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Question

50 mL of 0.1 M HCl and 50 mL of 2.0 M NaOH are mixed. The pH of the resulting solution is -


Consider log(2)=0.30


A
1.3
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B
4.2
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C
12.7
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D
11.7
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Solution

The correct option is C 12.7
Millimoles of HCl=0.1×50=5 mmol
Millimoles of NaOH=0.2×50=10 mmol
1 mmol of HCl is neutralized by 1 mmol of NaOH
Resulting solution contains 5 mmol of NaOH in (50 + 50) = 100 mL of solution
[OH]=5×103100×103=5×102 M
[OH]=5×102M

[H+]=KwOH=10145×102=2×1013
pH=log [H+]=log(2×1013)
=(13log 2)=130.30=12.7pH=12.7

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