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
To calculate the Pb2+ and IO3- concentrations when the mixture comes to equilibrium, we can use the stoichiometry of the balanced equation and the given initial concentrations.
1. Calculate the number of moles of lead (II) nitrate (Pb(NO3)2):
Pb(NO3)2 = (0.05 L)(0.05 M) = 0.0025 mol
2. Calculate the number of moles of sodium iodate (NaIO3):
NaIO3 = (0.04 L)(0.2 M) = 0.008 mol
3. Determine the limiting reagent:
In this case, NaIO3 is in excess, so it is not limiting the reaction. Pb(NO3)2 is limiting.
4. Calculate the number of moles of Pb(IO3)2 formed:
Pb(IO3)2 formed = 0.0025 mol (1 mol Pb(NO3)2 produces 1 mol Pb(IO3)2)
5. Calculate the remaining moles of NaIO3 after the reaction:
NaIO3 remaining = 0.008 mol - 2(Pb(NO3)2) = 0.008 mol - 2(0.0025 mol) = 0.008 mol - 0.005 mol = 0.003 mol
6. Calculate the concentration of IO3- in the solution:
(IO3-) = 0.003 mol / 0.090 L = 0.0333 M
Therefore, the concentrations of Pb2+ and IO3- in a solution saturated with Pb(IO3)2 and with the excess of 0.003 mol / 0.090 L (or 0.0333 M) of IO3- are 0 M and 0.0333 M, respectively.
Finally, we can use the Ksp expression to check if our equilibrium concentrations are consistent with the given Ksp value:
Ksp = (Pb2+)(IO3-)2 = (0 M)(0.0333 M)2 = 0
Since the Ksp is significantly smaller than the given value (2.6 x 10-13), it indicates that the reaction did not reach equilibrium and more Pb(IO3)2 would need to dissolve to reach saturation.