Question
Calculate the molar solubility of Cu(OH)2, Ksp = 2.2 × 10–20, in 0.52 M NH3. Don't forget to use the complexation reaction Cu2+ + 4 NH3 ⇌ Cu(NH3)42+, K = 5.0 × 1013
What does he mean by the complexation reaction??? So confused, please help!!!
What does he mean by the complexation reaction??? So confused, please help!!!
Answers
DrBob222
What the problem means is that the solubility is increased due to the formation of the Cu(NH3)4^+2 ion and in fact that concn is greater than that of the Cu^+2 by itself. Here is what I would do.
Cu(OH)2 ==> Cu^+2 + 2OH^-
Cu^+2 + 4NH3 ==> Cu(NH3)4^+2
Write the Ksp expression for Cu(OH)2.
Write the K for the complex. That is
K = [Cu(NH3)4^+2]/(Cu^+2)(NH3)^4 and I would include the NH3 equilibrium also.
That is NH3 + HOH ==> NH4^+ + OH^-
Write Kb for NH3.
I would do an ICE chart for Cu(OH)2 in which Cu(OH)2 = S (for solubility)
S = (Cu^+2)+[Cu(NH3)4^+2]
You also know that total NH3 = 0.52 and that is
0.52 = (NH4^+) + NH3 + [Cu(NH3)4^+2].
You solve these two equations together with the 3 above you wrote (Ksp, K, and Kb) for S. I worked through the problem and obtained about 0.0027 M for solubility if I didn't make an error somewhere.
Cu(OH)2 ==> Cu^+2 + 2OH^-
Cu^+2 + 4NH3 ==> Cu(NH3)4^+2
Write the Ksp expression for Cu(OH)2.
Write the K for the complex. That is
K = [Cu(NH3)4^+2]/(Cu^+2)(NH3)^4 and I would include the NH3 equilibrium also.
That is NH3 + HOH ==> NH4^+ + OH^-
Write Kb for NH3.
I would do an ICE chart for Cu(OH)2 in which Cu(OH)2 = S (for solubility)
S = (Cu^+2)+[Cu(NH3)4^+2]
You also know that total NH3 = 0.52 and that is
0.52 = (NH4^+) + NH3 + [Cu(NH3)4^+2].
You solve these two equations together with the 3 above you wrote (Ksp, K, and Kb) for S. I worked through the problem and obtained about 0.0027 M for solubility if I didn't make an error somewhere.