The 𝐾sp of cerium (III) iodate, Ce(IO3)3 , is 1.40×10−11 . Calculate the molar solubility, 𝑠 , of this compound.

Question

s= ? M

Can someone show me the steps as well...

Answer 1

Set up the Ksp relation. Ksp = [Ce³⁺] [IO₃^1-]³ as Ce(IO₃)₃ ====> Ce³⁺ +

3 IO₃^1-.

1.4 x 10^-11 = s x (3s)³, s is the solubility because if one mole of Ce(IO₃)₃ dissolved you would get one mole of Ce³⁺ ions. 3s is because the IO₃^1- concentration is 3 times the Ce³⁺. Now solve for s.

27 x s⁴ = 1.4 x 10^-11

s⁴ = 1.4 x 10^-11 / 27 = 5.19 x 10^-13

s = 8.49 x 10^-4 M