Table of Contents
Solubility Equilibria
When an only sparingly soluble ionic solid is in contact with water, dissolution does not go to completion. Instead, a dynamic equilibrium is established between the undissolved solid and the ions in solution. This is called a solubility equilibrium.
Consider a generic, slightly soluble salt $M_pX_q$ (where $M$ is a cation and $X$ is an anion). The dissolution process is written as a heterogeneous equilibrium:
$$
M_pX_q(s) \rightleftharpoons p\,M^{z+}(aq) + q\,X^{z'-}(aq)
$$
At equilibrium, the rate at which ions leave the crystal equals the rate at which they re-attach. The concentration of dissolved ions then becomes constant (at a given temperature and for given conditions). This defines the solubility of the substance under those conditions.
Because the activity of a pure solid is treated as constant, only the dissolved ions appear in the equilibrium expression. This leads naturally to the concept of the solubility product.
The Solubility Product $K_\text{sp}$
Definition
The solubility product constant $K_\text{sp}$ is the equilibrium constant associated with the dissolution of a sparingly soluble solid in water. For
$$
M_pX_q(s) \rightleftharpoons p\,M^{z+}(aq) + q\,X^{z'-}(aq)
$$
the solubility product is
$$
K_\text{sp} = [M^{z+}]^p [X^{z'-}]^q
$$
where the square brackets denote equilibrium molar concentrations of the ions in a saturated solution at a specified temperature.
Key points:
- The solid does not appear in the expression.
- $K_\text{sp}$ is temperature-dependent.
- A larger $K_\text{sp}$ means the salt is more soluble (under otherwise identical conditions).
Examples of Dissolution Equilibria and $K_\text{sp}$ Expressions
- Silver chloride:
$$
\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)
$$
$$
K_\text{sp} = [\text{Ag}^+][\text{Cl}^-]
$$
- Calcium fluoride:
$$
\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\,\text{F}^-(aq)
$$
$$
K_\text{sp} = [\text{Ca}^{2+}][\text{F}^-]^2
$$
- Barium sulfate:
$$
\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)
$$
$$
K_\text{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]
$$
From $K_\text{sp}$ to Solubility
Molar Solubility in Pure Water
The molar solubility $s$ of a salt is the number of moles of solid that dissolve per liter to form a saturated solution under specified conditions.
For a simple 1:1 salt:
1:1 Electrolytes (e.g. AgCl, BaSO$_4$)
For a salt $MX$:
$$
MX(s) \rightleftharpoons M^+(aq) + X^-(aq)
$$
Let $s$ be the molar solubility in pure water. Then at equilibrium:
- $[M^+] = s$
- $[X^-] = s$
Hence:
$$
K_\text{sp} = [M^+][X^-] = s \cdot s = s^2
$$
so
$$
s = \sqrt{K_\text{sp}}
$$
1:2 Electrolytes (e.g. CaF$_2$)
For
$$
\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\,\text{F}^-(aq)
$$
Let $s$ be the molar solubility. Then:
- $[\text{Ca}^{2+}] = s$
- $[\text{F}^-] = 2s$
The solubility product is:
$$
K_\text{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s\,(2s)^2 = 4s^3
$$
Thus:
$$
s = \sqrt[3]{\frac{K_\text{sp}}{4}}
$$
2:3 Electrolytes (e.g. Al$_2$(S0$_4$)$_3$)
For
$$
\text{Al}_2(\text{SO}_4)_3(s) \rightleftharpoons 2\,\text{Al}^{3+}(aq) + 3\,\text{SO}_4^{2-}(aq)
$$
Let $s$ be the molar solubility. Then:
- $[\text{Al}^{3+}] = 2s$
- $[\text{SO}_4^{2-}] = 3s$
Hence:
$$
K_\text{sp} = (2s)^2 (3s)^3 = 4 \cdot 27 \cdot s^5 = 108\,s^5
$$
so:
$$
s = \sqrt[5]{\frac{K_\text{sp}}{108}}
$$
These relations are valid for solubility in pure water and when no other equilibria significantly affect ion concentrations.
Solubility vs. Solubility Product
- Solubility is a concentration (e.g. mol/L or g/L).
- $K_\text{sp}$ is a constant characterizing the equilibrium.
A salt with a smaller $K_\text{sp}$ has a smaller molar solubility (in pure water, ignoring side effects).
The Ionic Product and Precipitation
To decide whether a precipitate will form in a given solution, it is useful to compare:
- The ionic product $Q_\text{sp}$ (also called the reaction quotient for the dissolution reaction)
- The solubility product $K_\text{sp}$
For a salt $M_pX_q$:
$$
Q_\text{sp} = [M^{z+}]^p [X^{z'-}]^q
$$
using the actual current concentrations in solution (which may or may not be at equilibrium).
Comparison:
- If $Q_\text{sp} < K_\text{sp}$: the solution is unsaturated; more solid can dissolve.
- If $Q_\text{sp} = K_\text{sp}$: the solution is saturated; equilibrium is established.
- If $Q_\text{sp} > K_\text{sp}$: the solution is supersaturated; precipitation is thermodynamically favored until $Q_\text{sp}$ decreases to $K_\text{sp}$.
This is the central tool for predicting whether mixing two ionic solutions will cause a precipitate to form.
Common-Ion Effect and Solubility
The solubility of a sparingly soluble salt decreases when one of its ions is already present in the solution from another source. This is known as the common-ion effect and is a direct consequence of the law of mass action and Le Châtelier’s principle (already introduced elsewhere).
Example: AgCl in NaCl Solution
Dissolution:
$$
\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)
\quad;\quad
K_\text{sp} = [\text{Ag}^+][\text{Cl}^-]
$$
- In pure water, $[\text{Ag}^+] = [\text{Cl}^-] = s$.
- In a solution already containing chloride ions, e.g. $[\text{Cl}^-] = c_0$ from NaCl, only a small extra amount $s$ of $\text{Cl}^-$ is added from dissolving AgCl.
Assuming $c_0 \gg s$, we approximate $[\text{Cl}^-] \approx c_0$. Then:
$$
K_\text{sp} \approx [\text{Ag}^+]\,c_0
$$
so:
$$
[\text{Ag}^+] \approx \frac{K_\text{sp}}{c_0}
$$
The molar solubility of AgCl has been reduced from $\sqrt{K_\text{sp}}$ (in pure water) to roughly $K_\text{sp}/c_0$ in the presence of the common ion Cl$^-$.
Common-ion effect is important in:
- Selective precipitation in qualitative analysis
- Controlling solubility in industrial processes
- Many natural systems (e.g. solubility of minerals in seawater vs. pure water)
Influence of pH on Solubility
Many sparingly soluble salts contain anions that can react with $\text{H}^+$ or $\text{OH}^-$, such as:
- Carbonate: $\text{CO}_3^{2-}$
- Sulfite: $\text{SO}_3^{2-}$
- Hydroxide: $\text{OH}^-$
- Phosphate: $\text{PO}_4^{3-}$
If an anion is protonated (or partially converted into another species) at low pH, its concentration in the dissolution equilibrium decreases. The system then tends to dissolve more solid to restore $K_\text{sp}$. Thus:
- For salts containing basic anions, lowering pH generally increases solubility.
- For salts containing acidic cations (e.g. some metal hydroxides), raising pH can increase solubility, while intermediate pH may favor precipitation.
Example: CaCO$_3$ and Acid
Dissolution equilibrium:
$$
\text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq)
\quad;\quad
K_\text{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}]
$$
Carbonate also takes part in acid–base equilibria:
$$
\text{CO}_3^{2-} + \text{H}^+ \rightleftharpoons \text{HCO}_3^-
$$
Addition of $\text{H}^+$ (lowering pH) converts $\text{CO}_3^{2-}$ to $\text{HCO}_3^-$, reducing $[\text{CO}_3^{2-}]$. The dissolution equilibrium responds by dissolving more CaCO$_3$ to replenish $\text{CO}_3^{2-}$, increasing the apparent solubility of the solid.
In practice, this explains, for example:
- Dissolution of limestone by acidic rainwater
- Removal of limescale with acidic cleaning agents
Note that a detailed quantitative treatment requires combining $K_\text{sp}$ with acid–base equilibrium constants from acid–base chapters.
Competing and Complex-Forming Equilibria
Solubility can be greatly altered by additional equilibria in solution involving the same ions. Important cases:
- Complex formation (coordination complexes)
- Precipitation of a different solid containing one of the ions
- Redox reactions changing the oxidation state of a metal ion
Complex Formation and Increased Solubility
Metal ions often form complexes with suitable ligands (e.g. NH$_3$, CN$^-$, EDTA). When a complex forms, the free metal ion concentration decreases, shifting the dissolution equilibrium to the right (more solid dissolves) to maintain $K_\text{sp}$.
Example: AgCl and NH$_3$
Dissolution:
$$
\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)
$$
Complexation with ammonia:
$$
\text{Ag}^+(aq) + 2\,\text{NH}_3(aq) \rightleftharpoons [\text{Ag(NH}_3)_2]^+(aq)
$$
The complexation equilibrium removes free $\text{Ag}^+$ from solution. To maintain $K_\text{sp}$, more AgCl dissolves, sometimes to the point where no visible precipitate remains.
A quantitative description requires combining $K_\text{sp}$ with complex formation constants, which are treated in coordination chemistry.
Selective Precipitation
Selective precipitation uses differences in $K_\text{sp}$ values to separate ions from mixtures. By carefully adjusting conditions (such as adding a reagent or changing pH), one ion is precipitated while others remain in solution.
Basic principle:
- Add a precipitating agent that forms sparingly soluble salts with several ions.
- Increase the concentration of the precipitating agent gradually.
- The salt with the smallest $K_\text{sp}$ (and thus lowest solubility) precipitates first.
- Once that ion is largely removed, further changes can cause other ions to precipitate.
Analyzing such processes relies on comparing the ionic product $Q_\text{sp}$ for each potential precipitate with the corresponding $K_\text{sp}$ under the evolving conditions.
This concept is central in qualitative inorganic analysis and in many industrial purification processes.
Temperature Dependence of Solubility Products
Like other equilibrium constants, $K_\text{sp}$ depends on temperature:
- If dissolution is endothermic, increasing temperature usually increases solubility and thus $K_\text{sp}$.
- If dissolution is exothermic, increasing temperature often decreases solubility and $K_\text{sp}$.
A full quantitative treatment of this dependence is linked to thermodynamics and the temperature dependence of equilibrium constants. For solubility equilibria, one should always consider the specified temperature for tabulated $K_\text{sp}$ values.
Summary of Key Relationships
- Solubility equilibrium of a sparingly soluble salt:
$$
M_pX_q(s) \rightleftharpoons p\,M^{z+}(aq) + q\,X^{z'-}(aq)
$$
- Solubility product:
$$
K_\text{sp} = [M^{z+}]^p [X^{z'-}]^q
$$
- Ionic product:
$$
Q_\text{sp} = [M^{z+}]^p [X^{z'-}]^q
$$
- $Q_\text{sp} < K_\text{sp}$: unsaturated, more solid can dissolve.
- $Q_\text{sp} = K_\text{sp}$: saturated, equilibrium.
- $Q_\text{sp} > K_\text{sp}$: supersaturated, precipitation favored.
- Molar solubility can be expressed in terms of $K_\text{sp}$ for a given dissolution stoichiometry (e.g. $s = \sqrt{K_\text{sp}}$ for a simple 1:1 salt in pure water).
- Solubility is modified by:
- Common-ion effect (decreases solubility)
- pH changes, if the ions participate in acid–base equilibria
- Complex formation and other side reactions (often increase or decrease solubility)
All these phenomena are direct applications of the law of mass action to heterogeneous equilibria involving a solid phase and dissolved ions.