Step-by-step Solution:

1. Analyze the Titration Data

   The volumes of NaOH solution used in the five experiments are 12.5 mL, 10.5 mL, 9.0 mL, 9.0 mL, and 9.0 mL. In a titration, multiple readings are taken to ensure accuracy. The first reading is often a rough estimate and subsequent readings that are close to each other (concordant) are used for calculation. The readings 9.0 mL, 9.0 mL, and 9.0 mL are concordant. Therefore, the average volume of NaOH solution used for the titration is: $V_{\text{NaOH}} = \frac{9.0 + 9.0 + 9.0}{3} = 9.0 \text{ mL}$

2. Write the Balanced Chemical Equation

   The reaction is a neutralization reaction between oxalic acid ($H_2C_2O_4$), a diprotic acid, and sodium hydroxide (NaOH), a monoprotic base. The balanced chemical equation is: $H_2C_2O_4 (aq) + 2NaOH (aq) \rightarrow Na_2C_2O_4 (aq) + 2H_2O (l)$

3. Determine the Stoichiometry

   From the balanced equation, 1 mole of oxalic acid reacts completely with 2 moles of sodium hydroxide. The stoichiometric coefficient for oxalic acid ($n_{\text{acid}}$) is 1. The stoichiometric coefficient for sodium hydroxide ($n_{\text{base}}$) is 2.

4. Apply the Titration Formula

   At the equivalence point, the relationship between the acid and base is given by the formula: $\frac{M_{\text{acid}} \times V_{\text{acid}}}{n_{\text{acid}}} = \frac{M_{\text{base}} \times V_{\text{base}}}{n_{\text{base}}}$ Where:

   • $M_{\text{acid}}$ = Molarity of oxalic acid = 0.10 M
   • $V_{\text{acid}}$ = Volume of oxalic acid = 5.00 mL
   • $n_{\text{acid}}$ = 1
   • $M_{\text{base}}$ = Molarity of NaOH (to be calculated)
   • $V_{\text{base}}$ = Volume of NaOH = 9.0 mL
   • $n_{\text{base}}$ = 2

5. Calculate the Molarity of NaOH

   Substitute the known values into the formula: $\frac{0.10 \text{ M} \times 5.00 \text{ mL}}{1} = \frac{M_{\text{NaOH}} \times 9.0 \text{ mL}}{2}$ Now, solve for $M_{\text{NaOH}}$: $M_{\text{NaOH}} = \frac{0.10 \times 5.00 \times 2}{9.0}$ $M_{\text{NaOH}} = \frac{1.0}{9.0}$ $M_{\text{NaOH}} = 0.1111... \text{ M}$ Rounding to two significant figures (consistent with the input data like 0.10 M and 9.0 mL), we get: $M_{\text{NaOH}} \approx 0.11 \text{ M}$

Thus, the concentration of the NaOH solution is 0.11 M.