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Basic Buffer Action

Question

$50$ % neutralisation of a solution of formic acid $(K_{a} = 2 \times 10^{- 4})$ with $N a O H$ would result in a solution having a hydrogen ion concentration of:

2 \times 10^{- 4}

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3.69

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4.0

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1.85

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Solution

The correct option is C $3.69$
Formic acid partial neutralization with $N a O H$ result in an acidic buffer.
Initial concentration of acid be $C .$
So 50% is neutralized. so let $[S a l t] = \frac{C}{2}$
Remaining acid will be $[A c i d] = C - \frac{C}{2} = \frac{C}{2}$
From Henderson's equation,
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]} p H = 3.69 + l o g \frac{C / 2}{C / 2} p H = 3.69$

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