3A. Some people have an allergic reaction to the food preservative sulfite (SO3

Question

3A. Some people have an allergic reaction to the food preservative sulfite (SO3 2- ), which can be measured by instrumental methods or by a redox titration: Initially, a solution of I3 - was prepared by dissolving 1.100 g of KIO3, 7.000 g of KI, and 1.0 mL of 6 M H2SO4 in a 100 mL volumetric flask and diluting to the mark. Write the reaction that occurs when the sulfuric acid is added to potassium iodate and potassium iodide.

3B. Write a balanced equation for the reaction that occurs between the I3 - and the sulfite in the wine

3C. 5.00 mL of the solution described in 3A is added to 50.00 mL of the wine and all of the sulfite reacts. The excess I3 - is then titrated with 0.05111 M NasSsO3 and 13.55 mL are required to reach a starch-indicator end point. What is the concentration of sulfite in the wine in mol/L.

3D. When should the starch-indicator be added to the titration solution – at the start or just before the equivalence point is reached? Justify your answer.

Explanation / Answer

3A): When H2SO4 is added to potassium iodate and potassium iodide solution I2 is liberated. The balanced chemical equation is;

8I-(aq) + IO3-(aq) + 3H2SO4(aq) -------- > 3 I3-(aq) + 3SO42-(aq) + 3H2O ---------- (1)

3B): The balanced equation for the reaction of I3-(aq) and SO32-(aq) is

SO32-(aq) + I3-(aq) + H2O ------- > SO42-(aq) + 2H+(aq) + 3 I-(aq) ------- (2)

3C): Moles of KIO3 in 100 mL of the solution = mass / molar mass = 1.100 g / 214.0 g/mol = 0.00514 mol

Hence moles of KIO3 in 5.00 mL of the solution = 0.00514 mol x (5mL/100mL) = 0.000257 mol  IO3-(aq)

From equation (1) moles of I3-(aq) formed by 0.000257 mol  IO3-(aq)

=  0.000257 mol  IO3-(aq) x [ 3 mol  I3-(aq) / 1 mol IO3-(aq) ] = 0.000771 mol  I3-(aq)

The balanced equation for the reaction of excess I3-(aq) with Na2S2O3 is

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

Moles of S2O32-(aq) reacted with excess I3-(aq) = MxV =  (0.05111 mol/L x 0.01355 L) = 0.0006925 mol

Hence moles of excess  I3-(aq) = 0.0006925 mol S2O32-(aq) x [1 mol I3-(aq) / 2 mol S2O32-(aq) ] = 0.000346 mol

Hence moles of I3-(aq) reacted =  0.000771 mol -  0.000346 mol = 0.000425 mol I3-(aq)

From reaction (2), moles of SO32-(aq) in 50.00 mL of wine = 0.000425 mol SO32-(aq)

Hence concentration of sulphite in wine = (0.000425 mol / 50 mL) x (1000 mL/ 1L) = 8.5x10-3 mol /L (answer)

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