A student fills a burette with a sodium hydroxide solution of unknown concentration…

Step 1. The three titres lie within $0.10\ \text{mL}$ of each other. The mean is $\dfrac{24.75 + 24.80 + 24.85}{3} = 24.80\ \text{mL}$.

Step 2. $n(\text{H}^+) = 2 \times 0.0500\ \text{mol/L} \times 0.02000\ \text{L} = 2.00\ \text{mmol}$.

Step 3. The sodium hydroxide neutralises exactly this amount: $n(\text{NaOH}) = 2.00\ \text{mmol}$. $c = \dfrac{2.00\ \text{mmol}}{24.80\ \text{mL}} = 0.0806\ \text{mol/L}$.