i. Define degree of dissociation. Derive Ostwald's dilution law for the CH3COOH.

i.

Consider an equilibrium of weak acid CH₃COOH that exists in solution partly as the undissociated species CH₃COOH and partly H⁺ and CH₃COO^- ions. Then

\[CH_3COOH_{(aq)} \rightleftharpoons H^+_{(aq)} + CH_3COO^-_{(aq)}\]

ii.

The acid dissociation constant is given as:

\(K_a = \frac{{[H^+][CH_3COO^-]}}{{[CH_3COOH]}}\) ....(1)

iii.

Suppose 1 mol of acid CH₃COOH is initially present in volume V dm³ of the solution. At equilibrium, the fraction dissociated would be \(\alpha\), where \(\alpha\) is the degree of dissociation of the acid. The fraction of an acid that remains undissociated would be (1 - \(\alpha\)).

iv.

Thus, at equilibrium \([CH_3COOH] = (1 - \(\alpha\))/V\) mol dm^-3,

\([H^+] = [CH_3COO^-] = \(\alpha\)/V\) mol dm^-3

v.

Substituting these in equation (1),

\(K_a = (\alpha / V)(\alpha / V) /((1 - \(\alpha\)) / V) = \alpha^2/(1 - \(\alpha\) V)\) ....(2)

vi.

If c is the initial concentration of CH₃COOH in mol dm^-3 and V is the volume in dm³ mol^-1 then c = 1/V. Replacing 1/V in equation (2) by c,

we get

\(K_a = (\alpha^2c)/(1 - \(\alpha\))\) ....(3)

vii.

For the weak acid CH₃COOH, \(\alpha\) is very small, or (1 - \(\alpha\)) \(\approx\) 1. With this, equation (2) and (3) become:

\(K_a = \(\alpha^2/V\)\) and \(K_a = \(\alpha^2c\) ....(4)

\(\alpha = \sqrt{K_a/c}\) or \(\alpha = \sqrt{K_aV}\) ....(5)

Equation (5) implies that the degree of dissociation of a weak acid (CH₃COOH) is inversely proportional to the square root of its concentration or directly proportional to the square root of the volume of the solution containing 1 mol of the weak acid.