If the ionization constant of hypochlorous acid (HOCl) is $2.5 \times 10^{-5}$, the pH of $1.0 \mathrm{M}$…

1. 3.3
2. 2.3
3. 4.3
4. 3.0

Solution

$$ \begin{aligned} & \text { Given, ionisation constant }\left(K_aight)=2.5 \times 10^{-5} \\ & \text { and molarity }(c)=1.0 \mathrm{M} \\ & \because K_a=C \alpha^2 \\ & \qquad \begin{aligned} \therefore & =\sqrt{\frac{K_a}{C}}=\sqrt{\frac{2.5 \times 10^{-5}}{1}} \\ & =0.5 \times 10^{-3} \\ \text { or } \quad \alpha & =5 \times 10^{-3} \end{aligned} \end{aligned} $$ or Also, for dissociation of $\mathrm{HOCl}$, the required relation is as follows
$$ (\because C=1 \mathrm{~m}) $$ $$ \begin{aligned} & \therefore \quad\left[\mathrm{H}^{+}ight]=\alpha=5 \times 10^{-4} \\ & \text { and } \mathrm{pH}=-\log \left[\mathrm{H}^{+}ight] \\ & =-\log \left[5 \times 10^{-3}ight] \\ & =-\log 5+3 \log 10 \\ & =-0.7+3 \\ & \mathrm{pH}=2.3 \\ & \end{aligned} $$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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