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pH of a Solution

The pH of a...

Question

The $p H$ of a solution obtained by mixing $100 m l$ of $0.2 M C H_{3} C O O H$ is with $100 m l$ of $0.2 M N a O H$ would be:
[Note : $p K_{a}$ for $C H_{3} C O O H = 4.74$ and $log 2 = 0.301)$ ].

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 $8.87$
$C H_{3} C O O H + N a O H \to C H_{3} C O O N a + H_{2} O$

Moles of $C H_{3} C O O H = 0.2 \times 100 \times 1 = 20 m m o l$
Moles of $N a O H = 0.2 \times 100 \times 1 = 20 m m o l$

So there is complete neutralisation
$[s a l t] = C = \frac{20}{100 + 100} = \frac{20}{200} = 0.1 M$

$p H = 7 + \frac{1}{2} p K_{a} + \frac{1}{2} {log}_{10} C$

$= 7 + \frac{1}{2} (4.74) + \frac{1}{2} log (0.1)$

$= 7 + 2.37 - \frac{1}{2} {log}_{10} 10$

$p H = 9.37 - 0.5 = 8.87$

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