Q.Calculate $\mathrm{[H_3O^+]}$ in 0.1 mol $\mathrm{dm^3}$ solution of acetic acid. Given : $K_a\ \mathrm{[CH_3COOH]} = 1.8 \times 10^{-5}$

[!NOTE]
The statement is printed exactly as in the textbook, including its own "0.1 mol dm$^{3}$" -- the superscript minus is missing in the print. The concentration is $0.1\ \mathrm{mol\ dm^{-3}}$, which is what the solution uses.

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 …