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

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6-2.Equilibrium-II (Ionic Equilibrium)

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

A

$2 \times 10^{-4}$

B

$10^{-8}$

C

$5 \times 10^{-11}$

D

$5 \times 10^{-12}$

Solution

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

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

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

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

Standard 11

Chemistry

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