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$5 \times 10^{-12}$
$=2 \times 10^{-4}$
$\left[\mathrm{H}^{+}\right] \times\left[\mathrm{OH}^{-}\right]=\mathrm{Kw}=10^{-14}$
$\therefore \quad\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{2 \times 10^{-4}}=5 \times 10^{-11}$
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