Question

42.3mL of 0.285M hydrochloric acid wee used to titrate a sodium hydroxide solution having a concentration of 0.714M. What volume of base was needed?

Answer 1

The reaction of the titration of sodium hydroxide solution by hydrochloric acid is:

HCl + NaOH$\rightarrow$ H₂O + NaCl.

As you see, 1 mol of HCl reacts with 1 mol of NaOH. Therefore, the number of the moles of NaOH at the equivalence point equals the number of the moles of HCl added:

$n(NaOH) = n(HCl)$ .

The volume of base needed is its number of the moles divided by its concentration:

$V (NaOH) = \frac{n(NaOH)}{c(NaOH)}$ .

Using a similar expression for the number of the moles of hydrochloric acid, we finally get:

$V(NaOH) = \frac{n(HCl)}{c(NaOH)}= \frac{c(HCl)V(HCl)}{c(NaOH)}$

$V(NaOH)= \frac{0.285\text{ M}·42.3\text{ mL}}{0.714 \text{ M}} = 16.9$ mL.

Answer: 16.9 mL of base was needed for the titration.