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Arrhenius Acid

To neutralize...

Question

To neutralize $25 m l$ of $0.25 M N a_{2} C O_{3}$ solution how much volume of $0.5 M H C l$ is required

12.5 m l

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25 m l

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37.5 m l

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50 m l

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Solution

The correct option is D $25 m l$
By law of equivalence:

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

$n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}$

$2 \times 0.25 M \times 25 = 1 \times 0.5 M \times V_{2}$

$V_{2} = \frac{2 \times 0.25 M \times 25}{0.5 M} = 25 m l$

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