Understanding Acid and Base Dissociation Constants

Expert reviewed • 22 November 2024 • 6 minute read

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

Acid Dissociation Constants (Ka)

Basic Principles

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

HA(aq)+H2O(l)⇌A(aq)−+H3O(aq)+\text{HA}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{A}^-_{(aq)} + \text{H}_3\text{O}^+_{(aq)}

The acid dissociation constant (Ka) is expressed as:

Ka=[H3O+][A−][HA]K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}

Important Assumptions

  • For weak acids with small Ka values:

    • The initial acid concentration [HA] approximately equals its equilibrium concentration
    • The change in [HA] during ionization is negligible

  • Water's self-ionization contribution to [H₃O⁺] is negligible

  • For polyprotic acids (with multiple H⁺ to donate):

    • Subsequent ionizations are usually much weaker than the first
    • Later Ka values can often be ignored if much smaller

For example, hydrogen sulfide (H₂S) shows this behavior:

First ionization: H2S(aq)+H2O(l)⇌HS(aq)−+H3O(aq)+Ka1=8.9×10−8\text{H}_2\text{S}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{HS}^-_{(aq)} + \text{H}_3\text{O}^+_{(aq)} \quad K_{a1} = 8.9 \times 10^{-8}

Second ionization: HS(aq)−+H2O(l)⇌S(aq)2−+H3O(aq)+Ka2=1.3×10−14\text{HS}^-_{(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{S}^{2-}_{(aq)} + \text{H}_3\text{O}^+_{(aq)} \quad K_{a2} = 1.3 \times 10^{-14}

Calculating [H⁺] from Ka - Worked Example

Problem: Calculate [H⁺] for a solution made by dissolving 0.0400 moles of hypochlorous acid (HOCl) in 500.0 mL of water, given Ka = 3.5 × 10⁻⁸.

Solution:

  • Write the equilibrium equation: HOCl(aq)⇌H(aq)++OCl(aq)−\text{HOCl}_{(aq)} \rightleftharpoons \text{H}^+_{(aq)} + \text{OCl}^-_{(aq)}

  • Calculate initial concentration: [HOCl]initial=0.0400 mol0.5000 L=0.0800 M[\text{HOCl}]_\text{initial} = \frac{0.0400 \text{ mol}}{0.5000 \text{ L}} = 0.0800 \text{ M}

  • Set up ICE table:

    Species[HOCl][H⁺][OCl⁻]
    Initial0.080000
    Change-x+x+x
    Equilibrium0.0800-xxx

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

  • Since Ka is small, assume x is negligible compared to 0.0800: 3.5×10−8=x20.08003.5 \times 10^{-8} = \frac{x^2}{0.0800} x=0.0800×3.5×10−8x = \sqrt{0.0800 \times 3.5 \times 10^{-8}} [H+]=5.3×10−5 mol/L[H^+] = 5.3 \times 10^{-5} \text{ mol/L}

Base Dissociation Constants (Kb)

When a base (B) dissolves in water, it establishes an equilibrium:

B(aq)+H2O(l)⇌HB(aq)++OH(aq)−\text{B}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{HB}^+_{(aq)} + \text{OH}^-_{(aq)}

The base dissociation constant (Kb) is expressed as:

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

For example, the carbonate ion acts as a base:

CO32−(aq)+H2O(l)⇌HCO3−(aq)+OH−(aq)\text{CO}_3^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{HCO}_3^-(\text{aq}) + \text{OH}^-(\text{aq})

With the corresponding Kb expression:

Kb=[OH−][HCO3−][CO32−]K_b = \frac{[\text{OH}^-][\text{HCO}_3^-]}{[\text{CO}_3^{2-}]}

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Acid and Base Dissociation Constants

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