Question

In the titration of total acid with sodium hydroxide, NaOH, it was determined that 36.03mL of 0.0980M NaOH was needed to neutralize all the acid in a 10.0mL aliquot of a powdered drink mix solution. In a second titration, it was determined that were 1.45 x 10^(-5) moles of ascorbic acid in a 10.0mL aliquot of the same powdered drink mix solution. Calculate the moles of citric acid in the 10.0mL aliquot of powdered drink mix solution.

So confused.. Thank you.

Answer 1

I believe ascorbic acid is a diprotic acid which I will label as H2A.
Total mols NaOH used = M x L = approx 0.00353 mols NaOH.
The solution contained 1.45E-5 mols ascorbic and that will use 2*1.45E-5 = 2.90E-5 mols NaOH.
That leaves 0.003531-2.90E-5 = 0.00350 for mols NaOH to titrate the citric acid in the 10.0 mL sample. Since citric acid is a triprotic acid, mols citric acid will be 1/3 mols NaOH = 0.00350/3 = ?
Check my work.

Answer 2

If you had to make a pitcher of punch with a powdered drink mix, explain how this lab would allow making this drink in the least amount of time