Question
pH of 0.08mol dm–3 HOCl solution is 2.85. Calculate its ionisation constant.
✓
Answer
pH of HOCl = 2.85
But, $-\text{pH}=\log\text{[H}^+]$
$\therefore-2.85=\log\text{[H}^+]$
$\Rightarrow\overline{3}.15=\log\text{[H}^+]$
$\Rightarrow\log\text{[H}^+]=1.413\times10^{-3}$
For weak mono basic acid $[\text{H}^+]=\sqrt{\text{K}_\text{a}\times\text{C}}$
$\Rightarrow\text{K}_\text{}a=\frac{[\text{H}^+]^2}{\text{C}}=\frac{(1.413\times10^{-3})^2}{0.08}$
$=24.957\times10^{-6}=2.4957\times10^{-5}$
But, $-\text{pH}=\log\text{[H}^+]$
$\therefore-2.85=\log\text{[H}^+]$
$\Rightarrow\overline{3}.15=\log\text{[H}^+]$
$\Rightarrow\log\text{[H}^+]=1.413\times10^{-3}$
For weak mono basic acid $[\text{H}^+]=\sqrt{\text{K}_\text{a}\times\text{C}}$
$\Rightarrow\text{K}_\text{}a=\frac{[\text{H}^+]^2}{\text{C}}=\frac{(1.413\times10^{-3})^2}{0.08}$
$=24.957\times10^{-6}=2.4957\times10^{-5}$
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