A solution of hydrochloric acid of unknown concentration was titrated with 0.11 M NaOH. If a 150-mL sample of the HCl solution required exactly 12 mL of the NaOH solution to reach the equivalence point, what was the pH of the HCl solution?

Question

Quizplus · Accepted Answer

At the equivalence point, moles of acid = moles of base.

The number of moles of NaOH used can be calculated as:
moles of NaOH = Molarity of NaOH x Volume of NaOH used (in L)
moles of NaOH = 0.11 M x 0.012 L
moles of NaOH = 0.00132 mol

Since HCl and NaOH react in a 1:1 ratio, the number of moles of HCl in the 150 mL sample can be calculated as:
moles of HCl = moles of NaOH used
moles of HCl = 0.00132 mol

The concentration of HCl can then be calculated as:
Molarity of HCl = moles of HCl / Volume of HCl (in L)
Molarity of HCl = 0.00132 mol / 0.150 L
Molarity of HCl = 0.0088 M

The pH of a 0.0088 M HCl solution can be calculated as:
pH = -log[H+]
pH = -log[0.0088]
pH = 2.06

Therefore, the pH of the HCl solution is 2.06. The correct answer is D.

A Solution of Hydrochloric Acid of Unknown Concentration Was Titrated