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Question

Calculate the pH of 0.2 M NaOCl solution.
Ka(HOCl)=3.5×108

A
10.01
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B
12.53
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C
8.63
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D
9.53
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Solution

The correct option is A 10.01
NaOCl is a salt of weak acid HOCl and strong base NaOH
pKa=log(Ka)pKa=log(3.5×108)pKa=(80.54)=7.46pH=7+12(pKa+logC)pH=7+12(7.46+log(0.2))pH=7+12(7.460.7)pH=10.38

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