a Using the Kp value for Cu(OH)2 ( 1.6 x 10-10) and the overall formation constant for Cu(NH3)42+ ( 1.0 x 1013), calculate the value for the equilibrium constant for the following reaction: Cu(OH)2(8)+4NH3 (aq) K = 1.6x10^-6 Cu(NH3)42+ (aq) + 20H (aq)

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Chapter16: Solubility And Complex Ion Equilibria
Section: Chapter Questions
Problem 63E: In the presence of CN, Fe3+ forms the complex ion Fe(CN)63. The equilibrium concentrations of Fe3+...
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a Using the
Kep value for
Cu(OH)2 (
1.6 x 10-19) and the overall formation constant for
Cu(NH3)2+ (
1.0 x 1013), calculate the value for the equilibrium constant for the following reaction:
Cu(OH)2(8)+4NH, (aq)
K = = 1.6x10^-6
Cu(NH3)42+ (aq) + 20H(aq)
buse the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of Cu(OH)2 in 4.8 M NH3. In 4.8 M NH3 the
concentration of OH is 0.0093 M.
Solubility =
mol/L
Transcribed Image Text:a Using the Kep value for Cu(OH)2 ( 1.6 x 10-19) and the overall formation constant for Cu(NH3)2+ ( 1.0 x 1013), calculate the value for the equilibrium constant for the following reaction: Cu(OH)2(8)+4NH, (aq) K = = 1.6x10^-6 Cu(NH3)42+ (aq) + 20H(aq) buse the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of Cu(OH)2 in 4.8 M NH3. In 4.8 M NH3 the concentration of OH is 0.0093 M. Solubility = mol/L
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