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 S2- 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,
Ka = ([H+ ][CN-])/[HCN] or [H+ ] (Ka [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+] = (Ka [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.
Example 7: Determine the concentration of hydroxyl ions in 0.4 M NH4OH solution having (i) no ammonium chloride, (ii) 5.35 g of NH4Cl in a litre of the solution. Ionization constant of NH4OH is 1.8 × 10-5.
Solution: (i) Let 'α' be the degree of dissociation of NH4OH is absence of NH4Cl.
α = √(Kb/c)
So, [OH-] = C α = √(Kbc) = √(1.8 * 10-5 * 0.4)
= 2.68 × 10-3 mol L-1
(ii) In presence of NH4Cl
[NH-4] = 5.35/53.5 = 0.1M and [NH4OH] = 0.4M
So, [OH-] =(Kb [NK4OH])/[NH4-]=(1.8×10-3×0.4)/0.1
= 7.2 × 10-5 mol L-1
Example 8: When 0.100 mole of ammonia, NH3, is dissolved in sufficient water to make 1.0 L solution, the solution is found to have a hydroxide ion concentration of 1.34 × 10-3 M. Calculate Kb for ammonia.
NH3 + H2O ↔ NH+4
At equilibrium (0.100 -1.34 ×10-3) M 1.34×10-3 M
= 0.09866 M + OH-
1.34 × 10-3 M
Kb = [NH4+ ][OH- ]/[NK3 ] = (1.34×10-3 × 1.34×10-3)/0.09866
Example 9: Ka for HA is 4.9 ×10-8. After making the necessary approximation, calculate for its decimolar solution
(a) % dissociation
(b) H+ ion concentration.
Solution: For a weak electrolyte.
α = √K/C = √((4.9×10-8)/0.1)
= 7 × 10-4
% dissociation = 100 × α = 100 × 7 × 10-4
= 7 ×10-2
HA ↔ H+ + A-
C(-1-α) Cα Cα
[H+] = C × α = 0.1 ×7 × 10-4 = 7 × 10-5 mol L-1
Example 10: Nicotinic acid (Ka = 1.4 ×10-5) is represented by the formula HNiC. Calculate its per cent dissociation in a solution which contains 0.10 mole of nicotine acid per 2 litre of solution.
Solution: Initial concentration of the nicotinic acid = 0.10/2 = 0.05molL-1
HNiC ↔ H+ + NiC-
Equilibrium conc. (0.05-x) x x
As x is very small, (0.05 - x), can be taken as 0.05
Ka = [H+ ][NiC- ]/[HNiC] = (x × x)/0.05
or x2 = (0.05) × (1.4 × 10-5)
or x = 0.83 × 10-3 mol L-1
% dissociation = (0.83×10-3)/0.05×100 = 1.66
Alternative method: Let α be the degree of dissociation
NHiC ↔ H+ + NiC-
At equilibrium 0.05 (1 -α) 0.05α 0.05α
Ka = (0.05α×0.05α)/0.05(1-α)
As α is very small, (1-α) → 1.
So, 1.4 × 10-5 = 0.05 α2
or α√((1.4×10-5)/0.05=1.67×10-2 )
Per cent dissociation = 100 × α = 100 × 1.67 ×10-2
Example 11: At 30o C the degree of dissociation of 0.066 M HA is 0.0145. What would be the degree of dissociation of 0.02 M solution of the acid at the same temperature?
Solution: Let the ionization constant of the acid be Ka.
Degree of dissociation at 0.66 M concentration = 0.0145.
Applying α = √Kα/C
0.0145 = √Kα/0.066 ..... (i)
Let the degree of dissociation of the acid at 0.02 M concentration be α1.
α1 = √Kα/0.02 ...... (ii)
Dividing Eq. (ii) by Eq. (i)
or α1 = 0.0145 × 1.8166 = 0.0263
Example 12: A solution contains 0.1 M H2S and 0.3 M HCl. Calculate the concentration of S2- and HS- ions in solution. Given and for H2S are 10-7 and 1.3 × 10-13 respectively.
Solution: H2S ↔ H+ + HS-
Kα1=[H+ ][HS- ]/[H2 S] ....... (i)
Further HS- = H+ + S2-
Kα2=[H+ ][HS2- ]/[H2 S] ....... (ii)
Multiplying both the equations
Kα1×Kα2=([H+ ]2 [HS2- ])/[H2S]
Due to common ion, the ionization of H2S is suppressed and the [H+] in solution is due to the presence of 0.3 M HCl.
Putting the value of [S2-] in Eq. (ii)
1.3×10-13 = (0.3×1.44×10-20)/([HS-])
or [HS-] (0.3×1.44×10-20)/(1.3×10-13)= 3.3×10-8 M