Lecture Notes on Strong Bases and pH Calculations

Key Concepts

• Strong Bases: Strong bases dissociate completely in solution to form hydroxide ions.
  • Example: Potassium hydroxide (KOH), a Group 1A metal hydroxide.
  • Other Group 1A examples: Lithium hydroxide (LiOH), Sodium hydroxide (NaOH).
  • Group 2A Metal Hydroxides also considered strong bases: Calcium hydroxide (Ca(OH)₂), Strontium hydroxide (Sr(OH)₂).

Problem 1: Sodium Hydroxide (NaOH)

• Objective: Calculate the initial concentration of NaOH given a pH of 13.00.

Steps:

1. Dissolution Equation:

   • NaOH (solid) → Na⁺ (aq) + OH⁻ (aq)
   • Balanced equation indicates [OH⁻] = [NaOH] initial concentration.

2. Calculate [OH⁻] Using pH:

   • At 25°C, pH + pOH = 14.00.
   • Given pH = 13.00:
     • pOH = 14.00 - 13.00 = 1.00.
   • pOH = -log[OH⁻]:
     • 1.00 = -log[OH⁻].
     • Solve: [OH⁻] = 10^(-1.00) = 0.10 M.

3. Conclusion:

   • Initial [NaOH] = 0.10 M.

Problem 2: Calcium Hydroxide (Ca(OH)₂)

• Objective: Find the pH of a 0.0010 M Ca(OH)₂ solution.

Steps:

1. Dissolution Equation:

   • Ca(OH)₂ (solid) → Ca²⁺ (aq) + 2OH⁻ (aq)
   • Mole ratio: 1:1 for Ca(OH)₂ to Ca²⁺, 1:2 for Ca(OH)₂ to OH⁻.

2. Concentration Calculations:

   • [Ca(OH)₂] = 0.0010 M results in [Ca²⁺] = 0.0010 M.
   • [OH⁻] = 2 × 0.0010 M = 0.0020 M.

3. Calculate pH:

   • Calculate pOH:
     • pOH = -log[OH⁻] = -log(0.0020) ≈ 2.70.
     • Maintain 2 significant figures.
   • Calculate pH:
     • pH + pOH = 14.00.
     • pH = 14.00 - 2.70 = 11.30.

Conclusion

• Key Takeaways:
  • Understanding the dissociation of strong bases in water is crucial.
  • Use of pH and pOH relationships to determine concentrations of solutions.
  • Handling significant figures is important in chemical calculations.