Calculate the $p H$ at the equivalence point of the titration between 0.1 M $C H_{3} C O O H$ (25 mL) with 0.05 M $N a O H$ .
(Given $p K_{a}$ for $C H_{3} C O O H = 4.74$ ; log(3)=0.48).

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Solution

Milli equivalents of $C H_{3} C O O N a$ formed = $0.1 \times 25$
$C H_{3} C O O H + N a O H$ $⟶$ $C H_{3} C O O N a + H_{2} O$

At the equivalnce point,
Milli equivalents of $N a O H$ = Milli equivalents of $C H_{3} C O O H$
$0.05 \times V = 0.1 \times 25$
V= 50 mL
The total volume of the solution = 75 mL
Concentration of $C H_{3} C O O N a$ = $\frac{0.1 \times 25}{75} = \frac{1}{30}$
$p H = 7 + \frac{1}{2} p K_{a} + \frac{1}{2} l o g C$
(for a salt of W.A. + S.B.)
$= 7 + \frac{1}{2} \times 4.74 + \frac{1}{2} l o g \frac{1}{30}$
$= 8.63$

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