Answer to Question #218528 in General Chemistry for Deere

Question

For this lab we had 9 trials

flask 1 solution 100 mL of 0.200 M KI made from a stock solution of KI that is 2.00M

flask 2 solution 100 mL of 0.100 M (NH4)2S2O8 made from a stock solution of (NH4)2S2O8 that is 1.00 M

given 100 mL of 0.0050 M Na2S2O3 (timing reagent)

50 mL of 0.20 M KCl

starch solution

experiment 1

flask 1

7.50 mL of 0.200 M KI

5.00 mL 0.00060 M S2O32-

one drop starch solution

add all listed above to one flask 1

determine the concentration of KI , (NH4)2S2O8,and Na2S2O3 when flask 1 and 2 are mixed together . The total volume for each reaction is 20.0 mL

Answer 1

The reaction of oxidation of potassium iodide by persulphate ammonium proceeds according to the equation:

2KI + (NH4)2S2O8 = I2 + K2SO4 + (NH4)2SO4 (1)

occurs with release of iodine by the total second order (first order for each component). The limiting stage of the process is the formation of an intermediate product – ion itselft:

I– + S2O82– = SO42– + ISO4– , (2)

(2)

which decomposes by the reaction:

ISO4– + I– = I2 + SO42– . (3)

The formation of iodine can be seen by the blue starch solution. Starch is a very sensitive reagent for molecular iodine. Therefore even the low speed response when merging components, the solution rapidly becomes blue. To delay the appearance of color, the system adds a fixed amount of sodium thiosulfate, which reacts with the iodine released by the reaction:

I2 + 2Na2S2O3 = Na2S4O6 + 2NaI. (4)

The volume of the solution in the first flask are equal. 40 + 20 = 60 ml.

The number of moles of KI in solution No. 1 is

40*0,16/1000 = 0,0064 mol

The number of moles of Na2S2O3 in the solution No. 1 is

20*0,0055/1000 = 0,00011 mol

The volume of the solution in the second flask is $ 35 + 5 = 40 ml

The number of moles (NH4)2S2O8 in the solution No. 2 is

5 * 0,12/1000 = 0,00060 mol

After mixing the solutions 1 and 2 begin reactions 1 – 4.

From each mole of М8 according to the reaction (1) produces 1 mol of I2

To recovery (and discoloration) 1 mol of I2 in equation(4) consumes 2 mol of Na2S2O3.

Thus, the recovery of 1 mol (NH4)2S2O8 consumes 2 mol of Na2S2O3 Na2S2O3. Then to restore 0,00060 mol mol (NH4)2S2O8 is spent 0,0012 mol Na2S2O3 – i.e. 0,00011 mol of thiosulfate present in solution, this is not enough.

Total volume is 7.50+5.00+7.50 =20.00 (+ 1 drop but I guess they don't count that).

You have 7.50 mL of 0.200 M KI so the concn in the mixed solution is (0.00750 L x 0.200 = 0.0015 moles and that divided by 0.0020 L) = 0.75 M KI. All of the others are done the same way. If you are comfortable with millimoles, you can make it much simpler by 7.50 x 0.200 = 1.5 millimoles and that divided by 20 mL = 0.075 M KI. Even simpler is to treat it as a dilution problem like so.

0.200 M x (7.50 mL/20.0 mL) = 0.075 M KI.