a. Using the Ksp for Cu(OH)2 (1.6 10-19) and the overall formation constant for Cu(NH3)42+ (1.0
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Cu(OH)2(s) + 4NH3(aq) ⇌ Cu(NH3) 42+(aq) + 2OH–(aq)
b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/ L) of Cu(OH)2 in 5.0 M NH3. In 5.0 M NH3, the concentration of OH- is 0.0095 M.
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