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Introduction and Degree of Hydrolysis

The pH of a 0...

Question

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 $log 2 = 0.301$ ]

A

2.65

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B

4.35

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C

4.65

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D

5.35

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Solution

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

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Similar questions

p H

0.02 M N H_{4} C l

solution will be:
[Given:

K_{b} (N H_{4} O H) = 10^{- 5}

log 2 = 0.301

]

p H

0.02 M N H_{4} C l

solution will be:
[Given:

K_{b} (N H_{4} O H) = 10^{- 5}

log 2 = 0.301

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10^{- 4.5} = 3.162 \times 10^{- 5}

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M g

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Assume constant volume

M g = 24.3 g / m o l (log 3 = 0.47, log 2 = 0.301)

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Introduction and Degree of Hydrolysis

Standard XIII Chemistry

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