Question

Consider the titration of $\mathrm{NaOH}$ solution versus $1.25M$ oxalic acid solution. At the endpoint following burette readings were obtained.

1. $4.5mL$
2. $4.5mL$
3. $4.4mL$
4. $4.4mL$
5. $4.4mL$

If the volume of oxalic acid taken was $10.0ml$. then the molarity of the $\mathrm{NaOH}$ solution is ____$M$. (Rounded-off to the nearest integer)

Answer 1

Acid-base titration:

1. Titration is the method by which the concentration of a base or acid can be determined by neutralizing it with an acid or base of known concentration.
2. The acid or base of a known concentration is known as a standard solution.
3. It is given that the concentration and volume of oxalic acid is $1.25M$ and $10mL$ respectively.
4. The five volumes of $\mathrm{NaOH}$ solution used in neutralizing the oxalic acid solution are also given, from which the average volume of $\mathrm{NaOH}$ solution used is calculated to be $4.4mL$.
5. We also know for titration that: ${\mathrm{M}}_{\mathrm{eq}}\mathrm{of}\mathrm{NaOH}={\mathrm{M}}_{\mathrm{eq}}\mathrm{of}\mathrm{oxalic}\mathrm{acid}$.
6. So to determine the concentration of $\mathrm{NaOH}$ solution used: $M_1\times V_1\times 1=M_2\times V_2\times 2$ where $M_1$ and $M_2$ are the concentration of $\mathrm{NaOH}$ and oxalic acid respectively and $V_1$ and $V_2$ are volumes of $\mathrm{NaOH}$ and oxalic acid respectively.
7. Placing the known values in the equation above: $$\begin{gathered} M_1\times V_1\times 1=M_2\times V_2\times 2 \\ {M}_1\times 4.4\times 1=1.25\times 10\times 2 \\ {M}_1=\frac{25}{4.4} \\ {M}_1=5.68M\approx 6M \end{gathered}$$
8. Therefore, the molarity of the $\mathbf{NaOH}$ solution is $\mathbf{6}\mathbf{}\mathit{M}$.