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Salt of Weak Acid and Weak Base

A 100 m L s...

Question

A $100 m L$ solution of $0.1 M C H_{3} C O O H$ is titrated with $0.1 M N a O H$ , calculate the $p H$ at $25 %$ completion and $75 %$ completion of the titration. $(p k_{a} = 4.74)$

4.26, 5.22

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5.22, 5.26

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4.74, 4.26

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5.00, 4.00

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Solution

The correct option is D $4.26, 5.22$
$C H_{3} C O O H + N a O H \to C H_{3} C O O N a + H_{2} O$
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$
When $25 %$ acid is dissociated, then at equilibrium $25 %$ of salt is formed and $75 %$ of acid is left.
$p H = 4.74 + l o g \frac{[0.25]}{[0.75]} = 4.74 + (- 0.477)$
$p H = 4.26$
When dissociation is $75 %$ , then at equilibrium $25 %$ acid is left and $75 %$ salt is formed.
$p H = 4.74 + l o g \frac{[0.75]}{[0.25]} = 4.74 + 0.477$
$p H = 5.22$

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