The ksp for Ce(IO3)3 is 3.2E-10. What is Ce^3+ concentration in a solution prepared by mixing 60ml of 0.025M Ce^3+ with 40ml of 0.045M IO3^-?

Answer 1

millimols Ce^3+ = mL x M = 60 x 0.025 = 1.5
mmols IO3^- = 40 x 0.045 = 1.8
Qsp = (Ce^3+)(IO3^-)^3 = approx 9E-8 but you can do the more exact number. Note that M = mmols/mL. which means Ksp will be exceeded and a ppt will form. Now we must find the limiting reagent.
Ce^3+ + 3IO3^- ==> Ce(IO3)3
1.5.....1.8
Therefore, IO3^- is the limiting reagent and all of it will be used to form a ppt. The (Ce^3+) will be what is left behind. Can you take it from here.

Answer 2

I understand the solution above but if the limiting reagent is 1.8 that means to get the [Ce^3+]= 1.5-1.8= 0.3mols/ 0.10L wherein 0.1L is the total volume. Is this correct? Thanks.