Calculate the equilibrium vale of Pb +2 (aq) in 1.00 L of saturated PbCl2 solution to which 0.250 mole of NaCl has been added. 60.0 ml of 0.0300 M NaCl 60.0mL of 0.0500 M Pb(NO3)2

PbCl2(s) ==> Pb^2+ + 2Cl-

Ksp =(Pb^+2)(Cl-)^2
All of the problems are an example of the common ion effect as well as Le Chatelier's principle.
NaCl ==> Na+ + Cl-
Set up an ICE chart for the PbCl2.
(Pb2+) = x
(Cl-) from the PbCl2 = x
Cl- from the NaCl is 0.25 M (since 0.250 mole was placed in 1L).
Therefore, (x)(x+0.25) = Ksp. Substitute Ksp and solve for x. You will have a quadratic equation unless you make the simplifying assumption that x + 0.25 = 0.25.

To calculate the equilibrium value of Pb+2 (aq) in the given solution, we need to use the concept of ionization of PbCl2 and the common ion effect.

Step 1: Write the balanced equation for the ionization of PbCl2:
PbCl2 ⇌ Pb+2 (aq) + 2 Cl- (aq)

Step 2: Determine the initial concentration of Pb+2 (aq) before any ions from NaCl and Pb(NO3)2 are added.

Since we have 0.250 moles of NaCl added to 1.00 L of saturated PbCl2 solution, the concentration of Cl- ions initially available from NaCl is:
Concentration of Cl- = (0.250 moles)/(1.00 L) = 0.250 M

Step 3: Calculate the concentration of Cl- ions contributed by Pb(NO3)2.

You mentioned adding 60.0 mL (0.0600 L) of 0.0500 M Pb(NO3)2. Since Pb(NO3)2 dissociates into one Pb+2 ion and two NO3- ions, the initial concentration of Cl- ions contributed by Pb(NO3)2 is:
Concentration of Cl- = (0.0600 L) x (0.0500 M) x (2) = 0.0060 M

Step 4: Determine the new concentration of Cl- ions after the addition of NaCl and Pb(NO3)2.

The total concentration of Cl- ions in the solution is the sum of the initial concentration of Cl- ions from NaCl and Pb(NO3)2:
Total concentration of Cl- = 0.250 M + 0.0060 M = 0.256 M

Step 5: Apply the common ion effect to determine the equilibrium value of Pb+2.

The common ion effect states that the presence of a common ion reduces the solubility of a slightly soluble salt. In this case, the common ion is Cl-.

At equilibrium, the concentration of Pb+2 (aq) will be determined by the solubility product constant (Ksp) of PbCl2. The Ksp expression for PbCl2 is:
Ksp = [Pb+2]eq [Cl-]eq^2

Since PbCl2 is saturated, we assume that the amount dissolved is equal to the initial concentration, which is also equal to the concentration of Pb+2 ions at equilibrium.

Substituting the values into the Ksp expression:
Ksp = [Pb+2]eq (0.256 M)^2

Rearranging the equation to solve for [Pb+2]eq:
[Pb+2]eq = Ksp / (0.256 M)^2

The value of Ksp for PbCl2 is 1.6 x 10^-5 (mol/L)^2.

Now, substitute the values into the equation:
[Pb+2]eq = (1.6 x 10^-5) / (0.256)^2

Calculate the equilibrium concentration of Pb+2 using a calculator:
[Pb+2]eq ≈ 2.81 x 10^-3 M

Therefore, the equilibrium value of Pb+2 (aq) in the given solution is approximately 2.81 x 10^-3 M.