Concentration Of The Citric Acid Solution - Titration Data

Question

In this titration, NaOHNaOH is the titrant, and citric acid is the analyte.

Please enter your titration data below. Then, with the data on MNet, determine the volume of the analyte, citric acid, at the equivalence point on your graphs. Finally, calculate the molarity of the citric acid, using the molarity of sodium hydroxide you calculated earlier i.e. 0.0119714879 M.

Trial 1Trial 2Trial 3	(Optional)
Volume of titrant,	NaOHNaOH (mL)	43.35	39.66	39.07
Volume of analyte,	citric acid (mL)	2.80	2.41	3.40
Molarity of citric	acid solution (M)

Average Molarity Of Citric Acid

UnroundedRounded
Average molarity of citric acid (M)
Actual molarity of citric acid (M)	-

Standard Deviation Of Citric Acid Molarities

What is the standard deviation of your molarity?

Answer 1

Citric acid is a triprotic acid, meaning it has 3 titratable hydrogens. Let us represent this as H₃A.

H₃A + 3NaOH ==> 3H₂O + Na₃A (where Na₃A is trisodium citrate)

moles NaOH used in trial 1 = 43.35 mls x 1 L / 1000 ml x 0.01197 mols / L = 5.189x10^-4 mols NaOH

moles citric acid in trial 1 = 5.189x10^-4 mols NaOH x 1 mol H₃A / 3 mol NaOH = 1.730x10^-4 moles

molarity of citric acid in trial 1 = 1.730x10^-4 moles / 2.80 ml x 1000 ml / L = 0.0618 M

moles NaOH in trial 2 = 39.66 mls x 1 L / 1000 ml x 0.01197 mol / L = 4.747x10^-4 moles NaOH

moles citric acid = 4.747x10^-4 moles x 1 mol acid / 3 mol NaOH = 1.582x10^-4 moles acid

molarity of citric acid in trial 2 = 1.582x10^-4 moles / 0.00241 L = 0.0657 M

moles NaOH in trial 3 = 39.07 ml x 1 L / 1000 ml x 0.01197 mol / L = 4.677x10^-4 mols NaOH

moles citric acid = 4.677x10^-4 mols NaOH x 1 mol citric acid/3 mol NaOH = 1.559x10^-4 moles acid

molarity of citric acid in trial 3 = 1.559x10^-4 moles acid / 0.00340 = 0.0458 M

Calculate mean and standard deviation from the three trials.