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$50 \mathrm{~mL}$ of $\mathrm{H}_2 \mathrm{O}$ is added to $50 \mathrm{~mL}$ of $1 \times 10^{-3} \mathrm{M}$ barium hydroxide solution. What is the $\mathrm{pH}$ of the resulting solution?
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Verified Answer
The correct answer is:
$11.0$
In water, barium hydroxide is hydrolysed as follows.
$$
\begin{aligned}
\mathrm{Ba}(\mathrm{OH})_2 \rightleftharpoons & \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-} \\
\text {conc. of } \mathrm{Ba}^{2+} & =1 \times 10^{-3} \mathrm{M} \\
\text { conc. of }\left[\mathrm{OH}^{-}\right] & =2 \times 1 \times 10^{-3} \mathrm{M} \\
& =2 \times 10^{-3} \mathrm{M} \\
\text { pOH } & =-\log \left[\mathrm{OH}^{-}\right] \\
& =-\log \left(2 \times 10^{-3}\right) \\
& =2.69
\end{aligned}
$$
$\begin{aligned} \mathrm{pH}+\mathrm{pOH} & =14 \\ \mathrm{pH} & =14-\mathrm{pOH} \\ & =14-2.69 \\ & =11.3 \\ & \approx 11.0\end{aligned}$
$$
\begin{aligned}
\mathrm{Ba}(\mathrm{OH})_2 \rightleftharpoons & \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-} \\
\text {conc. of } \mathrm{Ba}^{2+} & =1 \times 10^{-3} \mathrm{M} \\
\text { conc. of }\left[\mathrm{OH}^{-}\right] & =2 \times 1 \times 10^{-3} \mathrm{M} \\
& =2 \times 10^{-3} \mathrm{M} \\
\text { pOH } & =-\log \left[\mathrm{OH}^{-}\right] \\
& =-\log \left(2 \times 10^{-3}\right) \\
& =2.69
\end{aligned}
$$
$\begin{aligned} \mathrm{pH}+\mathrm{pOH} & =14 \\ \mathrm{pH} & =14-\mathrm{pOH} \\ & =14-2.69 \\ & =11.3 \\ & \approx 11.0\end{aligned}$
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