Q.19 Derive Ostwald’s dilution law for weak acid. Obtain relation between solubility product and its solubility for Al(OH)3.
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Answer:
Ostwald’s dilution law: Consider a weak monobasic acid HA of initial concentration \(c\) mol L−1 and degree of dissociation \(\alpha\):
| HA | H+ | A− | |
|---|---|---|---|
| Initial | \(c\) | 0 | 0 |
| At equilibrium | \(c(1-\alpha)\) | \(c\alpha\) | \(c\alpha\) |
\(K_a = \dfrac{[H^+][A^-]}{[HA]} = \dfrac{c\alpha \cdot c\alpha}{c(1-\alpha)} = \dfrac{\alpha^2 c}{1-\alpha}\)
For a weak acid \(\alpha \ll 1\), so \(1 - \alpha \approx 1\): \(K_a = \alpha^2 c\), i.e. \(\alpha = \sqrt{\dfrac{K_a}{c}}\).
Thus the degree of dissociation of a weak electrolyte is inversely proportional to the square root of its concentration (directly proportional to the square root of dilution). This is Ostwald’s dilution law.
Al(OH)3: If its molar solubility is \(S\) mol L−1:
Al(OH)3(s) ⇌ Al3+(aq) + 3OH−(aq), with \([Al^{3+}] = S\) and \([OH^-] = 3S\)
\(K_{sp} = [Al^{3+}][OH^-]^3 = S(3S)^3 = 27S^4\), so \(S = \left(\dfrac{K_{sp}}{27}\right)^{1/4}\)
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