Calculate the molar solubility of Cu(OH)2, Ksp = 2.2 × 10^–20, in 0.87 M NH3. Don't forget to use the complexation reaction Cu2+ + 4 NH3 ⇌ Cu(NH3)42+, K = 5.0 × 10^13.
The Attempt at a Solution
I wasn't sure where to begin. I figured that because Cu reacts with NH3, more of the Cu(OH)2 would "disappear" thus having a higher solubility. So I did an ICE table for Cu2+ + 4 NH3 ⇌ Cu(NH3)4 and set it up as x/(.87-x)^5=5 * 10^13, but the answers were non-real. I'm at a loss for where to begin, now. I know I need to do something with the Cu + NH3 reaction, too (otherwise, why would we even be given that information?), but I don't really know what.