0.630 g of $H X_{2} C X_{2} O X_{4} \cdot 2 H X_{2} O$ (molar mass 126 g/mol) is dissolved in water to make 100 mL. A 10.0 mL portion needs 13.0 mL of NaOH solution for complete neutralisation. Find the molarity of the NaOH (three decimal places).

Numerical answer type, answer in M, tolerance 0.0006.

Answer: 0.077 M

Oxalic acid in 100 mL = 0.630/126 = 5.00 mmol, so a 10.0 mL portion holds 0.500 mmol, i.e. 0.500

\times

2 = 1.00 meq (dibasic). NaOH meq = M

\times

13.0 = 1.00, so M = 0.077 M.