Need chem help. Have final tomorrow.

The Ksp of Ca(IO3)2 is 7.1x10^-7 at 298 K. Calculate the solubility (g/L) of calcium iodate in

(a) 0.060 mol/L Ca(NO3)2
(b) 0.060 mol/L NaIO3

I have a final tomorrow and I still don't get this one.

Ah, the common ion effect.

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

7.1 x 10^-7 = (0.060) (x)^2

x = 0.00344 M (this is the iodate concentration, which is twice the amount of Ca(IO3)2 that dissolves.

solubility in mol/L Ca(IO3)2 is 0.00172 M

convert to g/L

0.00172 mol/L times 389.872 g/mol = 0.67 g/L

----------------------

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

7.1 x 10^-7 = (x) (0.060)^2

x = 0.000197 M

This is the Ca^2+ conc, which is 1:1 with calcium iodate

convert to g/L

0.000197 mol/L times 389.872 g/mol = 0.077 g/L

More common ion effect:

http://www.chemteam.info/Equilibrium/Com…

More Ksp stuff:

http://www.chemteam.info/Equilibrium/Equ…

Best wishes on the final.