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(a) Define:

(i) Common ion effect with example

(ii) Solubility product

(iii) Buffer solution.

(b) Calculate degree of ionization of 0.1 mol/lit of acetic acid, given Ka for CH3COOH = 1.8 x 10–5 .

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(a) (i) Common ion effect: The effect by which the ionization of one electrolyte is suppressed by the presence of a common ion.

Example: A mixture of CH3COOH and CH3COONa

CH3COOH (aq) ⇌ CH3COO + H+ (aq)

(Weak electrolyte)

CH3COONa → CH3COO + Na+ (aq)

(Strong electrolyte) Common ion

The degree of ionisation of acetic acid is suppressed by the addition of a common ion containing electrolyte i.e., sodium acetate.

(ii) Solubility product: The solubility product of a sparingly soluble salt at a given temperature may be defined as the product of the concentration of its ions in the saturated solution, with each concentration term raised to the power equal to the number of times the ion occurs in the equation representing the dissociation of the electrolyte.

(iii) Buffer solution: A buffer solution is the solution which can resist the change in pH on addition of small amount of acid or base. The ability of buffer solution to resist change in pH on addition of acid or base is called buffer action.

(b) Concentration = 0.1 moles/liter

Ka = 1.8 × 10–5

Ka = Ca2

1.8 × 10–5 = 0.1 × a2

\(\sqrt{{1.8\times10^{-5}}\over0.1}\) = a

a = 0.0134

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