A solution is prepared by mixing 50.0 mL of 0.41 M Pb(NO3)2 with 50.0 mL of 1.8 M KCl.?

Question

Calculate the concentrations of Pb2+ and Cl2 at equilibrium. Ksp for PbCl2(s) is 1.6 10-5.
Pb2+ =?
Cl2=?

Answer 1

Imagine the solutions are well mixed before precipitation:
Pb(NO3)2: 0.205M in 100mL, or 0.0205 moles.
KCl: 0.9M in 100mL, or 0.09 moles.
The reaction now takes place:
Pb(NO3)2 + 2KCl ==> 2KNO3 + PbCl2
Obviously 0.205M Pb(NO3)2, in comparison to 0.9M KCl, is the limiting reactant. Let the final [Pb++] left in the solution be X M, that is, only 0.1X mole [Pb++] left in the solution. Thus we precipitate out (0.0205 - 0.1X) mole Pb++, which requires 2(0.0205 - 0.1X) of Cl- to precipitate out together. That means the left Cl- mole number to be:
0.09 - 2(0.0205 - 0.1X) = 0.09 - 0.041 + 0.2X = 0.049+0.2X. that means the left Pb++ and Cl- concentration in 100mL solution is:
[Pb++] = X M, and
[Cl-] = 10(0.049+0.2X) = 0.49+ 2X (M) = 0.49
Hence we get the final equation by puting data in Ksp:
Ksp = 1.6x10^-5 = [Pb++]*[Cl-]*[Cl-]
= X(0.49)^2
X = Ksp/0.49^2 = 6.66x10^-5 (M)
So [Pb2+] = 6.66x10^-5 M, and [Cl-] = 0.49 M