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Question: The pH of 0.02 M NH<sub>4</sub>Cl(aq.) (pK<sub>b</sub> = 4.73) is equal to –...

The pH of 0.02 M NH4Cl(aq.) (pKb = 4.73) is equal to –

A

3.78

B

4.73

C

5.48

D

7.00

Answer

5.48

Explanation

Solution

pH = 12\frac{1}{2} [pKw – pKb –log C]

= 12\frac{1}{2} [14 – 4.73 – log 0.02]

= 12\frac{1}{2} [14 – 4.73 + 1.698] = 5.48