A 1.2431 -g alloy sample was analyzed for its copper content iodometrically. The alloy was dissolved in hot, concentrated nitric acid and prepared for analysis. The addition of excess KI(aq) solution precipitated the copper as cuprous iodide and liberated elemental iodine according to the following reaction:

Question

Cu2+(aq) + I–(aq) → CuI(s) + I2(aq)

The liberated iodine was titrated against a standard thiosulfate solution until the disappearance of the blue starch-iodine indicator complex according to

I2(aq) + 2S2O32–(aq) → 2I–(aq) + S4O62–(aq)

Calculate the percent copper in the sample if 26.41 mL of a 0.2300 M thiosulfate solution was required to react with the liberated iodine.

Answer 1

mols S2O3^-2 = M x L
Convert moles S2O3^-2 to moles I2, then to moles Cu.
g Cu = moles Cu x atomic mass Cu

%Cu = (mass Cu/mass sample)*100 =

Answer 2

S2O3^-2..... why is it to the power of -2?

Answer 3

how do you Convert moles S2O3^-2 to moles I2

Answer 4

It is not to the power. We can't write superscripts easily on these boards so we write the carat to denote that a superscript follows. If we had the time we could do it right but few of us take the time and students don't take the time either. Thiosulfate should be
S₂O₃^-2. the -2 shows the charge on the ion.

Answer 5

Cu (+ HNO3) ==> Cu^+2 but not balanced.
2Cu^+2 + 4I^- ==>2CuI + I2
I2 + 2S2O3^-2 ==> 2I^- + S4O6^-2

2 moles Cu = 2 mole Cu^+2 = 1 mole I2 = 2 moles S2O3^-2

Therefore, 1 mole Cu metal = 1 mole S2O3^-