A 250. mL sample of 0.0328M HCl is partially neutralized by the addition of 100. mL of 0.0245M NaOH. Find the concentration of hydrochloric acid in the resulting solution.
A) 0.00700 M
B) 0.0164 M
C) 0.0383 M
D) 0.0230 M
E) 0.0575 M

Question

Quizplus · Accepted Answer

The moles of HCl initially present are 0.250 L * 0.0328 M = 0.0082 moles. The moles of NaOH added are 0.100 L * 0.0245 M = 0.00245 moles. The remaining moles of HCl after neutralization are 0.0082 - 0.00245 = 0.00575 moles. The total volume of the solution is 250 mL + 100 mL = 350 mL = 0.350 L. The concentration of HCl is 0.00575 moles / 0.350 L = 0.0164 M.

A 250 ML Sample of 0