Question

(a) Potassium iodate solution was prepared by dissolving 1.022 g of KIO3 (FM 214.00) in a 500 mL volumetric flask. Then 50.00 mL of the solution was pipetted into a flask and treated with excess KI (2 g) and acid (10 mL of 0.5 M H2SO4). How many millimoles of I3− are created by the reaction

b) The triiodide from part (a) reacted with 37.57 mL of Na2S2O3 solution. What is the concentration of the Na2S2O3 solution?

(c) A 1.223 g sample of solid containing ascorbic acid and inert ingredients was dissolved in dilute H2SO4 and treated with 2 g of KI and 50.00 mL of KIO3 solution from part (a). Excess triiodide required 14.30 mL of Na2S2O3 solution from part (b). Find the weight percent of ascorbic acid (FM 176.13) in the unknown.
HINT: Think "back-titration

Answer 1

(a) Potassium iodate solution was prepared by dissolving 1.022 g of KIO3 (FM 214.00) in a 500 mL volumetric flask. Then 50.00 mL of the solution was pipetted into a flask and treated with excess KI (2 g) and acid (10 mL of 0.5 M H2SO4). How many millimoles of I3− are created by the reaction

a. mol IO3 - = (1.022 g)(1 mol/214.00 g) = 0.004776 mol

Reaction

IO3 - + 8I - → 3I3 - + 3H2O

50/500 mL taken or 0.4776 mmol

Thus, 1.433 mmol I3 - is formed.

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b. Titration Reaction

I³ - + 2S₂O3 2- → 3I- + S₄O₆^2-

So… [S2O3 2- ] = (2)(.001433)/(0.03766 L) = 0.07609 M

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c. A + H2O → DA + 2H+ + 2e- E 0 = 0.390 V

I3 - + 2e- → 3I- E 0 = 0.535

V A + H2O + I3 - → DA + 2H+ + 3I-

14.33 mmol I3 - (or added to the ascorbic acid, 50 out of 500 mL)

(14.22 mL)(0.07609 M) = 1.082 mmol S2O3 2- added.

It will react with 0.5410 mmol I3 - .

Therefore, 1.433 – 0.5410 = 0.892 mmol I3 - reacted with 0.892 mmol of ascorbic acid.

(0.892 mmol)(176.13 mg/mmol)/1000 = 0.157 g A

Wt % = (0.157/1.223)∙100 = 12.8 %

d) must add indicator right before the endpoint.