The $p H$ of a $0.02 M N H_{4} C l$ solution will be: $[Given K_{b} (N H_{4} O H) = 10^{- 5} and log 2 = 0.301] .$

4.65

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5.35

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4.35

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2.65

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Solution

The correct option is B $5.35$
$N H_{4} C l$ is a salt of strong acid and weak base. Hence the formula for calculating its $p H$ can be written as
$p H = \frac{1}{2} [p K_{w} - p K_{b} - log C]$
$K_{b} = 10^{- 5}$
$p K_{b} = - l o g K_{b}$
$p K_{b} = 5$
$p H = \frac{1}{2} [14 - p K_{b} - log C] p H = 7 - \frac{5}{2} - \frac{1}{2} (log 2 \times 10^{- 2}) p H = 7 - \frac{5}{2} - \frac{1}{2} \times 0.301 + 1 p H = 7 - \frac{5}{2} - 0.1505 + 1 p H = 5.35$
Hence, option (b) is correct.

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p H

0.02 M N H_{4} C l

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