Determining relative solubility primarily involves comparing the solubility product constant ($K_{sp}$) values of different compounds, though specific conditions and calculation methods are crucial for an accurate assessment.
Understanding Solubility and Ksp
Solubility refers to the maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature. For sparingly soluble ionic compounds, this equilibrium is described by the solubility product constant ($K_{sp}$). The $K{sp}$ is an equilibrium constant that represents the product of the concentrations of the dissolved ions, each raised to the power of its stoichiometric coefficient in the balanced dissolution equation. A smaller $K{sp}$ generally indicates lower solubility, and a larger $K_{sp}$ indicates higher solubility.
Comparing $K_{sp}$ Values Directly
As stated in the reference, for compounds that dissolve to produce the same number of ions, we can directly compare their $K_{sp}$ values to determine their relative solubilities.
- Rule: When comparing ionic compounds that dissociate into the same total number of ions (e.g., AX type salts like AgCl and BaSO₄, both producing two ions; or AX₂ type salts like CaF₂ and Mg(OH)₂, both producing three ions), the compound with the larger $K_{sp}$ value is more soluble.
Example: Comparing AgCl and AgBr
Let's consider two silver halides, both of which dissociate into two ions (one Ag⁺ and one halide ion):
- AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) ; $K_{sp}$ (AgCl) = $1.8 \times 10^{-10}$
- AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq) ; $K_{sp}$ (AgBr) = $5.0 \times 10^{-13}$
Since both compounds produce two ions, we can directly compare their $K_{sp}$ values. $1.8 \times 10^{-10}$ is larger than $5.0 \times 10^{-13}$. Therefore, AgCl is more soluble than AgBr.
Calculating and Comparing Molar Solubility (s)
When compounds produce a different number of ions, directly comparing their $K{sp}$ values can be misleading. In such cases, it's necessary to calculate the molar solubility (s) for each compound from its $K{sp}$ and then compare these s values. Molar solubility (s) represents the number of moles of solute that dissolve per liter of solution.
Steps to Calculate Molar Solubility:
- Write the balanced dissolution equilibrium equation for the compound.
- Set up an ICE (Initial, Change, Equilibrium) table or use stoichiometry to express ion concentrations in terms of 's'.
- Substitute these expressions into the $K_{sp}$ expression and solve for 's'.
Example: Comparing AgCl and PbCl₂
-
AgCl: (2 ions: Ag⁺, Cl⁻)
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
$K{sp} = [Ag⁺][Cl⁻] = (s)(s) = s²$
$s = \sqrt{K{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$ M -
PbCl₂: (3 ions: Pb²⁺, 2Cl⁻)
PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq)
$K{sp} = [Pb²⁺][Cl⁻]² = (s)(2s)² = 4s³$
$s = \sqrt[3]{K{sp}/4}$
Given $K_{sp}$ (PbCl₂) = $1.7 \times 10^{-5}$
$s = \sqrt[3]{(1.7 \times 10^{-5})/4} = \sqrt[3]{4.25 \times 10^{-6}} = 1.62 \times 10^{-2}$ M
Comparison:
Molar solubility of AgCl ($1.34 \times 10^{-5}$ M) is much smaller than the molar solubility of PbCl₂ ($1.62 \times 10^{-2}$ M). Therefore, PbCl₂ is significantly more soluble than AgCl, even though its $K{sp}$ value ($1.7 \times 10^{-5}$) is much larger than that of AgCl ($1.8 \times 10^{-10}$). This clearly demonstrates why direct $K{sp}$ comparison is only valid when the number of ions is the same.
Summary Table: $K_{sp}$ vs. Molar Solubility (s)
| Compound | Dissociation Type | $K_{sp}$ Expression | Molar Solubility (s) from $K_{sp}$ | Example $K_{sp}$ | Calculated Molar Solubility (s) |
|---|---|---|---|---|---|
| AX | A⁺ + X⁻ | $s²$ | $s = \sqrt{K_{sp}}$ | $1.8 \times 10^{-10}$ (AgCl) | $1.34 \times 10^{-5}$ M |
| AX₂ | A²⁺ + 2X⁻ | $4s³$ | $s = \sqrt[3]{K_{sp}/4}$ | $1.7 \times 10^{-5}$ (PbCl₂) | $1.62 \times 10^{-2}$ M |
| A₂X | 2A⁺ + X²⁻ | $4s³$ | $s = \sqrt[3]{K_{sp}/4}$ | $2.1 \times 10^{-8}$ (Ag₂CrO₄) | $1.74 \times 10^{-3}$ M |
| AX₃ | A³⁺ + 3X⁻ | $27s⁴$ | $s = \sqrt[4]{K_{sp}/27}$ | $2.0 \times 10^{-19}$ (Al(OH)₃) | $2.94 \times 10^{-6}$ M |
Other Factors Influencing Relative Solubility
Beyond intrinsic $K_{sp}$ values, several external factors can influence the relative solubility of compounds in a given solution:
- Temperature: Solubility generally increases with increasing temperature for most solids (endothermic dissolution). However, some compounds (exothermic dissolution) may become less soluble at higher temperatures.
- Common Ion Effect: The solubility of a sparingly soluble salt decreases significantly when a common ion (an ion already present in the solution) is added. This shifts the equilibrium towards the undissolved solid, according to Le Chatelier's Principle.
- pH: For salts containing basic anions (e.g., hydroxide, carbonate, fluoride) or acidic cations, solubility is often pH-dependent.
- For example, metal hydroxides (like Mg(OH)₂) are more soluble in acidic solutions because H₃O⁺ ions react with OH⁻ ions, removing them from solution and shifting the equilibrium to the right.
- Similarly, salts of weak acids (e.g., CaCO₃, CaF₂) become more soluble in acidic solutions as the acid protonates the anion, reducing its concentration.
- Complex Ion Formation: The solubility of many metal salts can increase dramatically in the presence of ligands that can form stable complex ions with the metal cation. For example, AgCl is sparingly soluble in water, but its solubility increases significantly in the presence of ammonia (NH₃) due to the formation of the soluble complex ion [Ag(NH₃)₂]⁺.
By considering both the $K_{sp}$ values (and the stoichiometry of dissolution) and these environmental factors, one can accurately determine the relative solubility of various ionic compounds.
