n a volumetric analysis experiment, a solution of sodium oxalate (Na 2C 2O 4) in acidic solution is titrated with a solution of potassium permanganate (KMnO 4) according to the following balanced chemical equation: 2KMnO4(aq) + 8H2SO4(aq) + 5Na2C2O4(aq) → 2MnSO4(aq) + 8H2O(l) + 10CO2(g) + 5Na2SO4(aq) + K2SO4(aq) It required 40.0 mL of 0.0335 M KMnO4 to reach the endpoint. What mass of Na2C2O4 was present initially?

mNa₂C₂O₄ = 0.4489 g

Explanation:

This is a problem of stechiometry. in this case, we need to find the mole ratio between the permanganate solution and the oxalate. This can be seen in the balanced reaction which is the following:

2KMnO₄ + 8H₂SO₄ + 5Na₂C₂O₄---> 2MnSO₄ + 8H₂O + 10CO₂ + 5Na₂SO₄ + K₂SO₄

According to this balanced reaction, we have a mole ratio of 2:5 between permanganate and oxalate, so, the first thing to do here is calculate the moles of the permanganate needed to reach the endpoint of this titration:

n = M * V

Replacing we have:

nKMnO₄ = 0.0335 mol/L  * 0.040 L = 0.00134 moles

With these moles, and the above mole ratio, we can determine the moles of the oxalate:

2 moles KMnO₄ > 5 moles Na₂C₂O₄

0.00134 moles > X

X = 0.00134 * 5 / 2

nNa₂C₂O₄ = 0.00335 moles

Finally, with the molecular mass of Na₂C₂O₄, we can determine the mass of the oxalate:

Molecular weights:

Na: 23 g/mol;   C: 12 g/mol;    O: 16 g/mol

MMNa₂C₂O₄ = (23*2) + (2*12) + (4*16) = 134 g/mol

mNa₂C₂O₄ = nNa₂C₂O₄ * MM

mNa₂C₂O₄ = 0.00335 * 134

Hope this helps

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explanation: