[g1192]

How would one solve the following? How do you set up the ice table?

What happens when you add 0.1L of 0.01M NH3(aq) to 0.1L of 0.01M Fe2+ solution.
Ksp of Fe(OH)2 = 7.9 x 10-15

[g1192]

Fe²⁺_(aq) + 2OH^-_(aq)   Fe(OH)_2(s)

OH- comes from NH₃ +H₂O    NH4⁺ + OH^-

So when you add NH₃, I know that more OH^- is produced. However we are only given the above Ksp to work with. How do I incorporate the NH₃?

Thanks

[Borek]

You need Kb. Could be you are expected to get it from tables.

[g1192]

NH₃   +   H₂O     NH4⁺   +   OH^-    Kb= 1.8 x 10^-5

I: 0.005M of NH3 ; 0M of NH4+ ; 0M of OH-
C: -x                 ; +x             ;  +x
E: 0.005-x          ; x               ;  x

Kb = x²/0.005
x= 3 x 10^-4M

Fe(OH)₂    Fe²⁺   +   OH^-     Ksp of Fe(OH)2 = 7.9 x 10-15

Q = [Fe²⁺][OH^-]²
Q = (0.005)(0.005)²
Q=1.25 x 10^-7

Since Q is greater than K, a precipitate of Fe(OH)₂ will be formed.

Is this right?

Thanks

[rabolisk]

Why did you assume that [OH] is .0005M in the second equilibrium?

[g1192]

Oops. It should be this:

Q = [Fe²⁺][OH^-]²
Q = (0.005)(3 x 10^-4)²
Q= 4.5 x 10^-10

Since Q is greater than K, a precipitate of Fe(OH)₂ will be formed.

Better?