If the ionization constant of hypochlorous acid (HOCl) is $2.5 \times 10^{-5}$, the pH of $1.0 \mathrm{M}$ of its solution is $(\log 5=0.7)$

• A

  3.3
• Correct answer
  B

  2.3
• C

  4.3
• D

  3.0

Solution

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

Hence, the correct answer is option: 'B'.