Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained.
(i) 4.5 mL (ii) 4.5 mL (iii) 4.4 mL (iv) 4.4 mL (v) 4.4 mL
If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is ______ M. (Rounded-off to the nearest integer)

First, we find the mean burette reading from the five readings: $$(4.5 + 4.5 + 4.4 + 4.4 + 4.4)/5 = 22.2/5 = 4.44$$ mL. This is the mean volume of NaOH used.

Oxalic acid ($$H_2C_2O_4$$) is a diprotic acid, so the reaction with NaOH is: $$H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$$.

The millimoles of oxalic acid taken are: $$10.0 \times 1.25 = 12.5$$ mmol.

From the stoichiometry, the millimoles of NaOH required are: $$2 \times 12.5 = 25.0$$ mmol.

The molarity of NaOH is: $$M = \frac{25.0}{4.44} = 5.63$$ M.

Rounding to the nearest integer, the molarity of the NaOH solution is $$\boxed{6}$$ M.