Solubility Equilibria: Concepts, Calculations, and Applications

Solubility and Solubility Equilibria

Definition of Solubility

Solubility (s) is the maximum quantity of a solute that can dissolve in a certain quantity of solvent at a specified temperature or pressure. It is a fundamental concept in chemistry that determines how much of a substance can be dissolved before the solution becomes saturated.

• Common units:

  • Grams of solute per liter of solution (g/L)

  • Grams of solute per gram of solvent (g/g)

  • Moles of solute per liter of solution (mol/L)

Effect of Temperature on Solubility

For most solid solutes, increasing temperature increases solubility. This relationship can be visualized in solubility curves, which plot solubility (g solute per 100 g H2O) versus temperature (°C).

• Example: The solubility of KNO3 increases sharply with temperature, while NaCl shows only a slight increase.

Types of Solutions Based on Solubility

• Saturated Solution: Contains the maximum amount of solute that can dissolve at a given temperature. Any additional solute will not dissolve.

• Unsaturated Solution: Contains less solute than the maximum amount; more solute can still be dissolved.

• Supersaturated Solution: Contains more solute than can normally dissolve; it is unstable and can precipitate solute if disturbed. This can occur when a saturated solution is carefully cooled.

Solubility Equilibrium

Dynamic Equilibrium in Saturated Solutions

When the concentration of a solute reaches its solubility limit, an equilibrium is established between the dissolved and undissolved solute:

• Solute (s) ↔ Solute (aq)

At equilibrium, the rate at which solute dissolves equals the rate at which it precipitates.

Solubility Product Constant (Ksp)

Definition and Expression

The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound in water. It quantifies the extent to which a compound can dissolve.

• For the salt BaSO4:

$\mathrm{BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)}$

$K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}]$

where [Ba2+] and [SO42−] are the equilibrium concentrations of the ions.

Writing Ksp Expressions for Various Salts

• For CaCrO4:

  • $\mathrm{CaCrO_4(s) \rightleftharpoons Ca^{2+}(aq) + CrO_4^{2-}(aq)}$

  • $K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{CrO_4^{2-}}]$

• For PbCl2:

  • $\mathrm{PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)}$

  • $K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2$

• For Ca3(PO4)2:

  • $\mathrm{Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)}$

  • $K_{sp} = [\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2$

Solubility Product vs. Solubility

Ksp is not the same as solubility, but the two are related. Knowing one allows calculation of the other, provided the dissolution stoichiometry is known.

• Solubility is usually given as molar solubility (mol/L) or mass solubility (g/L).

• Ksp is the product of the equilibrium concentrations of the ions, each raised to the power of its coefficient in the balanced equation.

Calculations Involving Ksp and Solubility

Calculating Solubility from Ksp

• Example: Calculate the molar solubility of PbI2 in water at 25°C, given $K_{sp} = 7.1 \times 10^{-9}$.

• General steps:

  1. Write the dissolution equation and Ksp expression.

  2. Let the molar solubility be s.

  3. Express ion concentrations in terms of s and substitute into the Ksp expression.

  4. Solve for s.

For PbI2: $\mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)}$ $K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^-}]^2$ Let $[\mathrm{Pb^{2+}}] = s$, $[\mathrm{I^-}] = 2s$ $K_{sp} = s \times (2s)^2 = 4s^3$ Solve for $s$.

Calculating Ksp from Solubility

• Given the solubility of a salt, use the stoichiometry of dissolution to find ion concentrations, then substitute into the Ksp expression.

• Example: The solubility of CaC2O4 at 25°C is 0.0061 g/L. Calculate its Ksp.

Comparing Solubilities of Different Compounds

Comparing Ksp values is only valid if the cation:anion ratio is the same for both compounds. Otherwise, direct comparison can be misleading.

• Example: Which is more soluble, Mg(OH)2 ($K_{sp} = 2.06 \times 10^{-13}$) or FeCO3 ($K_{sp} = 3.07 \times 10^{-11}$)?

  • Calculate the molar solubility for each using their respective Ksp expressions.

Factors Affecting Solubility

The Common-Ion Effect

The presence of a common ion in solution decreases the solubility of a salt due to Le Châtelier's principle. Adding an ion already present in the equilibrium shifts the dissolution equilibrium to the left, reducing solubility.

• For XY(s) ↔ X+(aq) + Y−(aq), adding X+ or Y− decreases solubility.

• Example: The solubility of CaF2 in pure water vs. in a solution containing 0.100 mol/L NaF.

Effect of pH on Solubility

The solubility of salts containing basic anions increases in acidic solutions, while salts with acidic cations are more soluble in basic solutions.

• For a salt with a basic anion (e.g., Mg(OH)2):

  • $\mathrm{Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)}$

  • In acidic solution, [OH−] is reduced, shifting equilibrium right, increasing solubility.

  • In basic solution, [OH−] is high, shifting equilibrium left, decreasing solubility.

pH relationships:

• Neutral solution: $[H_3O^+] = [OH^-] = 1.0 \times 10^{-7}$ at 25°C

• Acidic solution: $[H_3O^+] > [OH^-]$

• Basic solution: $[OH^-] > [H_3O^+]$

• $[H_3O^+][OH^-] = 1.0 \times 10^{-14}$ at 25°C

• pH = $-\log [H_3O^+]$

Predicting Precipitation: The Reaction Quotient (Q) and Ksp

Will a Precipitate Form?

The reaction quotient, Q, is calculated using initial ion concentrations. Compare Q to Ksp to predict precipitation:

• If Q = Ksp: The system is at equilibrium; the solution is saturated.

• If Q < Ksp: More solid will dissolve until equilibrium is reached.

• If Q > Ksp: The salt will precipitate until equilibrium is reached.

Example Problems

• Given [Pb2+] = 0.015 M and [I−] = 0.0023 M, will PbI2 precipitate? $K_{sp} = 6.6 \times 10^{-9}$

• Given [Ni2+] = 0.65 M and [OH−] = 0.0053 M, will Ni(OH)2 precipitate? $K_{sp} = 2.0 \times 10^{-15}$

Calculate Q for each and compare to Ksp to determine if precipitation occurs.

Summary Table: Types of Solutions

Type of Solution	Description	Stability
Saturated	Contains the maximum amount of solute that can dissolve at a given temperature	Stable
Unsaturated	Contains less solute than the maximum; more can dissolve	Stable
Supersaturated	Contains more solute than can normally dissolve; excess can precipitate if disturbed	Unstable

Key Equations

• General Ksp expression for a salt $A_xB_y$:

  • $A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$

  • $K_{sp} = [A^{y+}]^x [B^{x-}]^y$

• Relationship between Ksp and solubility (for 1:1 salts):

  • If $K_{sp} = s^2$, then $s = \sqrt{K_{sp}}$

  • For other stoichiometries, set up the equilibrium table accordingly.

Additional info: For more complex salts, the relationship between Ksp and solubility involves solving higher-order equations based on the stoichiometry of dissolution.