Q13 / 160 · TS EAMCET 2019 (04 May Shift 2) · Chemistry
The concentration in $M$ of $\mathrm{OH}^{-}$in $0.001 \mathrm{M}$ $\mathrm{H}_2 \mathrm{SO}_4$ is
- A$1 \times 10^{-13}$
- B$0.5 \times 10^{-12}$
- Correct answerC$5 \times 10^{-12}$
- D$0.5 \times 10^{-13}$
Solution
$\mathrm{H}_2 \mathrm{SO}_4(0.001 \mathrm{M}) \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-}$
Concentration of $\mathrm{H}^{+}=2 \times 0.001 \mathrm{M}=0.002 \mathrm{M}$
$$
\begin{aligned}
& \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log (0.002) \\
& \mathrm{pH}=3-0.3010=2.6990
\end{aligned}
$$
From, $\mathrm{pH}+\mathrm{pOH}=14$
$$
\mathrm{pOH}=14-26990=11.3010
$$
$$
\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]
$$
or
$$
11.3010=-\log \left[\mathrm{OH}^{-}\right]
$$
or
$$
\left[\mathrm{OH}^{-}\right]=5 \times 10^{-12}
$$
Hence, the correct answer is option: 'C'.