In the redox persulfate-iodide experiment, a 100 mL reaction mixture is prepared for one of the runs as follows. 0.200 M KI 20 mL 0.200 M KNO3 15 mL 0.25% starch 5 mL 0.01 M Na2S2O3 10 mL 0.1 M EDTA 1 drop water to make 65 mL 0.200 M (NH4)2S2O8 35 mL a. How many moles of S2O32- would react when the solution turns dark blue? b. How many moles of S2O82- would react when the solution turns dark blue? c. What would be the initial rate of reaction if this reaction mixture took 200 seconds to turn dark blue?

a) The number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.

b) The number of moles of  S₂O₈2- that would react to produce the blue color will be 0.00005 M.

c) rate = 2.5 x 10-6M/L/s

Explanation:

a)

The reactions taking place in this experiment are represented by the ionic equations,

S₂O₈ 2- + 2I- > 2SO₄2- + I₂ (1)

2 S₂O₃2- + I₂ > S₄O₆ +2 I-  (2)

The persulphate ions react with the iodide ions to produce free iodine which is in turn reduced by the thiosulphate ions to produce iodide ions again. This reaction proceeds till all the thiosulphate ions are used up. Therefore the rate of the reaction will be the rate at which iodine is formed and used up.

When there is free iodine the reaction mixture,  the solution gives a dark blue coloration. This happens when all the thiosulphate ions are used up.

The volume of sodium thiosulphate ( Na₂S₂O3) solution added to the reaction vessel = 10ml

Molarity of sodium thiosulphate ( Na₂S₂O3) solution = 0.01M

Number of moles of ( Na2S2O3) = 0.01ml x 10M /1000ml = 0.0001M (molarity x volume in L)

The number of moles of thiosulphate reacted will be equal to the number of moles taken since the reaction proceeds till all the thiosulphate is consumed.

Hence, the number of moles of thiosulphate ( S₂O₃2-) that reacted to turn the solution dark blue will be 0.0001M.

b)

To calculate S₂O₈ 2-

The permanent blue color is produced once all the thiosulphate ions are used up and persulphate reacts with iodide ions to produce iodine, so, the number of moles of persulphate ions will be equal to the number of moles of iodine formed

According to the stoichiometry of equation 1.

1 mole of  S₂O₈ 2-produces 1 mole of iodine.

According to the stoichiometry of equation 2,

1mole of iodine produced consumes 2 moles of  S₂O₃2-

The number of moles of  S₂O₃2- taken = 0.0001M.

2 moles of   S₂O₃2- is equivalent to 1 mole I₂

therefore

0.0001 mole of   S₂O₃2- = 0.0001/2 = 0.00005M of I₂

Since stoichimetrically,

1 mole of  S₂O₈2- is equivalent to 1 mole I2, the number of moles of  S₂O₈2- that would react to produce the blue color will be 0.00005 M.

c)

The initial reaction rate is given by

rate =change in concentration of persulphate ion [S₂O₈2-] / time

rate = change in Concentraion of I₂ / t

since initial concentration of I₂ = 0.

rate = Concentraion of I₂/ t

The concentration of I₂ = number of moles of iodine / total volume of solution in L

= 0.00005M/ 0.1L = 0.0005M/L (Volume of the solution = 100ml = 0.1L)

rate = Concentraion of I₂ / t

= 0.0005 /200s

rate = 2.5 x 10-6M/L/s

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