Question

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

• A. 3.3
• B. 2.3
• C. 4.3
• D. 3.0

Answer 1

Answer: (B)

Given, ionisation constant $\left(K_{a}\right)=2.5 \times 10^{-5}$ and molarity $(c)=1.0\, M$

$\because K_{a}=C \alpha^{2} (\because \alpha<<1)$

$\therefore \alpha=\sqrt{\frac{K_{a}}{C}}=\sqrt{\frac{2.5 \times 10^{-5}}{1}}$

$=0.5 \times 10^{-3}$

or $\alpha=5 \times 10^{-3}$

Also, for dissociation of $HOCl$, the required relation is as follows



$(\because C=1\, m )$

$\therefore \left[ H ^{+}\right] =\alpha=5 \times 10^{-4}$

and $pH =-\log \left[ H ^{+}\right]$

$=-\log \left[5 \times 10^{-3}\right]$

$=-\log 5+3 \log 10$

$=-0.7+3$

$pH =2.3$