What is the solubility of Cd(IO3)2 in water and in 0.20 M CdCl2 at 25 degrees C? Ksp of Cd(IO3)2= 2.5*10^-8 (at 25 C)

Question

anonymous · Answer

Any ideas @sweetburger ? :)

anonymous · Answer

Oh oops, yep

sweetburger · Answer

what exactly are we looking for molar solubility?

anonymous · Answer

Yeah I think so

sweetburger · Answer

Well given the ksp I guess i can help you find the molar solubility in water 
so we have that the Ksp=[Cd][IO3]^2
                             Ksp= [x][2x]^2 
                             2.5*10^-8=4x^3 
so divide the ksp by 8 and then take the cube root of that number.

sweetburger · Answer

I don't know does the Ksp change based upon the solute?

sweetburger · Answer

or is your question saying that .2M Cadium Chloride is already in the water and were trying to determine the solubility with the Cadium Chlorida already there.

anonymous · Answer

It just says to determine the solubility of Cd(IO3)2 in water and 0.20 M CdCl2+ Ksp of Cd(IO3)2 is 2.5*10^-8 at 25 degrees C

jfraser · Answer

It's a common ion effect problem. Instead of the Cd dissolving impure water with an initial concentration of zero, the starting concentration of the cadmium comes from the cadmium chloride. Plug the concentrations into the KSP formula

jfraser · Answer

That should be in pure water

sweetburger · Answer

Thanks! @JFraser