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Question

The pH of a solution obtained by mixing 100 mL of 0.2 M CH3COOH with 100 mL of 0.2 M NaOH will be: (pKa for CH3COOH=4.74)

A
4.74
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B
8.87
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C
9.1
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D
8.87
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Solution

The correct option is D 8.87
0.2 M CH3COOH 0.2 M NaOH each 100 mL is taken here.

Milimoles of CH3COOH=0.2×100=20 mmol
Milimoles of CH3COOH=0.2×100=20 mmol

CH3COOH+NaOHCH3COONa+H2O
20 + 20 0 + 0
0 + 0 20 + 20
So there is complete neutralisation
[salt]=20200=0.1

Using the formula,
pH=7+12log pKa+log C
pH=7+12log pKa+12log[salt]


putting the values,
pH=7+12log (4.74)+12log(0.1)
pH=7+2.370.5
pH=8.87

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