A solution is prepared by mixing 50.0 mL of 0.33 M Pb(NO3)2 with 50.0 mL of 1.2 M KCl.?

Question

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

Answer 1

Pb(NO3)2 dissociates 100% with a stoichiometry
1 Pb(NO3)2 : 1 Pb(+2), thus the initial concentration of Pb(+2) is equal to the initial concentration of Pb(NO3)2.
The latter is C(Pb+2) = M1V1/(V1+V2) = 0.33*0.050 / (0.050+0.050) = 0.165 M

Similarly C(Cl-)= M2V2/(V1+V2) = 1.2*0.050/0.1 =0.6 M

.. .. .. .. .. .. PbCl2 <=> Pb(+2) + 2Cl-
Initial .. .. .. .. .. .. .. .. .. 0.165 .. .. 0.6
Precipitate .. .. .. .. .. .. .. x .. .. .. .. 2x
At equil .. .. .. .. .. .. .. 0.165-x .. 0.6-2x

Ksp =[Pb(+2)][Cl-]^2 = (0.165-x)(0.6-2x)^2

Let's assume that 0.6 >> x so that 0.6-x= 0.6
Then the equation is simplified to
(0.165-x)*(0.6)^2= 1.6*10^-5 =>
x=0.1649 so we can't do any simplifications

(0.165-x)(0.6-2x)^2 = 1.6*10^-5 =>
(0.165-x)* (0.36-2.4x+4x^2)= 1.6*10^-5 =>
0.0594 -0.396x+0.66x^2 -0.36x+2.4x^2-4x^3 =1.6*10^-5 =>

-4x^3 +3.06x^2 -0.756x +0.059384=0

Use the free on-line tool at http://www.1728.com/cubic.htm
to solve the equation

The only acceptable solution is
x= 0.16478

Thus [Pb+2]= 0.165 -0.16478 =2.2*10^-4 M
[Cl-] = 0.6-2x= 0.6-2*0.16478 = 0.27044 =0.27 M

Answer 2

That's way too hard for us dummies out here..