Question

The $pH$ of $0.5M$aqueous solution of $HF$ $(K_a=2\times {10}^{-4})$ is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

$K_a=c\times {\alpha }^2$

$2\times {10}^{-4}=0.5\times {\alpha }^2$

$\alpha =2\times {10}^{-2}$

$\left[H^{+}\right]=c\times \alpha$

$\therefore \left[H^{+}\right]=0.5\times 2\times {10}^{-2}={10}^{-2}$

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

$=-log\left[{10}^{-2}\right]$

$=2\times \left(log10\right)$

$=2$

Hence, option B is correct.