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(A) Using the Ksp for Cu(OH)2 (1.6 x 10^-19) and the overall formation constant for Cu(NH3)4...

(A) Using the Ksp for Cu(OH)2 (1.6 x 10^-19) and the overall formation constant for Cu(NH3)4 (1.0 x 10^13), calculate a value for the equilibrium constant for the reaction: Cu(OH)2(s) + 4NH3(aq) <===> Cu(NH3)4(aq) + 2OH-(aq)

(B) Use the value of the equilibrium constant you calculated in Part A to calculate the (approximate) solubility (in mols/liter, M) of Cu(OH)2 in 5.0 M NH3. In 5.0 M NH3 the concentration of OH- is 9.5 x 10^-3 M. (Although not strictly correct you may assume that s, the solubility, is small, and can be neglected in the calculation.) Also, calculate a more exact value of the solubility by assuming that s is NOT SMALL, i.e., cannot be neglected.

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