A student titrates 25.0 mL of H2SO4 with 0.100 M NaOH. If it takes 50.0 mL of NaOH to reach the endpoint, how many moles of H2SO4 were present in the titrate?

• A. 0.00500 mol
• B. 0.00250 mol
• C. 0.0100 mol
• D. 0.0250 mol

Answer: (A)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between sulfuric acid (H\_2SO\_4) and sodium hydroxide (NaOH): \[\mathrm{H_2SO_4 + 2\ NaOH \rightarrow Na_2SO_4 + 2\ H_2O}\]
2. Identify the volume and molarity of NaOH used in the titration. Given volume is 50.0 mL (which should be converted to liters) and molarity is 0.100 M. Calculate the moles of NaOH using the formula: \[\text{moles NaOH} = M \times V = 0.100\ \mathrm{mol/L} \times 0.0500\ \mathrm{L}\]
3. Use the stoichiometric ratio from the balanced equation to relate moles of NaOH to moles of H\_2SO\_4. Since 2 moles of NaOH react with 1 mole of H\_2SO\_4, calculate moles of H\_2SO\_4 as: \[\text{moles H_2SO_4} = \frac{\text{moles NaOH}}{2}\]
4. Substitute the calculated moles of NaOH into the equation to find the moles of H\_2SO\_4 present in the 25.0 mL sample.
5. Interpret the result as the number of moles of H\_2SO\_4 originally present in the titrated volume.