1. Calculate the concentration of IO−3 in a 5.52 mM Pb(NO3)2 solution saturated with Pb(IO3)2 . The 𝐾sp of Pb(IO3)2 is 2.5×10−13 . Assume that Pb(IO3)2 is a negligible source of Pb2+ compared to Pb(NO3)2 .

Question

2. A different solution contains dissolved NaIO3 . What is the concentration of NaIO3 if adding excess Pb(IO3)2(s) produces a Pb2+ concentration of 5.40×10−6 M ?

Answer 1

................Pb(IO3)2 ==> Pb^2+ + 2IO3^-
I.....................solid...........0..............0
C....................solid...........x..............2x
E.....................solid...........x..............2x
Ksp = 2.5E-13 = (Pb^2+)(IO3^-)^2
The problem tells you that (Pb^2+) is 5,52E-3 M, you want to calculate (IO3^-) in this solution AND THAT we can ignore the 2x contribution from the solubility of Pb(IO3)2. So
Ksp = 2.5E-13 = (Pb^2+)(IO3^-)^2
2.5E-13 = (5.52E-3)(IO3^-)^2
Solve for (IO3^-) = ?

2. A separate post.

Answer 2

................Pb(IO3)2 ==> Pb^2+ + 2IO3^-
I.....................solid...........0..............0
C....................solid...........x..............2x
E.....................solid...........x..............2x

The problem tells you that you have a solution of NaIO3 (remember it's completely ionized; i.e., 100%) so the amount of Pb(IO3)2 SOLID that dissolves will have (Pb^2+) = 5.4E-6 M. So if we ignore any extra IO3^- that might come from Pb(IO3)2 (whatever amount it will be small), then
Ksp = 2.5E-13 = (Pb^2+)(IO3^-)^2
Plug in the (Pb^2+) and solve for (IO3^-).
Post your work if you get stuck.