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The pH of 0.02 M is equal to
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a
3.78
b
4.73
c
5.48
d
7.00
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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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