Overview
Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature and pressure. Understanding solubility helps predict precipitation reactions and solution behavior.
Key Terms
Term Definition Saturated Contains maximum dissolved solute Unsaturated Contains less than maximum solute Supersaturated Contains more than maximum (unstable) Soluble Dissolves readily Insoluble Does not dissolve appreciably
Factors Affecting Solubility
1. Nature of Solute and Solvent
"Like dissolves like"
Polar solutes dissolve in polar solvents
Nonpolar solutes dissolve in nonpolar solvents
Solute Solvent Solubility NaCl (ionic) Water (polar) High Oil (nonpolar) Water (polar) Low Oil (nonpolar) Hexane (nonpolar) High
2. Temperature
For Most Solids:
Solubility increases with temperature
Endothermic dissolution
For Gases:
Solubility decreases with temperature
Exothermic dissolution
3. Pressure (Gases Only)
Henry's Law:
C = k H × P C = k_H \times P C = k H × P Where:
C C C k H k_H k H P P P
Solubility Rules for Ionic Compounds
Soluble Compounds
Ions Exceptions All Group 1 (Na⁺, K⁺, etc.) None All NH₄⁺ None All NO₃⁻ None All CH₃COO⁻ (acetate) None Most Cl⁻, Br⁻, I⁻ Ag⁺, Pb²⁺, Hg₂²⁺ Most SO₄²⁻ Ca²⁺, Sr²⁺, Ba²⁺, Pb²⁺, Ag⁺
Insoluble Compounds
Ions Exceptions Most OH⁻ Group 1, Ba²⁺, Sr²⁺, Ca²⁺ (slightly) Most S²⁻ Group 1, Group 2, NH₄⁺ Most CO₃²⁻ Group 1, NH₄⁺ Most PO₄³⁻ Group 1, NH₄⁺
Solubility Product (K s p K_{sp} K s p
For a sparingly soluble salt:
M a X b ( s ) ⇌ a M n + ( a q ) + b X m − ( a q ) \text{M}_a\text{X}_b(s) \rightleftharpoons a\text{M}^{n+}(aq) + b\text{X}^{m-}(aq) M a X b ( s ) ⇌ a M n + ( a q ) + b X m − ( a q ) K s p = [ M n + ] a × [ X m − ] b K_{sp} = [\text{M}^{n+}]^a \times [\text{X}^{m-}]^b K s p = [ M n + ] a × [ X m − ] b Example: Silver Chloride
AgCl ( s ) ⇌ Ag + ( a q ) + Cl − ( a q ) \text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) AgCl ( s ) ⇌ Ag + ( a q ) + Cl − ( a q ) K s p = [ Ag + ] [ Cl − ] = 1.8 × 10 − 10 K_{sp} = [\text{Ag}^+][\text{Cl}^-] = 1.8 \times 10^{-10} K s p = [ Ag + ] [ Cl − ] = 1.8 × 1 0 − 10 Calculating Molar Solubility
If s s s
K s p = ( s ) ( s ) = s 2 K_{sp} = (s)(s) = s^2 K s p = ( s ) ( s ) = s 2 s = K s p = 1.8 × 10 − 10 = 1.3 × 10 − 5 M s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5} \text{ M} s = K s p = 1.8 × 1 0 − 10 = 1.3 × 1 0 − 5 M More K s p K_{sp} K s p
CaF₂
CaF 2 ( s ) ⇌ Ca 2 + ( a q ) + 2 F − ( a q ) \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) CaF 2 ( s ) ⇌ Ca 2 + ( a q ) + 2 F − ( a q ) K s p = [ Ca 2 + ] [ F − ] 2 = 3.9 × 10 − 11 K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = 3.9 \times 10^{-11} K s p = [ Ca 2 + ] [ F − ] 2 = 3.9 × 1 0 − 11 Let s s s [ Ca 2 + ] = s [\text{Ca}^{2+}] = s [ Ca 2 + ] = s [ F − ] = 2 s [\text{F}^-] = 2s [ F − ] = 2 s
K s p = ( s ) ( 2 s ) 2 = 4 s 3 K_{sp} = (s)(2s)^2 = 4s^3 K s p = ( s ) ( 2 s ) 2 = 4 s 3 s = K s p 4 3 = 2.1 × 10 − 4 M s = \sqrt[3]{\frac{K_{sp}}{4}} = 2.1 \times 10^{-4} \text{ M} s = 3 4 K s p = 2.1 × 1 0 − 4 M Fe(OH)₃
Fe(OH) 3 ( s ) ⇌ Fe 3 + ( a q ) + 3 OH − ( a q ) \text{Fe(OH)}_3(s) \rightleftharpoons \text{Fe}^{3+}(aq) + 3\text{OH}^-(aq) Fe(OH) 3 ( s ) ⇌ Fe 3 + ( a q ) + 3 OH − ( a q ) K s p = [ Fe 3 + ] [ OH − ] 3 K_{sp} = [\text{Fe}^{3+}][\text{OH}^-]^3 K s p = [ Fe 3 + ] [ OH − ] 3 Let s s s
K s p = ( s ) ( 3 s ) 3 = 27 s 4 K_{sp} = (s)(3s)^3 = 27s^4 K s p = ( s ) ( 3 s ) 3 = 27 s 4 Common Ion Effect
Adding a common ion decreases solubility.
Example
Solubility of AgCl in 0.10 M NaCl:
K s p = [ Ag + ] [ Cl − ] = 1.8 × 10 − 10 K_{sp} = [\text{Ag}^+][\text{Cl}^-] = 1.8 \times 10^{-10} K s p = [ Ag + ] [ Cl − ] = 1.8 × 1 0 − 10 [ Cl − ] = 0.10 + s ≈ 0.10 ( s is small ) [\text{Cl}^-] = 0.10 + s \approx 0.10 \quad (s \text{ is small}) [ Cl − ] = 0.10 + s ≈ 0.10 ( s is small ) [ Ag + ] = s [\text{Ag}^+] = s [ Ag + ] = s s × 0.10 = 1.8 × 10 − 10 s \times 0.10 = 1.8 \times 10^{-10} s × 0.10 = 1.8 × 1 0 − 10 s = 1.8 × 10 − 9 M s = 1.8 \times 10^{-9} \text{ M} s = 1.8 × 1 0 − 9 M Solubility decreased from 1.3 × 10 − 5 1.3 \times 10^{-5} 1.3 × 1 0 − 5 1.8 × 10 − 9 1.8 \times 10^{-9} 1.8 × 1 0 − 9
Predicting Precipitation
Compare ion product (Q) to K s p K_{sp} K s p
Condition Result Q < K s p Q < K_{sp} Q < K s p Unsaturated, no precipitate Q = K s p Q = K_{sp} Q = K s p Saturated, equilibrium Q > K s p Q > K_{sp} Q > K s p Supersaturated, precipitate forms
Example
Will PbI₂ precipitate if [Pb²⁺] = 0.001 M and [I⁻] = 0.01 M?
K s p ( PbI 2 ) = 8.5 × 10 − 9 K_{sp}(\text{PbI}_2) = 8.5 \times 10^{-9} K s p ( PbI 2 ) = 8.5 × 1 0 − 9 Q = [ Pb 2 + ] [ I − ] 2 = ( 0.001 ) ( 0.01 ) 2 = 1 × 10 − 7 Q = [\text{Pb}^{2+}][\text{I}^-]^2 = (0.001)(0.01)^2 = 1 \times 10^{-7} Q = [ Pb 2 + ] [ I − ] 2 = ( 0.001 ) ( 0.01 ) 2 = 1 × 1 0 − 7 Q > K s p Q > K_{sp} Q > K s p
pH and Solubility
Salts containing basic anions are more soluble in acidic solutions:
Mg(OH) 2 , CaCO 3 , CaF 2 , PbS \text{Mg(OH)}_2, \text{CaCO}_3, \text{CaF}_2, \text{PbS} Mg(OH) 2 , CaCO 3 , CaF 2 , PbS Example: CaCO₃ in acid:
CaCO 3 ( s ) + 2 H + ( a q ) → Ca 2 + ( a q ) + H 2 O ( l ) + CO 2 ( g ) \text{CaCO}_3(s) + 2\text{H}^+(aq) \rightarrow \text{Ca}^{2+}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g) CaCO 3 ( s ) + 2 H + ( a q ) → Ca 2 + ( a q ) + H 2 O ( l ) + CO 2 ( g ) Complex Ion Formation
Complex ions increase solubility of some salts.
Example: AgCl in NH₃
AgCl ( s ) + 2 NH 3 ( a q ) → [ Ag(NH 3 ) 2 ] + ( a q ) + Cl − ( a q ) \text{AgCl}(s) + 2\text{NH}_3(aq) \rightarrow [\text{Ag(NH}_3\text{)}_2]^+(aq) + \text{Cl}^-(aq) AgCl ( s ) + 2 NH 3 ( a q ) → [ Ag(NH 3 ) 2 ] + ( a q ) + Cl − ( a q ) The formation of the complex ion drives dissolution.