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Answer to Question #57565 in General Chemistry for Sarah

Question #57565
For this lab we had 6 trials where we had flask one contain 0.0055 M Na2S2O3 and 0.16M KI, 0.1M EDTA (only 1 drop) and starch. Flask 2 contained 0.12M (NH4)2S2O8 and DI Water. Each trial had different volumes used.
For example trial one had the following:

Flask1:
40mL of 0.16M KI
20mL of 0.0055M Na2S2O3
one drop of 0.1M EDTA
1.000g of starch

Flask 2:
5mL of 0.12M (NH4)2S2O8
35mL of DI Water

The 2 flasks were later mixed until a reaction (change of color) was observed.
Here are the instructions for the calculation:
For each of the 6 trials find the initial [I-]. Use the equation M1V1=M2V2 where M1 and V1 are the initial molarity and volume and M2 and V2 are the molarity and volume AFTER mixing but BEFORE reacting.
Remember that when substances are mixed, the concentrations of all the substances are lowered.

Help?!?
Expert's answer
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.
When all the thiosulphate is used up, you will see a blue color iodobromine solution. If the solution does not become blue, whereas the concentration do not meet the claims.
Answer: If the solution does not become blue, whereas the concentration do not meet the claims.

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