Question

A titration of 0.1 M NaOH into 1.2 L of HCl was stopped once the pH reached 7 (at 25C). If 0.4 L of NaOH needed to be added to achieve this pH, what was the original concentration of the sample of HCl? Enter your answer in normal notation (no exponents) and out to the 1/1000th place if the answer is less than 1. (answer in the form 0.00x; enter no units, although the number should correspond to Molarity)

Answer 1

Answer:

0.033 M

Explanation:

Let's consider the neutralization reaction between NaOH and HCl.

NaOH + HCl → NaCl + H₂O

0.4 L of 0.1 M NaOH were used. The reacting moles of NaOH are:

0.4 L × 0.1 mol/L = 0.04 mol

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 0.04 moles.

0.04 moles of HCl are in 1.2 L. The molarity of HCl is:

M = 0.04 mol / 1.2 L = 0.033 M