizes partially to produce S2- ions, which precipitate the basic radicals of this group as their corresponding sulfides (CuS, PbS, CdS, etc.).

The solubilityproduct ofthe metallic sulphides of group-2 [Ksp(CuS) = 8.0 X IO-37, rip(PbS) = 3.0 X 10-28, Ksp(CdS) = 1.0 x lO’27, ATsp(HgS) = 2 X 10-53,

Ksp (Bi₂S₃) = 1.6 X10-72] are much smaller as compared to those of the sulfides of the subsequent groups. So, when H₂S gas is passed through the solution consisting of different basic radicals in the presence of HC1, the product of molar concentrations of the corresponding ions of each of the sulfides of this group exceeds its corresponding solubility product. Hence, only the sulfides of this group are precipitated.

It is important to note that the degree of ionization of weak acid H₂S in the presence of HC1 decreases due to the common ion (H+) effect. As a result, except for group- 2, the solubility products of sulfides of the subsequent groups are not exceeded. Therefore, except for the sulfide salts of group-2, sulfides of all other groups are not precipitated.

Dynamic Equilibrium Concept

Another notable point is that among the sulfides of group-2, the solubility products of CdS and PbS are sufficiently high. So, as a result of the common ion effect, if the concentration of S₂– ions decreases to a large extent, CdS and PbS are not precipitated. It has been experimentally proved that if H₂S gas is passed through the solution using 0.3(N) HC1 solution, then all the sulfides of this group are completely precipitated, but the precipitation of the sulfides of other groups does not occur.

Group 3A Basic radicals: Fe3+, Al3+, Cr3+ Group reagent: NH4OH solutions presence of NH4C1 The basic radicals of this group are precipitated as their corresponding hydroxides [Fe(OH)3, Al(OH)3, Cr(OH3) ]. The solubility products of the hydroxides of the basic radicals of group 3A are much smaller than those of the hydroxides belonging to the subsequent groups.

So, when NH4OH is added to the solution of different basic radicals in the presence of NH4C1, only the solubility products of the hydroxides of group-3A are exceeded, resulting in the preferential precipitation of the hydroxides of group-3A basic radicals. The hydroxides of the basic radicals of subsequent groups are not removed by this way of precipitation

Dynamic Equilibrium Concept

Here, it is to be noted that the degree of ionization of weak base NH₃ in the presence of NH4C1 is decreased due to the common ion (NH4-) effect. As a result, the concentration of OH ions decreases to such an extent that except for group-3A, the solubility product of metallic hydroxides of the subsequent groups is not exceeded. Therefore, except for group-IIIA, metallic hydroxides of other groups are not precipitated.

Group 3 Basic radicals: Co2+, Ni2+, Mn2+, Zn2+

Group reagent: H₂S in the presence of NH3 For precipitation of the basic radicals of this group as their corresponding sulfides [CoS, ZnS, MnS, NiS], H₂S gas is passed through the test solution in the presence of NH3.

In the presence of OH- ions produced by the dissociation of NH₄OH, the equilibrium involving the dissociation of the weak acid H2S gets shifted to the right, forming sulfide ions (S2-) to a greater extent [H₂S(a<7) + 2H₂O(Z) 2H₃O+(aq) + S2-(aq)].

Due to a sufficient increase in the concentration of S2- ions, the solubility products of the sulfides of group-3B are exceeded. Hence, the basic radicals of this group are precipitated as sulfides.