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Henderson Hassleback Equation

In a basic bu...

Question

In a basic buffer, $0.0025$ mole of $N H_{4} C l$ and $0.15 m o l e$ of $N H_{4} O H$ are present. The $p H$ of the solution will be:
$(p K_{a}) = 4.74$

A

11.04

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B

10.24

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C

6.62

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D

5.48

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Solution

The correct option is A $11.04$
From Henderson's equation,

$p O H = p K_{b} + l o g \frac{[s a l t]}{[b a s e]}$

$= 4.74 + l o g \frac{0.0025}{0.15}$

$= 4.74 + l o g \frac{1}{60}$
$= 2.96$

$p H = 14 - 2.96$
$= 11.04$

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