If 50 mL of 0.5 M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is ______ g.

Solution

Chemistry ma’am · 4 steps

Oxalic acid gives two H⁺ per molecule, so each mole of it uses up two moles of NaOH.

1. Moles of acid. Moles of oxalic acid = 0.5 mol L⁻¹ × 0.050 L = 0.025 mol.

   H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O

2. Moles of NaOH. The acid is dibasic, so the 25 mL of NaOH contains 2 × 0.025 = 0.05 mol. That makes the NaOH solution 0.05/(0.025 L) = 2 M.
3. Scale to 50 mL. 50 mL of this 2 M solution holds 2 × 0.050 = 0.1 mol NaOH. With molar mass 40 g mol⁻¹, the mass is 0.1 × 40 = 4 g.
4. So. The amount of NaOH in 50 mL of the solution is 4 g.

The common slip: 2. Treating oxalic acid as monobasic, or giving the mass in only the 25 mL that was titrated, gives 2 g instead of 4 g.

The idea to keep. In acid–base titrations, always count the replaceable H⁺ or OH⁻ per molecule first. Then watch which volume the question finally asks about.

Answer from the NTA final answer key. Solution by PrepWiser.