calculate the solubility of PbI2 in 0.0123M CaI2. Ksp=6.81x10^-9
..........PbI2 ==> Pb^2+ + 2I^-
I.........solid....0........0
C.........solid....x........2x
E.........solid....x........2x
......CaI2 ==> Ca^2+ + 2I^-
I...0.0123.....0........0
C..-0.0123....0.0123..2*0.0123
E.......0.....0.0123...0.0246
Ksp - (Pb^2+)(I^-)^2
(Pb^2+) = x
(I^-) = 2x from PbI2 and 0.0246 from CaI so ((I^-) = x+0.0246
Substitute and solve for x = (Pb^2+) = (PbI2) in solution.
To calculate the solubility of PbI2 in the presence of CaI2, we need to use the concept of the common ion effect. The common ion effect states that the solubility of a compound decreases when a common ion is added to the solution.
In this case, CaI2 contains the common ion "I-" which can potentially affect the solubility of PbI2. The solubility product constant (Ksp) of PbI2 is given as 6.81x10^-9.
Let's assume that the solubility of PbI2 in the presence of CaI2 is "s" mol/L. Since PbI2 dissociates into one Pb2+ ion and two I- ions, the concentrations of these ions can be expressed as "s" and "2s" respectively.
For PbI2:
PbI2 ⇌ Pb2+ + 2I-
The concentration of I- ions from CaI2 is 0.0123 M, and we should consider these ions when calculating the solubility of PbI2.
Now, we can write down the expression for the solubility product constant (Ksp) using the concentrations of the ions present in the solution:
Ksp = [Pb2+][I-]^2
Substituting the concentrations:
6.81x10^-9 = s * (0.0123 + 2s)^2
Simplifying the equation:
6.81x10^-9 = s * (0.0123 + 2s)^2
Expanding and rearranging the equation:
6.81x10^-9 = s * (0.0123^2 + 2 * 0.0123 * s + 4s^2)
Now, solve this quadratic equation for "s".