Lecture Notes: Solving Basic Acid-Base Problems in Chemistry

Key Formulas and Concepts

• pH Calculation

  • Formula: ( \text{pH} = -\log[\text{H}_3\text{O}^+] )
  • Estimate pH Range:
    • Look at the exponent of the hydronium ion concentration.
    • If the number is greater than 1, pH is in the lower range of the exponent.
    • Example: For ( 2.5 \times 10^{-4} ), the pH is approximately between 3 and 4.

• pOH and pH Relationship

  • Formula: ( \text{pH} + \text{pOH} = 14 ) (at 25°C)
  • To find pH from pOH: ( \text{pH} = 14 - \text{pOH} )

• Acidity and Basicity

  • pH < 7: Acidic
  • pH = 7: Neutral
  • pH > 7: Basic

• Hydroxide Ion Concentration

  • Formula: ( \text{pOH} = -\log[\text{OH}^-] )
  • To find ([\text{OH}^-]): ( [\text{OH}^-] = 10^{-\text{pOH}} )_

Example Problems and Solutions

1. Calculating pH from Hydronium Ion Concentration

   • Example: ( [\text{H}_3\text{O}^+] = 2.5 \times 10^{-4} )
   • Calculation: ( \text{pH} = 3.6 )

2. Finding pH from pOH

   • Example: ( \text{pOH} = 4.5 )
   • Calculation: ( \text{pH} = 14 - 4.5 = 9.5 )
   • Basic solution since pH > 7.

3. Hydroxide Ion Concentration from pOH

   • Example: ( \text{pOH} = 3.8 )
   • Calculation: ([\text{OH}^-] = 1.58 \times 10^{-4} \text{ M} )

4. Calculating pOH from Hydronium Ion Concentration

   • Example: ( [\text{H}_3\text{O}^+] = 4.2 \times 10^{-3} )
   • Steps:
     • Calculate ( \text{pH} = 2.377 )
     • ( \text{pOH} = 14 - 2.377 = 11.623 )

5. Finding pKa and pKb

   • Given ( \text{K}_a = 1.8 \times 10^{-5} )
   • Calculation: ( \text{pK}_a = 4.745 )
   • ( \text{pK}_b = 14 - 4.745 = 9.255 )

6. Hydroxide Concentration from Hydronium Ion Concentration

   • Given ( [\text{H}_3\text{O}^+] = 7.1 \times 10^{-2} )
   • Steps:
     • ( \text{pH} = 1.149 )
     • ( \text{pOH} = 12.851 )
     • ([\text{OH}^-] = 1.41 \times 10^{-13} \text{ M} )

7. Calculating Kb from pKa

   • Given ( \text{pK}_a = 3.7 )
   • Calculation: ( \text{pK}_b = 10.3 )
   • ( \text{K}_b = 5.01 \times 10^{-11} )

8. pH from Hydroxide Concentration

   • Given ([\text{OH}^-] = 0.015 )
   • Steps:
     • ( \text{pOH} = 1.824 )
     • ( \text{pH} = 12.176 )_

Review of Important Formulas

• pH and pOH Relationships

  • ( \text{pH} = -\log[\text{H}^+] )
  • ( \text{pOH} = -\log[\text{OH}^-] )
  • ( \text{pH} + \text{pOH} = 14 )

• Concentration Calculations

  • ([\text{H}^+] = 10^{-\text{pH}} )
  • ([\text{OH}^-] = 10^{-\text{pOH}} )
  • ([\text{H}_3\text{O}^+] \times [\text{OH}^-] = 1 \times 10^{-14} )

• Acid and Base Constants

  • ( \text{pK}_a = -\log \text{K}_a )
  • ( \text{pK}_b = -\log \text{K}_b )
  • ( \text{K}_a \times \text{K}b = 1 \times 10^{-14} )

These notes cover the basic equations and examples for solving acid-base problems in chemistry. Understanding these principles is essential for tackling more complex chemistry problems involving acids and bases.