Redox Reaction Titration Lab Question

Question

How do I calculate the purity of a sample of impure oxalic acid if 0.4006 g of this material requires a titration with 28.62 mL of a potassium permanganate solution, 1.000 mL of which contains 5.980 mg of KMnO4?

Answer 1

Following up on Barry M's answer, I can offer the following.

Balanced equation for the titration of oxalic acid (C₂H₂O₄) with potassium permanganate (KMnO₄) is:

2KMnO₄ + 3H₂SO₄ + 5C₂H₂O₄ ==> K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂ (taking place in acidic solution)

First, we will find the moles (mmoles) KMnO₄ used: 5.980 mg KMnO₄ x 1 mmol/158.03 mg = 0.03784 mmoles

0.03794 mmoles/ml x 28.62 ml = 1.0830 mmoles KMnO₄ used

Next, we find the moles (mmoles) of C₂H₂O₄ equivalent to 1.0830 mmoles KMnO₄:

1.0830 mmol KMnO₄ x 5 mmol C₂H₂O₄ / 2 mmol KMnO₄ = 2.7075 mmoles C₂H₂O₄

Converting mmoles C₂H₂O₄ to milligrams of C₂H₂O₄, and then to grams, we have...

2.7075 mmoles x 90.03 mg/mmole = 243.76 mg C₂H₂O₄

Converting to grams, we have 243.76 mg x 1g/1000 mg = 0.24376 g of oxalic acid (C₂H₂O₄)

To obtain percent purity, we divide the mass of oxalic acid by the original mass of the sample:

% purity = 0.24376 g/0.4006 g (x100%) = 60.85% pure

Answer 2

First, get the balanced equation.

Then get the moles of KMnO4 used in the titration. (I prefer to work in millimoles, mmol, = mL X M, where M is the molarity). In this problem, the value of M of KMnO4 is derived from the information that 1.000 mL contains 5.980 mg of KMnO4, so 1 L has 5.980 g. Convert that to mol, and since M = mol/L, you'll have M also.

Next, get the mol of oxalic acid used in the titration based on the mol KMnO4 and the ratio of coefficients of oxalic acid to KMnO4. Convert mol oxalic acid to g oxalic acid, and then divide by the 0.4006 g. Multiiply by 100 to get % purity of oxalic acid.