Acid and Base Dissociation Constants (Ka and Kb)

Acid Dissociation Constant (Ka)

• Definition: Measures how much acid molecules dissociate in a water solution.

• Expression: For an acid HA, dissociating in water:

  HA \leftrightharpoons H^+ + A^-

  K_a = \frac{[H^+][A^-]}{[HA]}

• Interpretation:

  • Larger $K_a$ value: Stronger acid (more dissociation).
  • Smaller $K_a$ value: Weaker acid (less dissociation).

Using Data to Calculate Ka

• Use equilibrium concentrations (final data), not initial data.
• Example: If $[H^+] = 0.2$, $[A^-] = 0.2$, and $[HA] = 0.8$, then use these in the $K_a$ expression.

Exercises

1. Write dissociation reactions and $K_a$ expressions:

   • Nitrous acid (HNO_2):

     HNO_2 \leftrightharpoons H^+ + NO_2^-

     K_a = \frac{[H^+][NO_2^-]}{[HNO_2]}

   • Acetic acid (CH_3COOH):

     CH_3COOH \leftrightharpoons H^+ + CH_3COO^-

     K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}

   • Phosphoric acid (H_3PO_4):

     H_3PO_4 \leftrightharpoons H^+ + H_2PO_4^-

     K_{a1} = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]}

     • Only the first proton dissociation is considered here for $K_{a1}$.

2. Identify the weakest acid:

   • Compare $K_a$ values; the smallest $K_a$ indicates the weakest acid.
   • Example: Acetic acid $K_a = 1.8 \times 10^{-5}$ (weak acid).

Base Dissociation Constant (Kb)

• Definition: Measures how much base molecules dissociate in a water solution.

• Expression: For a base B, dissociating in water:

  B + H_2O \leftrightharpoons BH^+ + OH^-

  K_b = \frac{[BH^+][OH^-]}{[B]}

• Interpretation:

  • Larger $K_b$ value: Stronger base.
  • Smaller $K_b$ value: Weaker base.

• Example: Formate ion ($HCOO^-)$:

  HCOO^- + H_2O \leftrightharpoons HCOOH + OH^-

  K_b = \frac{[HCOOH][OH^-]}{[HCOO^-]}

Ion Product Constant (Kw)

• Expression:

  K_w = [H^+][OH^-] = 1.0 \times 10^{-14} at 25°C.

• Implications:

  • Neutral solution: $[H^+] = [OH^-] = 1.0 \times 10^{-7}$ (pH 7).
  • Acidic solution: $[H^+] > [OH^-]$.
  • Basic solution: $[H^+] < [OH^-]$.

Example Calculations

1. Given: $[OH^-] = 1.0 \times 10^{-5}$ at 25°C.

2. Find: $[H^+]$.

3. Solution:

   [H^+] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-5}} = 1.0 \times 10^{-9}

4. Interpretation:

   • Higher $[OH^-]$ than $[H^+]$: Basic solution.

5. Neutral Solution Criteria: $[H^+] = [OH^-] = 1.0 \times 10^{-7}$ at 25°C.