Titration question! Calculate percentage of unknown in a mixture?

a mixture is known to contain oxalic acid h2c2o4 a diprotic acid as well as other non acidic compounds. an 8.00g sample of this mixture was dissolved in a 250.0ml volumetric flask and a 25.0ml aliquot of this solution was titrated with 0.200M potassium hydroxide and required 36.0ml to reach the end point. calculate the percentage of the oxalic acid in the mixture?Titration question! Calculate percentage of unknown in a mixture?
1. calculate the number of moles of KOH used to reach the end poing: mol(KOH) = 0.2*(36.0/1000) = 0.0072mols

2. KOH will react with oxalic acid in a 2:1 ratio. Therefore you know that the number of moles of oxalic acid present in the sample is: mols(oxalic acid) = 1/2 * 0.0072 = 0.0036mols. This will be the number of moles present in a 25ml aliquote. 3. Transform the numbr of moles of the acid in grams: mass(oxalic acid) = mol weight * mols = 90*0.0036 = 0.324g.

4. You can use ratios to calculate how many grams of oxalic acid where in 250ml of solution: g(in 25) / 25 = g(in 250)/250

hence g(in 250) = g(in 25) * 250 / 25 = 3.24 g.

5. Calculate the percentage by dividing the mass of oxalic acid in the mixture by the total mass of the mixture:

%(oxalic acid) = (g(oxalic acid)/mass tot) *100 =5 (3.24/8)*100 = 40.5%



hope it helps!