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Base Dissociation Constants (Kb)

Key Concepts

  • Kb, the base dissociation constant or base ionisation constant, is an equilibrium constant that refers to the dissociation, or ionisation, of a base.

  • For the reaction in which the Arrhenius base, BOH, dissociates to form the ions OH- and B+:
    BOH OH- + B+
    Kb = [OH-][B+]
    [BOH]

    For a Brönsted-Lowry base:
    B + H2O BH+ + OH-
    Kb = [OH-][BH+]
    [B]

    The concentration of water is absorbed into the value of Kb

  • Kb provides a measure of the equilibrium position
  1. if Kb is large, the products of the dissociation reaction are favoured

  2. if Kb is small, undissociated base is favoured.

  • Kb provides a measure of the strength of a base
  1. if Kb is large, the base is largely dissociated so the base is strong

  2. if Kb is small, very little of the base is dissociated so the base is weak.

  • The degree to which a base dissociates can be represented as a percentage:
    % dissociation (ionization) = [OH- at equilibrium] ÷ [base initial] x 100

    If %dissociation ≈ 100%, the base is a strong base

    If %dissociation is small, the base is a weak base

Example : Calculating [OH-], pOH and %dissociation for a Strong Base

Calculate the [OH-], pOH and %dissociation in 0.10 mol L-1 NaOH(aq) at 25oC.
  1. Write the base dissociation equation:
    NaOH OH- + Na+

  2. Calculate the initial and equilibrium concentrations of the species present:

    NaOH is a strong base, it completely dissociates to form OH- and Na+
      NaOH OH- + Na+
    initial concentrations (M) 0.10   0   0
    Equilibrium concentrations (M) 0.10 - 0.10 = 0   0.10   0.10

    [OH-] = 0.10M

  3. Calculate pOH : pH = -log[OH-]

    pOH = -log[0.10] = 1

  4. Calculate %dissociation:

    %dissociation = [OH-]/[base initial] x 100

    [OH-] = 0.1M

    [NaOH initial] = 0.1M

    %dissociation = 0.1/0.1 x 100 = 100%

Example : Calculating [OH-], pOH and %dissociation for a Weak Base

Calculate the [OH-], pOH and %dissociation in 0.40 mol L-1 NH3(aq). (Kb = 1.8 x 10-5 at 25oC)
  1. Write the base dissociation equation:
    NH3 + H2O NH4+ + OH-

  2. Write the equilibrium expression for the base dissociation:
    Kb = [NH4+][OH-]
    [NH3]
    1.8 x 10-5 = [NH4+][OH-]
    [NH3]

  3. Calculate the initial and equilibrium concentrations of the species present:

    let x = moles of NH3 that dissociate to form NH4+ and OH-
      NH3 + H2 O NH4+ + OH-
    initial concentrations (M) 0.40   0   0
    Equilibrium concentrations (M) 0.40 - x   x   x

    Since NH3 is a weak base (Kb is small), it dissociates only slightly, x will be very small compared to 0.40

    So, at equilibrium, [NH3] ≈ 0.40M

  4. Substitute the concentration values into the expression for the base dissociation:
    Kb = [NH4+][OH-]
    [NH3]
    1.8 x 10-5 = [x][x]
    [0.40]

    1.8 x 10-5 = x2 ÷ 0.40

    x2 = 1.8 x 10-5 x 0.40 = 7.2 x 10-6

    x = √7.2 x 10-6 = 2.7 x 10-3

    [OH-] = x = 2.7 x 10-3 mol L-1

  5. Calculate pOH:

    pOH = -log[OH-]

    pOH = -log[2.7 x 10-3] = 2.6

  6. Calculate %dissociation:

    %dissociation = [OH-]/[base initial] x 100

    [OH-] = 2.7 x 10-3

    [NH3 initial] = 0.40M

    %dissociation = 2.7 x 10-3/0.40 x 100 = 0.68%

Example : Calculating [H+] and pH for a Weak Base at 25oC

Calculate the [H+] and pH for a 0.62M aqueous ammonia solution.
Kb = 1.8 x 10-5 at 25oC.
  1. Write the base dissociation equation:
    NH3 + H2O NH4+ + OH-

  2. Write the equilibrium expression for the base dissociation:
    Kb = [NH4+][OH-]
    [NH3]
    1.8 x 10-5 = [NH4+][OH-]
    [NH3]

  3. Calculate the initial and equilibrium concentrations of the species present:

    let x = moles of NH3 that dissociate to form NH4+ and OH-
      NH3 + H2O NH4+ + OH-
    initial concentrations (M) 0.62   0   0
    Equilibrium concentrations (M) 0.62 - x   x   x

    Since NH3 is a weak base (Kb is small), it dissociates only slightly, x will be very small compared to 0.62

    So, at equilibrium, [NH3] ≈ 0.62M

  4. Substitute the concentration values into the expression for the base dissociation:
    Kb = [NH4+][OH-]
    [NH3]
    1.8 x 10-5 = [x][x]
    [0.62]

    1.8 x 10-5 = x2 ÷ 0.62

    x2 = 1.8 x 10-5 x 0.62 = 1.1 x 10-5

    x = √1.1 x 10-5 = 3.3 x 10-3

    [OH-] = x = 3.3 x 10-3 mol L-1

  5. Calculate the concentration of H+:

    At 25oC, Kw, the equilibrium constant for the dissociation of water, is 10-14

    ie, [H+][OH-] = 10-14

    [H+] = 10-14/[OH-]

    Substitute in the value for [OH-]:

    [H+] = 10-14/3.3 x 10-3 = 3 x 10-12M

  6. Calculate pH:

    pH = -log[H+]

    pH = -log[3 x 10-12] = 11.5

Practice Questions
For AUS-e-TUTE members:
  1. Click on the Kb drill link:
    Kb drill
  2. Enter your username and password if prompted.
  3. Click the "New Question" button to begin the drill.
  4. Worked solutions are provided if you need some help!

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Weak Base Dissociation Constants

baseformulaKb
methylamine CH3NH2 4.4 x 10-4
ammonia NH3 1.8 x 10-5
hydroxylamine NH2OH 9.1 x 10-9
phosphine PH3 1.0 x 10-14
 

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