An impure sample of (COOH)2 · 2 H2O that has a mass of 2.9 g was dissolved in water and titrated with standard NaOH solution. The titration required 32.4 mL of 0.144 mo- lar NaOH solution. Calculate the percent (COOH)2 · 2 H2O in the sample.

Answer 1

After dissolving the oxalic acid into water, the problem reduces to 1mole Oxalic Acid + 2moles Sodium Hydroxide => 1mole Sodium Oxalate + 2moles Water. In this, moles of NaOH used > moles of OxA neutralized by 2-times. To equate moles of OxA and NaOH => 2(moles OxA) = 1(moles NaOH).

Then, 2 x moles Oxalic Acid neutralized = Molarity of Sodium Hydroxide x Volume of NaOH(liters)

2(moles OxA) = (0.144M)(.0324L) =>
moles OxA neutralized = [(0.144M)(.0324L)/2]mole OxA = 0.0023 mole OxA.Then, grams of OxA = (0.0023 mole OxA)(90 gms OxA/mole) = 0.21gms Oxa.

%OxA neutralized = (0.21g/2.90g)100% = 7.2%