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Acidic Buffer Action

2 g of acetat...

Question

$2 g$ of acetate ions in the form of sodium acetate are added to $500 mL$ of a $0.10 M$ acetic acid solution of $p K_{a} (C H_{3} C O O H) = 4.74$ . What is the $p H$ of the final solution?

4.56

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4.74

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5.01

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5.70

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Solution

The correct option is A $4.56$
No of mmoles of acetate ion $= \frac{2 \times 1000}{60} = 33.33 m m o l e s$ also, 50 mmoles of $C H_{3} C O O H$ are present in the solution.
$p H = p K_{a} + l o g \frac{[C H_{3} C O O^{-}]}{C H_{3} C O O H]}$
$p H = 4.74 + l o g \frac{33.33}{50}$
$p H = 4.74 + l o g 2 - l o g 3 = 4.564$

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2 g

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