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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium3 Marks
Question
pH of $0.08 \mathrm{mol} \mathrm{~dm}^{-3} \mathrm{HOCl}$ solution is 2.85 . Calculate its ionisation constant.
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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 $[\text{H}^+]=\sqrt{\text{K}_\text{a}\times\text{C}}$
$\Rightarrow\text{K}_\text{}a=\frac{[\text{H}^+]^2}{\text{C}}=\frac{(1.413\times10^{-3})^2}{0.08}$
$=24.957\times10^{-6}=2.4957\times10^{-5}$

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