Consider a solution that contains Ag+ , Ba2+, and Pb2+ each at a concentration of .2M
a) You add NaCl until the concentration of Cl- is 5.0 x 10^-3 M. A white precipitate forms. How do you determine whether that precipitate was AgCl or PbCl2?
b) If you separated the supernatant from above and treated it with 3.0 M HCl would a precipitate of PbCl2 be observed? Justify answer mathmatically.
c) if you took the supernatant from above and treated it with 3 M Na2SO4, would you expect a precipitate to form? What would be the identity of the precipitate?
What values do you have for Ksp PbCl2 and AgCl. I can look them up in my text but the numbers probably will not the same as in your text. I would rather use your numbers.
pbcl2 ksp=1.7x10^-5
and agcl ksp = 1.8 x 10^-10
a.
Ksp PbCl2 = (Pb^2+)(Cl^-)^2
Qrxn = (0.2)(5E-3)^2 = 5E-6 and this is smaller than Ksp; therefore, no PbCl2 ppts.
Ksp AgCl = (Ag^+)(Cl^-)
Qrxn = (0.2)(5E-3) = 1E-3 which is larger than Ksp; therefore, AgCl ppts.
b.
The total chloride is now 3M + 5E-3M. I will ignore the small 5E-3. So the 0.2M Ag will use up part of the 3M HCl to leave about 2.7M HCl. We will get a ppt and it will be the rest of the AgCl. Will PbCl2 come down too?
(Pb^2+)(Cl^2-) = (0.2)(2.7)^2 = about 1.5 which exceeds Ksp for PbCl2. Yes a ppt will form.
c. Will BaSO4 ppt?
Q = (Ba^2+)(SO4^2-) = (0.2)(.3) = ?. Is Ksp for BaSO4 smaller or larger
a) To determine whether the precipitate formed is AgCl or PbCl2, you can utilize the solubility product constant (Ksp) values of both compounds. AgCl has a lower Ksp value than PbCl2, which means it is less soluble and more likely to form a precipitate.
First, calculate the molar solubility of AgCl and PbCl2 at the given concentration of Cl- (5.0 x 10^-3 M) using their respective Ksp values:
For AgCl:
AgCl dissociates into Ag+ and Cl-
AgCl ⇌ Ag+ + Cl-
Ksp = [Ag+][Cl-]
Given: [Cl-] = 5.0 x 10^-3 M
Ksp = (0.2)(5.0 x 10^-3) = 1.0 x 10^-3
For PbCl2:
PbCl2 dissociates into Pb2+ and 2Cl-
PbCl2 ⇌ Pb2+ + 2Cl-
Ksp = [Pb2+][Cl-]^2
Given: [Cl-] = 5.0 x 10^-3 M
Ksp = (0.2)(5.0 x 10^-3)^2 = 5.0 x 10^-6
Comparing the two Ksp values, we can see that the Ksp for AgCl (1.0 x 10^-3) is higher than that of PbCl2 (5.0 x 10^-6). Therefore, if a precipitate forms, it is more likely to be AgCl rather than PbCl2.
b) To determine whether a precipitate of PbCl2 would form when the supernatant is treated with 3.0 M HCl, we can again use the solubility product constant (Ksp).
For PbCl2:
PbCl2 ⇌ Pb2+ + 2Cl-
Ksp = [Pb2+][Cl-]^2
If the concentration of Cl- is increased by adding 3.0 M HCl, the new concentration of Cl- is the sum of the initial Cl- concentration (5.0 x 10^-3 M) and the concentration of HCl (3.0 M):
[Cl-] = 5.0 x 10^-3 + 3.0 = 3.005 M
Now, compare this concentration of Cl- with the Ksp value of PbCl2 (5.0 x 10^-6):
[Cl-]^2 = (3.005)^2 = 9.030025
Ksp (PbCl2) = 5.0 x 10^-6
Since the concentration of Cl- squared (9.030025) is greater than the Ksp value of PbCl2 (5.0 x 10^-6), a precipitate of PbCl2 would indeed be expected to form when the supernatant is treated with 3.0 M HCl.
c) To determine whether a precipitate would form when the supernatant from above is treated with 3 M Na2SO4, we need to consider the solubility rules for sulfates.
In general, most sulfates are soluble except for those of barium (Ba2+) and lead (Pb2+).
Since Pb2+ is already present in the solution, adding Na2SO4 would result in the formation of a white precipitate, which is PbSO4.
Therefore, if the supernatant is treated with 3 M Na2SO4, a precipitate of PbSO4 is expected to form.