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
$\mathrm{HA}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-}$
$\mathrm{HA}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-}$
$\Rightarrow \frac{\left[\mathrm{H}^{+}\right]}{[\mathrm{HA}]} \times 100=2$
$\Rightarrow \frac{\left[\mathrm{H}^{+}\right]}{0.01} \times 100=2$
$\Rightarrow\left[\mathrm{H}^{+}\right]=\frac{0.02}{100}=2 \times 10^{-4} \mathrm{M}$
$\therefore\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{2 \times 10^{-4}}=0.5 \times 10^{-10} \mathrm{m}$
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