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.
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.
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To determine the balanced net ionic equation that shows why the solubility of Ca(OH)2(s) increases in the presence of a strong acid, we first need to understand the reaction between Ca(OH)2 and the acid.
When Ca(OH)2 (calcium hydroxide) reacts with a strong acid such as HCl (hydrochloric acid), it undergoes an acid-base reaction. The base, Ca(OH)2, reacts with the acid, HCl, to form calcium chloride (CaCl2) and water (H2O). The balanced equation for this reaction is:
Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
To write the net ionic equation, we remove the spectator ions (ions that are present on both sides of the equation and do not participate in the reaction). In this case, the Ca2+ and Cl- ions are spectator ions. Thus, the net ionic equation is:
2OH- + 2H+ -> 2H2O
This equation shows that the hydroxide ions (OH-) from Ca(OH)2 react with the hydrogen ions (H+) from the acid, which results in the formation of water. This reaction consumes the OH- ions, and since the solubility of Ca(OH)2 is based on the concentration of OH- ions, by consuming OH- ions in the reaction, more Ca(OH)2 will dissolve in the presence of the acid.
Now, to calculate the equilibrium constant for this reaction, we can use the concentration of the products and reactants. However, since Ca(OH)2 is sparingly soluble, its concentration is very low and mostly remains in the solid state. Thus, we cannot directly calculate the equilibrium constant.
However, if we know the concentration of the strong acid (HCl), we can assume it is in excess, which means it completely ionizes in water, giving a high concentration of H+ ions. In this case, the concentration of OH- ions can be considered negligible compared to the concentration of H+ ions. As a result, the net ionic equation simplifies to:
H+ + OH- -> H2O
In this simplified equation, the equilibrium constant is the autoionization constant of water, which is known as Kw and is equal to 1.0 x 10^-14 at 25°C.
Therefore, in the presence of a strong acid, the solubility of Ca(OH)2 increases due to the reaction of OH- ions with H+ ions to form water. The equilibrium constant for this simplified reaction is the autoionization constant of water (Kw).
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