posted by mel on .
OKOK I know that I have posted this question but I just don't understand the last part.
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. (10 marks)
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.
How do I determine the mols Pb(IO3)2 pptd and mols NaIO3 remaining.
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.
So, you have a 90 mL solution that is saturated with Pb(NO3)2 [it is insoluble] and it has an excess of 0.003 mols NaIO3. So the NEW problem now reads,
calculate (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 = (Pb^+2)(IO3^-)^2 = 2.6E-13
The (Pb^+2) is determined from Ksp.