Solubility Product Principle and Applications

Solubility-Product Constant (Ksp)

Definition and Expression

The solubility-product constant (Ksp) is an equilibrium constant for the dissolution of a sparingly soluble ionic compound. It represents the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the balanced equation.

• General form: For a salt AB that dissociates as AB(s) → A+(aq) + B-(aq), the Ksp expression is $K_{sp} = [A^+][B^-]$.

• Example: For BaSO4: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$, $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.

• Note: The concentration of the solid is not included in the Ksp expression.

Writing Ksp Expressions

• Ag3PO4: $Ag_3PO_4(s) \rightleftharpoons 3Ag^+(aq) + PO_4^{3-}(aq)$, $K_{sp} = [Ag^+]^3[PO_4^{3-}]$

• AgCl: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$, $K_{sp} = [Ag^+][Cl^-]$

• Ag2S: $Ag_2S(s) \rightleftharpoons 2Ag^+(aq) + S^{2-}(aq)$, $K_{sp} = [Ag^+]^2[S^{2-}]$

Solubility and Molar Solubility

Solubility is the amount of a substance (in grams) that dissolves to form a saturated solution. Molar solubility is the number of moles of solute dissolving to form one liter of saturated solution.

• To convert solubility to Ksp:

  1. Convert solubility (g/L) to molar solubility (mol/L) using molar mass.

  2. Determine the equilibrium concentrations of ions based on the stoichiometry.

  3. Calculate Ksp as the product of ion concentrations.

Example: The molar solubility of silver chromate (Ag2CrO4) is $6.7 \times 10^{-5}$ M. The Ksp expression is $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$.

The Common-Ion Effect

Principle and Application

The common-ion effect describes the decrease in solubility of an ionic compound when a solution already contains one of the ions present in the compound. This is an application of Le Châtelier’s principle.

• Adding a common ion shifts the equilibrium toward the solid, decreasing solubility and causing precipitation.

• Example: Adding NaF to a solution of CaF2 increases [F-], shifting equilibrium to form more CaF2 solid.

• Removing a common ion (e.g., by adding acid to remove F- as HF) increases solubility.

Formation of Complex Ions

Effect on Solubility

The formation of complex ions can increase the solubility of otherwise insoluble salts. Complex ions are formed when a metal ion reacts with ligands (such as ammonia).

• Example: Addition of ammonia to AgCl forms the complex ion $Ag(NH_3)_2^+$, removing Ag+ from solution and favoring the dissolution of AgCl.

• Overall reaction: $AgCl(s) + 2NH_3(aq) \rightleftharpoons Ag(NH_3)_2^+(aq) + Cl^-(aq)$

Selective Precipitation

Separation of Ions

Selective precipitation is a technique used to separate ions based on differences in their solubility products. Ions with lower Ksp values precipitate first.

• Example: In a mixture of Zn2+ and Cu2+, CuS (Ksp = $6 \times 10^{-37}$) precipitates before ZnS (Ksp = $2 \times 10^{-25}$).

• Adding HCl to a solution containing Ag+ and Cu2+ causes AgCl to precipitate (Ksp = $1.8 \times 10^{-10}$), while Cu2+ remains in solution.

Calculations and Examples

Solubility and Ksp Calculations

• Example: The molar solubility of PbCl2 is calculated from its Ksp ($1.7 \times 10^{-5}$).

• Study Question: Calculate the molar solubility for zinc hydroxide (Ksp = $4.5 \times 10^{-17}$ at 25°C).

• Study Question: The molar solubility of barium sulfate is $1.1 \times 10^{-5}$ M. What is the Ksp?

Precipitation Criteria

• For a salt AB: $Q = [A^+][B^-]$

• If $Q > K_{sp}$, precipitation occurs.

• If $Q = K_{sp}$, equilibrium exists.

• If $Q < K_{sp}$, more solid dissolves.

Additional Problems and Answers

Solubility Product Expressions

• Correct Ksp expressions must use ion concentrations raised to their stoichiometric powers.

• Incorrect examples: $K_{sp}(PbCl_2) = [Pb]^2[Cl_2]$ (should be $[Pb^{2+}][Cl^-]^2$), $K_{sp}(MgF_2) = [Mg^{2+}][F^-]$ (should be $[Mg^{2+}][F^-]^2$), $K_{sp}(Ag_2CrO_4) = [Ag^+]^4[CrO_4^{2-}]^2$ (should be $[Ag^+]^2[CrO_4^{2-}]$).

Sample Calculations

• Lead(II) sulfate: Ksp = $1.8 \times 10^{-8}$, molar solubility = $1.3 \times 10^{-4}$ M.

• MgF2: Molar solubility = $1.2 \times 10^{-3}$ M, Ksp = $6.91 \times 10^{-9}$.

Summary Table: Solubility Product Expressions

Salt	Ksp Expression
CaF2	$K_{sp} = [Ca^{2+}][F^-]^2$
Ag2CrO4	$K_{sp} = [Ag^+]^2[CrO_4^{2-}]$
FeS	$K_{sp} = [Fe^{2+}][S^{2-}]$
PbCl2	$K_{sp} = [Pb^{2+}][Cl^-]^2$
BaSO4	$K_{sp} = [Ba^{2+}][SO_4^{2-}]$

Key Points for Exam Preparation

• Understand how to write Ksp expressions for various salts.

• Be able to convert solubility (g/L) to molar solubility and then to Ksp.

• Apply the common-ion effect and Le Châtelier’s principle to solubility equilibria.

• Recognize the role of complex ion formation in increasing solubility.

• Use selective precipitation to separate ions based on their Ksp values.

• Perform calculations involving Ksp, molar solubility, and precipitation criteria.