Acid-Base Chemistry: Solving Basic Problems

Key Concepts and Formulas

• pH and pOH

  • pH = -log[H₃O⁺]
  • pOH = -log[OH⁻]
  • At 25°C, pH + pOH = 14 (due to kW = 1 x 10⁻¹⁴)

• Hydronium and Hydroxide Concentrations

  • [H₃O⁺] = 10^(-pH)
  • [OH⁻] = 10^(-pOH)
  • [H₃O⁺] x [OH⁻] = 1 x 10⁻¹⁴

• Acid and Base Dissociation Constants

  • pKa = -log(Ka)
  • pKb = -log(Kb)
  • pKa + pKb = 14
  • Ka x Kb = 1 x 10⁻¹⁴

Example Problems and Solutions

1. Calculating pH from Hydronium Ion Concentration

   • Given: [H₃O⁺] = 2.5 x 10⁻⁴ M
   • Estimate pH: Look at exponent (-4), pH ~ 4
   • Exact: pH = -log(2.5 x 10⁻⁴) = 3.6

2. Calculating pH from pOH

   • Given: pOH = 4.5
   • Formula: pH = 14 - pOH = 9.5
   • Solution type: Basic (pH > 7)

3. Calculating Hydroxide Ion Concentration from pOH

   • Given: pOH = 3.8
   • Formula: [OH⁻] = 10^(-pOH) = 1.58 x 10⁻⁴ M

4. Calculating pOH from Hydronium Ion Concentration

   • Given: [H₃O⁺] = 4.2 x 10⁻³ M
   • pH = -log([H₃O⁺]) = 2.377
   • pOH = 14 - pH = 11.623

5. Calculating pKa and pKb from Ka

   • Given: Ka = 1.8 x 10⁻⁵
   • pKa = -log(Ka) = 4.745
   • pKb = 14 - pKa = 9.255

6. Calculating Hydroxide Ion Concentration from Hydronium Ion Concentration

   • Given: [H₃O⁺] = 7.1 x 10⁻² M
   • pH = -log([H₃O⁺]) = 1.149
   • pOH = 14 - pH = 12.851
   • [OH⁻] = 10^(-pOH) = 1.41 x 10⁻¹³ M

7. Calculating Kb from pKa

   • Given: pKa = 3.7
   • pKb = 14 - pKa = 10.3
   • Kb = 10^(-pKb) = 5.01 x 10⁻¹¹

8. Calculating pH from Hydroxide Ion Concentration

   • Given: [OH⁻] = 0.015 M
   • pOH = -log([OH⁻]) = 1.824
   • pH = 14 - pOH = 12.176

Summary

• Familiarize yourself with basic formulas for calculating pH, pOH, [H₃O⁺], and [OH⁻].
• Understanding the relationship between acidity (pH), basicity (pOH), and the dissociation constants (pKa, pKb) is crucial.
• Practice using these calculations to solve various acid-base problems in chemistry.