48.8 mL of a 1.42 M NaOH solution is titrated with a 1.86 M HCl solution. What is the final volume of the solution when the NaOH has been completely neutralized by the HCl? (Assume the volumes are additive.)
A) 37.3 mL
B) 86.1 mL
C) 63.9 mL
D) 49.6 mL
E) 113 mL

Question

Quizplus · Accepted Answer

The final volume of the solution is the sum of the volumes of the NaOH and HCl solutions used. First, calculate the moles of NaOH: $1.42 \, \text{mol/L} \times 0.0488 \, \text{L} = 0.069296 \, \text{mol}$. Since the reaction between NaOH and HCl is 1:1, the moles of HCl needed are the same. To find the volume of HCl needed: $0.069296 \, \text{mol} \div 1.86 \, \text{mol/L} = 0.03726 \, \text{L} = 37.26 \, \text{mL}$. The final volume is the sum of the volumes of NaOH and HCl: $48.8 \, \text{mL} + 37.26 \, \text{mL} = 86.06 \, \text{mL}$, which rounds to 86.1 mL.

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