Another concentration question??

Question

A solution is made by dissolving 2.015 g of potassium dichromate (VI) in 250.0 mL of water. A 25 mL portion is added to an excess of iodide ions and some dilute sulfuric acid. The iodine produced is then reacted with sodium thiosulfate solution, 36.20 mL of the thiosulfate solution are need for reaction. Calculate the concentration of the thiosulfate solution.

thanks, and im noticing answerers having different answers, mostly toward the end of the answer.

So I might have to add to my question, asking for the reason why they would be different. Hopefully you can check back on this question for updates.

Thanks, I need the help.

Answer 1

1K2Cr2O7 + 6KI + 7H2SO4 --> Cr2(SO4)3 + 4K2SO4 + 3 I2 + 7H2O

which simplifies to:

1K2Cr2O7  --> 3 I2    (a 1:3 ratio)

--------------------------------------...

1 I2 + 2 Na2S2O3 --> 2 NaI + NaS4O6   (a 1:2 ratio)

------------------------

which means that:

1K2Cr2O7  --> (3 I2 ) --->  6 Na2S2O3  (a 1:6 ratio)

===========================

find moles:

2.015 g @ 294.19 g/mol = 0.0068493 moles K2Cr2O7 was dissolved in 250.0 ml of water

since only 25 ml was used,(1/10th of the K2Cr2O7solution), it means only 0.0006849 moles of K2Cr2O7 was used, (1/10th of the total moles of K2Cr2O7)

0.0006849 mol K2Cr2O7 @ 6 mol Na2S2O3/1mol K2Cr2O7 = 0.0041096 moles of Na2S2O3 reacted

now for the molarity of the 36.20 ml Na2S2O3 solution:

0.0041096 mol Na2S2O3 / 0.03620litres = 0.1135 Molar

your answer is: 0.1135Molar Na2S2O3