What volume (in mL) of 0.150 M NaOH (aq) is required to neutralize 25.0 mL of 0.100 M H2SO4 (aq)?

• A. 16.7 mL
• B. 66.7 mL
• C. 75.0 mL
• D. 33.3 mL

Answer: (D)

Neutralization here hinges on the stoichiometry between a diprotic acid and a strong base. Each mole of H2SO4 requires two moles of NaOH to neutralize its two protons.

First find the moles of acid present: 25.0 mL of 0.100 M H2SO4 equals 0.0250 L × 0.100 mol/L = 0.00250 mol H2SO4.

Because two NaOH molecules are needed per one H2SO4 molecule, the moles of NaOH required are 2 × 0.00250 = 0.00500 mol NaOH.

The NaOH solution is 0.150 M, so the volume needed is 0.00500 mol ÷ 0.150 mol/L = 0.0333 L = 33.3 mL.

Thus, about 33.3 mL of 0.150 M NaOH is required to neutralize the 25.0 mL of 0.100 M H2SO4.