Question

If 200 mL of a standard 0.15 M NaOH solution is required to neutralize 30 ml of $H_{2}S{O}_4$, what is the molarity of acid?

• A. 1.0 M
• B. 2.0 M
• C. 0.5 M
• D. 1.5 M

Answer 1

The correct option is C 0.5 M

We have,
200 mL of a 0.15 M NaOH solution and 30 mL of $H_{2}S{O}_4$solution of unknown molarity.

Let’s assume the molarity of $H_{2}S{O}_4$ solution to be M.
For complete neutralization,
Number of millimole of $H^{+}$ ions = Number of millimole of $O{H}^{-}$ ions
Since, 1 mol of $H_{2}S{O}_4$ produces 2 mol of $H^{+}$ ions. i.e. $H_{2}S{O}_4$ is a diprotic acid
$1\times 200\times 0.15=2\times 30\times M$
M = 0.5
Therefore the molarity of the acid solution is 0.5 M