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Basic Buffer Action

The number of...

Question

The number of grams/weight of $N H_{4} C l$ required to be added to $3 l i t r e s$ of $0.01 M N H_{3}$ to prepare the buffer of $p H = 9.45$ at temperature $298 K$ ( $K_{b}$ for $N H_{3}$ is $1.85 \times 10^{- 5})$ .

A

0.354 g m

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B

4.55 g m

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C

0.455 g m

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D

3.55 g m

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Solution

The correct option is A $0.354 g m$
$∵ p O H = p K_{b} + log \frac{[N H_{4} C l]}{[N H_{3}]}$
$p O H = - log K_{b} + log \frac{[N H_{4} C l]}{[N H_{3}]}$
$(∵ p K_{b} = - log K_{b})$
Also, $p O H = 14 - p H = 14 - 9.45 = 4.55$
and $log \frac{[N H_{4} C l]}{[N H_{3}] . K_{b}} = log 3 \approx 0.470$
Thus, $log 10^{14} - (log 10^{9} + log 3) \approx log \frac{[N H_{4} C l]}{[N H_{3}] . K_{b}}$
$\Rightarrow log \frac{10^{14}}{10^{9} \times 3} = log \frac{[N H_{4} C l]}{[N H_{3}] . K_{b}}$
$\Rightarrow \frac{10^{14}}{10^{9} \times 3} \approx \frac{[N H_{4} C l]}{[N H_{3}] . K_{b}}$
or $\frac{10^{14} \times [N H_{3}] . K_{b}}{10^{9} \times 3}$
$= [N H_{4} C l] \approx \frac{10^{14}}{10^{9} \times 3} \times 0.01 \times 1.85 \times 10^{- 5}$
$= [N H_{4} C l] \approx 0.354 g m$ .

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