Understanding Acid and Base Dissociation Constant Calculations

Expert reviewed • 22 November 2024 • 5 minute read

The dissociation of acids and bases in water is fundamental to understanding chemical equilibria. This article explores how to calculate and apply dissociation constants for both acids (Ka) and bases (Kb).

Acid Dissociation Constant (Ka)

When an acid (HA) dissolves in water, it establishes an equilibrium:

HA(aq)+H2O(l)⇌A(aq)−+H3O(aq)+HA_{(aq)} + H_2O_{(l)} \rightleftharpoons A^-_{(aq)} + H_3O^+_{(aq)}

The acid dissociation constant (Ka) is expressed as:

Ka=[H3O+][A−][HA]K_a = \frac{[H_3O^+][A^-]}{[HA]}

Important Assumptions for Ka Calculations

  • For weak acids with small Ka values:

    • The change in initial acid concentration is negligible
    • [H₃O⁺] equals [A⁻] at equilibrium
    • Water self-ionization is negligible

  • For polyprotic acids:

    • Subsequent ionizations are typically much smaller than the first

    • Example: Hydrogen sulfide (H₂S)

      First ionization: H2S(aq)+H2O(l)⇌HS(aq)−+H3O(aq)+Ka1=8.9×10−8H_2S_{(aq)} + H_2O_{(l)} \rightleftharpoons HS^-_{(aq)} + H_3O^+_{(aq)} \quad K_{a1} = 8.9 \times 10^{-8}

      Second ionization: HS(aq)−+H2O(l)⇌S(aq)2−+H3O(aq)+Ka2=1.3×10−14HS^-_{(aq)} + H_2O_{(l)} \rightleftharpoons S^{2-}_{(aq)} + H_3O^+_{(aq)} \quad K_{a2} = 1.3 \times 10^{-14}

Solving Ka Problems

Practice Question 1

Given: - HOCl (hypochlorous acid) - Ka = 3.5 × 10⁻⁸ - 0.0400 moles HOCl in 500.0 mL water

Find [H⁺] from Ka.

1. Write equilibrium equation:

HOCl(aq)⇌H(aq)++OCl(aq)−HOCl_{(aq)} \rightleftharpoons H^+_{(aq)} + OCl^-_{(aq)}

  • Calculate initial concentration: [HOCl]initial=0.0400 mol0.500 L=0.0800 M[HOCl]_{initial} = \frac{0.0400\text{ mol}}{0.500\text{ L}} = 0.0800\text{ M}

  • Set up ICE table [Image-ICE-Table]

  • Apply Ka expression: 3.5×10−8=(x)(x)0.0800−x3.5 \times 10^{-8} = \frac{(x)(x)}{0.0800 - x}

  • Assuming x is negligible: x=0.0800×3.5×10−8=5.3×10−5 Mx = \sqrt{0.0800 \times 3.5 \times 10^{-8}} = 5.3 \times 10^{-5}\text{ M}

Base Dissociation Constant (Kb)

For a base (B) in water:

B(aq)+H2O(l)⇌HB(aq)++OH(aq)−B_{(aq)} + H_2O_{(l)} \rightleftharpoons HB^+_{(aq)} + OH^-_{(aq)}

The base dissociation constant (Kb) is expressed as:

Kb=[OH−][HB+][B]K_b = \frac{[OH^-][HB^+]}{[B]}

Example: Carbonate Ion as a Base

For carbonate ion in water:

CO3(aq)2−+H2O(l)⇌HCO3(aq)−+OH(aq)−\text{CO}_{3(aq)}^{2-} + \text{H}_{2}\text{O}_{(l)} \rightleftharpoons \text{HCO}_{3(aq)}^{-} + \text{OH}_{(aq)}^{-}

Kb=[OH−][HCO3−][CO32−]K_b = \frac{[OH^-][HCO_3^-]}{[CO_3^{2-}]}

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Acid and Base Dissociation Constants

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