Question

25.0cm3 of a 0.10moldm-3 solution of sodium hydroxide was titrated against a solution of hydrochloric acid of unknown concentration, 27.3cm3 of the acid was required. What was the concentration of the acid

Answer 1

The chemical reaction is;

$NaOH + HCl \rightarrow NaCl + H_2O$

From the reaction above and information given, it can be inferred that;

$\begin{aligned} n_A \textsf{(number of moles of acid used) }&= 1\\ n_B \textsf{(number of moles of base used) }&= 1 \end{aligned}$ $\begin{aligned} \\ V_A \textsf{(Volume of acid used) }&= 27.3cm^3\\ V_B \textsf{(Volume of base used) } &= 25.0 cm^3\\ C_A \textsf{(Concentration of acid) } &= \ ?\\ C_B \textsf{(Concentration of base) } &= 0.10\ moldm^{-3} \end{aligned}$

Where A(acid) = HCl

and B(Base) = NaOH

Using the formula;

$\dfrac{C_AV_A}{C_BV_B} = \dfrac{n_A}{n_B}$

$C_A = \dfrac{n_A × C_BV_B}{n_B × V_A}$

$C_A = \dfrac{1 × 0.10 × 25}{1 × 27.3}$

$C_A = 0.0916\ moldm^{-3}$

$\therefore \textsf{The concentration of the acid }= 0.0916\ moldm^{-3}$