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
- 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.
- 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.
- % 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.
- In this solution, a dynamic equilibrium is established between the dissolved gas and the gas over the liquid surface.
- 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.
- 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:
- 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.
- Since an irreversible reaction undergoes completion, the reactants participating the reaction in equivalent amounts are completely exhausted.
- 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.
⇒ \(\begin{aligned}
& \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) ; \mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \\
& \mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) ; \mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)
\end{aligned}\)
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.
- 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).
- 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.
- 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.
- 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.
- Eventually a moment comes when the rate of the forward reaction becomes equal to the rate ofthe backward reaction.
- 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
- 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 CO2, 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, H2, 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, H2, 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 KP and KX
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, PA, PB, PD, and PE 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 PA = xA xP, PB = xB xP, PD = xD xP, and PE = xE 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 KC and KX
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 KP and Kk.
Relation Between KP, KC & KX:
KP = Kc(RT)Δn = KX(P)Δn When An = 0 for a reaction [e.g.,H2(g) +I2(g)⇌2HI(g)
or, N2(g) + O2(g) 2NO(g)], ⇌then, KP= KC = 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 KP and KC are not influenced by pressure. If the temperature remains constant, the increase or decrease in pressure does not alter the values of KP and KP. 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 QP.
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 ∞.
- 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.
- 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.
- 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.
- 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—
- ΔG is negative, the reaction is spontaneous as written.
- ΔG is positive, the reaction Is non-spontaneOus as written.
- 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
- 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.
- 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.
- 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
- 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.
- As a result, the reaction will no longer remain in the state of equilibrium.
- 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.
- 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.
- 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
- 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.
- As a result, the reaction will no longer remain in the state of equilibrium.
- 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.
- 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.
- 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.
- 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
- 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.
- As a result, the reaction will no longer exist in the state of equilibrium.
- 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.
- 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.
- 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.
- 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
- 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.
- As a result, the reaction will no longer remain in the equilibrium state.
- 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.
- 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.
- 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.
- 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)
- 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:
- 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.
- 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.
- 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).
- 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:
- At constant temperature, if the pressure of a reaction system at equilibrium is decreased, then the reaction will no longer remain at equilibrium.
- 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.
- 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.
- NaOH(ag) Na+(ag) + OH-(aq)
- KOH(ag)→ K+(aq) + OH-(aq)
- Ca(OH)2(aq)→ Ca2+{aq) + 20H-(aq)
- 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.
- 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.
- The acidic or basic properties of a substance that is insoluble in water cannot be explained by this theory.
- 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.
- 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.
- 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:
- Acid: An acid is a substance that can donate a proton (or H+ ion)
- 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 ;
- CH3COOOH(aq)+ H2O(l) ⇌ H3O+(aq)+CH3COO-(aq)
- H2SO4(aq) + 2H2O(l)⇌ 2H3O(aq)+SO2-4(aq)
- NH+4{aq) H- H2O(/)⇌ H3O+(aq) +NH3(aq)
- [Fe(H2O)6]3+(H2O) + H2O(l)⇌
- [Fe(H2O)5OH]2(aq)+ H3O+(aq)
- 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:
- CH3COO-(aq) + H2O(l)⇌ CH3COOH(aq) + OH-(aq)
- NH3(aq) +H2O(l)⇌NH+4(aq) + OH-(aq)
- 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.
- 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.
- 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.
- 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.
- 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:
- This concept provides no idea regarding the relative strengths of acids and bases.
- This theory contradicts the general concept of acids by considering BF3 A1C13, and simple cations as acids.
- 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.
- 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 Kw.
therefore Kw=[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 Kw(25C)=10-14]
PKw = pKw = -log10Kw At 25°C,Kw=10-14
Therefore, at 25°C, pKw = -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:
- With a decrease or Increase In (lie concentration of OH- ions In the solution, the pOH of the solution Increases or decreases respectively
- 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, pKw = 14. Hence, at this temperature, the pH scale extends from 0 to 14. On the other hand, at 100°C, pKw = 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:
- 0.01(N)
- HC1
- 0 0.05(M) H2SO4
- 0 0.001(N)
- 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
∴ PKa+PKb = PKw
Dynamic Equilibrium Concept
At 25°C, pKw = 14. So, at 25°C, pKa + pKb = 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:
- 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.
- 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.
- 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:
- 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.
- 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.
- 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:
- At a particular temperature, the solubility product of a sparingly soluble salt has a fixed value. Its value changes with temperature variation.
- 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 AxBy: 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.
