Acid dissociation constant

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  1. What is the acid/base dissociation constant?
  2. From Kw to Ka and Kb.
  3. The Ka and Kb expressions.
  4. Relating Ka, Kb and Kw using conjugate pairs.
  1. Use the relationship between the Ka and Kw expressions to find an unknown Kb value.
    1. Ethanoic acid, or acetic acid CH3COOH is a weak acid with Ka = 1.4*10-5.1
      1. Write the formula of its conjugate base.
      2. Find the Kb for its conjugate base using the relationship between Ka/Kb and Kw.
    2. Explain using the Ka/Kb expression why Ka and Kb values are not normally used when studying strong acids and bases.

      1 Source for Ka acid dissociation constants: ATKINS, P. W., & DE PAULA, J. (2006). Atkins' Physical chemistry. Oxford, Oxford University Press.
Topic Notes

In this lesson, we will learn:

  • To recall the equilibrium ionization expressions for weak acids and bases.
  • How to relate Ka and Kb for conjugate acid/base pairs.
  • How to calculate concentration of aqueous ions in weak acid/base solutions.


  • In Autoionization of water, we looked at the equilibrium:

    2 H2O(l)  ⇌ \, \rightleftharpoons \, H3O+ (aq) + OH- (aq)

    The equilibrium constant was expressed as:

    Kw = [H3O+] [OH-] = 1 * 10 -14 at 25oC

    We saw the effect of adding strong acids/bases to equilibrium concentrations of water and dissolved ions. That was straightforward because strong acids and bases experience 100% dissociation in water.
    We also saw in
  • In strong and weak acids and bases that weak acids and bases do not experience 100% dissociation. This makes expressions for their dissociation in water more complicated.

  • Every weak acid has an acid dissociation constant, Ka and a weak base a base dissociation constant, Kb. These are equilibrium constants showing how much the acid/base dissociates when dissolved in water (aqueous solution).
    For a weak acid HX dissolved in water:

    Ka = [H3O+][X−][HX]\frac{[H_3O^{+}] [X^{-}]}{[HX]}

    For a weak base B dissolved in water:

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

    • Just like in the ionization of water and other equilibria; this is an equilibrium constant expression. This means that the higher the Ka/Kb value, the greater the degree of dissociation (because the concentrations of the dissociated ions, in the numerator, are larger values) and therefore the stronger the acid or base.
    • Remember that for strong acids and bases Ka/Kb values are not normally used. This is because in the Ka expression, their [HX] or [B] is equal to or almost zero due to complete dissociation, so the values are incredibly large.
    • For strong acids, pKa is used instead of Ka. pKa is the negative logarithm of the Ka value and is more appropriate to use for strong acids, instead of the extremely large Ka values they have.

  • In Conjugate acids and bases, we learned in a conjugate pair that a stronger conjugate acid will have a weaker the conjugate base. This relationship affects the concentration of aqueous ions and therefore affects the Ka and Kb values for a conjugate acid/base pair!
    Consider the equations for the conjugate pair acid-base pair HX and X-:

    Conjugate acid: \qquad HX + H2O  ⇌ \, \rightleftharpoons \, X- + H3O+ \qquad

    Ka = [H3O+][X−][HX]\frac{[H_3O^{+}] [X^{-}]}{[HX]}

    Conjugate base: \qquad X- + H2O  ⇌ \, \rightleftharpoons \, HX + OH- \qquad

    Kb = [HX][OH−][X−]\frac{[HX] [OH^{-}]}{[X^{-}]}

    • Both equations depend on [X-] and [HX] so Ka and Kb themselves can be related, and terms cancelled out:

      buffer solutions weak acid or base ph = pka

      As you can see, the result is the product of [H3O+] and [OH-] which is the expression for Kw. Therefore for a conjugate pair:

      Ka * Kb = Kw