Question

A 1.0567 gram tablet of vitamin C is ground with a mortar and pestle. 0.9975 grams of the resulting powder are added to a 500.0 mL volumetric flask, dissolved in water and then diluted to volume. A 25.0 mL aliquot of the solution is used for the titration with a 0.0210 M solution. If it takes 9.55 milliliters of the triiodide solution to reach the endpoint of the titration, how many milligrams of the ascorbic acid (MW 176.12 g/mol) were in the original tablet?

Answer 1

The reaction between Vitamin C and triiodide is given as :

C6H8O6 + I3^- + H2O = C6H6O6 + 3I^- + 2H⁺

So there reaction says , that equimolar concentration of both the reagents are used.

So using the formula M1V1 = M2V2

M1 x 25.0 = 0.0210 x 9.55

M1 = 0.008022 M

M = no. of moles of solute / volume of solution in L

0.008022 = n / 0.500

n = 0.004011 moles

So 0.004011 moles of ascorbic acid is present in 0.9975 grams of powder

The number of moles of acid present in 1.0567 gram tablet will be (1.0567 x 0.004011) / 0.9975

= 0.00425 moles

The amount in milligrams = 0.00425 x 176.12 x 1000 = 748.51 milligrams