Question

Please check through my work, not 100% sure I took the rightsteps to doing this or not.

A saturated aqueous solution of slightly soluble salt, bariumiodate [ Ba(IO3)2] was prepared and titrated with sodiumthiosulfate. Based on the data tabulated below, calculate theindicated quantities in questions 1-5.

(Column A)Volume of saturated Barium Iodate solution (filteredsolution).
(Column B) Initial (before titration ) volume of the thiosulfate inthe burette.
(Column C) Final (after titration) volume of the thiosulfate in theburette.
(Column D) Molarity of titrant (thiosulfate solution). 1,2,3 aretests.
------A------B--------C---------D
---[ mL ]--[mL]--- [mL]---[mol/L]
1) [ 1.0 ]--[0.0]---[6.5]---[0.01]
2) [ 2.0 ]--[6.8]---[20.1]--[0.01 ]
3) [ 3.0 ]--[20.3]- [46.8]--[0.01]


Organize your calculations using the steps below, but firstdetermine and write the proper chemical equation. Ba(IO3)2 +Na2S2O3---> ??

1mol IO3^- reacts with 6mol S2O3^2-

1. Calculate the number of moles of thiosulfate used in eachtitration: n S2O3^2- =
1. 0.01x0.00 = 0
2. 0.01x0.0068 = 0.000068 = 6.8e^-5
3. 0.01x0.0203 = 0.000203 = 2.03e^-4
2. Calculate the number of moles of Iodate titrated: n IO3^-=
1. 0/6 = 0
2. 6.8e^5/6 = 0.0000113 = 1.13e^-5
3. 2.03e^-4/6 = 0.0000338 = 3.38e^-5

3. Calculate the molar concentartion of Iodate ions: [IO3^2-]
1. 0/0.001 = 0
2. 1.13e^-5/0.002 = 0.00565 = 5.65e^-3
3. 3.38e^-5/0.003 = 0.01126 = 1.13e^-2

4. Calculate the average molar concentration of Iodate ions: Avg[IO3^2-] =
0 + 0.00565 + 0.01226 = 0.01695/3 0.00565 = 5.65e^-3

5. Calculate Ksp for the dissolution of Ba (IO3)2 : Ksp =

Ksp = [Ba^2+][IO3^-]^2