20 mL of 0.125 M potassium iodide solution is mixed with 10 mL of 0.1 g L–1 sodium thiosulfate (Na2S2O3.5H2O) solution and 0.2 g of solid iodine indicator is added.

To this solution is added 20 mL of 0.025 M peroxodisulfate solution.

After 46 seconds a blue colour appears, indicating that sufficient I2 has been formed from the reaction between KI and peroxodisulfate to consume all the thiosulfate present. The final temperature is 18 °C.

1. The following prelab questions are designed to guide you through calculations and processes that will enable you to use experimental data to i) determine the rate of the reaction, ii) derive the rate law and iii) determine the rate constant, k. You will do this during the laboratory using your own
experimental results.

a) What is the initial concentration of iodide in the reaction mixture?

b) What is the initial concentration of peroxodisulfate in the reaction mixture?

c) What is the number of moles of thiosulfate consumed after 46 seconds of the reaction?

d) What is the number of moles of iodine formed after 46 seconds of the reaction?

e) What is the change in concentration of iodine (I2) formed after 46 seconds of the reaction?

1 answer

20 mL of 0.125 M potassium iodide solution is mixed with 10 mL of 0.1 g L–1 sodium thiosulfate (Na2S2O3.5H2O) solution and 0.2 g of solid iodine indicator is added. To this solution is added 20 mL of 0.025 M peroxodisulfate solution.
So total volume is 20 mL of 0.125 M KI + 10 mL of 0.1 g/L Na2S2O3.5H2O + 20 mL of 0.025 M Na2S2O8 = 50 mL.

a) What is the initial concentration of iodide in the reaction mixture?
0.125 M KI x (20/50) = ? M
b) What is the initial concentration of peroxodisulfate in the reaction mixture?
0.025 M Na2S2O8 x (20/50) = ? M

c) What is the number of moles of thiosulfate consumed after 46 seconds of the reaction?
All of the Na2S2O3 has been consumed. mols Na2S2O3.5H2O = g/molar mass = 0.1 g/L = 0.1/molar mass Na2S2O3.5H2O = ?
Then mols Na2S2O3.5H2O = M x L = M x 0.010 L = ?


d) What is the number of moles of iodine formed after 46 seconds of the reaction?
mols I2 formed = 1/2 mols KI initially - 1/2 mold Na2S2O3 used. That's 1/2 * moles KI from part a - 1/2 * mols Na2|S2O3 from part c.

e) What is the change in concentration of iodine (I2) formed after 46 seconds of the reaction?
(I2) initially - (I2) when all Na2S2O3 is used.