Hello, I would greatly appreciate some help with the following problem, as I'm not sure how to proceed.

Question

I'm trying to find solubility (Ksp) of Cu(IO3)2 from titration experiment.

The following equations were provided:

Cu(IO3)2 --> Cu^2+ + 2IO3^-
Ksp = [Cu^2+][IO3^-]^2
2Cu^2+ + 5I^- --> 2Cul + I^-
IO3^-+8I^- + 6H3O^+ --> 3I3^- + 9H2O
I3^- + 2S2O3^2- --> 3I^- + S4O6^2-
I3^- + 2S2O3^2- --> 3I^- + S4O6^2-

I pipetted 10.00mL of saturated Cu(IO3)2 solution into a 125mL flask and added ~10mL of distilled water, 0.8 grams of KI and 1.0mL of glacial acetic acid. The sample was titrated with Na2S2O3. I found that I needed 14mL of Na2S2O3 for the first titration titration trial.

Could someone please let me know how I should go about finding Ksp for Cu(IO3)2? I would greatly appreciate any help.

Answer 1

Also, I forgot to mention that the molarity of Na2S2O3 was previously established to be 0.032 M.

Answer 2

Yes, you need M S2O3^- but surely you have more places than that in the answer.And I wouldn't think 14 mL is close enough. I would think something like four places(that is 14.22 mL). Also I think you have a typo; I believe the first equation after Ksp is not right.That should be
2Cu^2+ + 5I^- ==> 2CuI + I3^-. Do you know what's going on here?
Cu^2+ is reduced to Cu^+ and ppts as CuI while I^- is oxidized to I2 (I2 + I^- ==> I3^-). The liberated I2 is then titrated with standard S2O3^-.
mols S2O3 = M x L = ? Then convert from mols S2O3^- to mols Cu^2+.
?mols S2O3^2- x (1 mol I3^-/2 mols S2O3^2-) x (2 mol Cu^2+/1 mol I3^-) = x mols Cu^2+.
That gives you the mols Cu in 10 mL of the saturated solution and mols x molar mass = grams in 10 mL. Correct that for grams in 1L if you want grams. I would leave it in mols so you will have mols/L and can then solve for Ksp.

Answer 3

Thank you much Dr. Bob!