posted by Jasleen .
a solution is 0.10M in free NH3, 0.10M in NH4Cl, and 0.015M in [cu(NH3)4]. Will Cu(OH)2 (s) precipitate from this solution? Ksp of Cu(OH)2 is 2.2810^-20
I would write the reaction for the formation of Cu(NH3)4^+2. Now write the expression for Kformation for this reaction. You have Cu(NH3)4^+2 and NH3; the only unknown is (Cu^+2). Solve for that.
Now write the expression for NH3 + HOH ==> NH4^+ + OH^-
Write Kb expression.
Kb = (NH4^+)(OH^-)/(NH3).
You know (NH4^+) from the NH4Cl, you know NH3 from the problem, calculate (OH^-).
Now see if (Cu^+2)(OH^-)^2 exceeds Ksp.
[note:since the problem says that the FREE ammonia is 0.10 M, I assume we don't need to make a correction for the amount of NH3 used up forming the copper ammine complex.]
i did all the steps you said, however the final answer i get is 1.36*10^-11
the answer is percipitate does not form