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Electrolytes

100 mL of a 0...

Question

$100 m L$ of a $0.5 M C H_{3} C O O N a (a q)$ is titrated against a $0.5 M H C l (a q)$ . Calculate the $p H$ of the solution initially i.e. when no $H C l$ has been added.
Given: $p K_{a} (C H_{3} C O O H) = 4.74$

6.58

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8.24

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7.96

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9.22

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Solution

The correct option is D 9.22
When only $100 m L$ of a $0.5 M C H_{3} C O O N a (a q)$ is present. It will show simple salt hydrolysis.
The formula to calculate the $p H$ of a salt of strong base and weak acid is :
$p H = \frac{1}{2} (p K_{w} + p K_{a} + log C) p H = \frac{1}{2} (14 + 4.74 + log 0.5) p H = \frac{1}{2} (18.74 - 1 + log 5) p H = \frac{1}{2} (18.44) p H = 9.22$

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