Solubility and Equilibrium question

  • #1

Homework Statement


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.


Homework Equations


Ksp=[A]^m^n
Keq=[product]/[reactants]


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.
 

Answers and Replies

  • #2
Borek
Mentor
28,825
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Try to assume concentration of ammonia doesn't change. Note that OH- concentration is a function of ammonia concentration as well.
 

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