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Salt of Weak Acid and Strong Base

A 50 mL solut...

Question

A $50 m L$ solution of $0.2 M C H_{3} C O O H$ is titrated against $0.2 M N a O H$ . Find the $p H$ of the solution when $50 m L$ of $N a O H$ is added.
$p K_{a} (C H_{3} C O O H) = 4.74$

8.87

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9.44

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5.66

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10.52

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Solution

The correct option is A 8.87
mmol of $C H_{3} C O O H = 50 \times 0.2 = 10$
mmol of $N a O H = 50 \times 0.2 = 10$

$C H_{3} C O O H (a q) + N a O H (a q) \to C H_{3} C O O N a (a q) + H_{2} O (l) I n i t i a l l y : 101000 F i n a l : 001010$
Since only salt of weak acid and strong base i.e. $C H_{3} C O O N a$ is left, it will undergo hydrolysis.
$Concentration = \frac{Moles}{Total Volume}$

$[C H_{3} C O O N a] = [C H_{3} C O O^{-}] = \frac{10}{100} = 0.1 M$

$C H_{3} C O O^{-} (a q) + H_{2} O (l) ⇌ C H_{3} C O O H (a q) + O H^{-} (a q)$
$p H = 7 + \frac{1}{2} (p K_{a} + l o g C) p H = 7 + \frac{1}{2} (4.74 + l o g (0.1)) p H = 7 + \frac{1}{2} (4.74 - 1) p H = 7 + 1.87 p H = 8.87$
Theory :
Phenolphthalein indicator is used for weak acid and strong base titrations

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