Solubility and Solubility Product (Ksp) Study Notes

Dissolution is defined as the process in which an ionic solid dissolves in a polar liquid. This process can be represented by the chemical equation: $AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq)$ , where " $(aq)$ " stands for aqueous.
Precipitation is the process in which ions leave a solution and regenerate an ionic solid.
A precipitate is the solid substance formed during the process of precipitation.
The precipitation process is represented by the reverse reaction: $AgCl(s) \leftarrow Ag^{+}(aq) + Cl^{-}(aq)$ .
Solubility is defined as the maximum quantity of a substance that will dissolve in a given solvent.
A Saturated Solution is a solution that contains as much solute as can possibly be dissolved under the existing conditions of temperature and pressure.
A Supersaturated Solution is a solution that contains more solute than can be possibly dissolved. These solutions are considered unstable, and the excess solid will precipitate out of the solution.

Solubility Equilibrium occurs when the rate of dissolution and the rate of precipitation are equal.
The equilibrium state is represented as: $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$ .
Directional rates in equilibrium: - Dissolution: The forward reaction rate. - Precipitation: The reverse reaction rate.
Examples of solubility equilibrium equations include: - Example 1: $NaCl(s) \rightleftharpoons Na^{+}(aq) + Cl^{-}(aq)$ - Example 2: $CaCl_2(s) \rightleftharpoons Ca^{2+}(aq) + 2Cl^{-}(aq)$
Guidelines for writing dissolution or precipitation reactions: - Ionic compounds must be broken up into their constituent ions according to their chemical formulas. - The equation must be balanced so that both sides are electronically neutral (charge conservation).

Categorization of Ionic Compounds by solubility ( $S$ ): - Soluble: If solubility S > 0.1\,M. - Slightly soluble: If solubility $S$ is between $0.01\,M$ and $0.1\,M$ . - Insoluble: If solubility S < 0.01\,M.
Comprehensive Solubility Rules: 1. All compounds containing group 1 ions ( $Li^{+}$ , $Na^{+}$ , $K^{+}$ , $Rb^{+}$ , $Cs^{+}$ ) or ammonium ( $NH_4^{+}$ ) are soluble. 2. All compounds containing the nitrate ion ( $NO_3^{-}$ ) and the acetate ion ( $CH_3CO_2^{-}$ ) are soluble, with the specific exception of Silver Acetate ( $AgCH_3CO_2$ ). 3. All compounds containing halide ions ( $Cl^{-}$ , $Br^{-}$ , $I^{-}$ ) are soluble, EXCEPT when paired with $Ag^{+}$ , $Pb^{2+}$ , and $Hg_2^{2+}$ . 4. All compounds containing the sulfate ion ( $SO_4^{2-}$ ) are soluble, EXCEPT when paired with $Ag^{+}$ , $Pb^{2+}$ , and $Hg_2^{2+}$ , or with the alkaline earth metals $Ca^{2+}$ , $Sr^{2+}$ , and $Ba^{2+}$ . 5. All compounds containing the hydroxide ion ( $OH^{-}$ ) are insoluble, EXCEPT when paired with $Ca^{2+}$ , $Sr^{2+}$ , and $Ba^{2+}$ , group 1 ions, or ammonium ions. 6. All compounds containing any other anions (not specified above) are insoluble, EXCEPT when paired with group 1 or ammonium ions.
If a compound is classified as insoluble according to these rules, it means it will precipitate.

Procedure for predicting the formation of a precipitate: 1. Identify the ions present in the reactants. 2. Switch the partners (cation-anion swap) to identify the potential products and see which ions precipitate based on solubility rules. 3. Write a balanced net ionic equation specifically for the formation of the precipitate.
Practical Application Examples: 1. Mixing Silver Nitrate ( $AgNO_3$ ) with Sodium Chloride ( $NaCl$ ). 2. Mixing Calcium Chloride ( $CaCl_2$ ) and Ammonium Carbonate ( $(NH_4)_2CO_3$ ). 3. Mixing Barium Hydroxide ( $Ba(OH)_2$ ) and Iron (III) Sulfate ( $Fe_2(SO_4)_3$ ).

The equilibrium expression applies to the dissolution of a solid: $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$ .
The Solubility Product ( $K_{sp}$ ) for this process is: $K_{sp} = [Ag^{+}][Cl^{-}]$ .
Because $AgCl$ is a pure solid, and concentrations of pure solids and liquids are considered constant, the concentration of the solid ( $[AgCl]$ ) is incorporated into the equilibrium constant $K_{sp}$.
Interpretation of $K_{sp}$ : A very small value of $K_{sp}$ indicates that the substance is not very soluble in $H_2O$ .
Examples of writing $K_{sp}$ expressions: - (a) For $Fe(OH)<em>3(s)$ : $K</em>{sp} = [Fe^{3+}][OH^{-}]^3$ - (b) For $Ag_2CrO_4(s)$ : $K_{sp} = [Ag^{+}]^2[CrO_4^{2-}]$ - (c) For $PbI_2(s)$ : $K_{sp} = [Pb^{2+}][I^{-}]^2$

Example: A sample of $CaSO_4(s)$ in pure water at $25^{\circ}C$ with $K_{sp} = 2.4 \times 10^{-5}$ . - Determine concentration of $Ca^{2+}$ and $SO_4^{2-}$ at equilibrium.
Saturated Solutions at $25^{\circ}C$ : - Dissolved ion concentrations for $Mg(OH)<em>2$ where $K</em>{sp} = 8.9 \times 10^{-12}$ . - Dissolved ion concentrations ( $Ca^{2+}$ and $F^{-}$ ) for $CaF_2$ where $K_{sp} = 3.9 \times 10^{-11}$ .

Solubility ( $S$ ) units are typically given in $g/100\,mL$ or Molarity ( $M$ ).
Steps to calculate $K_{sp}$ : 1. Write the balanced equation for dissolving the solid. 2. Write the $K_{sp}$ expression. 3. Convert solubility from $g/100\,mL$ to $M$ if necessary. 4. Determine the concentration for ions at equilibrium based on stoichiometry. 5. Substitute the concentrations into the $K_{sp}$ expression.
Calculation Examples: 1. Solubility of $BaSO_4$ at $293\,K$ is $2.4 \times 10^{-4}\,g/100\,mL$ . Calculate $K_{sp}$ . 2. Solubility of $CaF_2$ is $1.5\,g/100\,mL$ . Calculate $K_{sp}$ .

Steps to find solubility: 1. Write the balanced equation for solubility equilibrium. 2. Write the $K_{sp}$ expression. 3. Assign variables (e.g., $x$ or $S$ ) for the molar concentrations of ions. 4. Substitute variables into the $K_{sp}$ expression and solve for the variable. 5. The result is the molar solubility of the solid. 6. Convert the solubility into the specific required unit (e.g., $g/100\,mL$ ).
Example: Lead (II) sulfide ( $PbS$ ) has $K_{sp} = 7.9 \times 10^{-28}$ . Determine solubility in $g/100\,mL$ .

The solubility of a substance decreases when one of its constituent ions is already present in the solution from another source.
Comparison Example with Silver Chloride ( $AgCl$ ), $K_{sp} = 1.8 \times 10^{-10}$ : - Calculate molar solubility in pure water. - Calculate molar solubility in tap water containing $0.0010\,M\,Cl^{-}$ .

Steps for prediction: 1. Identify the expected precipitate based on the combination of reactants. 2. Write the balanced equation for dissolving that expected precipitate. 3. Write the $K_{sp}$ expression for that precipitate. 4. Determine the Trial Product (Reaction Quotient, $Q$ ) value. If equal volumes of two solutions are mixed, the concentrations of the original solutions must be divided by two to account for the volume doubling. 5. Compare the trial product ( $Q$ ) to the known $K_{sp}$ value. 6. Comparison Logic: - If Q > K_{sp}, a precipitate forms. - If Q < K_{sp}, no precipitate forms.
Application Examples: - Mixing $0.010\,M\,MgCl_2$ and $0.020\,M\,Na_2CO_3$ . Given $K_{sp}$ for the precipitate is $4.0 \times 10^{-5}$ . - Mixing $0.0020\,M\,AgNO_3$ and $0.020\,M\,Na_2SO_4$ . Given $K_{sp}$ for the precipitate is $1.2 \times 10^{-5}$ .