Question

Volume of $0.1 \, \text{M}$ $\text{H}_{2}\text{S}\text{O}_{4}$ required to neutralize $30 \, \text{m}\text{L}$ of $0.2 \, \text{N}$ $\text{N}\text{a}\text{O}\text{H}$ is

• A. $30 \, mL$
• B. $15 \, mL$
• C. $40 \, mL$
• D. $60 \, mL$

Answer 1

Answer: (A)

$0.1 \, \text{M} \, \, \text{o}\text{f} \, \, \text{H}_{2}\text{S}\text{O}_{4} \, \Rightarrow 0.2 \, \text{N} \, \, \text{o}\text{f} \, \, \text{H}_{2}\text{S}\text{O}_{4}$

$\therefore $ $\text{N}_{1}\text{V}_{1}=\text{N}_{2}\text{V}_{2}$ $\left[\text{N}_{1} = 0.2 \, \text{N} \, \text{H}_{2} \text{SO}_{4}\right]$

$0.2 \, \times \, \text{V}_{1}=30 \, \times \, 0.2$

$\therefore $ $\text{V}_{1}=30 \, \text{m}\text{L}$