Question

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

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

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

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

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

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



$(\because C=1m)$

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

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

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

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

$=-0.7+3$

$pH=2.3$