The pH of a solution obtained by mixing $100 m l$ of $0.2 M$ $C H_{3} C O O H$ with 100 ml of 0.2 M $N a O H$ would be:
$(p K_{a}$ for $C H_{3} C O O H = 4.74)$

4.74

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8.87

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9.10

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8.57

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Solution

The correct option is B $4.74$
$p H = p K_{a} + l o g \frac{B a s e}{A c i d} p H = 4.74 + 0 p H = 4.74$

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