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The $\mathrm{pH}$ of a buffer solution made by mixing $25 \mathrm{~mL}$ of $0.02 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ and $25 \mathrm{~mL}$ of $0.2 \mathrm{M}$ $\mathrm{NH}_4 \mathrm{Cl}$ at $25^{\circ}$ is $\left(\mathrm{pK}\right.$ of $\mathrm{NH}_4 \mathrm{OH}=4.8$ )
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Verified Answer
The correct answer is:
$8.2$
As a mixture of $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{NH}_4 \mathrm{Cl}$ acts as a basic buffer, so its pH must be basic, (i.e., greater than 7), hence the answer must be 2 nd. It can also be find by calculations :
$$
\begin{aligned}
\mathrm{pOH} & =\mathrm{p} K_b+\log \frac{[\text { salt }]}{[\text { base }]} \\
& =4.8+\log \frac{0.2 \mathrm{M}}{0.02 \mathrm{M}} \\
& =4.8 \times \log 10 \\
\mathrm{pOH} & =4.8+1=5.8 \\
\mathrm{pH} & =14-\mathrm{pOH}=14-5.8=8.2
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{pOH} & =\mathrm{p} K_b+\log \frac{[\text { salt }]}{[\text { base }]} \\
& =4.8+\log \frac{0.2 \mathrm{M}}{0.02 \mathrm{M}} \\
& =4.8 \times \log 10 \\
\mathrm{pOH} & =4.8+1=5.8 \\
\mathrm{pH} & =14-\mathrm{pOH}=14-5.8=8.2
\end{aligned}
$$
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