Question

How many mL of $1M$ $H_{2}S{O}_4$ is required to neutralise $10mL$ of $1M$ $NaOH$ solution?

• A. 2.5
• B. 5.0
• C. 10.0
• D. 20.0

Answer 1

Answer: (B)

Molecular weight of $H_{2}S{O}_4=98$
Equivalent weight of $H_{2}S{O}_4=\frac{98}{2}=49$
Molarity $\times$ mol. wt. $=$ normality $\times$ eq. wt.
$1\times 98=N_1\times 49N_1=\frac{98}{49}=2$
Similarly, for $NaOH1N=1M{N}_2=1$
Normality equation, $\underset{\text{ Acid }}{N_1{V}_1}=\underset{\text{ Base }}{N_2{V}_2}$
$2\times V_1=1\times 10V_1=5mL$
So, $5mL1M{H}_{2}S{O}_4$
is required to neutralize $10mL$ of $1MNaOH$ solution.