Question

Assume that you mixed 20.00 mL of 0.040 M KI with 20.00 mL of 0.060 M (NH4)2S2O8 , 10.00 mL of 0.00070 M Na2S2O3 , and a few drops of starch. The point of mixing sets time = 0.

The Initial Concentrations are:

[KI] = 0.016M

[(NH4)2S2O8] = 0.024M

[Na2S2O3] = 0.00014M

Oxidation-Reduction Reaction between iodide and peroxydisulfate: 2I- + S2O8^2- -> 2SO4^2- + I2

Iodine Clock Method reaction: 2S2O3^2- +I2 -> 2I^- + S4O6^2-

Question: The basis of the "Method of Initial Rates" used in this experiment, is that the initial concentrations of reagents are essentially unchanged during the measurement time period. Calculate the % of the initial (NH4)2S2O3 that has reacted when the blue colour appears.

Answer 1

The important point in iodine clock reaction is that thiosulfate is completely consumed by the iodine generated in the first part of the reaction so that the decrease in concentration of thiosulfate = -[(final concentration) – (initial concentration)] = (initial concentration) = 0.00014 M.

The iodine reacts when starch when thiosulfate gets consumed.

The total volume of the solution = (20 + 20 + 10) mL = 50 mL.

Millimoles of thiosulfate = (50 mL)*(0.00014 M) = 0.007 mmole.

As per the balanced stoichiometric equation (equation 2, reaction of thiosulfate with iodine),

2 mole thiosulfate = 1 mole iodine.

Therefore, millimoles of iodine = (0.007 mmole thiosulfate)*(1 mole iodine/2 mole thiosulfate) = 0.0035 mmole.

Again, millimoles of peroxydisulfate = (50 mL)*(0.024 M) = 1.2 mmole.

As per stoichiometric equation (equation 1, reaction peroxydisulfate to produce iodine),

1 mole iodine = 1 mole peroxydisulfate.

Therefore, millimoles of peroxydisulfate consumed = (0.0035 mmole iodine)*(1 mole peroxydisulfate/1 mole iodine) = 0.0035 mmole.

Molar concentration of peroxydisulfate that has reacted = (0.0035 mmole)/(50 mL) = 0.00007 M.

Percent of initial concentration of (NH₄)₂S₂O₈ that has reacted when the blue color appears = (0.00007 M)/(0.024 M)*100 = 0.2917% ≈ 0.292% (ans).

Note: I believe the question asks for the percentage of (NH₄)₂S₂O₈ that has reacted since 100% Na₂S₂O₃ reacts during the reaction.