Question

How do I find the molar solubility of this compound? The Ksp of cerium (III) iodate, Ce(IO3)3, is 1.40 × 10-11.

Answer 1

Well, you solve the solubility equilibrium...and get `S_"cerium iodate"=8.49xx10^-4*mol*L^-1`

Explanation:

Well, first off we write the solubility expression to inform our calculations....

`Ce(IO_3)_3(s) stackrel(H_2O)rightleftharpoonsCe^(3+) + 3IO_3^-`

And `K_"sp"=[Ce^(3+)][IO_3^(-)]^3=1.40xx10^-11`

And we can express each ionic concentration in terms of the solubility of the lanthanide salt...i.e. `[Ce^(3+)]=S` and clearly `[IO_3^(-)]=3S`... Capisce?

And so `K_"sp"=Sxx(3S)^3=27S^4`...

and thus `S=""^(4)sqrt(K_"sp"/27)=""^(4)sqrt((1.40xx10^-11)/27)`

Now to find my calculator....`S=8.49xx10^-4*mol*L^-1`

And this gives a solubility in grams per litre of ....`664.83*g*mol^-1xx8.49xx10^-4*mol*L^-1=0.564*g*L^-1`..