Question

 

Molarity of Na2S2O3 = 0.0258 M

Volume Na2S2O3 = 23.23 mL

Molarity IO3- = 4.30 x 10^-2 M

Solubility product constant expression for Ca(IO3)2 : [Ca2+][IO3-]^2

Trying to figure out:

1) What is the molar solubility of Ca(IO3)2 in mol/L?

2) Solubility of Ca(IO3)2 in g/L?

3) Solubility product constant (Ksp) value for Ca(IO3)2?

Help on this would be greatly appreciated! I am sincerely stuck and really don't know how to proceed from this point. thanks in advance!

Answer 1

3) Calculations for solubility product ( Ksp ) value of Ca (IO3)2

The formula for calcium Iodate is Ca (IO3)2 , represents that for every 2 moles of IO3^- ion , the number of moles of Ca ^2+ present = 1 mole / L

when molarity (M) of IO3^2- is given as 4.30 x 10^-2 M the Number of moles of Ca^2+ = ( 4.30 x 10^-2 ) / 2

.................................................................................................................................................. = 0.0215 moles / L

Now Ksp of Ca ( IO3 )2 = [ Ca²⁺ ] [ IO3 ^- ]²

Substituting the values we get,

Ksp = ( 0.0215 ) ( 0.043 )²

...... = 3.9754 x 10^-5  

The net reaction between iodate and thiosulphate is

IO3^- (aq ) + 6S2O3^2- (aq) + 6H3O^+ (aq)-----------> I^-(aq) + 3S4O6^2- (aq) + 9H2O (l)

The stoichiometry of the reaction represents that-

Moles of S2O3^2- = M S2O3 2- x L S2O3 ^2- = 0.0258 x .02323

.............................................................................. = 5.9934 x 10^-4 mole

Again , as per equation above it takes 1 mole of IO3^- to react with 6 moles of S2O3^2-

So, 0.00059934 moles of S2O3^2- x (1mole of IO3^- / 6 moles of S2O3^2- )

....................................................................................................= 9.988x 10^{- 5} moles of IO3^-

Therefore the molar solubility of Ca`(IO3)2 = 9.99 x 10^-5 moles per Litre

2 ) Therefore solubility of Ca(IO3)2 in gms / Litre = 9.99 x 10^-5 x 390

...............................................................................= 0.0389 gms / L