Ncert Exercises & Solutions

pH of 0.08 mol dm^-3 HOCl solution is 2.85. Calculate its ionisation constant.

pH of HOCl = 2.85

But - pH = \log [H^{+}]

∴ - 2.85 = \log [H^{+}]

\Rightarrow \bar{3} . 15 = \log [H^{+}]

\Rightarrow [H^{+}] = 1.413 \times 10^{- 3}

For weak monobasic acid [H^{+}] = \sqrt{K_{a} \times C}

\Rightarrow K_{a} = \frac{{[H^{+}]}^{2}}{C} = \frac{{(1.413 \times 10^{- 3})}^{2}}{0.08}

= 24.957 \times 10^{- 6} = 2.4957 \times 10^{- 5}