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Acidic Buffer Action

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Question

Calculate approximate $p H$ of the resultant solution formed by titration of $25 m L$ of $0.4 M N a_{2} C O_{3}$ with $50 m L$ of $0.025 M H C l$ .

[Given that: $p K_{a_{1}} = 6.4$ and $p K_{a_{2}} = 10.3$ for $H_{2} C O_{3}$ ]

5.92

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11.14

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6.4

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5.88

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Solution

The correct option is B $11.14$
The number of millimoles of sodium carbonate $= 25 \times 0.4 = 10$ millimoles.
The number of millimoles of $H C l$ $= 50 \times 0.025 = 1.25$ millimoles.
Out of 10 millimoles of sodium carbonate, 1.25 millimoles will react with 1.25 millimoles of $H C l$ to from 1.25 millimoles of sodium bicarbonate.
8.75 millimoles of sodium carbonate remain unreacted.
An acidic buffer solution is obtained in which acid is sodium bicarbonate and salt is sodium carbonate.
The expression for the pH of the solution is $p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]} = 10.3 + l o g \frac{8.75}{1.25} = 11.14$

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