Question

The purity of a sample of sodium oxalate, Na₂C₂O₄, is determined by titrating with a standard solution of KMnO₄. If a 0.5116-g sample requires 35.62 mL of 0.0400 M KMnO₄ to reach the titration’s end point, what is the %w/w Na₂C₂O₄ in the sample.

Answer 1

Question:

The purity of a sample of sodium oxalate, Na₂C₂O₄, is determined by titrating with a standard solution of KMnO₄. If a 0.5116-g sample requires 35.62 mL of 0.0400 M KMnO₄ to reach the titration’s end point, what is the %w/w Na₂C₂O₄ in the sample.

Solution:

The reaction of the titration is:

16 H⁺(aq) + 2 MnO₄^-(aq) + 5 C₂O₄^2-(aq)  →  2 Mn²⁺(aq) + 8 H₂O(l) + 10 CO₂(g)

One can see that 5 moles of oxalate anion react with 2 moles of permanganate:

$n(C_2O_4^{2-}) = \frac{5}{2}n(MnO_4^-)$ .

The number of the moles of the permanganate used to reach the titration's end point is:

$n(MnO_4^-) = cV = 0.0400·0.03562 = 0.00142$ mol.

Therefore, the number of the moles of oxalate in the sample is:

$n(C_2O_4^{2-}) = \frac{5}{2}·0.00142 = 0.003562$ mol.

The molar mass of Na₂C₂O₄ is 134 g/mol. Therefore, the mass of sodium oxalate in the sample was:

$m(Na_2C_2O_4) = nM = 0.003562·134 = 0.477$ g

The mass percentage of the sodium oxalate in the sample is:

$w = \frac{0.477}{0.5116}·100\% =93.3\%$ .

Answer: the mass percentage of the sodium oxalate in the sample is 93.3%.