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Question

A solution is prepared by adding 750.0 mL of 4.00 x
10-3 M Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 M KIO3.
Will Ce(IO3)3 (Ksp= 1.9 x 10-10) precipitate from this
solution?

Answer 1

First of all you need to find the concentration of the respective ions in the final solution. Then you check if the product
[Ce+3]*[IO3(-)]^3 is bigger or smaller than the Ksp (you will have a precipitate forming only if the product is bigger than the Ksp)

mole Ce+3 =mole Ce(NO3)3 =M1*V1, when you mix the two solutions the number of moles will be the same but the concentration changes due to the change in volume. Thus

[Ce+3]= M1V1/(V1+V2)= 0.004*750/(750+300) =2.86*10^-3 M

similarly

[IO3(-)] =M2V2/(V1+V2) = 0.02*300/(750+300) = 5.71*10^-3 M

[Ce+3]*[IO3(-)]^3 =(2.86*10^-3)*(5.71*10^-3)^3 = 5.32*10^-10 > Ksp so you will have precipitation