Understanding pH in Salt Solutions

Aug 14, 2024

Lecture on pH of Salt Solutions

Overview

  • Goal: Find the pH of different salt solutions.
  • Focus: Sodium acetate and ammonium chloride solutions.

Sodium Acetate Solution

  • Components in Solution:
    • Sodium ions (Na+) and acetate anions (CH3COO-).
    • Sodium ions do not react with water.
    • Acetate anion acts as a base, reacts with water.
  • Reaction:
    • Acetate anion (CH3COO-) takes a proton from water (H2O) to form acetic acid (CH3COOH) and hydroxide ion (OH-).
  • Initial Concentrations:
    • 0.25 molar sodium acetate = 0.25 molar acetate anion.
    • Initial product concentrations are 0.
  • Equilibrium:
    • Change in concentration denoted by 'X'.
    • Acetate: 0.25 - X, Acetic Acid: X, Hydroxide: X.
  • Equilibrium Expression:
    • Kb = (CH3COOH)(OH-) / (CH3COO-).
    • Solve for Kb using Ka = 1.8 x 10^-5 (Ka x Kb = Kw = 1.0 x 10^-14).
    • Kb = 5.6 x 10^-10.
  • Assumption:
    • X is much smaller than 0.25, thus 0.25 - X ≈ 0.25.
  • Calculate X:
    • Solve X² = 1.4 x 10^-10, resulting in X = 1.2 x 10^-5.

Calculating pH

  • Hydroxide Ion Concentration (X): 1.2 x 10^-5.
  • pOH: -log(1.2 x 10^-5) = 4.92.
  • pH Calculation: pH = 14 - pOH = 9.08.
  • Nature of Solution: Basic.

Ammonium Chloride Solution

  • Components in Solution:
    • Ammonium ions (NH4+) and chloride ions (Cl-).
    • Chloride ions do not react with water.
    • Ammonium ions act as an acid, donate a proton.
  • Reaction:
    • NH4+ donates a proton to water (H2O), forming hydronium ions (H3O+) and ammonia (NH3).
  • Initial Concentrations:
    • 0.050 molar ammonium chloride = 0.050 molar ammonium ions.
    • Initial product concentrations are 0.
  • Equilibrium:
    • Change denoted by 'X'.
    • Ammonium: 0.050 - X, Hydronium: X, Ammonia: X.
  • Equilibrium Expression:
    • Ka = (H3O+)(NH3) / (NH4+).
    • Use Kb for NH3 = 1.8 x 10^-5 to find Ka using Ka x Kb = Kw.
    • Ka = 5.6 x 10^-10.
  • Assumption:
    • X is much smaller than 0.050, thus 0.050 - X ≈ 0.050.
  • Calculate X:
    • Solve X² = 5.6 x 10^-10 x 0.050, resulting in X = 5.3 x 10^-6.

Calculating pH

  • Hydronium Ion Concentration (X): 5.3 x 10^-6.
  • pH: -log(5.3 x 10^-6) = 5.28.
  • Nature of Solution: Acidic.

Conclusion

  • Sodium acetate results in a basic solution with a pH of 9.08.
  • Ammonium chloride results in an acidic solution with a pH of 5.28.