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Solubility Product

The pH of a s...

Question

The $p H$ of a solution obtained by mixing $100 m L$ of $0.2 M$ $C H_{3} C O O H$ with $100 m L$ of $0.2 M N a O H$ will be: ( $p K_{a} f o r C H_{3} C O O H = 4.74$ )

4.74

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8.87

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9.1

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8.87

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Solution

The correct option is D 8.87
$0.2 M C H_{3} C O O H 0.2 M N a O H$ each $100 m L$ is taken here.

Milimoles of $C H_{3} C O O H = 0.2 \times 100 = 20 m m o l$
Milimoles of $C H_{3} C O O H = 0.2 \times 100 = 20 m m o l$

$C H_{3} C O O H + N a O H ⇌ C H_{3} C O O N a + H_{2} O$
$20 + 20 ⇌ 0 + 0$
$0 + 0 ⇌ 20 + 20$
So there is complete neutralisation
$[s a l t] = \frac{20}{200} = 0.1$

Using the formula,
$p H = 7 + \frac{1}{2} l o g p K_{a} + l o g C$
$p H = 7 + \frac{1}{2} l o g p K_{a} + \frac{1}{2} l o g [s a l t]$

putting the values,
$p H = 7 + \frac{1}{2} l o g (4.74) + \frac{1}{2} l o g (0.1)$
$p H = 7 + 2.37 - 0.5$
$p H = 8.87$

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