5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette using phenolphthalein indicator. The volume of NaOH required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution?

The burette readings for the five trials (given in the table) are first averaged.

Let the five volumes of $$NaOH$$ noted be (in mL): $$V_1,\,V_2,\,V_3,\,V_4,\,V_5$$.
Their mean volume is
$$\overline V = \frac{V_1+V_2+V_3+V_4+V_5}{5}\;{\text{mL}}$$

Substituting the numerical values supplied in the question, the average comes to
$$\overline V = 9.10\;{\text{mL}}$$

Convert this to litres for use in the molarity equation:
$$\overline V = 9.10\;\text{mL} = 9.10 \times 10^{-3}\;\text{L}$$

Now calculate the number of moles of oxalic acid $$\left(H_2C_2O_4\right)$$ present in the conical flask.
Given: volume of oxalic acid solution $$V_{\text{acid}} = 5.00\;\text{mL} = 5.00 \times 10^{-3}\;\text{L}$$, and its molarity $$M_{\text{acid}} = 0.10\;M$$.

Moles of oxalic acid:
$$n_{\text{acid}} = M_{\text{acid}}\times V_{\text{acid}}$$
$$\phantom{n_{\text{acid}}}=0.10\;{\text{mol L}^{-1}}\times 5.00\times10^{-3}\;{\text{L}}$$
$$\phantom{n_{\text{acid}}}=5.0\times10^{-4}\;\text{mol}$$

Oxalic acid is diprotic. One mole of $$H_2C_2O_4$$ reacts with two moles of $$NaOH$$:
$$H_2C_2O_4 + 2\,NaOH \rightarrow Na_2C_2O_4 + 2\,H_2O$$

Therefore, moles of $$NaOH$$ required are
$$n_{\text{base}} = 2\times n_{\text{acid}} = 2\times 5.0\times10^{-4} = 1.0\times10^{-3}\;\text{mol}$$

Let $$M_{\text{base}}$$ be the molarity of the $$NaOH$$ solution. Using $$n = M\,V$$,

$$M_{\text{base}} = \frac{n_{\text{base}}}{V_{\text{base}}} = \frac{1.0\times10^{-3}\;\text{mol}}{9.10\times10^{-3}\;\text{L}}$$

$$M_{\text{base}} = 0.1099\;\text{mol L}^{-1} \approx 0.11\;M$$

Hence, the concentration of the $$NaOH$$ solution is

0.11 M.