In one such titration exercise, the following results were obtained for the volume of the acid used. Complete the table and determine the mean titre.

Titre = Final Burette Reading - Initial Burette Reading

• Approximate: $21.8 \text{ cm}^3 - 0.0 \text{ cm}^3 = 21.8 \text{ cm}^3$
• Accurate 1: $21.5 \text{ cm}^3 - 0.5 \text{ cm}^3 = 21.0 \text{ cm}^3$
• Accurate 2: $43.3 \text{ cm}^3 - 22.3 \text{ cm}^3 = 21.0 \text{ cm}^3$

\begin{tabular}{|l|c|c|c|}
\hline
& Approximate & Accurate 1 & Accurate 2 \ \hline
Final Burette Reading & $21.8 \text{ cm}^3$ & $21.5 \text{ cm}^3$ & $43.3 \text{ cm}^3$ \ Initial Burette Reading & $0.0 \text{ cm}^3$ & $0.5 \text{ cm}^3$ & $22.3 \text{ cm}^3$ \ Titre & $\mathbf{21.8 \text{ cm}^3}$ & $\mathbf{21.0 \text{ cm}^3}$ & $\mathbf{21.0 \text{ cm}^3}$ \ \hline
\end{tabular}

The accurate titres are $21.0 \text{ cm}^3$ and $21.0 \text{ cm}^3$. The approximate titre is usually discarded.

Concentration of NaOH ($C_b$) = $0.1 \text{ M}$
Volume of NaOH ($V_b$) = $25 \text{ cm}^3 = 0.025 \text{ dm}^3$
Volume of HCl ($V_a$) = Mean titre = $21.0 \text{ cm}^3 = 0.021 \text{ dm}^3$
From the balanced equation, the mole ratio of HCl to NaOH ($n_a:n_b$) is $1:1$.

Where $C_a$ is the concentration of HCl.

Rounding to three significant figures: