A basic buffer solution contains 0.3 M NH _4 OH and 0.2 M ( NH _4… - Vidyadip

Question

A basic buffer solution contains $0.3 M NH _4 OH$ and $0.2 M \left( NH _4\right)_2 SO _4$. If $K _{ b }$ for $NH _4 OH$ at a certain temperature is $2 \times 10^{-5}$, what is the $pH$ of the solution?

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Answer

$\begin{aligned}
& \text { Given: }\left[ NH _4 OH \right]=0.3 M \\
& {\left[ NH _4{ }^{+}\right]=2 \times 0.2=0.4 M } \\
& p K_{ b }=-\log _{10} K_{ b }=-\log _{10} 2 \times 10^{-5}=\overline{5} .3010 \\
& =4.6990 \\
& pOH = p K_b+\log \frac{\left[ NH _4^{+}\right]}{\left[ NH _4 OH \right]} \\
& =4.6990+\log \frac{0.4}{0.3} \text {} \\
& =4.6990+0.1249 \\
& =4.8239 \\
& pH + pOH =14 \\
& \therefore pH =14- pOH =14-4.8239=9.1761 \\
&
\end{aligned}$
$pH =9.1761$

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