Solubility Equilibria and the Ksp Constant

Solubility Equilibria and the Solubility Product

• When a typical ionic solid dissolves in water, it separates into cations and anions.
• For Example, CaF₂(s) + H₂O → Ca²⁺(aq) + 2F⁻(aq), and the ions formed are Ca²⁺ and F⁻.
• When solid salt is initially added, no ions are present; as dissolution occurs, the ionic concentration increases.
• The dissolution reaction and its reverse happen simultaneously, Ca²⁺(aq) + 2F⁻(aq) → CaF₂(s).
• Eventually solution achieves a state of saturation, also known as dynamic equilibrium symbolized as CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq).
• The equilibrium expression is Ksₚ = [Ca²⁺][F⁻]².
• The solubility product constant (Ksp) is the equilibrium expression constant for the dissolution of an ionic solid in water, also known as the solubility.
• Solubility equilibrium remains unaffected by the presence of excess solid formed and the size of present particles.
• Solubility represents an equilibrium position, while the solubility product is an equilibrium constant, having only fixed value for a given solid at a given temperature.
• Bismuth sulfide (Bi₂S₃) has a solubility of 1.0 × 10⁻¹⁵ mol/L at 25°C, allowing for the calculation of the Ksp value.
• Nickel(II) hydroxide (Ni(OH)₂), with approximately 0.14 g dissolving per liter of water at 20°C, allows for the calculation of the Ksp at this temperature.
• Copper(II) iodate, Cu(IO₃)₂, has a Ksp value of 1.4 × 10⁻⁷ at 25°C, allowing for the calculation of its solubility at the mentioned temperature.
• To calculate solubility for: Ag₃PO₄ where Ksₚ = 1.8 × 10⁻¹⁸
• To calculate solubility for: CaCO₃ where Ksₚ = 8.7 × 10⁻⁹
• Exercise caution when using Ksₚ values to predict the relative solubilities of a range of salts.
• Salts being compared produce the same number of ions, Ksₚ values can be used to determine relative solubilities.
• Relative solubilities cannot be predicted using Ksₚ values when salts being compared produce differing numbers of ions.
• Comparing AgI (s) Ksₚ = 1.5 x 10⁻¹⁶, CuI (s) Ksₚ = 5.0 x 10⁻¹², and CaSO₄ (s) Ksₚ = 6.1 x 10⁻⁵;
• Comparing CuS (s) Ksₚ = 8.5 x 10⁻⁴⁵, Ag₂S (s) Ksₚ = 1.6 x 10⁻⁴⁹, and Bi₂S₃ (s) Ksₚ = 1.1 x 10⁻⁷³.

Common Ion Effect

• The solubility of a solid decreases when the solution already contains ions common to the solid.
• To calculate solid CaF₂ solubility (Ksₚ = 4.0 × 10⁻¹¹) in 0.025-M NaF solution.

Solubility and pH

• Solubility decreases and the equilibrium shifts to the left from an Increase in pH , example reaction Mg(OH)₂(s) → Mg⁺²(aq) + 2(OH⁻)(aq).
• Solubility increases and the equilibrium shifts to the right from a Decrease in pH.
• If the anion X⁻ is a reliable base, the salt MX will show an increase in solubility within an acidic solution.
• AgCl maintains the same solubility in acid as it does in pure water as the Cl⁻ ion is a weak base where no HCl molecules form.
• Adding H⁺ ions to a solution containing Cl⁻ ions does not affect the concentration of Cl⁻ ions of the solubility of the chloride salt.

Precipitation and Qualitative Analysis

• Ion product (Q) is defined similarly to the expression as to Ksₚ for a given solid, with initial concentrations used instead of the equilibrium concentrations.
• It can be predicted whether precipitation occurs by considering the relationship between Q and Ksₚ.
• If Q > Ksₚ, precipitation occurs and continues until concentrations reach the point where they satisfy Ksₚ.
• If Q < Ksₚ, precipitation won't occur.
• When adding 750.0 mL of 4.00 × 10⁻³ M Ce(NO₃)₃ to 300.0 mL of 2.00 × 10⁻² M KIO₃ , you can determine whether Ce(IO₃)₃ (Ksₚ = 1.9 × 10⁻¹⁰ will precipitate or not.

Calculating Equilibrium Concentrations after Precipitation

• First, determine whether a product is formed when the solutions mix, and after that calculate the concentration of ions in the mixed solution to determine Q.
• Second, if a reaction appears to be going to completion when two solutions are mixed, conduct stoichiometric calculations before equilibrium ones.
• Third, allow the system to adjust to equilibrium, and determine the concentrations of ions in the solution.
• Fourth, substitute the expressions derived in step 3 to determine Ksₚ.
• When mixing 150.0 mL of 1.00 × 10⁻² M Mg(NO₃)₂ and 250.0 mL of 1.00 × 10⁻¹ M NaF to form a solution, calculating the concentrations of Mg²⁺ and F⁻ at equilibrium with solid MgF₂ (Ksₚ = 6.4 × 10⁻⁹) is possible.

Selective Precipitation

• Selective precipitation is a method used to separate mixtures of metal ions within aqueous solutions.
• It involves using a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture.
• When NaCl is added to a solution containing Ba²⁺ and Ag⁺ ions, AgCl precipitates as a white solid, whilst BaCl₂ is soluble and Ba² ions remain in the solution.
• When I- is added gradually to a solution containing 1.0 × 10⁻⁴ M Cu⁺ and 2.0 × 10⁻³ M Pb²⁺ it can be determined whether Pbl₂ (Ksₚ = 1.4 × 10⁻⁸) or Cul (Ksₚ = 5.3 × 10⁻¹²) precipitate first.
• Specify the concentration of I- necessary to begin precipitation of each salt.
• If a solution is 1 × 10⁻⁴ M in NaF, Na₂S, and Na₃PO₄, the order of precipitation can be determined as a source of Pb²⁺ is gradually added.
  • Ksp(PbF₂) = 4 × 10⁻⁸
  • Ksp(PbS) = 7 × 10⁻²⁹ -Ksp[Pb₃(PO₄)₂] = 1 × 10⁻⁵⁴

Equilibria Involving Complex Ions

• A complex ion is a charged species consisting of a metal ion surrounded by ligands.
• Ligands act as Lewis bases.
• The molecule or ion possesses a lone pair that can be donated to an empty orbital on a metal ion creating a covalent bond.
• Coordination number refers to the amount of ligands attached to the metal ion.

Formation(Stability) Constant

• Equilibrium constant pertains to each step of complex ion formation through the addition of an individual ligand to a metal ion or a complex ion within an aqueous solution.
• Typically, the ligand's total concentration exceeds the overall concentration of the metal ion.
• When mixing 150.0 mL of 1.00 × 10⁻³ M AgNO₃ with 200.0 mL of 5.00 M Na₂S₂O₃ it can calculated: the concentrations of Ag⁺, Ag(S₂O₃)⁻, and Ag(S₂O₃)₂³⁻.
• The stepwise formation equilibria proceeds as:
  • Ag⁺ + S₂O₃⁻ ⇌ Ag(S₂O₃)⁻, K₁ = 7.4 × 10⁸
  • Ag(S₂O₃)⁻ + S₂O₃⁻ ⇌ Ag(S₂O₃)₂³⁻, K₂ = 3.9 × 10⁴