Question

Please! This is urgent!

I don't know how to do this.

Determine the solubility at 25°C of Cd(IO3)2 in water and in 0.20 M CdCl2. Ksp of Cd(IO 3) 2 is 2.5 × 10 –8 at 25°C.

Please help!

Answer 1

.In water:...........................Cd(IO3)2 ==> Cd^2 + 2 IO3^-
I..............................solid..............0..............0
C............................solid...............x.............2x
E............................solid...............x.............2x

Ksp = (Cd^2+)(IO3^-)^2
2.5E-9 = (x)(*2x)^2. Solve for x = solubility.

In 0.2M CdCl2:
Remember CdCl2 is comp;letelyionized; therefore,
..................CdCl2 ==> Cd^2+ + 2Cl^-
I..................0.20............0..............0
C...............-0.20...........0.20.........0.40
E...................0............0.20..........0.40

Then you can see (Cd^2+) = 0.20 + x ; i.e.(0.20 from CdCl2 and x from Cd(IO3)2
And (IO3^-) = 2x
Plug those values into the Ksp expression as above and solve for x. Thjat will be the solubility. This problem shows the common ion effect; that is, when a common ion is present (Cd^2+ in this case), the solubility is decreased. Post your work if you get stuck.