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Solubility and Equilibrium question

  1. Apr 1, 2012 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    Ksp=[A]^m^n
    Keq=[product]/[reactants]


    3. 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.
     
  2. jcsd
  3. Apr 1, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    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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