Q.

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

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a

3.78

b

4.73

c

5.48

d

7.00

answer is C.

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Detailed Solution

Ammonium chloride is salt of strong acid and weak base, solution is acidic.pH=12  pKw- pKb - log C = 12 14-4.73-log 0.02 = 12 14-4.73+ 1.698 = 5.48
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The pH of 0.02 M NH4Cl (aq.)   pKb = 4.73 is equal to