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pH the Power of H

50 mL of 0.2 ...

Question

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCI. If $p K_{b}$ , of ammonia solution is 4.75, the pH of the mixture will be :

A

3.75

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B

9.25

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C

8.25

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D

4.75

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Solution

The correct option is C 9.25
Number of millimoles of $N H_{3} = 50 m L \times 0.2 M = 10 m m o l$
Number of millimoles of $H C l = 25 m L \times 0.2 M = 5 m m o l$
5 mmol $N H_{3}$ will react with 5 mmol $H C l$ to form 5 mmol $N H_{4} C l$
$10 - 5 = 5$ mmol $N H_{3}$ will remain.
$p O H = p K_{b} - l o g \frac{[N H_{4} C l]}{[N H_{3}]}$
$p O H = 4.75 - l o g \frac{5 / V}{5 / V}$
$p O H = 4.75$
$p H = 14 - p O H = 14 - 4.75 = 9.25$

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