Calculate the solubility of Pbso4 in a water sample, which contain 1.4x10-2M Cacl2?
Given : Ksp for Pbs04= 1.6x10-8M2
Ksp for Cas04= 2.6x10-5M2
I think the idea here is that CaSO4 is so much more soluble than PbSO4 that the CaCl2/CaSO4 equilibrium will determine the sulfate concentration and that in turn will determine the solubility of PbSO4.
.........CaSO4 ==> Ca^2+ + SO4^2-
I........solid.....0........0
C........solid.....x........x
E........solid.....x........x
..........CaCl2 ==> Ca^2+ + 2Cl^-
I........0.014.......0........0
C.......-0.014.....0.014....2*0.014
E.........0........0.014....0.028
Ksp CaSO4 = (Ca^2+)(SO4^2-)
(Ca^2+) = x from CaSO4 + 0.014 from CaCl2. (SO4^2-) is x. Plug those into Ksp expression for CaSO4 and solve for SO4. Note that you may need to solve a quadratic but I can't tell just by looking. Then go to the PbSO4 part of the problem.
.............PbSO4 --> Pb^2+ + SO4^2-
I ...........solid.....0........0
C............solid.....x........x
E............solid.....x........x
Ksp PbSO4 = (Pb^2+)(SO4^2-)
(Pb^2+) = x = solubility
(SO4^2-) = x from PbSO4 + the number for sulfate from CaSO4. Solve for (Pb^2+). Again, you may need to consider a quadratic here.
To calculate the solubility of PbSO4 in a water sample containing CaCl2, we need to consider the common ion effect. The presence of CaCl2 in the water sample will increase the concentration of Ca2+ ions, which will affect the solubility of PbSO4.
First, let's consider the dissociation equation for PbSO4:
PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq)
The solubility product constant (Ksp) for PbSO4 is given as 1.6x10^-8M^2, which represents the equilibrium constant for the dissociation reaction:
Ksp = [Pb2+][SO42-]
Now, we need to consider the effect of the CaCl2 on the solubility of PbSO4. The Ca2+ ions from CaCl2 will react with the SO42- ions from PbSO4 and form an insoluble precipitate of CaSO4.
The dissociation equation for CaSO4 is:
CaSO4(s) ⇌ Ca2+(aq) + SO42-(aq)
The Ksp for CaSO4 is given as 2.6x10^-5M^2, which represents the equilibrium constant for the dissociation reaction:
Ksp = [Ca2+][SO42-]
Since both PbSO4 and CaSO4 share the SO42- ion, their solubilities are related. Assuming x is the solubility of PbSO4, the concentration of SO42- will be x.
Using the given Ksp for PbSO4, we can write:
Ksp for PbSO4 = [Pb2+][SO42-]
1.6x10^-8M^2 = [Pb2+][x]
Using the given Ksp for CaSO4, we can write:
Ksp for CaSO4 = [Ca2+][SO42-]
2.6x10^-5M^2 = [Ca2+][x]
Since the concentration of SO42- is the same for both equations, we can equate the expressions for [SO42-]:
[Pb2+] = [Ca2+]
Now, we know that the concentration of CaCl2 is 1.4x10^-2M, which is equal to the concentration of Ca2+ because each CaCl2 molecule dissociates into one Ca2+ ion.
Therefore, [Ca2+] = 1.4x10^-2M and [Pb2+] = 1.4x10^-2M.
Finally, to calculate the solubility of PbSO4 in the water sample, we substitute the value of [Pb2+] into the equation for Ksp of PbSO4:
1.6x10^-8M^2 = (1.4x10^-2M)(x)
Now, solving for x, the solubility of PbSO4:
x = (1.6x10^-8M^2) / (1.4x10^-2M)
x ≈ 1.14x10^-7M
The solubility of PbSO4 in the water sample, which contains 1.4x10^-2M CaCl2, is approximately 1.14x10^-7M.