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Basic Buffer Action

If 50 mL of...

Question

If $50 m L$ of $0.2 M K O H$ is added to $40 m L$ of $0.5 M H C O O H$ . The $p H$ of the resulting solution is: $(K_{a} = 1.8 \times 10^{- 4})$

3.74

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5.6

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7.5

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3.4

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Solution

The correct option is A 3.74
$H C O O H + K O H ⟶ H C O O H + H_{2} O$

After reaction, it forms Buffer solution
$[H C O O H] = \frac{10}{90} [H C O O K] = \frac{10}{90}$

$p H = p K_{a} + log \frac{[s a l t]}{[a c i d]}$

$p H = p K_{a}$

$p H = 4 - log (1)$ ( since [Salt] = [Acid] )

$p H = 3.74$

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