A 25.00 mL sample of 0.100 M sulfuric acid is titrated with sodium hydroxide. If 34.36 mL of NaOH is required to reach the equivalence point, what is the concentration of the NaOH solution?

Sulfuric acid ($\text{H}_2\text{SO}_4$) is a diprotic acid, meaning it releases two $\text{H}^+$ ions. Sodium hydroxide ($\text{NaOH}$) is a monoprotic base, releasing one $\text{OH}^-$ ion.

From the balanced equation, 1 mole of $\text{H}_2\text{SO}_4$ reacts with 2 moles of $\text{NaOH}$.

Concentration of acid ($M_a$) = $0.100 \text{ M}$
Volume of acid ($V_a$) = $25.00 \text{ mL} = 0.02500 \text{ L}$
Volume of base ($V_b$) = $34.36 \text{ mL} = 0.03436 \text{ L}$
Concentration of base ($M_b$) = ?

At the equivalence point of a neutralization reaction, the moles of acid and base are related by their stoichiometric coefficients:

Where $n_a$ is the stoichiometric coefficient of the acid (1 for $\text{H}_2\text{SO}_4$) and $n_b$ is the stoichiometric coefficient of the base (2 for $\text{NaOH}$).

Multiply both sides by 2:

Solve for $M_b$:

The concentration of the acid ($0.100 \text{ M}$) has 3 significant figures. The volumes ($25.00 \text{ mL}$ and $34.36 \text{ mL}$) have 4 significant figures. The result should be rounded to 3 significant figures.

The concentration of the base solution is $\text{0.146 M}$.