Question

Vitamin C (ascorbic acid) from foods can be measured by titration with I3-If 29.41 mL of I3- solution are required to react with 0.1970 g of pure ascorbic acid, what is the molarity of the I3- solution? A vitamin C tablet containing ascorbic acid plus inert binder was ground to a powder, and 0.4242 g was titrated by 31.63 mL of I3-. How many moles of ascorbic acid are present in the 0.4242 g sample? Find the weight percent of ascorbic acid in the tablet.

I got .08 mol/L for the first part, is that correct? Then I'm not sure where to go from there?

Answer 1

Solution :-

Mass of pure ascorbic acid = 0.1970 g

volume of I3- solution = 29.41 ml = 0.02941 L

moles of pure ascorbic acid = mass / molar mass

                                       = 0.1970 g / 176.12 g per mol

                                       = 0.001119 mol

mole ratio of the ascorbic acid to I3^- is 1 : 1

therefore moles of I3^- = 0.001119 mol

molarity of the I3^- = moles / volume in liter

                            = 0.001119 mol / 0.02941 L

                            = 0.038 M

now lets calculate the moles of the I3^- reacted in the second part

moles= molarity * volume in liter

        = 0.038 mol per L * 0.03163 L

        = 0.001202 mol ascorbic acid

now lets calculate the mass of ascorbic acid

mass of ascorbic acid = moles * molar mass

                                = 0.001202 mol * 176.12 g per mol

                                = 0.21169 g

now lets find the mass percent of the ascorbic acid

mass % = (mass of ascornbic acid / mass of tablet )*100%

             =(0.21169 g / 0.4242 g)*100%

             = 49.9 %

Therefore the mass percent of the ascorbic acid in the tablet is 49.9 %