In: Chemistry
#3 R&D Exp. 26
Should Cu(OH)2 precipitate in 3 M NH3 if the original [Cu2+] is 0.1 M ( the Kb for ammonia is 1.6x10-5)?
Please show work and explain a little of how you got this result
In the above problem volumes are not mentioned, so we will consider that volume will remian almost constant even after mixing the two.
Now
NH3 + H2O --> NH4+ + OH-
initial
3M
0
0
equilibrium
3-x
x
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