Wondering If I did this correctly so far
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.
Now if this is correct what do I do next?
So far so good.
NaIO3 is is excess; therefore, determine mols Pb(IO3)2 pptd and mols NaIO3 remaining.(All of the Pb(NO3)2 will be consumed.) Then it becomes a common ion problem with Ksp.
You will have solid Pb(IO3)2 in a saturated solution of Pb(IO3)2 that also contains xx mols NaIO3. Calculate Pb^+2.
To determine the moles of Pb(IO3)2 precipitated and the moles of NaIO3 remaining, you need to use the stoichiometry of the balanced equation.
From the balanced equation, we can see that 1 mole of Pb(NO3)2 reacts to form 1 mole of Pb(IO3)2. You initially have 2.5 x 10^-3 moles of Pb(NO3)2, so you will have the same amount of moles of Pb(IO3)2 precipitated.
Next, we need to determine the remaining moles of NaIO3. From the balanced equation, 2 moles of NaIO3 react to form 1 mole of Pb(IO3)2. You initially have 8.0 x 10^-3 moles of NaIO3, so you will have (8.0 x 10^-3 moles NaIO3) / 2 = 4.0 x 10^-3 moles of NaIO3 remaining.
Now, since Pb(IO3)2 is a sparingly soluble salt, it will form a saturated solution. This means that the concentration of Pb^2+ ions (the Pb2+ concentration) in the solution will be equal to the moles of Pb(IO3)2 precipitated divided by the total volume of the solution.
To calculate the Pb2+ concentration, you need to add the volumes of the two solutions together. In this case, the volumes are 50.0 mL and 40.0 mL, which add up to 90.0 mL or 0.0900 L.
The Pb2+ concentration can be calculated as follows:
Pb2+ concentration = (moles of Pb(IO3)2 precipitated) / (total volume of solution)
Pb2+ concentration = (2.5 x 10^-3 moles) / 0.0900 L = 2.778 x 10^-2 M
Lastly, to determine the IO3- concentration, you can use the fact that the Pb(IO3)2 will dissociate to form 2 moles of IO3- ions for every mole of Pb(IO3)2. Since you know the moles of Pb(IO3)2 precipitated, you can multiply that by 2 to calculate the moles of IO3- ions.
IO3- concentration = 2 x (moles of Pb(IO3)2 precipitated) / (total volume of solution)
IO3- concentration = 2 x (2.5 x 10^-3 moles) / 0.0900 L = 5.556 x 10^-2 M
Therefore, the Pb2+ concentration is 2.778 x 10^-2 M and the IO3- concentration is 5.556 x 10^-2 M when the mixture comes to equilibrium.