How would you determine the % wt of ascorbic acid in a vitamin c tablet when you

Question

How would you determine the % wt of ascorbic acid in a vitamin c tablet when you are doing a red-ox titration using titrant of 7.8 g thiosulfate in 450 mL of freshly boiled water with 0.15 g of sodium carbonate. The weight of the actual tablet was 0.6915 g but the amount of transfered powder after crushing the tablet was 0.6483 g. For this titration, approximately 34 mL of the titrant was used. The powder was dissolved in 60 mL of 0.5 M sulfuric acid. 2 g of solid KI was added along with 25 mL of the standard KIO3 solution (standard was formed with 2.5 g of solid reagent dissolved in 250 mL volumetric flask). 2 mL of starch indicator was added before the end point.

Explanation / Answer

Write down the chemical equations for the reactions taking place:

IO3- + 8 I- + 6 H+ -------> 3 I3- + 3 H2O ……(1)

Ascorbic Acid + I3- + H2O ------> Dehydroascorbic acid + 3 I- + 2 H+ …..(2)

I3- + 2 S2O32- -----> 3 I- + S4O62- ……(3)

As per the balanced reactions above, we have

1 mole IO3- = 3 moles I3-

1 mole I3- = 1 mole Ascorbic Acid

1 mole I3- = 2 moles S2O32-

Start by finding out the molarity of S2O32- prepared. Na2CO3 is used to stabilize the pH of the solution and plays no role in the molarity of the S2O32-.

Molar mass of S2O32- = 110.117 g/mol.

Moles of S2O32- taken = (7.8 g)/(110.117 g/mol) = 0.0708 mole.176.13

Molarity of S2O32- prepared = (0.0708 mole)/[(450 mL)*(1 L/1000 mL)] = 0.1573 mol/L = 0.1573 M.

Find out the molarity of KIO3 solution taken.

Molar mass of KIO3 = 214.001 g/mol.

Moles of KIO3 taken = (2.5 g)/(214.001 g/mol) = 0.01168 mole.

Molarity of KIO3 = (0.01168 mole)/[(250 mL)*(1 L/1000 mL)] = 0.0467 M

This is the information that we require.

What is essentially happening here is that ascorbic acid reacts with I3- and the excess I3- is titrated with S2O32-.

Moles of S2O32- required for the titration = (34 mL)*(1 L/1000 mL)*(0.1573 mol/L) = 5.3482*10-3 mole.

Moles of I3- titrated = (5.3482*10-3 mole S2O32-)*(1 mole I3-/2 moles S2O32-) = 2.6741*10-3 mole.

Moles of KIO3 taken = (25 mL)*(1 L/1000 mL)*(0.0467 mol/L) = 1.1675*10-3 mole.

Moles of I3- generated = (1.1675*10-3mole KIO3)*(3 moles I3-/1 mole KIO3) = 3.5025*10-3 mole.

Moles of I3- reacted with ascorbic acid = (3.5025*10-3 – 2.6741*10-3) mole = 8.284*10-4 mole.

Moles of ascorbic acid reacted with I3- = (8.248*10-4 mole ascorbic acid)*(1 mole ascorbic acid/1 mole I3-) = 8.248*10-4 mole.

Find out the mass of ascorbic acid contained in 8.248*10-4 mole ascorbic acid.

Molar mass of ascorbic acid = 176.13 g/mol.

Therefore, mass of ascorbic acid = (8.248*10-4 mole)*(176.13 g/mol) = 0.1459 g.

Next calculate the percentage of ascorbic acid in the tablet = (mass of ascorbic acid)/(mass of tablet)*100 = (0.1459 g)/(0.6915 g)*100 = 22.38% (ans).

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