[spunkylulu]

I've been trying to solve this problem for about a week now and my professor is of no help.  I really don't even know where to start and would appreciate some guidance! 

To a solution of Pb2+ with a concentration of 3.87 x 10-8 M is added ammonium hydroxide with a concentration of  [OH-] of 1 x 10-8 M.  Will a precipitate form which is Pb(OH)2? Ksp for Pb(OH)2 is 1.2 x 10-15?

What will be the Pb2+ concentration in Part a if 0.1 M Na2CO3 is added.  (assume no change in solution volume).

[Borek]

To a solution of Pb2+ with a concentration of 3.87 x 10-8 M is added ammonium hydroxide with a concentration of  [OH-] of 1 x 10-8 M.  Will a precipitate form which is Pb(OH)2? Ksp for Pb(OH)2 is 1.2 x 10-15?

Looks like just plug and chug, where is the problem?

[spunkylulu]

To a solution of Pb2+ with a concentration of 3.87 x 10-8 M is added ammonium hydroxide with a concentration of  [OH-] of 1 x 10-8 M.  Will a precipitate form which is Pb(OH)2? Ksp for Pb(OH)2 is 1.2 x 10-15?

Looks like just plug and chug, where is the problem?

What do I plug in chug into what formula?  That's what I don't understand...how to approach the problem.

[Borek]

What is the formula for Ksp of the Pb(OH)₂?

[spunkylulu]

What is the formula for Ksp of the Pb(OH)₂?

Isn't it: Ksp = [Pb(OH)₂][NH4]² / [Pb] [NH₄OH]₂ or is that for the whole reaction and the Ksp of Pb(OH)₂ = [Pb][OH]²

[Borek]

the Ksp of Pb(OH)₂ = [Pb][OH]²

That's it.

[spunkylulu]

Okay so I got "no" because Q < K.  To determine the concentration of Pb²⁺ when .1M Na₂CO₃, won't I just divide the Ksp by .1M?  I found the Ksp to be 1.49 x 10^-15 given the solubility of lead carbonate in water as 3.87 X 10^-8M

[Borek]

Looks reasonably correct to me.

Ask your teacher how to prepare ammonia solution with pOH 8, that's really something.