If 20ml of 0.4 N NaOH solution completely neutralise 40ml of a dibasic acid, the molarity of the acid solution is:

0.1 M

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0.2 M

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0.3 M

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0.4 M

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Solution

The correct option is A $0.1 M$
For neutralization, no. of equivalent should be same,

$N_{1} V_{1} = N_{2} V_{2}$

$0.4 \times 20 = N_{2} \times 40$

$N_{2} = \frac{0.4 \times 20}{40}$

$N_{2} = 0.2 N$

n-factor of acid $= 2 \to Because of dibasic acid [2 acidic-hydrogen]$

$M_{2} = \frac{N_{2}}{2}$

$M_{2} = \frac{0.2}{2} = 0.1 M$

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