The concentration of an unknown acid solution with general formula \( \mathrm{H}_{3} \mathrm{~A} \) is to be determined by titration with a 0.850 M KOH solution. Determine the concentration of the acid solution given that 38.34 mL of KOH solution were needed to fully neutralize 15.00 mL of the acid solution. (A) 6.52 M (B) 0.996 M (C) 0.111 M (D) 0.724 M (E) 2.17 M

To determine the concentration of the unknown acid \( \text{H}_3\text{A} \), we can use the concept of stoichiometry in acid-base reactions. The balanced reaction between \( \text{H}_3\text{A} \) and \( \text{KOH} \) is: \[ \text{H}_3\text{A} + 3 \text{KOH} \rightarrow \text{K}_3\text{A} + 3 \text{H}_2\text{O} \] From this equation, we can see that one mole of \( \text{H}_3\text{A} \) reacts with three moles of \( \text{KOH} \). First, let's calculate the moles of \( \text{KOH} \) used in the titration: \[ \text{Moles of KOH} = \text{Concentration} \times \text{Volume} = 0.850 \, \text{mol/L} \times 0.03834 \, \text{L} = 0.0326 \, \text{mol} \] Since the ratio of \( \text{KOH} \) to \( \text{H}_3\text{A} \) is 3:1, the moles of \( \text{H}_3\text{A} \) is: \[ \text{Moles of H}_3\text{A} = \frac{\text{Moles of KOH}}{3} = \frac{0.0326}{3} = 0.01087 \, \text{mol} \] Next, we can find the concentration of the acid solution using the volume of the acid solution used (15.00 mL = 0.01500 L): \[ \text{Concentration of H}_3\text{A} = \frac{\text{Moles of H}_3\text{A}}{\text{Volume of acid solution}} = \frac{0.01087 \, \text{mol}}{0.01500 \, \text{L}} = 0.724 \, \text{M} \] Thus, the concentration of the acid solution is \( \text{Answer: (D) 0.724 M} \). So, the correct answer is (D) 0.724 M.