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Write a balanced net ionic equation to show why the solubility of Cu(OH)2(s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid. Use the pull-down boxes to specify states such as (aq) or (s).

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Answer:

2.2×10^8

Explanation:

Cu(OH)2(s)<---------> Cu^2+(aq) + 2OH^-(aq) Ksp=2.2 x 10 ^-20

2H3O^+(aq) + 2OH^-(aq) <-------> 4H2O(l). Kw= 1×10^14

Cu^2+(aq) + 4H2O(l) <--------> [Cu(H2O)4]^2+(aq)

Overall ionic reaction:

Cu(OH)2(s) +2H3O^+(aq) <---------> [Cu(H20)4]^2+(aq)

Equilibrium constant for the reaction: Ksp×Kw= 2.2 x 10 ^-20 × (1/(1×10^-14))^2

Keq= 2.2×10^8

Kw= ion dissociation constant of water