Question

You have 25 mL of an aqueous HCl solution in a conical flask. By titration (with a pH meter), 10 mL of 0.4 M NaOH is needed to neutralise the solution. (a) What pH reading will the neutralised solution give? [1 mark] (b) What is the molar concentration of the aqueous HCl solution (to 2 d.p.)? [3 marks]

Answer 1

(a) The pH is an important quantity for the acid-base titration. When we determine an acid we may talk about the solution with pH < 7, and when it concerns a base the pH > 7. The process of neutralisation occurs exactly at pH = 7, when the amount acidic $H_3O^+$ is equal to the amount of basic $OH^-$

(b) The molar concentration of $HCl$ can be determined from the following equation:

$n(NaOH)=n(HCl)$

$c(NaOH) \times V(NaOH) = c(HCl) \times V(HCl)$

where c and V are the molar concentration and volume of corresponding samples.

Deducing the molar concentration will give following:

$c(HCl)=\frac{c(NaOH) \times V(NaOH)}{V{HCl}} = \frac{0.4M \times 10.0 ml}{25.0 ml} = 0.16M$