# chemistry check 2

posted by on .

I know you've read this question too many times already but I just want to make sure I did the last part (Ksp) correctly.

50.0 ml of a 0.0500 M solution of lead (II) nitrate is mixed with 40.0 ml of a 0.200 M solution of sodium iodate at 25°C. Calculate the Pb2+ and IO3- concentrations when the mixture comes to equilibrium. At this temperature the Ksp for lead (II) iodate is 2.6 x 10-13.

The balanced equation
Pb(NO3)2 + 2NaIO3 ==> Pb(IO3)2 + 2NaNO3
Pb(NO3)2 = (0.05L)(0.05M) = 2.5 x 10-3 mols.
NaIO3 = (0.04L)(0.2M) = 8.0 x 10-3 mols.
NaIO3 is in excess.
Pb(NO3)2 - 0.0025 mols
NaIO3 = 0.008 mols.

Pb(IO3)2 formed = 0.0025 mols [1 mol Pb(NO3)2 produces 1 mol Pb(IO3)2.]
NaIO3 remaining after the reaction is 0.008 - 2[Pb(NO3)2] = 0.008 - 2(0.0025) = 0.008 - .005 = 0.003 mols. (IO3-) = 0.003 mols/0.090 L = 0.0333 M.

Therefore 90 mL solution is saturated with Pb(NO3)2 [it is insoluble] and it has an excess of 0.003 mols NaIO3.
Therefore (Pb+2) and (IO3-) in a solution saturated with Pb(IO3)2 and 0.003 mols/0.090 L (or 0.0333 M) in IO3-.

Ksp = (Pb2+)(IO3-)2 = 2.6 x 10-13