0.01 , M ,HA(aq.) is 2 % ionized, [OH^-] of solution is :- - Vidyadip

MCQ

$0.01\, M \,HA(aq.)$ is $2\%$ ionized, $[OH^-]$ of solution is :-

A
$2 \times 10^{-4}$
B
$10^{-8}$
✓
$5 \times 10^{-11}$
D
$5 \times 10^{-12}$

✓

Answer

Correct option: C.

$5 \times 10^{-11}$

c
$\left[\mathrm{H}^{+}\right]_{\mathrm{wea} \mathrm{k} \text { acid }}=\mathrm{C} \alpha=0.01 \times\left(\frac{2}{100}\right)$

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

$\left[\mathrm{H}^{+}\right] \times\left[\mathrm{OH}^{-}\right]=\mathrm{Kw}=10^{-14}$

$\therefore \quad\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{2 \times 10^{-4}}=5 \times 10^{-11}$

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