Question

What volume of 0.3 M sulphuric acid is required to exactly neutralize 200 ml of 0.5 M NaOH?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$2NaOH+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2H_{2}O$
2 mol of NaOH react with 1 mol of $H_{2}S{O}_4$.
$\therefore$ $\frac{M_1{V}_1}{M_2{V}_2}=\frac{n_1}{n_2}=\frac{0.3\times V}{0.5\times 200}=\frac{1}{2}$
Hence, V = 167 ml