Hello, I would greatly appreciate some help with the following problem, as I'm not sure how to proceed.
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.
Also, I forgot to mention that the molarity of Na2S2O3 was previously established to be 0.032 M.
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.
Thank you much Dr. Bob!
To find the solubility product constant (Ksp) for Cu(IO3)2, there are a few steps you need to follow based on the provided information and equations:
1. Write the balanced chemical equation for the reaction that occurs between Cu(IO3)2 and Na2S2O3. From the given equations, we can see that the reaction is between I3^- and S2O3^2- ions.
I3^- + 2S2O3^2- → 3I^- + S4O6^2-
2. Calculate the moles of Na2S2O3 used in the titration. You know the volume (14 mL) and concentration of Na2S2O3 is typically given, so use these values to calculate the number of moles.
3. Use stoichiometry to determine the moles of I3^- ions present in the reaction. From the balanced equation, you can see that 1 mol of I3^- is formed for every 2 moles of Na2S2O3 used.
4. Calculate the moles of Cu(IO3)2 in the 10 mL sample. Because Cu(IO3)2 → Cu^2+ + 2IO3^-, you know that 1 mol of Cu(IO3)2 produces 1 mol of Cu^2+ ions.
5. Calculate the concentration of Cu^2+ ions in the 10 mL sample. Divide the moles of Cu^2+ ions by the volume of the sample (in liters) to get the concentration.
6. Calculate the concentration of IO3^- ions in the 10 mL sample. The ratio of Cu(IO3)2 to Cu^2+ and IO3^- is 1:1:2. Therefore, the concentration of IO3^- ions is double the concentration of Cu^2+ ions.
7. Square the concentration of IO3^- ions to get the concentration of [IO3^-]^2.
8. Substitute the values obtained for [Cu^2+] and [IO3^-]^2 into the Ksp expression:
Ksp = [Cu^2+][IO3^-]^2
9. Solve for Ksp.
Following these steps should allow you to find the solubility product constant (Ksp) for Cu(IO3)2.