In the above problem volumes are not mentioned, so we will
consider that volume will remian almost constant even after mixing
the two.
Now
               Â
NH3 + H2O --> NH4+ + OH-
initial           Â
3MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â
0
equilibrium    Â
3-x          Â
x           Â
x
Kb = [NH4][OH-]/[NH3] = 1.6X10^-5 = [x][x]/[3-x]
Its a weak base so x<<1
1.6X10^-5 = x^2 / 3
x^2 = 48X10^-6
x = 6.92X10^-3 = [OH-]
Now Ksp of copper hydroxide is
Cu(OH)2 <-----> Cu2+ + 2OH- Ksp = 4.8X10^-20
And Ionic product = [Cu+2][OH-] = 0.1 X 6.92X10^-3 = 6.92X10^-4
IP >>Ksp
so precipitate will form
