What is the molar solubility of PbCl2 in a solution of 0.23 M CaCl2? Ksp = 1.6 10-5 for PbCl2.

idk how to do this because of the Cl2

It's a common ion and you just add it in.

PbCl2==>Pb^+2 + 2Cl^-
So Ksp = (Pb^+2)(Cl^-)^2
(Pb^+2) = x
(Cl^-) = 2x (from the PbCl2 solubility + 2*0.23 from the CaCl2.
Plug in the numbers and solve for x.
You can avoid a quadratic equation by making the simplifying assumption that x + 0.46 = 0.46 (but since the solubility of PbCl2 is high (ksp is relatively large) that may lead to an error. At any rate, you can try it an see if there is much error or you can go ahead and solve the quadratic equation since most calculators handle that easily now.

To find the molar solubility of PbCl2, we can use the concept of the common ion effect. The common ion effect states that the solubility of a salt is reduced when a common ion is present in the solution.

In this case, we have a solution of 0.23 M CaCl2. The Ca2+ ion is the common ion, which can potentially affect the solubility of PbCl2. To determine the molar solubility, we need to calculate the concentration of Pb2+ ions in the presence of Ca2+ ions.

The balanced chemical equation for the dissolution of PbCl2 is:

PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)

The solubility product constant (Ksp) for PbCl2 is given as 1.6 x 10^-5, which is equal to the product of the Pb2+ and Cl- ion concentrations.

Using the equilibrium expression for Ksp, we can write:

Ksp = [Pb2+][Cl-]^2

We need to express the concentration of the Cl- ions in terms of Pb2+ concentration:

[Cl-] = 2[Pb2+]

Substituting this into the Ksp expression:

Ksp = [Pb2+] * (2[Pb2+])^2
Ksp = [Pb2+] * 4[Pb2+]^2
Ksp = 4[Pb2+]^3

Now, we know that the concentration of Ca2+ ions is 0.23 M, which is equal to the concentration of Cl- ions. Therefore,

[Cl-] = 0.23 M

We substitute this into the [Pb2+] expression:

0.23 M = 2[Pb2+]
[Pb2+] = 0.23/2 M
[Pb2+] = 0.115 M

Therefore, the molar solubility of PbCl2 in a solution of 0.23 M CaCl2 is approximately 0.115 M.

To determine the molar solubility of PbCl2 in a solution of 0.23 M CaCl2, you need to consider the effect of the common ion (Cl-) on the solubility of PbCl2. The presence of CaCl2 will affect the equilibrium of PbCl2 dissolving in water.

The solubility product constant (Ksp) for PbCl2 is given as 1.6 x 10^-5. This expression represents the equilibrium constant for the dissociation of PbCl2 into Pb2+ and 2Cl- ions in a saturated solution.

In this case, the additional Cl- ions from CaCl2 will shift the equilibrium of PbCl2 dissolving in water to the left, reducing its solubility. To find the molar solubility of PbCl2, you can follow these steps:

1. Write the balanced equation for the dissociation of PbCl2:
PbCl2 ⇌ Pb2+ + 2Cl-

2. Set up an ICE (initial, change, equilibrium) table for the dissociation of PbCl2.
Let "x" be the molar solubility of PbCl2 in moles per liter (M).

Initial:
PbCl2: Pb2+: 2Cl-:
0.023 M 0 M 0 M

Change:
PbCl2: Pb2+: 2Cl-:
- x + x + 2x

Equilibrium:
PbCl2: Pb2+: 2Cl-:
0.023 M - x x 2x

3. Write the expression for the solubility product constant (Ksp) and substitute the equilibrium concentrations:
Ksp = [Pb2+] * [Cl-]^2
= (x) * (2x)^2
= 4x^3

4. Substitute the known concentration of [Cl-] from the 0.23 M CaCl2 solution into the expression for Ksp.
Ksp = 1.6 x 10^-5
1.6 x 10^-5 = 4x^3

5. Solve the equation for x to find the molar solubility of PbCl2.
x^3 = (1.6 x 10^-5) / 4
x^3 = 4 x 10^-6
x = (4 x 10^-6)^(1/3)

By solving this equation, you can find the molar solubility of PbCl2 in the solution of 0.23 M CaCl2.

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