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Introduction to Solubility
Solubility describes the extent to which a solute dissolves in a solvent under specified conditions, usually at a particular temperature and pressure. A solution forms when solute particles become dispersed among solvent particles through interactions between them.
In a solution of NaCl in water, the ionic crystal separates into hydrated ions:
NaCl(s) → Na⁺(aq) + Cl⁻(aq)
Water molecules surround the ions, stabilizing them through ion–dipole attractions. The resulting solution is homogeneous because the ions are distributed throughout the solvent.
Solubility is different from the rate of dissolving. Solubility describes how much solute can dissolve at equilibrium, whereas the rate describes how quickly dissolution occurs. Stirring and crushing a solid into smaller particles can increase the rate of dissolving without necessarily increasing its equilibrium solubility.
Types of Solutions
A solution is unsaturated when it contains less dissolved solute than the equilibrium amount possible under the specified conditions. Additional solute can dissolve.
A saturated solution contains the maximum equilibrium concentration of dissolved solute under the specified conditions when excess undissolved solute is available. At equilibrium, dissolution and crystallization occur at equal rates.
For a sparingly soluble ionic compound, this equilibrium can be represented as:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
At equilibrium, AgCl continues to dissolve and recrystallize at equal rates. The concentrations of dissolved ions remain constant, even though particles continue to exchange between the solid and solution.
A supersaturated solution contains more dissolved solute than is normally present at equilibrium under the same conditions. It is unstable or metastable and may crystallize when disturbed or when a suitable crystal is introduced.
Supersaturated solutions can be prepared by dissolving additional solute at a higher temperature and then cooling the solution carefully. Crystallization occurs when the solution can no longer maintain the excess dissolved solute.
Factors Affecting Solubility
Intermolecular Forces and the Nature of the Solute and Solvent
Solubility depends strongly on the interactions among solute particles, solvent particles, and solute–solvent pairs. Dissolution requires separating solute particles and making room for them in the solvent, while forming new interactions between the solute and solvent.
A useful guideline is that substances with similar types of intermolecular interactions tend to dissolve well in one another.
Polar solvents, such as water, generally dissolve ionic and polar substances effectively. Nonpolar solvents, such as hexane, generally dissolve nonpolar substances more effectively.
For example, NaCl dissolves readily in water because water molecules stabilize the separated ions. Hexane does not dissolve NaCl effectively because its nonpolar molecules cannot stabilize the ions sufficiently.
Likewise, nonpolar substances such as oils have limited solubility in water because mixing them does not provide sufficiently favorable interactions to compensate for disrupting water’s hydrogen-bonding network.
This guideline is useful but not absolute. Solubility depends on the overall energetic and entropic changes associated with dissolution, not simply on whether a substance is classified as polar or nonpolar.
Temperature and the Solubility of Solids
The effect of temperature on the solubility of a solid depends on the energetics of dissolution.
For many solid solutes, solubility increases as temperature increases. If dissolving a substance absorbs heat, increasing temperature often favors the dissolved state. For other substances, solubility may decrease as temperature rises.
The relationship can be understood using Le Châtelier’s principle when the dissolution process has a significant heat effect. If dissolution is endothermic, heat can be treated as a reactant, so increasing temperature generally favors dissolution. If dissolution is exothermic, heat behaves as a product, so increasing temperature generally favors the solid state.
These are equilibrium-based predictions; the actual solubility trend depends on the specific substance.
Solubility curves show how the maximum amount of a solute that dissolves in a given amount of solvent changes with temperature. A point below a curve represents an unsaturated solution, a point on the curve represents saturation, and a point above the curve represents a supersaturated solution if the excess solute remains dissolved.
Pressure and the Solubility of Gases
Pressure has a particularly important effect on gas solubility. At constant temperature, increasing the partial pressure of a gas above a liquid generally increases the concentration of that gas dissolved in the liquid.
This relationship is described by Henry’s law:
C = kP
where C is the equilibrium concentration of dissolved gas, P is the partial pressure of that gas above the solution, and k is Henry’s law constant for the specified gas–solvent system and temperature.
The exact numerical value and units of k depend on the convention used. In the form shown, k has units that make C = kP dimensionally consistent.
For example, carbon dioxide dissolves in a sealed soft drink under elevated CO₂ pressure. When the container is opened, the CO₂ partial pressure above the liquid decreases, reducing the equilibrium concentration of dissolved CO₂. The excess gas escapes as bubbles.
Temperature also affects gas solubility. For many gases dissolved in water, increasing temperature decreases solubility because dissolved gas molecules more readily escape into the gas phase.
Solubility of Ionic Compounds
Ionic compounds dissolve when solvent–ion interactions can sufficiently compensate for the separation of ions from the crystal lattice, together with the entropy changes associated with dissolution.
Water is an effective solvent for many ionic compounds because its molecules are polar. The oxygen end of a water molecule is partially negative and attracts cations, while the hydrogen ends are partially positive and attract anions.
These ion–dipole attractions produce hydration, in which water molecules surround dissolved ions.
However, not every ionic compound is highly soluble in water. Some compounds have strong lattice attractions that are difficult to overcome, and the overall balance of enthalpy and entropy may not favor extensive dissolution.
For example, AgCl is only sparingly soluble in water, whereas NaNO₃ is highly soluble. Both contain ions, but their different solid-state and solution interactions lead to different solubilities.
Solubility Equilibrium and the Solubility-Product Constant
For a sparingly soluble ionic compound, the equilibrium between the solid and its dissolved ions can be described using the solubility-product constant, Ksp.
Consider calcium fluoride:
CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)
Its solubility-product expression is:
Ksp = [Ca²⁺][F⁻]²
The pure solid is omitted from the expression because its activity is effectively constant. The exponents correspond to the stoichiometric coefficients of the dissolved ions.
For a given temperature, Ksp is constant for a specified equilibrium system. A larger Ksp often indicates greater solubility when comparing salts with the same dissolution stoichiometry, but Ksp values cannot always be compared directly to rank compounds with different ion ratios.
Calculating Molar Solubility from Ksp
Molar solubility, s, is the number of moles of a compound that dissolve per liter of solution to establish equilibrium. It is measured in mol/L.
For CaF₂ dissolving in pure water, let its molar solubility be s.
CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)
At equilibrium:
[Ca²⁺] = s
[F⁻] = 2s
Substituting into the Ksp expression:
Ksp = [Ca²⁺][F⁻]²
Ksp = (s)(2s)²
Ksp = 4s³
Therefore:
s = (Ksp / 4)^(1/3)
Once the Ksp value is provided, substituting it into this equation gives the molar solubility of CaF₂ in pure water, assuming the contribution from water itself is negligible.
For a compound such as AgCl:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag⁺][Cl⁻]
In pure water, if the molar solubility is s:
[Ag⁺] = s and [Cl⁻] = s
Therefore:
Ksp = s²
s = √Ksp
These examples show why the dissolution equation must be balanced before calculating solubility: the stoichiometric coefficients determine the ion concentrations and the powers in the Ksp expression.
The Common-Ion Effect
The common-ion effect occurs when adding an ion already present in a solubility equilibrium decreases the solubility of the sparingly soluble compound.
For example:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Adding NaCl increases the concentration of Cl⁻. According to Le Châtelier’s principle, the equilibrium shifts toward solid AgCl, reducing the equilibrium concentration of Ag⁺ compared with that in pure water.
The Ksp expression remains unchanged at the same temperature, but the equilibrium concentrations change to satisfy it.
This principle is useful for predicting precipitation and understanding why a compound may be less soluble in a solution containing one of its ions than in pure water.
Precipitation and the Reaction Quotient
A precipitate forms when dissolved ions produce a solid that is insufficiently soluble to remain entirely in solution.
The reaction quotient, Q, is calculated using the same concentration expression as Ksp but uses the current ion concentrations rather than necessarily using equilibrium concentrations.
For AgCl:
Q = [Ag⁺][Cl⁻]
Compare Q with Ksp:
- If Q < Ksp, the solution is unsaturated with respect to the solid, and no precipitation is required to reach equilibrium.
- If Q = Ksp, the solution is at equilibrium with the solid.
- If Q > Ksp, the solution is supersaturated with respect to the solid, and precipitation is thermodynamically favored until equilibrium is restored.
When mixing solutions, calculate the ion concentrations after accounting for dilution and the total mixed volume. Then evaluate Q to determine whether precipitation is expected.