Suppose a solution contains 0.21 M Pb2 and 0.41 M Al3 . Calculate the pH range that would allow Al(OH)3 to precipitate but not Pb(OH)2.

Answer 1

Pb(OH)2 ==> Pb^2+ + 2OH^-
Al(OH)3 ==> Al^3+ + 3OH^-

Calculate OH^- needed for a 0.21M Pb^2+ solution to ppt Pb(OH)2,
Calculate OH^- needed for a 0.41M Al^3+ solution to ppt Al(OH)3.
You will want the (OH^-) to be less than you calculate to prevent pptn of Pb(OH)2. Convert that OH to pH so you will want the pH to be greater than that. When you don't show the Ksp values you are using we can't calculate anything and hope to get the right answer since you are probably not using the same values for Ksp that we are. Textbooks differ in the tables they use.

Answer 2

Oops, sorry. The Ksp for Al(OH)3 is given as 4.6×10^–33 and the Ksp for Pb(OH)2 is given as 1.43×10^–20.

Answer 3

What's the best way to calculate the answer with those values? I don't understand.

Answer 4

Why do you never post a step by step solution so others can follow. instead its just brief explanation