Chemistry · Ch 6 — Equilibrium

Solubility Equilibria of Sparingly Soluble Salts

6.13

Solubility Equilibria of Sparingly Soluble Salts

The solubility of ionic solids in water spans an enormous range. Some salts, like calcium chloride, are so soluble they are hygroscopic — they absorb water vapour from the atmosphere. Others, such as lithium fluoride, dissolve so little that they are often called "insoluble." In reality, no salt is truly insoluble; every salt has some finite, measurable solubility.

What Determines Whether a Salt Dissolves?

Two opposing energy factors decide the fate of an ionic salt placed in a solvent:

Lattice enthalpy — the energy required to separate the ions from the crystal lattice. This is a positive quantity (energy must be supplied).
Solvation enthalpy — the energy released when the separated ions interact with solvent molecules. This is always negative (energy is released).

For dissolution to occur, the ion-solvent interactions must be strong enough to overcome the electrostatic forces holding the crystal together. In other words, the magnitude of the solvation enthalpy must exceed the lattice enthalpy.

Note

Solvation enthalpy depends heavily on the nature of the solvent. In a non-polar (covalent) solvent, solvation enthalpy is small and cannot overcome the lattice enthalpy. That is why ionic salts do not dissolve in non-polar solvents like benzene or hexane.

Classification by Solubility

Chemists group salts into three categories based on their molar solubility in water at room temperature:

Table unnumbered-solubility-categories-tableClassifying ionic salts by solubility at 298 K — Category I soluble (> 0.1 M), Category II slightly soluble (0.01–0.1 M), Category III sparingly soluble (< 0.01 M).

The textbook sorts salts into three categories by their solubility in water at 298 K:

Category	Solubility
Category I — Soluble	Solubility > 0.1 M
Category II — Slightly soluble	0.01 M < Solubility < 0.1 M
Category III — Sparingly soluble	Solubility < 0.01 M

Each salt has a characteristic solubility that depends on temperature. For most salts, solubility increases with temperature, but there are exceptions.

The Equilibrium in a Saturated Solution

When a sparingly soluble ionic salt is placed in water, it dissolves until the solution becomes saturated. At that point, a dynamic equilibrium is established between the undissolved solid salt and the ions in solution.

Consider a general sparingly soluble salt $A_xB_y$ that dissociates as:

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

The equilibrium constant for this heterogeneous equilibrium is called the solubility product constant, $K_{sp}$ . Since the concentration of a pure solid is constant (it does not appear in the equilibrium expression), we write:

K_{sp} = [A^{y+}]^x [B^{x-}]^y

The square brackets denote molar concentrations of the ions in the saturated solution.

Important

$K_{sp}$ is a constant at a given temperature. It depends only on the nature of the salt and the temperature — not on the amount of solid present or the volume of solution.

Relating $K_{sp}$ to Molar Solubility

Let $s$ be the molar solubility of the salt — the number of moles of salt that dissolve per litre of solution to form a saturated solution.

For the salt $A_xB_y$ :

Each mole of $A_xB_y$ that dissolves produces $x$ moles of $A^{y+}$ and $y$ moles of $B^{x-}$ .
Therefore, at equilibrium: $[A^{y+}] = xs$ and $[B^{x-}] = ys$ .

Substituting into the $K_{sp}$ expression:

K_{sp} = (xs)^x (ys)^y = x^x y^y s^{(x+y)}

This gives a direct relationship between $K_{sp}$ and molar solubility $s$ .

Tip

For a salt of the type AB (1:1 ratio, like AgCl), $x = 1$ and $y = 1$ , so $K_{sp} = s^2$ and $s = \sqrt{K_{sp}}$ .

For a salt of the type AB $_2$ (like PbCl $_2$ ), $x = 1$ and $y = 2$ , so $K_{sp} = 4s^3$ and $s = \sqrt[3]{K_{sp}/4}$ .

The Common Ion Effect on Solubility

The solubility of a sparingly soluble salt decreases when a soluble salt containing one of the same ions (a common ion) is added to the solution. This is a direct consequence of Le Chatelier's principle applied to the solubility equilibrium.

Consider the equilibrium for AgCl:

\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)

If we add NaCl (which provides Cl $^-$ ions), the concentration of Cl $^-$ increases. To maintain $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$ , the concentration of Ag $^+$ must decrease — which means more AgCl precipitates out of solution. The net effect is that the solubility of AgCl is reduced. …

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Solubility Equilibria of Sparingly Soluble Salts