Ksp and Solubility

Introduction

Calculating solubility using the solubility product constant (Ksp) is vital in areas such as environmental science and industrial chemistry. This measurement helps determine water quality and affects processes, including mineral extraction.

Concept of Equilibrium in Saturated Solutions

• Chemical Equilibrium: A dynamic condition where the rate at which the solute dissolves equals the rate of precipitation in a saturated solution.
• Saturated Solutions: These occur when no additional solute can dissolve under specific conditions, such as temperature and pressure.
• Dynamic Equilibrium: In this state, solute particles dissolve and precipitate at equal rates, maintaining a stable composition.
• Examples:
  • NaCl dissolving in water.
  • Chalk (calcium carbonate) in water, showcasing limited solubility.

Diagram: Illustrating Dynamic Equilibrium

Solubility Product Constant (Ksp)

• Ksp: The equilibrium constant for saturated solutions, measuring the maximum potential ion concentration.

Deriving the Ksp Expression

• Consider a generic ionic compound AB dissolving into A⁺ and B⁻. The Ksp is represented as: $\text{Ksp} = [A^+][B^-]$

Example Derivation – Lead Chloride (PbCl₂)

• Equilibrium Equation:
  • PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq)
• Ksp Expression:
  • $\text{Ksp} = [\text{Pb}^{2+}][\text{Cl}^-]^2$.
• Significance: The Ksp indicates the ion concentration in a solution.

Factors Affecting Solubility

Introduction to Factors Affecting Solubility

• Solubility: The capability of a solute to dissolve in a solvent, indicating how much solute can dissolve before the solution becomes saturated.
• Key Influences:
  • Enthalpy (ΔH) and Entropy (ΔS): Energy changes and disorder impacting solubility.
  • Polarity: The principle of 'like dissolves like.'
  • Hydration Energy: The energy released when ions are surrounded by water molecules.

Solvent and Solute Nature

• Hydration Spheres: Form around ions, enhancing solubility by stabilising dispersed ions.

Ion Separation and Energy Exchange

• Lattice Energy: The energy required to separate ions in a crystal lattice. Balancing with Hydration Energy influences solubility.

Temperature Effects

• Effect of Temperature:

  • Endothermic reactions benefit from increased temperature.
  • Exothermic reactions may decrease in solubility with heat.

• Illustrated example: More sugar dissolves in hot versus cold tea.

• Van't Hoff Equation: $\frac{d(\ln K)}{dT} = \frac{\Delta H}{RT^2}$

Common Ion Effect

• Common Ion Effect: Solubility reduction occurs when a solution already contains a common ion.

• Example: Adding NaCl to a solution of AgCl decreases AgCl's solubility due to the shared chloride ion.

Solubility Product Constant (Ksp) vs. Reaction Quotient (Q)

• Ksp: A threshold indicating saturation of ions in a solution.
• Q (Reaction Quotient): Measures the current ion product, compared to Ksp, to assess precipitation likelihood.

Decision-Making Procedure

• Calculate Q using initial ion concentrations and compare to Ksp:
  • Q < Ksp: Undersaturated (no precipitation).
  • Q = Ksp: Saturated equilibrium (no additional precipitate).
  • Q > Ksp: Precipitation occurs.

Essential Formulas for Ksp-related Calculations

• Ksp Expression: Defined by: $K_{sp} = [C]^c[D]^d$
  • Describes equilibrium in a saturated solution like $aA(s) \rightarrow cC(aq) + dD(aq)$.

Practice Problems

• Example 1: Calculate solubility for $\text{CaF}_2$ given $K_{sp} = 3.9 \times 10^{-11}$:

  • Equation: $\text{CaF}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$
  • Ksp Expression: $K_{sp} = [Ca^{2+}][F^-]^2 = s \times (2s)^2 = 4s^3$
  • Solving for $s$: $s = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} = 2.1 \times 10^{-4}$ mol/L

• Problem 2: Calculate the molar solubility of BaSO₄ in a 0.05 M solution of Na₂SO₄ given Ksp = 1.1 × 10⁻¹⁰.

  • Equation: $\text{BaSO}_4(s) \rightarrow \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$
  • Initial $[\text{SO}_4^{2-}]$ = 0.05 M (from Na₂SO₄)
  • Let $s$ = molar solubility of BaSO₄
  • At equilibrium: $[\text{Ba}^{2+}] = s$ and $[\text{SO}_4^{2-}] = 0.05 + s$
  • $K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s(0.05 + s)$
  • Since $s$ will be very small compared to 0.05: $K_{sp} \approx s \times 0.05$
  • $s = \frac{K_{sp}}{0.05} = \frac{1.1 \times 10^{-10}}{0.05} = 2.2 \times 10^{-9}$ mol/L

Exam Tips

• Emphasise the comparison between Q and Ksp in assessing potential precipitate formation.
• Consider temperature's effect on Ksp during calculations.