2 - solubility product
Solubility Equilibria
Review of Solubility Concepts
Solvent: The substance in which a solute dissolves.
Solute: The substance that is dissolved in a solvent.
Solution: A homogeneous mixture of solvent and solute.
Saturated: A solution that cannot dissolve any more solute at a given temperature.
Unsaturated: A solution that can still dissolve more solute.
Supersaturated: A solution that holds more dissolved solute than is required to reach equilibrium.
Solubility Curve: A graph that shows the relationship between solubility and temperature.
Solubility: The ability of a substance to dissolve in a solvent at a given temperature.
Dissociation: The separation of ions that occurs when an ionic compound dissolves.
Precipitation: The formation of a solid (precipitate) during a chemical reaction in a solution.
Ionization: The process of forming ions from neutral atoms or molecules.
Net Ionic Equation: A chemical equation that shows only the particles that participate in a reaction.
Solubility Chart: A chart that lists the solubility of various substances in specific solvents.
Factors Affecting Solubility: Include temperature, pressure, and the nature of solute and solvent.
Solubility Product (Ksp)
When a small amount of a salt (solute), AxBy, is placed in water (solvent):
It may dissolve completely, liberating ions Ay+ and Bx- into the solution.
Saturated Solutions: Once no more solute can dissolve despite being in contact with the solvent, the solution is saturated.
Dynamic Equilibrium: At saturation, the system reaches dynamic equilibrium where the rate of dissolving equals the rate of precipitation.
Rate of Dissolving vs. Precipitation
Rate of dissolving (Rdiss) and rate of precipitation (Rppt) can be represented as:
Rdiss: NaCl(s) ⇌ Na+(aq) + Cl-(aq)
The solution is in equilibrium when Rdiss = Rppt.
Initially, salt dissolves faster than its ions precipitate, causing a net movement toward dissolution.
Equilibrium Constant Expression
General representation:
AxBy(s) ⇋ xAy+(aq) + yBx-(aq)
Equilibrium Constant:
[ K_{eq} = \frac{[Ay^+]^x[Bx^-]^y}{[AxBy]} ]
Since the concentration of pure solid salt is constant, it leads to modified equilibrium constant:
[ K_{eq} \cdot [AxBy] = K_{sp} = [Ay^+]^x[Bx^-]^y ]
Ksp is known as the solubility product.
Constants in Saturated Solutions
For saturated solutions at equilibrium:
The product of the ion concentrations, each raised to their respective coefficients, is constant and termed Ksp.
All solubility equilibrium equations represent the dissolution of ionic solids into their ions.
Even after reaching equilibrium, dissolving and precipitation processes continue dynamically.
Example: Writing Ksp Expressions
CaCl2(s) ⇌ Ca2+(aq) + 2Cl-(aq)
Ksp = [Ca2+][Cl−]²
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO4³-(aq)
Ksp = [Ca2+]³[PO4³−]²
Factors Affecting Ksp
Ksp is dependent on the solubility of the salt:
Higher solubility leads to higher Ksp values.
Temperature Impact: As temperature increases, solubility generally increases, thus affecting Ksp.
According to Le Chatelier’s Principle, increasing temperature shifts equilibrium in the endothermic direction.
Factors Not Affecting Ksp
Adding or removing solid ionic substances does not affect equilibrium or Ksp since solid concentration is constant.
Changes in solution volume (addition of water or evaporation) change ion concentration but not Ksp.
Limitations of Ksp
Ksp values are reliable only for sparingly soluble salts:
Ions should not exceed 0.001 mol/L for valid Ksp values.
At higher solubility levels, Ksp may no longer remain constant.
Some salts dissolve through multiple steps, each with its own equilibrium constant.
If ions react with water post-dissolution (hydrolysis), Ksp calculations may become unreliable.
Calculating Ksp from Solubility Example #2
Calculating the concentrations and Ksp for calcium hydroxide with a solubility of 0.012 mol/L:
Determine [Ca2+] and [OH-].
Example #3: Calcium Phosphate
Solubility given as 2.21 x 10⁻⁴ g/L:
Calculate the molar concentrations of calcium ions ([Ca2+]) and phosphate ions ([PO4³-]).
Example #4: Lead(IV) Iodide
Given solubility is 0.85 g/100 mL, calculate Ksp = 6.1 x 10⁻⁹.
Calculating Solubility from Ksp
Solubility calculated in g/100 mL or g/L, but molar solubility is preferred for Ksp calculations.
Let molar solubility of salt = ‘s’ mol/L.
Example #5: Calculate molar solubilities for CuCO3 (Ksp = 2.5 x 10⁻¹⁰) and Ag2SO4 (Ksp = 1.7 x 10⁻⁵).
Example #6: Determining Ion Concentration
If the equilibrium constant for dissolving lead(II) iodide is 8.5 x 10⁻⁹, calculate the ion concentrations at equilibrium, where s = 1.3 x 10⁻³ mol/L.