Solubility and Solubility Equilibria (General Chemistry Study Notes)

Solubility and Solubility Equilibria

Introduction

This chapter covers the principles of solubility and solubility equilibria, focusing on the solubility product constant (Ksp), calculation of solubility, precipitation reactions, and factors affecting solubility such as the common ion effect and pH. These concepts are essential for understanding the behavior of slightly soluble salts in aqueous solutions.

Solubility Equilibria

A. Solubility Product Constant (Ksp)

The solubility product constant, Ksp, is the equilibrium constant for the dissolution of a slightly soluble ionic compound in water. It quantifies the extent to which a salt can dissolve to form its constituent ions.

• General Dissolution Reaction: For a salt such as AgCl: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$

• Solubility Product Expression: $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$

• Molar Solubility (s): Number of moles of solute dissolved in 1 L of a saturated solution (units: mol/L).

• Solubility (g/L): Number of grams of solute dissolved in 1 L of a saturated solution.

Example: Writing Ksp Expressions

• For AgF2: $\text{AgF}_2(s) \rightleftharpoons \text{Ag}^{2+}(aq) + 2\text{F}^-(aq)$ $K_{sp} = [\text{Ag}^{2+}][\text{F}^-]^2$

• For Mg2CO3: $\text{Mg}_2\text{CO}_3(s) \rightleftharpoons 2\text{Mg}^{2+}(aq) + \text{CO}_3^{2-}(aq)$ $K_{sp} = [\text{Mg}^{2+}]^2[\text{CO}_3^{2-}]$

• For Al3(PO4)2: $\text{Al}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Al}^{3+}(aq) + 2\text{PO}_4^{3-}(aq)$ $K_{sp} = [\text{Al}^{3+}]^3[\text{PO}_4^{3-}]^2$

B. Finding Ksp from Solubility

Given the solubility of a compound, you can calculate its Ksp by determining the equilibrium concentrations of its ions and substituting them into the solubility product expression.

• Steps:

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

  2. Determine ion concentrations at equilibrium based on the stoichiometry of the dissolution reaction.

  3. Substitute ion concentrations into the Ksp expression.

Example: CaSO4

• Given: Solubility = 0.67 g/L $\text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq)$ $K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}]$

Example: Ag2CrO4

• Given: Solubility = 0.044 g/L $\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq)$ $K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]$

C. Finding Solubility from Ksp

If Ksp is known, the molar solubility of a compound can be calculated by setting up an equilibrium table and solving for the concentration of ions at saturation.

• Steps:

  1. Write the balanced dissolution equation.

  2. Let the molar solubility be 's'.

  3. Express ion concentrations in terms of 's'.

  4. Substitute into the Ksp expression and solve for 's'.

Example: AgCl

• Given: $K_{sp} = 1.6 \times 10^{-10}$ $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ $K_{sp} = s^2$ Solve for $s$.

Example: CaF2

• Given: $K_{sp} = 3.9 \times 10^{-11}$ $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$ $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2$ Let $s$ = molar solubility, then $[\text{Ca}^{2+}] = s$, $[\text{F}^-] = 2s$ $K_{sp} = s \times (2s)^2 = 4s^3$ Solve for $s$.

D. Precipitation Reactions

Precipitation occurs when the product of the ion concentrations in solution exceeds the Ksp of a salt. The reaction quotient, Q, is used to predict whether a precipitate will form.

• Reaction Quotient: $Q = [\text{Ag}^+][\text{Cl}^-]$ (for AgCl)

• Comparison:

  Condition	Result
  $Q < K_{sp}$	Unsaturated solution; no precipitate forms
  $Q = K_{sp}$	Saturated solution; no precipitate forms
  $Q > K_{sp}$	Supersaturated solution; precipitate forms

Example: Will a Precipitate Form?

• Mixing solutions of NaF and Ca(NO3)2 to test for CaF2 precipitation. $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$ Calculate $Q$ and compare to $K_{sp}$.

Factors Affecting Solubility

A. Common Ion Effect

The common ion effect refers to the decrease in solubility of a slightly soluble salt when a solution already contains one of the ions present in the salt. This shifts the equilibrium to favor the undissolved solid.

• Example: $\text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)$ Adding AgNO3 (source of Ag+) or NaBr (source of Br-) decreases AgBr solubility.

• Assume the common ion concentration remains constant if it is much larger than the amount produced by the salt's dissolution.

Example: Molar Solubility of AgBr

• In pure water: Use $K_{sp}$ to solve for $s$.

• In 0.010 M NaBr: $[\text{Br}^-]$ is fixed at 0.010 M; solve for $[\text{Ag}^+]$ using $K_{sp}$.

Example: Molar Solubility of CaF2

• In 0.010 M Ca(NO3)2: $[\text{Ca}^{2+}]$ is fixed; solve for $[\text{F}^-]$.

• In 0.010 M NaF: $[\text{F}^-]$ is fixed; solve for $[\text{Ca}^{2+}]$.

B. Effect of pH on Solubility

The solubility of salts containing basic anions (anions of weak acids) increases as the solution becomes more acidic (lower pH). This is because the added acid reacts with the anion, removing it from solution and shifting the equilibrium toward more dissolution.

• Example: $\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq)$ In acidic solution, OH- is neutralized, increasing solubility.

• Soluble bases dissolve better in acidic solutions; soluble acids dissolve better in basic solutions.

• Salts with anions that do not hydrolyze (anions of strong acids) are unaffected by pH.

Example: Which Compounds Are More Soluble in Acidic Solution?

• CaCO3: Contains CO32-, a basic anion; solubility increases in acid.

• AgCl: Contains Cl-, an anion of a strong acid; solubility unaffected by pH.

• PbSO4: Contains SO42-, an anion of a strong acid; solubility unaffected by pH.

Additional info: The notes above expand on the brief points in the original slides, providing full academic context, definitions, and stepwise explanations for calculation and conceptual understanding.