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Buffer Solutions

pK b of N H 3...

Question

$p K_{b}$ of $N H_{3}$ is 4.74. The $p H$ when 100 mL of 0.01 M $N H_{3}$ solution is 50% neutralised by 0.01 M HCl is,

A

4.74

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B

2.37

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C

9.26

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D

9.48

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Solution

The correct option is C 9.26
At 50% neutralisation,
$[N H_{4}^{+}] = [N H_{3}]$
The solution is a buffer
$∴ p O H = p K_{b} + l o g \frac{[N H_{4}^{+}]}{[N H_{3}]}$ = 4.74 + log 1 = 4.74
pH = 14 - 4.74 = 9.26

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