Answer to Question #317359 in General Chemistry for Musonda 02

Question

Calculate the concentration in mol dm–3 and g dm–3, of a sodium ethanedioate (Na2C2O4) solution, 5.00 cm3 of which were oxidized in acid solution by 24.50 cm3 of a potassium manganate(VII) solution containing 0.05 mol dm–3.

Answer 1

Moles of Manganese (VII) ions used

mol MnO₄^– used in titration = 0.05 x 24.5 / 1000 = 0.001225,

Balanced equation for the reaction

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

Mole ratio MnO₄^–:Na₂C₂O₄ is 2:5

so mol Na₂C₂O₄ titrated = 0.001225 x 2.5 = 0.003063 in 5 cm³,

scaling up to 1 dm³,

molarity Na₂C₂O₄ = 0.003063 x 1000 / 5 = 0.613 mol dm^–3

M_r(Na₂C₂O₄) = 134,

so concentration = 0.613 x 134 = 82.1 g dm^–3