how many L of 0.12M HCl would have to be added to 120mL of a saturated soln of PbCl2..?

Question

question continued..
...in contact w/ 45g solid PbCl2 to reduce the Pb^2+ concentration to 0.0055 M? Ksp(PbCl2) = 1.7 * 10^-5

Answer 1

You have a common ion effect problem

PbCl2 <=> Pb+2 +2Cl-

Ksp=[Pb+2][Cl-]^2
You know that [Pb+2]=5.5*10^-3
Also from the reaction you know that the amount of Cl- coming from the PbCl2 is double the amount of Pb+2.
So if C is the concentration of Cl- coming from HCl, the amount of Cl- at equilibrium is [Cl-]=C+2*[Pb+2]

So Ksp= [Pb+2]* (C+2*[Pb+2])^2=>
if we assume that C>>2[Pb+2] then C+2*[Pb+2] =C and
Ksp=[Pb+2]C^2 =>1.7*10^-5= 5.5*10^-3 C^2 =>
C=55.6 * 10^-3 but 2*[Pb+2]=11*10^-3, so we can't do the approximation...

Let's name [Pb+2]=p for simplicity

Ksp=p(C+2p)^2 =>
Ksp/p= C^2+4pC+4p^2 =>
C^2+4pC+4p^2-Ksp/p=0 =>
C^2+2.2*10^-3 C -2.97*10^-3 =0

C=0.5*(-2.2*10^-3 + SQRT( (2.2*10^-3)^2 +4*2.97*10^-3) ) =>
C= 5.34*10^-2

If you added Va L of HCl, then
C=mole HCl/Vtotal = Ca*Va/(Va+0.12) =>
CVa+0.12C=CaVa =>
Va=0.12C/(Ca-C)= 0.12*0.0534/(0.12-0.0534) = 0.0962L or 96.2 ml