Write a balanced net ionic equation to show why the solubility of Ca(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.

Answer 1

It increases because H^+ of the acid reacts with the OH^- of the Ca(OH)2 and that shifts the equilibrium to the right which is to be more soluble.
Ca(OH)2 ==> Ca^2+ + 2OH^- but
2OH + 2H^+ ==> 2H2O so the OH^- is decreased and the reaction (which is solubility) must shift to the right to try and counterbalance this.

I don't understand the remaining parts of the question.

Answer 2

From table of thermodynamics constants, the equilibrium constant can be calculated using Thermodynamic constants for Free Energy of Formation(∆G⁰fmn) to calculate Free energy of reaction(∆G⁰Rxn). Substitute (∆G⁰fmn) values into Hess's Law Equation for Free Energy to get Free Energy of Reaction (∆G⁰-Rxn). Then use (∆G⁰Rxn)=-RTln(Ksp) => ln(Ksp) = -((∆G⁰Rxn)/(RT)). Solve for Ksp = e^-((∆G⁰Rxn)/(RT))

=>(∆G⁰Rxn) = [∑n·(∆G⁰fmn)Products] –[∑n·(∆G⁰fmn)Reactants].
=>(∆G⁰-Rxn) = -RTln(Ksp)

Ca(OH)₂(s) <=> Ca⁺²(aq) + 2OH⁻(aq)
(∆G⁰fmn)Ca(OH)₂(s) = -898.1Kj
(∆G⁰fmn)Ca⁺²(aq) = -553.6Kj
(∆G⁰fmn)2OH⁻(aq)=2(-157.25Kj)=-314.5Kj

(∆G⁰Rxn)=[∑n(∆G⁰fmn)P]-∑n(∆G⁰fmn)R]
=[(-553.6Kj)+(-314.5Kj)]-[(-898.1Kj)]
=+30.0Kj

+30.0Kj = -(0.008314Kj/mol-K)(298K)ln(Ksp)
=> ln(Ksp) = [(30.0)/-(0.008314)(298)]
=> Ksp = exp(-12.1) => Ksp = 5.5x10⁻⁶

Answer 3

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