For any slightly soluble salt placed in water, an equilibrium isestablished between the solid and ions in solution. Consider thefollowing reaction: Ca(IO3)2 (s) <----> Ca 2+ (aq) + 2IO3- a)Write the equilibrium constant expression for this reaction (Ksp).b) When calcium Iodate is added to water at 25 degrees celsius ,the equilibrium concentration of calcium ion was found to be0.011M. Calculate the value of Ksp for this salt.

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The solubility of lead(II) bromide, PbBr₂, in water is 4.34 g/L. In an aqueous solution, an equilibrium exists between the dissolved ions and the solid salt.

1) Write the solubility product expression for this equation.

2) Calculate the molar concentration and the K_sp (solubility product constant) for lead(II) bromide.

turquoisewhale289

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