• Equilibrium in physical and chemical processes
• Dynamic nature of equilibrium
• Law of mass action
• Equilibrium constant
• Factors affecting equilibrium
• Le Chatelier’s principle
• Ionic equilibrium
• Ionisation of acids and bases
• Strong and weak electrolytes
• Degree of ionization
• Concept of pH
• Hydrolysis of salts(elementary idea)
• Buffer solutions
• Solubility product
• Common ion effect (with illustrative example)

It is generally observed that many chemical reactions do not proceed to completion when they are carried out in a closed container. This implies that the reactants are not completely converted into products. Instead , after some time concentrations of the reactants do not undergo further decrease and the reaction appears to have stopped. This state of the system in which no further net change occurs is called a state of equilibrium.
In all processes which attain equilibrium , two opposing forces are involved. Equilibrium is attained when rates of the two opposing processes become equal.
If the opposing processes involve only physical changes, the equilibrium is called physical equilibrium. If the opposing processes are chemical reactions, the equilibrium is called chemical equilibrium.
In spite of a large number and variety of chemical reactions, their state of chemical equilibrium may be classified in three groups characterized by the extent to which the reactions proceed.
(i) The reactions that proceed nearly to completion and only negligible concentration of the reactants is left.
(ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage.
(iii) The reactions in which the concentrations of the reactants and products are comparable when the system is in equilibrium.
The equilibrium involving physical processes are referred to as physical equilibria. The physical equilibria involving change in state may be of the following three types.

When a crystalline solid is heated , the temperature at which solid-liquid equilibrium is attained under 1 atmospheric pressure is called the normal melting point or normal freezing point of the substance. If heat energy is added to a mixture of solid and liquid at equilibrium, the solid is gradually converted into liquid while the temperature remains constant. If a solid-liquid system at melting point is taken in a well-insulated container, then this constitutes a system in which solid is in dynamic equilibrium with the liquid. For example, let us consider ice and water at 273 K (melting point of ice) , taken in a perfectly insulated thermos flask. It may be noted that, the temperature as well as the masses of ice and water remains constant. This represents a dynamic equilibrium between ice and water.
ice ⇌water
H2O(s) ⇌ H2O(ℓ)
Since there is no change in mass of ice and water, the number of molecules going from ice into water is equal to the number of molecules of water going into ice.
Thus, at equilibrium,
Rate of melting = Rate of freezing
When a liquid is placed in an open container, it starts evaporating spontaneously. This is because a liquid is composed of molecules which are in constant rapid motion. The velocity distribution in a liquid is almost the same as in case of gases. All molecules of a liquid do not possess the same velocity. At all temperatures, a liquid possesses some molecules whose velocity and hence kinetic energy is much larger as compared to that of the rest of the molecules. When a molecule reaches the surface of the liquid , an inward pull due to surface tension acts on it. If the molecule possesses sufficient kinetic energy to overcome the inward pull due to surface tension, it comes out of the surface and collects over it (fig a).

Evaporation of a liquid in an open container
Such molecules constitute the vapour and the process is called vapourisation or evaporation.
When a liquid is taken in closed container having some free space over the liquid (Fig b)

Evaporation and condensation processes occurring simultaneously in a closed container leading to equilibrium.
When a liquid is taken in a closed container having some free space over the liquid ( Fig b ), the energetic molecules collect in the free space due to vaporisation. Some of the vapour molecules however get condensed and return back to the liquid. The rate of condensation depends upon the concentration of molecules in the vapour phase. The rate of evaporation remains constant throughout if the temperature is kept constant. Since the rate of condensation is proportional to the concentration of molecules in vapour phase, it slowly increases with time. After certain time, the rate of condensation becomes equal to rate of evaporation as shown graphically in Fig.

Attainment of liquid ⇌ Vapour equilibrium
As soon as the rate of evaporation becomes equal to the rate of condensation, a state of dynamic equilibrium is attained which can be shown as follows:

It is to be noted that in a liquid vapour system, equilibrium can be attained only when the liquid is taken in a closed container. In an open container, vapour disperse in the atmosphere and equilibrium cannot be attained.
In the state of equilibrium i.e., the state when the rate of evaporation and rate of condensation are exactly equal, a constant pressure of vapour is attained. The pressure at which the liquid and vapour phases exist together in equilibrium is called saturation vapour pressure or simply vapour pressure of the liquid at the same temperature.
The liquid vapour equilibrium as described above is dynamic one. The two processes occur simultaneously in opposite direction but with equal rate i.e., at equilibrium.
Rate of evaporation = Rate of condensation
Liquid–vapour or liquid-gas equilibrium can be illustrated by taking a small amount of water at room temperature in a evacuated vessel attached to a manometer as shown in Fig.

Fig Evaporation of water in a closed vessel (a) Initial stage (b) Equilibrium stage
Due to evaporation of water, a pressure is developed in the vessel and the mercury level in the manometer rises gradually. After some time , the mercury level in the manometer becomes steady (fig b) . This indicates that no more water is evaporating although liquid water is still present in the vessel. This shows that a state of equilibrium has been attained as shown below :
H2O(ℓ) ⇌H2O(g)
At this stage , the rate of evaporation of water molecules present in the liquid phase is equal to the rate of condensation of water molecules present in the gaseous phase (i.e., vapour phase).
01. Vapour pressure of water, acetone and ethanol at 293 K are 2.34, 12.36 and 5.85 kPa respectively. Which of these has the lowest and the highest boiling point ? At 293 K which of these evaporates least in a sealed container before equilibrium is established ?
Consider the systems where solids sublime to vapour phase. If we place solid iodine in a closed vessel, after some time the vessel gets filled up with violet vapour and the intensity of colour increases with time. After a certain time the intensity of colour becomes constant and at this stage equilibrium is attained. Hence solid iodine sublimes to give iodine vapour and the iodine vapour condenses to give solid iodine. The equilibrium can be represented as :
I2(solid) ⇌ I2(vapour)
Other examples showing this kind of equilibrium are :
Camphor(solid) ⇌Camphor(vapour)
NH4Cl(solid) ⇌ NH4Cl(vapour)
It is not possible to dissolve any amount of a solute in a given amount of solvent. For example, when we add sugar to water , the crystals of sugar keep on going into solution in the beginning. But after some time no more of sugar dissolves.
A solution in which no more solute can be dissolved is called a saturated solution. The amount of solute required to prepare a saturated solution in a given quantity of solvent is known as solubility of the solute at a particular temperature.
The saturated solution corresponds to the state of equilibrium. When we add sugar crystals to water, molecular vibration tends to dislodge molecules from the surface of crystals. The molecules of sugar which go into solution are free to move throughout the water. In the beginning, the rate at which molecules leave the crystal is much greater than the rate of their return. As the number of molecules in solution increases , the rate at which molecules return to crystal also increases. Soon a balance between the two rates i.e., the rate of dissolution and rate of precipitation is established and this corresponds to the state of equilibrium.

Attainment of solid-liquid equilibrium

Thus, in a saturated solution, a dynamic equilibrium exists between the molecules of sugar in the solid state and the molecules of sugar in solution.

The dynamic nature of equilibrium can be demonstrated by adding radioactive sugar into a saturated solution of non-radioactive sugar (Fig).

It is observed that the solution and rest of the non-radioactive sugar also become radioactive. This shows that even at equilibrium, the process of dissolution and precipitation are taking place. This means that equilibrium is dynamic in nature. However, at equilibrium :
Rate of dissolution = Rate of precipitation.
The gases can be dissolved in suitable liquids. At a given temperature, a liquid can dissolve only certain definite mass of the gas. This suggests that a state of equilibrium exists between molecules in the gaseous state and the molecules dissolved in the liquid. For example, when carbon dioxide is dissolved in soda water, the following equilibrium exists :
CO2(g) ⇌ CO2(solution)

At equilibrium the rate at which gas molecule pass into the solution becomes equal to the rate at which dissolved gas molecules come back into the gaseous phase. This is shown in Fig.

The solubility of a gas in liquid depends on the pressure. The effect of pressure on the solubility of a gas in a liquid is given by Henry's law.
Henry’s Law
The law can be stated as follows:
The mass of a gas that dissolves in a given mass of a solvent at constant temperature is proportional to to the pressure of the gas at equilibrium with the solution provided the gas does not undergo any chemical change during the formation of a solution.
If the mass m of a gas dissolves in a given mass of a solvent at a particular temperature under the equilibrium pressure p , then according to Henry’s Law, we have,
m  p
m = k p
where k is a proportionality constant.
Limitations of Henry’s Law
Following are the limitations of Henry’s law :
i) Henry’s law is applicable to ideal gases only. However, this law can be applied to real gases at low pressure when the real gases approach the behaviour of ideal gases.
ii) The law is not applicable to those gases which undergo a chemical change in solution. For example, the law cannot be applied to ammonia which reacts with water to form NH4OH.
iii) The law cannot be applied to those gases which dissociate into ions in solution. For example, the law cannot be applied to HCl gas because it dissociates into H+ and Cl ions in solution.
A soda water bottle contains CO2 gas dissolved in water at a higher pressure and posses the following equilibria:
CO2(g) ⇌ CO2(solution)
Since the soda water bottle is sealed when the gas is at high pressure, there is quite appreciable amount of carbon dioxide dissolved in water and the gas pressure above the solution is high. When the bottle is opened, the pressure of the gas above the solution decreases, therefore, the dissolved gas escapes(fizzes out) to attain new equilibrium state.
The solubility of a gas in a liquid decreases with increase in temperature.
02. 0.200 g of iodine is stirred in 100 mL of water at 288 K till equilibrium is reached. What will be the mass of iodine found in solution and the mass that is left undissolved ? After equilibrium is reached with 0.200 g of iodine in 100 mL of water, we add 150 mL of water to the system. How much iodine will be dissolved and how much will be left undissolved and what will be the concentration of iodine in solution ? [I2(aq)]at equilibrium = 1.1 x 104 mol L1 at 288 K
03. At 20C the solubility of N2 gas in water is 0.0150 g/L , when the partial pressure of the gas is 580 torr. Find the solubility of nitrogen in water at 200C when the partial pressure is 800 torr.
04. The solubility of sodium chloride in water is 6.150 mol/L at 20C . 80 g of sodium chloride are dissolved in 100 cm3 of water at 20C. How much sodium chloride is left undissolved ? After equilibrium is reached, an additional 50 cm3 of water is added to the system at the same temperature. Find the amount of NaCl present in the solution and in the undissolved state.
From the discussions of physical equilibria, the following points are noted :
i) In the case of Liquid ⇌ vapour equilibrium, the vapour pressure is a constant at a given temperature.
ii) For solid ⇌Liquid equilibrium , there is only one temperature(melting point) at 1 atm at which the two phases can coexist without any exchange of heat with the surroundings, the mass of the two phases remain constant.
iii) For dissolution of solids in liquids, the solubility is constant at a given temperature.
iv) For dissolution of gases in liquids the concentration of a gas in liquid is proportional to the pressure(concentration) of the gas over the liquid.
These four concentration-related findings are expressed in the TABLE below:
Some Features of Physical Equilibria
Process Conclusions
H2O() ⇌ H2O(g) PH2O constant at given temperature.
H2O(s) ⇌ H2O() Melting point is fixed at constant pressure
Solute(s) ⇌Solute(solution)
Sugar(s) ⇌Sugar(solution)
Concentration of solute in solution is constant at a given temperature.
Gas ⇌ Solution

CO2 (g) ⇌ CO2(aq)
[gas(aq)]/[gas(g)] is constant at a given temperature.

is constant at a given temperature.
General Characteristics of Equilibria Involving Physical Processes
From the physical equilibria discussed above the characteristics of the systems may be stated as follows:
1. The system has to be closed; that it should not gain matter from the surroundings nor lose matter to the surroundings.
2. There is a dynamic but stable condition. Two opposite processes occur at the same rate.
3. The measurable properties of the system remain constant, since the concentration of the substances remain constant.
4. When equilibrium is attained, there exists an expression involving concentration of reacting substances which reaches a constant value at a given temperature. The TABLE above lists the concentration-related expressions for certain physical processes.
5. The magnitude of the constant value of the concentration-related expressions is an indication of the extent to which the reaction proceeds before reaching equilibrium.
A large number of chemical reactions do not go to completion and attain a state of equilibrium after some time. The state of equilibrium involving a chemical system is referred to as chemical equilibrium.
Reversible and Irreversible Reactions
A reversible reaction is one in which the products can combine to give back the reactants i.e., a reaction which takes place in both forward and backward directions under specified conditions. The reaction is represented by a double arrow ( ⇌ ) between the reactants and products as in the form of equation. Reactants are written on the left hand side while the products on the right hand side of the equation as;
A + B ⇌ C + D
N2 + 3 H2 ⇌2 NH3 ; H =  92.5 kJ
An irreversible reaction is one in which products do not interact to give back reactants under the same conditions. It is represented by single arrow between the reactants and products. It is an equation pointing from the reactants towards the product as :
A + B  C + D
For example, when silver nitrate is added to sodium chloride solution , a white precipitate of AgCl is formed along with NaNO3. AgCl and NaNO3, will not interact to give back AgNO3and NaCl.
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
A reversible reaction can be made irreversible if one of the product (s) which is gaseous is allowed to escape out. This is why reversible reactions are done in a closed vessel.
Equilibria Involving Chemical systems
Consider a general reversible reaction:
A + B ⇌ C + D
In the initial state the concentrations of A and B are very high and that of C and D is very low. (suppose the time t = 0, i.e., at the beginning, the amount of C and D = 0). As the reaction proceeds, the amount of A and B become less , while that of C and D increases. At a particular time under specified conditions(in closed system), the rate of forward reaction becomes less and becomes equal to the rate of backward reaction . At this state the reaction does not stop. Both forward and backward reactions take place simultaneously with equal magnitude and reactions seems to be at a stand-still. At this time the concentration of reactants (A and B) and the concentration of products (C and D) are constant. This state of chemical system is called chemical equilibrium. Graphical representation is given below.

Diagram showing how equilibrium is established in a reversible reaction
Consider the combination of hydrogen and iodine in a closed glass container. At first the intense colour can be observed due to the evolution of large amount of HI. After a while the colour diminishes. This is due to the decomposition of HI to give back H2 and I2. At a particular instant, there will be neither diminishing of colour nor increase in colour. Colour becomes steady. This shows that as much hydrogen and iodine combine to form HI in a particular rate, an equivalent amount of HI decomposes in the same rate to give back hydrogen and iodine. If the reaction has stopped, there will not be a change in the intensity of the colour due to the non-availability of HI or by an increase in the production of HI. This change is not observed which means that both reactions are going on continuously though it has attained an equilibrium point. So the reaction is dynamic and not remaining static.
Characteristics of Chemical Equilibrium
The important characteristics of chemical equilibrium are :
(i) At equilibrium the concentration of each component of reactants and products becomes constant.
(ii) The rate of forward reaction will be equal to rate of backward reaction. So, a chemical equilibrium is said to be dynamic in nature.
(iii) A chemical equilibrium is established provided none of the products is allowed to escape, i.e., only in a closed system a chemical equilibrium can be attained and not in an open system.
(iv) Chemical equilibrium can be attained either from the direction of reactants or from the direction of products depending upon the conditions like temperature and pressure.
(v) A catalyst does not alter the state of equilibrium. It is due to fact that a catalyst influences both the forward and backward reaction to the same extent. It only helps to attain the equilibrium rapidly.
The Law of Mass Action
On the basis of observations of many equilibrium reactions, two Norwegian chemists Guldberg and Waage (1864) suggested a quantitative relationship between rates of reactions and the concentrations of the reacting species. This relationship is known as the Law of Mass Action. This law states that :
At constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of reacting species with each concentration term raised to the power equal to the numerical coefficient of that species in the chemical equation representing the chemical change.
The molar concentration i.e., active mass ( moles per litre) of a substance is expressed by enclosing the symbol or formula of the substance in square brackets. For example, molar concentration of A is written as [A] or CA.
Consider a hypothetical reaction :
A + B  Products
According to the law of mass action:\
Rate of reaction , r  [A] [B]
= k [A] [B]
where [A] and [B] are the concentrations of the reactants A and B respectively, k is a proportionality constant called rate constant.
For any reaction :
a A + b B  Products
The law of mass action may be written as :
Rate of forward reaction  [A]a [B]b
Application of Law of Mass Action To Chemical Equilibrium
By applying the Law of Mass Action to a reversible reaction in equilibrium, it is possible to derive a simple mathematical expression known as Law of Chemical Equilibrium .
Consider a simple reversible reaction ,
A + B ⇌ C + D
in which an equilibrium exists between the reactants A and B and the products C and D. According to the law of mass action,
Rate of forward reaction  [A] [B]
= kf [A] [B]
where kf is the rate constant for forward reaction. [A] and [B] are the molar concentrations of reactants A and B respectively.
Rate of backward reaction  [C] [D]
= kb [C] [D]
where kb is the rate constant for the backward reaction and [C] and [D] are the molar concentrations of products C and D respectively.
At equilibrium , the rates of two opposing reactions become equal. Therefore at equilibrium :
Rate of forward reaction = Rate of backward reaction
 k f[A][B] = k b[C] [D]

Since kf and kb are constants the ratio kf / kb is also constant and it is represented by KC.

The constant K is called Equilibrium constant.
For a general reaction of the type :

The expression for equilibrium constant can be written as :

The equilibrium constant, K may be defined as the ratio of the product of equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficient of the substance in balanced chemical equation.
Equation (1) and (2) are the expressions for the Law of Equilibrium.
The concentration ratio :

is called concentration quotient and is denoted by Q. According to the law of chemical equilibrium,
Q = K ; at equilibrium.
If an equilibrium involves gaseous species , then the concentrations in the concentration quotient may be replaced by partial pressures because at a given temperature, the partial pressure of gaseous component is proportional to its concentration. If the above mentioned reaction has all the gaseous species then :

The equilibrium constant Kp defined in terms of partial pressures is not the same as the equilibrium constant KC, defined in terms of molar concentrations.
Since P = C R T for an ideal gas,

Homogeneous and Heterogeneous Equilibria
The equilibria in which all the substances are present in the same phase are known as Homogeneous equilibria. For example,

On the other hand, equilibria in which the substances involved are present in different phases are called Heterogeneous equilibria. For example,

Equilibrium Expressions of Some Reactions
While writing equilibrium expressions certain conventions are followed. Let us illustrate these conventions with the help of some examples.
1. In case some solid is involved in the equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentration of all solids are taken as unity, i.e., [Solid] = 1. For example,

2. In case some liquid is in equilibrium with some gas or gases, the concentration of the pure liquid is taken as unity. For example,

3. For the equilibria in aqueous media if water is also involved, its concentration will not change appreciably because it is present in a large quantity ( =55.5 mol L1) and hence [H2O] is taken as constant. For example,

By convention, the concentration of solvent is taken as constant.
5. Write the equilibrium constant expressions for the following
reactions :

Characteristics of Equilibrium constant (K)
Some important characteristics of equilibrium constant are :
1. The equilibrium constant has a definite value for every chemical reaction at a particular temperature.
2. The value of equilibrium constant is independent of initial concentrations of reacting species.
For example,
The equilibrium constant for the reaction:

at 298 K is :

Whatever may be the value of the initial concentrations
of reactants, Fe3+ and SCN ions, the value of K
comes out to be 138 at 298 K.
3. For a reversible reaction, the equilibrium constant for backward reaction is inverse of equilibrium constant for the forward reaction.
For example,
Equilibrium constant for the combination between
hydrogen and iodine at 717 K is 49.

The equilibrium constant for the decomposition of
hydrogen iodide at the same temperature is
inverse of the above equilibrium constant.

4. The equilibrium constant is independent of the presence of catalyst. This is so because the catalyst does not affect the equilibrium state.
5. If the equilibrium is expressed in terms of concentration, it has different units for different reactions.
(a) If the number of moles of the products is same as the number of moles of reactants, K has no units i.e., dimensionless. Thus the reaction between hydrogen and iodine to form hydrogen iodide, K has no units as illustrated below:

(b) If the number of moles of the products is not the same as the number of moles of the reactants, K will have certain units depending upon the change in the number of moles. Thus, for the reaction between nitrogen and hydrogen to form ammonia, K would have units L2 mol2 as shown below:

Thus K will have the units of L2mol2.
(c) It may be noted that the value of K changes if the coefficients of various species in the equation representing equilibrium are multiplied by some number. For example, equilibrium constant for the equation:

is different from the equilibrium constant for the reaction:

The equilibrium constant for the two equations are related to each other as :

In general, if K1 is the equilibrium constant for a particular equation, then the equilibrium constant K2 for some new equation which is obtained by multiplying the first equation by 'n' is given by the following relation:

Relations between Equilibrium Constants for a General Reaction and its Multiples

Some of the applications of equilibrium constant are :
i) Predicting the extent of reaction on the basis of its magnitude.
ii) Predicting the direction of the reaction.
iii) Calculating equilibrium concentrations.
i) Prediction of Extent of Reaction
The value of equilibrium constant tells us about the extent to which the reactants are converted into the products before the equilibrium is attained. Larger the value of K, greater is the extent to which reactants are converted into products. On the other hand, if the value of K is small, it indicates that equilibrium is mainly in favour of reactants and that only a small fraction of reactants is converted into products. For example, consider the following reaction:

The expression for equilibrium constant of this reaction

Consider the reaction:

K for this reaction is given by the expression :

The large value of K indicates that at equilibrium, concentration of CO2 is quite high. In other words, this indicates that the reaction between CO and O2 to form CO2 is almost complete under the given conditions.
The value of K also gives us an idea about the relative stabilities of reactants and products. If the value of K is large, the products are more stable, whereas if K is small, the reactants are more stable.
These generalizations are illustrated in Fig.

Dependence of extent of reaction on Kc
ii) Predicting the direction of the Reaction
The equilibrium constant is used to find in which direction an arbitrary reaction mixture of reactants and products will proceed. For this purpose we calculate the reaction quotient Q. The reaction quotient is defined in the same way as the equilibrium constant (with molar concentrations to give QC or with parial pressures to give QP) at any stage of the reaction. For a general reaction :
a A + b B ⇌ c C + d D

Then , if QC > KC , the reaction will proceed in the direction of reactants (reverse reaction)
If QC = KC , the reaction mixture is already at equilibrium.
In the reaction H2(g) + I2(g) ⇌ 2 HI (g) , if molar concentrations of H2(g) , I2(g) and HI(g) are 0.1 mol/L, 0. 2 mol/L, and 0.4 mol/L respectively at 783 K, then the reaction quotient at this stage is :

KC for the reaction at 783 K is 46 and we find that QC < KC. The reaction, therefore move to right i.e., more H2(g) and I2(g) will react to form HI(g) and their concentration will decrease till QC = KC. The reaction quotient Qc is useful in predicting the direction of reaction by comparing the values of Qc and Kc Thus we can make following generalizations concerning the direction of the reaction. (Fig) Predicting the direction of the reaction.  If Qc < Kc , net reaction goes from left to right  If Qc > Kc , net reaction goes from right to left.
 If Qc = Kc , no net reaction occurs.
6. A mixture of H2, N2 and NH3 with molar concentrations 3 x 103 mol/L , 1.0 x 103 mol/L and 2.0 x 103 mol/L respectively was prepared at 500 K At this temperature, the value of KC for the reaction N2(g) + 3 H2(g) ⇌2 NH3(g) , is 61. Predict whether at this stage the concentration of NH3 will increase or decrease.
7. The value of Kc for the reaction
2 A ⇌ B + C
is 2 x 103 . At a given time , the composition of reaction mixture is [A] = [B] = [C] = 3 x 103 M. In which direction the reaction will proceed ?
iii) Calculations of Equilibrium Constants and
Equilibrium Concentrations
If the equilibrium concentrations of various reactants and products are known in a reaction, the equilibrium constant can be calculated. On the other hand, if the equilibrium constant is known, then the equilibrium concentrations can be calculated.
In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations , the following three steps shall be followed :
Step 1 : Write the balanced equation for the reaction.
Step 2 : Under the balanced equation, make a table that lists for each substance involved in the reaction.
(a) the initial concentration.
(b) the change in concentration on going to equilibrium and
(c) the equilibrium concentration.
In constructing the table, define x as the concentration (mol/L) of one of the substances that reacts on going to equilibrium, then use the stiochiometry of the reaction to determine the concentrations of the other substances in terms of x.
Step 3 : Substitute the equilibrium concentrations into equilibrium equation for the reaction and solve for x. If you have to solve a quadratic equation choose the mathematical solution that makes chemical sense.
Step 4 : Calculate the equilibrium concentrations from the calculated value of x.
Step 5 : Check your results by substituting them into the equilibrium equation.
8. 13.8 g N2O4 was placed in a 1 L reaction vessel at 400 K and allowed to attain equilibrium.
N2O4(g) ⇌ 2 NO2(g)
The total pressure at equilibrium was found to be 9.15 bar. Calculate Kc, Kp and partial pressure at equilibrium.
9. 3 mol of PCl3 kept in 1 L closed vessel was allowed to attain equilibrium at 380 K. Calculate the composition of the mixture at equilibrium . Kc = 1.80.
10. 2.0 moles of PCl5 were introduced in a 2 L flask and heated at 625 K to establish equilibrium when 60% of PCl5 was dissociated into PCl3 and Cl2. Find the equilibrium constant :
11. An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 675 K in a 5 litre volume contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodide. Calculate the equilibrium constant.
12. 1 mole of CO and 1 mol of H2O are taken in a 10 litre vessel and heated to 1200 K ; 40% of water reacts with carbon monoxide according to the equation :
H2O(g) + CO(g) ⇌ H2(g) + CO2(g)
Calculate the equilibrium constant for the reaction.
13. One mole of HI was heated in a closed vessel of 1 litre capacity. At equilibrium 0.2 mol was dissociated. Calculate the equilibrium constant.
14. The equilibrium constant of a reaction is found to be 0.612 at certain temperature. What will be the equilibrium constant for the reverse reaction
15. For a reversible reaction, the rate constant for forward reactions and backward reactions are 2.38 x 104 and 8.15 x 105 respectively. Calculate the equilibrium constant.
16. The following concentrations were obtained for the formation of NH3 from N2 and H2 atequilibrium at 500 K. [N2] = 1.5 x 102 M, [H2] = 3.0 x 102 M and [NH3] = 1.2 x 102 M. Calculate equilibrium constant.
17. PCl5, PCl3 and Cl2 are at equilibrium at 500 K and having concentration 1.59 M PCl3 , 1.59 M Cl2 and 1.41 M PCl5 . Calculate Kc for the reaction . (N 195 P 7.3)
PCl5 ⇌ PCl3 + Cl2
18. At equilibrium, the concentrations of N2 = 3.0 x 103 M , O2 = 4.2 x 103 M and NO = 2.8 x 103 M in a sealed vessel at 800 K. What will be Kc for the reaction.
N2(g) + O2(g) ⇌ 2 NO(g)
19. The value of Kp for the reaction
CO2(g) + C(s) ⇌ 2 CO(g)
is 3.0 at 1000 K . If initially PCO2 = 0.48 bar and PCO = 0 bar and pure graphite is present , calculate the equilibrium partial pressures of CO and CO2.
20. For the equilibrium
2 NOCl ⇌ CO2(g ) + H2(g)
The value of equilibrium constant Kc is 3.75 x 106 at 1069 K. Calculate the Kp for the reaction at this temperature.
The value of KC for a reaction does not depend on the rate of the reaction. However, it is directly related the thermodynamics of the reaction and in particular , to the change in Gibbs energy , G . If :
 G is negative , then the reaction is spontaneous and proceeds in the forward direction.
 G is positive , then the reaction is considered non-spontaneous. Instead , as reverse reaction would have a negative G , the products of forward reaction shall be converted to the reactants.
 G is zero, reaction has achieved equilibrium ; at this point , there is no longer any free energy left to drive the reaction.
A mathematical expression of this thermodynamic view of equilibrium can be described by following equation :
G = Gө + RT ln Q ……… (1)
where Gө is standard Gibbs energy.
At equilibrium, where G = 0 and Q = Kc, equation (1) becomes
G = Gө + RT ln K = 0
Gө =  RT ln K …… (2)

Taking antilog of both sides , we get,

Hence using the equation (3) , the reaction spontaneity can be interpreted in terms of the value of Gө.
 If Gө < 0 , then is positive and > 1 , making K > 1 , which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.
 If Gө > 0 , then is negative and < 1 , making K > 1 which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.
21. The value of Gө for the phosphorylation of glucose in glycolysis is 13.8 kJ mol1. Find the value of Kc at 298 K.
22. Hydrolysis of sucrose gives
Sucrose + H2O ⇌Glucose + Fructose
Equilibrium constant Kc for the reaction is 2 x 1013 at 300 K. Calculate atGө 300 K.
Factors Affecting Equilibrium Le-Chatelier's Principle
Le-Chatelier's principle states that ' If a system in equilibrium is subjected to a change of concentration, temperature or pressure, the equilibrium shifts in a direction that tries to cancel the effect of the change that is imposed'.
1. Effect of Change in Concentration
Consider the reaction involving the formation of ammonia :

If at equilibrium , more of H2 or N2 is added to the above reaction, equilibrium shifts in the forward direction resulting in the formation of more NH3. On the other hand, if NH3 is added then equilibrium shifts in backward direction.
Consider the reaction :

If the concentration of O2 is increased(or N2 increased), it increases the concentration of the reactants. To nullify the effect, the forward reaction increases or more nitric oxide will be formed. Thus the equilibrium shifts to the forward direction.
In general , it may be concluded that if the concentration of one or all the reactant species is increased, the equilibrium shifts in the forward direction and more of the products are formed. Alternately, if the concentration of one or all the product species is increased, the equilibrium shifts in the backward direction forming more of the reactants.
2. Effect of Pressure : The change of pressure has effect only on those equilibria which involve gaseous substances and proceed with a change in the number of moles of the gases. According to Le-Chatelier's principle, increase of external pressure should effect the equilibria in such a way as to reduce the pressure. This implies that the equilibrium will shift in the direction which has smaller number of moles of gaseous substances. This can be easily understood from the following equilibrium representing the formation of ammonia :

On increasing the pressure, the volume occupied by the system will decrease. As such there will be greater number of moles per unit volume. The effect of this change can be counter balanced if equilibrium shifts in the direction involving a decrease in number of moles. This can happen only if nitrogen and hydrogen combine to form ammonia. Thus increase in pressure in this case will be in favour of forward reaction.
Consider the equilibrium in the following reaction:

Here the backward reaction takes place with a decrease in the number of moles of gaseous species. As the pressure is increased, the volume decreases and the number of moles per unit volume increases. Therefore the equilibrium shifts in the backward direction. In this way, the number of moles of gases decreases and the effect of increase of pressure is counteracted to some extent. On the other hand , decrease in pressure shifts the equilibrium in favour of forward reaction.
Consider some reactions which do not involve any change in the number of moles of gaseous species :

In such cases, pressure does not have any effect on equilibrium.
Let us explain the effect of pressure on the solubility of gases in liquid solvents on the basis of Le-Chatelier's Principle. Consider the equilibrium involving the dissolution of CO2 in water. The equilibrium may be represented as follows :
CO2(g) ⇌ CO2(aq)
On increasing pressure of CO2 , the equilibrium shifts in the direction which results in decrease of pressure of CO2. The pressure of CO2 will be lowered only if CO2(g) dissolves in water to form CO2(aq). Thus the solubility of a gas in some liquid is directly proportional to the pressure of the gas in equilibrium with the solution.
3. Effect of Temperature
A chemical equilibrium involves two opposing reactions, one favouring products and other favouring the reactants. For example, consider the equilibrium representing the formation of ammonia.

In this equilibrium, the forward reaction is exothermic , while the backward reaction is endothermic. If the temperature is increased, the equilibrium will shift in the direction of endothermic reaction, which tends to undo the effect of added heat. Since the backward reaction is endothermic, so the equilibrium shifts in favour of backward reaction. In other words, it will result into lesser amount of ammonia. On the other hand, if temperature is decreased, the equilibrium will shift towards exothermic reaction. So low temperature favours the formation of ammonia.
In the case of reaction between nitrogen and oxygen, the forward reaction is endothermic, while backward reaction is exothermic.

According to Le-Chatelier's Principle, the increase in temperature will favour the forward(endothermic) reaction. This is because, the equilibrium shifts towards the direction in which heat is absorbed to relieve the system of the stress of added heat. On the other hand, decrease in temperature shifts the equilibrium in the direction of exothermic reaction, i.e., in favour of reactants in the above example.
We can explain the effect of temperature on the solubility of solids in liquids with the help of Le-Chatelier's principle. If during the dissolution heat is absorbed, solubility increases with increase in temperature. NH4Cl and NaNO3 are examples of such substances. On the other hand, substances such as CaCl2, Li2CO3, Li2SO4 and NaI liberate heat during dissolution. The solubility of such substances decreases with increase in temperature.
Dissolution of gases in liquids is always exothermic. Therefore, solubility of gases in liquids decreases with increase in temperature.
4. Effect of Catalyst
The presence of catalyst does not disturb the state of equilibrium because , it increases the rate of forward reaction as well as backward reaction to the same extent. It simply hastens the attainment of equilibrium.
Certain reactions such as the formation of ammonia from nitrogen and hydrogen cannot be carried out at high temperatures because high temperature, favours the backward reaction. So such reactions have to be carried out at low temperature. But at low temperature the rate of reaction becomes very slow and it takes very long time to attain equilibrium. In order to increase the rates of such reactions, generally , catalysts are used so that equilibrium is attained early. For example, in the case of the formation of ammonia, iron-molybdenum is used as catalyst.
5. Effect of Addition of Inert Gases
The effect of addition of inert gases (i.e., gas which does not react with any species involved in equilibrium) can be discussed under different conditions as described below.
(a) Addition of inert gas at constant volume : When inert gas is added to the equilibrium system at constant volume, it will cause the increase in total pressure of the system. But partial pressure of each of the reactant as well as product species will not be affected and will remain the same. Hence under these conditions, there will be no effect on the equilibrium.
(b) Addition of inert gas at constant pressure : When inert gas is added to the system at constant pressure, it will result in the increase in volume. As a consequence of this, the number of moles per unit volume of various reactants and products will decrease. To counterbalance this stress, the equilibrium will shift to the side where number of moles are increased.
For example, for an equilibrium :

The addition of inert gas at constant pressure will push
the equilibrium to the backward direction.
Similarly, for the equilibrium :

the addition of inert gas at constant pressure will push the equilibrium to the forward direction.
The effect of change of concentration, pressure and temperature has been summed up in the following table.

Stress Direction in which equilibrium shifts
Increase in the concentration of one or more reactants. Forward direction.
Increase in the concentration of one or more of products. Backward reaction
Increase in temperature Towards endothermic direction.
Decrease in temperature. Towards exothermic direction.
Decrease in pressure Towards larger number of gaseous moles.
Increase in pressure Towards lesser number of gaseous moles.
Addition of Catalyst No effect
Addition of inert gas :
a) At constant volume
b) At constant pressure
No effect
Towards larger number of moles of gaseous moles.

Electrolytes when dissolved in water splits up into charged particles called ions.. The process is called ionisation or dissociation. Certain electrolytes such as NaCl, KCl, HCl are almost completely ionised in solutions, whereas electrolytes such as NH4OH, CH3COOH etc. are weakly ionised. The electrolytes which are almost completely ionised in their solutions are called strong electrolytes. On the other hand, electrolytes which are weakly ionised in their solutions are called weak electrolytes. In case of solutions of weak electrolytes, the ions produced by dissociation of electrolyte are in equilibrium with undissociated molecules of the electrolyte. The equation for dissociation of strong electrolytes are written with only a single arrow directed to the right.
KCl(aq)  K+(aq) + Cl(aq)
NH4Cl(aq)  NH4+(aq) + Cl(aq)
On the other hand, equations for the dissociation of weak

According to Arrhenius concept :
An acid is a substance which can furnish hydrogen ions in its aqueous solution. A base is a substance which can furnish hydroxyl ions in its aqueous solution.
For example, substances such as HNO3, HCl, CH3COOH etc are acids, whereas substances such as NaOH , KOH , NH4OH etc. are bases, according to this concept.

Acids such as HCl and HNO3 which are almost completely ionised in aqueous solution are termed as strong acids whereas acids such as CH3COOH which are weakly ionised are called weak acids.
Similarly, bases which are almost completely ionised in aqueous solution are called Strong bases , for example, NaOH and KOH. The bases such as NH4OH are only slightly ionised are called weak bases.
According to Arrhenius theory , neutralisation of acids and bases is basically a reaction between H+ and OH ions in solutions.

Nature of Hydrogen ion in aqueous solutions
Hydrogen atom contain one proton and one electron. H+ ion is formed by loss of this electron. Therefore H+ ion is simply a proton. Charge density of this unshielded proton is very high. Therefore, it is not likely to exist as H+ ion. In an aqueous solution, H+ ion is considered to be present in hydrated form in combination with a water molecule as H3O+.
H+ + H2O  H3O+
H3O+ is called hydronium ion.
In 1923, Bronsted and Lowry independently proposed new definitions for acids and bases. They proposed that :
An acid is a substance that can donate a proton.
A base is a substance that can accept a proton.
These definitions are more general because according to these definitions even ions can behave as acids or bases. Moreover, these definitions are not restricted to reactions taking place in aqueous solutions only. In order to understand this concept of acids and bases , consider some specific example :

From the above equations, it is obvious that acid base reactions according to Bronsted-Lowry concept involve transfer of proton from the acid to a base. A substance can act as an acid only if another substance capable of accepting a proton is present.
Conjugate acid-base pairs
An acid after losing a proton becomes a base whereas a base after accepting the proton becomes an acid. For example, consider the reaction between water and ammonia as represented by the following equilibrium:

In the forward reaction, water donates a proton to ammonia(base) and acts as acid. In the reverse reaction NH4+ ions donate a proton to OH ions (base) and acts as acid. A base formed by the loss of proton by an acid is called conjugate base of the acid, whereas an acid formed by gain of a proton by the base is called conjugate acid of the base. In the above example, OH ion is the conjugate base of H2O and NH4+ ion is the conjugate acid of NH3. Acid- base pairs such as H2O  OH and NH4+ NH3 which are formed by loss or gain of proton are called conjugate acid-base pairs.
A strong acid would have large tendency to donate a proton. Thus, conjugate base of a strong acid would be a weak base. Similarly, a conjugate base of a weak acid would be a strong base.
Some more conjugate acidbase pairs has been given in the following equations :

It may be noted that in equation (1) H2O is behaving as a base whereas in equation (2) it s behaving as an acid. Similarly HCO3 ion in equation (3) acts as an acid and in equation (4) it acts as a base. Such substances which act as acids as well as bases are called amphoteric substances.
In both Arrhenius and Bronsted concepts, acids are sources of protons. Hence all Arrhenius acids are also Bronsted acids. However, there is a difference in the definition of bases. Arrhenius theory requires base to the source of OH ions in aqueous medium, but Bronsted theory requires base to be a proton acceptor. Hence Arrhenius bases may not be Bronsted bases. For example, NaOH is a base according to Arrhenius theory because it gives OH ions in aqueous solution, but NaOH does not accept proton as such. Hence it may not be classified as a base according to Bronsted theory.
23. Write the conjugate acids for the following Bronsted bases :
F , HSO4 , CO32 
24. Write the conjugate acid base of :
(i) HCO3 (ii) H2O (iii) HS (iv) NH3

25. In the reaction :

which reactant would be considered as a Bronsted
26. What will be the the conjugate bases for the following acid :
HF, H2SO4 and HCO3 ?
27. The species H2O, HCO3, HSO4 and NH3 can both act as
Bronsted acids and bases. For each give the corresponding
conjugated acid and base.
28. Which of the following are Lewis bases ?
H2O, BF3, H+, and NH3
29. What is the conjugate base of HCN ?
30. Which conjugate base is stronger CN or F ?
31. Write the conjugate acids for the following Bronsted bases :
NH2 , NH3 and HCOO.
Strengths of acids and bases
Strength of an acid is measured in terms of its tendency to lose proton whereas strength of a base is measured in terms of its tendency to accept proton. The conjugate base of a strong acid is a weak base.

On the other hand, conjugate base of a weak acid is a strong base.

A base is considered to be strong if it has great tendency to accept a proton. Therefore, conjugate acid of a strong base has little tendency to lose proton and hence is a weak acid.

On the other hand, conjugate acid of a weak base is a strong acid.

The strength of acids or bases is experimentally measured by determining its ionisation or dissociation constants.
Although Bronsted-Lowry theory was more general than Arrhenius theory of acids and bases , but failed to explain the acid base reactions which do not involve transfer of protons. For example it fails to explain how acidic oxides such as anhydrous CO2, SO2, SO3 etc. can neutralise basic oxides such as CaO, BaO etc. even in absence of solvent.
Lewis proposed a more general definition for acids and bases, which do not require the presence of protons to explain the acid-base behaviour.
Accoding to Lewis concept :
An acid is a substance which can accept a pair of electrons.
A base is a substance which can donate a pair of electrons.
Acid-base reactions according to this concept involve the donation of electron pair by a base to an acid to form a co-ordinate bond. Lewis bases can be neutral molecules such as

having one or more unshared pairs of electrons. , or anions such as CN , OH , Cl, etc.
Lewis acids are the species having vacant orbitals in the valence shell of one of its atoms. The following species can act as Lewis acids.
(a) Molecules having an atom with incomplete octet.
For example , BF3 and AlCl3.

(b) Simple cations.
For example H+, Ag+ etc.

( c) Molecules in which central atom has vacant orbitals
and may acquire more than octets of valence electrons.
Forexample, SiF4.

(d) Molecules containing multiple bonds. For example CO2, SO2, etc.

It may be noted that all Bronsted bases are also Lewis bases but all Bronsted acids are not Lewis acids. Lewis bases generally contain one or more lone pairs of electrons and therefore , they can also accept a proton (Bronsted base). Thus, all Lewis bases are also Bronsted bases. On the other hand, Bronsted acids are those which can give a proton, for example , HCl, H2SO4 . But they are not capable of accepting a pair of electrons . Hence , all Bronsted acids are not Lewis acids.

32. Classify the following species into Lewis acids and Lewis
bases and show how these act as such :
(a) HO (b) F c) H+ (d) BF3
33. Classify the following into acids and bases according to
Lewis concept:
SO3, CaO, OH, BF3, RNH2, S2, Ag+.
34. In the reaction:
SnCl4 + 2 Cl  [SnCl6]2
Which is Lewis acid and which one is the Lewis base ?
Acids like HCl, HNO3 and H2SO4 when dissolved in water dissociate completely and thus producing a large number of H+ ions. Hence these acids are called strong acids. On the other hand, acids like CH3COOH, HF, H2CO3, H3PO3 etc,. dissociate only to a small extent in the aqueous solution giving small amount of H+ ions and hence are called weak acids. Similarly bases like NaOH and KOH dissociate almost completely in the aqueous solution producing a large number of OH ions and are called strong bases, whereas bases like NH4OH, Ca(OH)2, Al(OH)3 etc dissociate only to a small extent in aqueous solution and are therefore , called weak electrolytes.
Strong acids dissociate almost completely in water and therefore the molar concentrations of H3O+ions in the solution is same as that of acid itself. But weak acids are not completely dissociated and relative strengths of weak acids can be compared in terms of their dissociation constants. For example, the dissociation equilibrium of an acid HA may be represented as :

Applying the law of Chemical equilibrium:

Since the concentration of water is very large and remains almost constant in solution, it can be combined to give another constant Ka.

Here Ka is called Dissociation constant of the acid.
The value of dissociation constant gives an idea about the relative strength of the acid. Larger the value of Ka ,greater is the concentration of H3O+ions and stronger is the acid. If dissociation constants of two acids are known, their relative strength can be compared. For example, consider the following examples:

= 1.80 x 105

Since Ka for CH3COOH is larger than Ka for HCN, acetic acid is a stronger acid than HCN.
It may be noted that the above expression of Ka is applicable to monobasic acids. The polybasic acids like H2S, H2CO3, H3PO4 etc. dissociate in stages and different stages of dissociation have different values of dissociation constants. For example, the dissociation stages of H2S , a dibasic acid are given as follows :

Similarly for tribasic acids like H3PO4 , we have three ionization constants. The values of ionization constants for some common polyprotic acids are given below.
It can be seen that higher order ionization constants (Ka2 , Ka3 ) are smaller than the lower order ionization constant (Ka1) of polybasic acid. The reason for this is that it is more difficult to remove a positively charged proton from a negative ion due to electrostatic forces. This can be seen in the case of removing a proton from the uncharged H2CO3 as compared from a negatively charged H2CO3. Similarly it is more difficult to remove a proton from a doubly charged HPO42 anion as compared to HPO4.
Polyprotic acid solutions contain a mixture of acids like H2A , HA, and A2 in case of a diprotic acid. H2A being a strong acid, the primary reaction involves the dissociation of H2A and H3O+ in solution comes mainly from the first dissociation step.
Factors Affecting Acid strength
The extent of dissociation of an acid depends on the strength and polarity of the HA bond.
In general , when strength of HA bond decreases , that is , the energy required to break the bond decreases. HA becomes a stronger acid. Also, when the HA bond becomes a stronger acid. Also, when the HA bond becomes more polar i.e., the electronegativity difference between the atoms H and A increases and there is marked charge separation, cleavage of bond becomes thereby increasing the acidity.
But it should be noted that while comparing elements in the same group of the periodic table, HA bond strength is a more important factor in determining acidity than its polar nature. As the size of A increases down the group, HA bond strength decreases and so the acid strength increases. For example,

Similarly, H2S is stronger acid than H2O.
But, for elements in the same row of the periodic table, HA bond polarity becomes the deciding factor for determining the acid strength. As the electronegativity of A increases , the strength of the acid also increases. For example,

Calculation of [H3O+] and degree of Dissociation
From the knowledge of Ka , it is possible to calculate hydronium ion concentration and degree of ionisation of a weak acid. Degree of dissociation or ionisation may be defined as the fraction of the total number of molecules an electrolyte (acid or base) which dissociates into ions. Thus ,

As an example, consider acetic acid. The following equation represents the ionisation of acetic acid in aqueous solution.

Suppose C moles of CH3COOH are dissolved per litre of the solution and  the degree of ionisation of CH3COOH, then at equilibrium the concentration of various species would be as follows :

Since for weak acids  is very small as compared to 1,  in the denominator can be neglected. The expression for Ka then becomes :

Knowing the value of Ka , it is possible to calculate the degree of ionisation of weak acid at any particular concentration C.

From the degree of ionisation, hydronium ion concentration can be calculated as :

The ionisation constant Kb for weak base BOH can be represented as follows :

(Concentration of water remains constant)
Smaller the value of ionisation constant for a base, weaker is the base.
If C is the molar concentration of base and  is its degree of dissociation, then the concentrations of each species at equilibrium are:

But for weak base ,  is very small so that ( 1   )  1

The relative strengths of acids and bases can be compared in terms of their dissociation or ionisation constants. The acid (or a base ) having a higher value of dissociation constant is stronger. But for a quantitative comparison, the concentration of H+ ions should be determined for both acids in solution of equal molarity. In general, strengths of two acids HA(1) and HA(2)

Where Ka(1) and Ka(2) are dissociation constants.
Similarly for two bases,

Ionisation constants of some acids
Acid Ka Acid Ka
Hydrofluoric acid 3.5 x 104 Benzoic acid 6.5 x 105
Nitrous acid 4.5 x 104 Formic acid 1.8 x 104
Niacin 1.5 x 105 Acetic acid 1.85 x 105
Hypochlorous acid 3.0 x 108 Hydrocyanic acid 4.9 x 1010
Phenol 1.3 x 1010
Ionisation constants of some bases
Base Kb Base Kb
Dimethyl amine 5.4 x 104 Triethyl amine 6.45 x 105
Ammonia 1.77 x 105 Quinine 1.10 x 106
Pyridine 1.77 x 109 Aniline 4.27 x 1010
Urea 1.3 x 1014
35. Calculate [H3O+] ion concentration of 0.01 M solution of an
acid HA. The dissociation constant of the acid is 1.6 x 105.
36. At 298 K a 0.1 M solution of acetic acid is 4 % ionised. What is the ionisation constant Ka for acetic acid.
37. Calculate the degree of ionisation of 0.01 M solution of HCN , if its Ka is 4.8 x 1010 . Also calculate the hydronium ion concentration.
38. Calculate the concentration of H3O+ ion in a mixture of 0.02 M acetic acid and 0.2 M sodium acetate. Ka of acetic acid is 1.8 x 105.
39. The acid dissociation constants of HCN, CH3COOH and HF are 7.2 x 10 10 , 1.80 x 105 and 6.7 x 104 respectively. Write their names in the increasing order of their strengths.
The conductivity measurements of water indicate that water is a weak electrolyte. Even in pure state water is weakly ionised to give H3O+(aq) and OH(aq) ions as shown under :
H2O(ℓ) + H2O(ℓ) ⇌H3O+(aq) + OH (aq)
Applying the law of equilibrium :

Since concentration of water is very high ( 55.5 M ) and only a small fraction of it undergo ionisation, therefore [H2O] may be taken as constant and may be combined with K to have another constant Kw .

The constant Kw is called ionic product of water. Its value is 1.008 x 1014 mol2 L2.
In pure water [ H3O+] and [ OH ] are equal. Therefore ,

Thus in pure water,
[H3O+] = [OH ] = 1.0 x 107 mol L1 at 298 K .
Since with increase in temperature dissociation of water increases, therefore, value of Kw increases as the temperature is increased. However, at all temperatures [H3O+] remains equal to [OH ] in pure water. The values of Kw at different temperatures are given in TABLE.
Temperature (K) Ionic Product (Kw)
273 0.113 x 1014
283 0.292 x1014
298 1.008 x 1014
313 2.917 x 1014
323 5.474 x 1014
373 7.50 x 1014


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