Q.ix. pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation constant.

For a weak acid HA that dissociates partially, $HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)$, the acid-dissociation constant is $K_a=\dfrac{[H^+][A^-]}{[HA]}$; for a weak base BOH, $BOH(aq)\rightleftharpoons B^+(aq)+OH^-(aq)$, the base-dissociation constant is $K_b=\dfrac{[B^+][OH^-]}{[BOH]}$. A larger Ka or Kb means a stronger (more dissociated) weak acid or base. Ostwald's dilution law connects this constant to the degree of dissociation $\alpha$ (the fraction of the electrolyte that has split into ions) and the initial concentration $c$: starting 1 mol of acid in V dm3 of solution and substituting the equilibrium amounts into the Ka expression gives the exact relation $K_a=\dfrac{\alpha^2c}{1-\alpha}$, which for a genuinely weak acid (small $\alpha$, so $1-\alpha\approx1$) simplifies to the very useful approximate form $K_a\approx\alpha^2c$, i.e. $\alpha\approx\sqrt{K_a/c}$. The same pair of relations, $K_b=\dfrac{\alpha^2c}{1-\alpha}\approx\alpha^2c$, holds for a weak base with Kb. This is the single relation that lets a chemist move freely between three quantities -- the dissociation …