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Henderson Hassleback Equation

If pKaNH4+=9....

Question

If $p K_{a} (N H_{4}^{+}) = 9.2$ , then the amount of $(N H_{4})_{2} S O_{4}$ to be added in 500 mL of 0.1 M $N H_{4} O H$ for the preparation of a buffer of pH=8.5 would be:
(Given: log5=0.7)

0.175 mol

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0.125 mol

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0.150 mol

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0.20 mol

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Solution

The correct option is B 0.125 mol
This will make a basic buffer.
So, according to the Henderson- Hasselbalch equation,
$p O H = p K_{b} + l o g \frac{[s a l t]}{[b a s e]}$
So, $5.5 = 4.8 + l o g \frac{n_{s a l t}}{0.05}$
$0.7 = l o g \frac{n_{s a l t}}{0.05}$
$\Rightarrow l o g \frac{(n_{s a l t})}{0.05} = l o g 5$
$\Rightarrow n_{s a l t} = 0.25 m o l$
Since $(N H_{4})_{2} S O_{4}$ includes 2 $N H_{4}^{+}$ ions, the amount of $(N H_{4})_{2} S O_{4}$ required will be 0.125 mol.

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