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Basic Buffer Action

50 mL of 0....

Question

$50 m L$ of $0.05 M N a_{2} C O_{3}$ is titrated against $0.1 M H C l$ . On adding $40 m L$ of $H C l$ , $p H$ of the solution will be
[Given for $H_{2} C O_{3}, p K_{a_{1}} = 6.35; p K_{a_{2}} = 10.33; log 3 = 0.477, log 2 = 0.30$ ]

6.35

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6.526

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8.34

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6.173

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Solution

The correct option is C $6.173$
The number of milliequivalents of sodium carbonate $= 50 \times 0.05 = 2.5$
The number of milliequivalents of HCl $= 40 \times 0.1 = 4$

$N a_{2} C O_{3} + H C l ⟶ N a C l + N a H C O_{3}$

$N a H C O_{3} + H C l ⟶ N a C l + H_{2} C O_{3}$
Thus, the resulting solution will contain 1.0 milliequivalent of $N a H C O_{3}$ and 1.5 milliequivalents of $H_{2} C O_{3}$ .
This is an acidic buffer solution.
$p H = p K_{a} + l o g \frac{[N a H C O_{3}]}{[H_{2} C O_{3}]} = 6.35 + l o g \frac{1.0}{1.5}$

$p H = 6.35 + l o g \frac{2.0}{3.0} = 6.35 + l o g 2.0 - l o g 3.0 = 6.35 + 0.30 - 0.477 = 6.35 - 0.177 = 6.173$

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