Consider a solution that contains Ag+ , Ba2+, and Pb2+ each at a concentration of .2M

Question

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?

Answer 1

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.

Answer 2

pbcl2 ksp=1.7x10^-5
and agcl ksp = 1.8 x 10^-10

Answer 3

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