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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

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detailed solution

Correct option is C

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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