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 37.54 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 14.37 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

Answer A

The equation for the given reaction …
IO₃⁻_(aq) + 8I⁻_(aq) + 6H⁺_(aq) → 3I₃⁻ + 3H₂O _(l)
1 mol IO₃⁻ will produce 3 mol I₃⁻
Molar mass of KIO₃ = 214.0 g/mol.
1.022 g KIO₃, which is 0.004776 mol.
we use 1/10 (50/500) of that in your reaction, so we use 0.4776mmol.
So, 1.433 mmol I₃⁻ is formed.
Answer B

The reaction given is

I₃⁻ + 2S₂O₃^2- → 3I^- + S₄O₆^2-

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

Answer C

A + H2O + I₃⁻ → DA + 2H⁺ + 3I^-

(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

Weight Percentage = (0.157/1.223)∙100 = 12.8 %