Question

When 20.0mL of 1.40M solution of calcium nitrate is mixed with 70.0mL of 0.234M potassium iodate, a precipitate Ca (IO3)2 is formed.
Calculate the concentrations of both NO3- and IO3- before mixing and after mixing.

Answer 1

(NO3^-) = 1.40 M x [2 nitrates/mol Ca(NO3)2] = ? before mixing.
(NO3^-) = 1.40 M x (20/90) x [2 nitrates/mol Ca(NO3)2] = ? after mixing. Remember nitrate ion is not involved in the rxn; it is a spectator ion.

(IO3^-) before mixing = 0.234 M.
After mixing is a lot of work to do it right and I don't know if your prof intended it to be done this way. Here are the two ways to do it; one long and one short. First, this is a limiting reagent (LR) problem with KIO3 being the LR. You have 20 x 1.4 = 28 mmols Ca(NO3)2 and 70 x 0.234 = 16.38 mmols KIO3 initially.
....Ca(NO3)2 + 2KIO3 =>Ca(IO3)2(s) + 2KNO3
I....28..........0.......0..........0
add............16.38.................
C..-16.38.....-16.38......8.19.........
E...11.62.........0.......8.19 solid.....

The easy, but incorrect way, of doing this is to say that IO3^- is zero from above BUT the Ca(IO3)3 ppt dissolves slightly and the solubility depends upon the concn of excess Ca^2+ since not all of that is used. Look up Ksp for Ca(IO3)2

Ksp = (Ca^2+)(IO3^-)^2
(Ca^2+) from above is 11.62 mmols/90 mL = ?
Substitute and solve for (IO3^-) and that is the (IO3^-) after mixing. Post any work if you have questions.

Answer 2

Thank you