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$20 m L o f 0.1 M H_{2} S O_{4}$ solution is added to 30 mL of $0.2 M N H_{4} O H$ solution. The pH of the resultant mixture is $[p K_{b} o f N H_{4} O H = 4.7]$

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answer is 9.

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Detailed Solution

$20 m l 0.1 M H_{2} S O_{4} \Rightarrow η_{H +} = 4 M m o l 30 m l 0.2 M N H_{2} O H \Rightarrow η_{N H_{4} O H} = 6 M m o l \begin{array}{l} N H_{4} O H + H^{+} N H_{4}^{+} + H_{2} O \\ \Rightarrow 6 4 0 0 \\ \Rightarrow 2 0 4 4 \end{array}$

So, solution is a basic buffer

$p O H = p K_{b} + \log \frac{N H_{4}^{+}}{N H_{4} O H}$

$p O H = p K_{b} + \log 4 / 2$

$p O H = 4.7 + \log 2$

$p O H = 4.7 + 0.3 = 5$

$p H + p O H = 14$

$p H = 14 - 5 = 9$

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