COMMON ION EFFECT:

        Let AB to the weak electrolyte. Considering its dissociation,

        AB ↔ A⁺ + B^-

        and applying law of mass action we have

         K = [A⁺][B^-]/[AB]

The equilibrium constant, K, has a definite value at any given temperature. If now another electrolyte furnishing the A⁺ and B^- ions be added to the above solution. It will increases the concentration of either A⁺ ions or B^- ions (whichever has been added) and in order that K may remain constant, the concentration of AB must increases, i.e., the equilibrium will shift to the left hand side.

In other words, the degree of dissociation of an electrolyte (weak) is suppressed by the addition of another electrolyte (strong) containing a common ion. This is termed as common ion effect. Acetic acid is a weak electrolyte and its ionization is suppressed in presence of a strong acid (H⁺ ion as common ion) or a strong salt like sodium acetate (acetate ion as common ion). Similarly, the addition of NH₄Cl or NaOH to NH₄OH solution will suppress the dissociation of NH₄OH due to common ion either NH^-₄ or OH^-.

As a result of common ion effect, the concentration of the ion not in common in two electrolytes, is decreased. The use of this phenomenon is made in qualitative analysis to adjust concentration of S^2- ions in second group and OH^- ion concentration in third group of analysis.

Example 6:  The ionization constant of HCN is 4 × 10^-10. Calculate the concentration of hydrogen ions in 0.2 M solution of HCN containing 1 mol L^-1 of KCN?

Solution:   The dissociation of HCN is represented as

                   HCN ↔ H⁺ + CN^-

                   Applying law of mass action,

                    K_a =   ([H⁺ ][CN^-])/[HCN] or [H⁺ ] (K_a [HCN])/[CN^- ]

In presence of strong electrolyte, the total CN^- concentration comes from KCN which undergoes complete dissociation. It is further assumed that dissociation of HCN is very-very small and the concentration of HCN can be taken as the concentration of undissociated HCN.

Thus, [HCN] = 0.2 M and [CN^-] = 1M

Putting these values in the expression

[H⁺] = (K_a [HCN])/([CN^- ]) = (4×10^-10×0.2)/1 = 8×10^-11 mol L^-1

[Note: When KCN is not present, the [H+] concentration is equal to √CK i.e., √(0.2*4*10^-10) = 8.94 *10^-8mol L^-1 . This shows that concentration of H⁺ ions fails considerably when KCN is added to HCN solution.