To what volume should 10 litre of 0.5M CH₃COOH (K_a = 1.8×10^–5) be diluted in order to double the hydroxide ion concentration :

${\left[{\mathrm{H}}^{+}\right]}_{\mathrm{i}}=\sqrt{{\mathrm{K}}_{\mathrm{a}}{\mathrm{C}}_{\mathrm{i}}}=3\times {10}^{-3}$

[OH^-] is doubled means [H⁺ ] becomes half of initial [H⁺ ]

${\left[{\mathrm{H}}^{+}\right]}_{\mathrm{f}}=\frac{3\times {10}^{-3}}{2}=1.5\times {10}^{-3}$

${\left[{\mathrm{H}}^{+}\right]}_{\mathrm{f}}=\sqrt{{\mathrm{K}}_{\mathrm{a}}{\mathrm{C}}_{\mathrm{f}}}\Rightarrow {\mathrm{C}}_{\mathrm{f}}=0.125$

Calucuting final volume

${\text{M}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}={\mathrm{M}}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{f}}\Rightarrow {\mathrm{V}}_{\mathrm{f}}=\frac{0.5\times 10}{0.125}=40 \mathrm{lit}$