What is the molar solubility of lead(II) chromate (ksp=1.8x10^-14) in 0.13 M potassium chromate?

Let X = solubility Pb(CrO4)2.

....Pb(CrO4)2 ==> Pb^2+ + 2CrO4^2-
.......X...........X.......2X

K2CrO4 is 100% dissociated.
.........K2CrO4 ==> 2K^+ + CrO4^2-
initial..0.13......0.......0
change..-0.13......+0.13....0.13
equil......0.......0.13......0.13

Ksp = (Pb^2+)(CrO4^2-)
For (Pb^2+) substitute X
For (CrO4^2-) substitute 2X for that from Pb(CrO4)2 + 0.13 from K2CrO4 for total of 0.13+2X.
Solve for X.

4.46x10^-11 g/l

To find the molar solubility of lead(II) chromate (PbCrO4) in 0.13 M potassium chromate (K2CrO4), we need to use the concept of common ion effect and the salt's solubility product constant (Ksp).

The balanced chemical equation for the dissociation of lead(II) chromate is:

PbCrO4(s) ⇌ Pb2+(aq) + CrO42-(aq)

The solubility product constant (Ksp) expression is:

Ksp = [Pb2+][CrO42-]

Since we are given the Ksp value for lead(II) chromate (1.8x10^-14) and the concentration of potassium chromate (0.13 M), we can use these values to calculate the molar solubility of PbCrO4.

Let's assume that "x" represents the molar solubility of PbCrO4. Therefore, the concentration of Pb^2+ ions and CrO4^2- ions in the solution will also be "x".

Using the Ksp expression, we can write:

Ksp = [Pb2+][CrO42-]
1.8x10^-14 = (x)(x)

Since the stoichiometric coefficient for both ions is 1, they will have the same concentration in the solution.

Now, let's substitute the concentration of potassium chromate (0.13 M) into the equilibrium expression to form the quadratic equation:

1.8x10^-14 = (0.13 - x)(x)

Solving this quadratic equation will give us the value of "x", which represents the molar solubility of PbCrO4.

To find "x", we can use a numerical method or approximate the value.

Using a numerical method, we can solve this quadratic equation to find the value of "x".

To find the molar solubility of lead(II) chromate (PbCrO4) in 0.13 M potassium chromate (K2CrO4), we need to use the common-ion effect and the solubility product constant (Ksp) of PbCrO4.

The solubility product constant (Ksp) is an equilibrium constant that represents the maximum amount of a solute that can dissolve in a solvent at a given temperature. In this case, the Ksp for PbCrO4 is given as 1.8x10^-14.

The common-ion effect states that the solubility of a slightly soluble salt is reduced by the presence of a common ion. In this case, the common ion is the chromate ion (CrO4^2-).

To determine the molar solubility, we need to consider the ionic equation for the dissociation of PbCrO4:
PbCrO4(s) ⇌ Pb2+(aq) + CrO4^2-(aq)

Let's assume the molar solubility of PbCrO4 is x. Consequently, the concentration of Pb2+ and CrO4^2- ions is also x.

Since we have a solution of potassium chromate (K2CrO4) with a concentration of 0.13 M, the concentration of CrO4^2- ion is 0.13 M(x), as each K2CrO4 molecule dissociates to provide two CrO4^2- ions.

Writing the solubility product expression for PbCrO4:
Ksp = [Pb2+][CrO4^2-]
Ksp = (x)(x)

Substituting the value of Ksp (1.8x10^-14) into the equation:
1.8x10^-14 = x^2

To solve this equation, take the square root of both sides:
x = √(1.8x10^-14)

Using a calculator, we find that x ≈ 1.34x10^-7 M

Therefore, the molar solubility of PbCrO4 in 0.13 M K2CrO4 is approximately 1.34x10^-7 M.