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Base Dissociation Constants (Kb)

Key Concepts

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Some Base Dissociation Constants (25oC)

The table below gives the value of the base dissociation constant, Kb, for aqueous solutions of different weak bases at 25°C:

baseformula Kb  
phosphine PH3 1.0 × 10-14 smaller Kb
hydroxylamine NH2OH 9.1 × 10-9
ammonia NH3 1.8 × 10-5
methanamine
(methylamine)
CH3NH2 4.4 × 10-4 larger Kb

Compare the values of the base dissociation constant for ammonia and methylamine:

Kb(ammonia) = 1.8 × 10-5 (smaller)

Kb(methanamine) = 4.4 × 10-4 (larger)

At 25°C, using solutions of the same concentration, for example 0.1 mol L-1:

(i) There will be more undissociated ammonia molecules than undissociated methylamine molecules.

(ii) There will be more hydroxide ions in the methylamine solution than in the ammonia solution, therefore

  • pOH of the methylamine solution will be lower than the pOH of the ammonia solution,
  • pH of the methylamine solution will be higher than the pH of the ammonia solution.

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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 25°C.

  1. Write the base dissociation equation:
    NaOH(aq) OH-(aq) + Na+(aq)
  2. Calculate the initial and equilibrium concentrations of the species present:
    NaOH is a strong base, it completely dissociates to form OH- and Na+

    Set up a R.I.C.E. Table as shown below to find the equilibrium concentration of each species in solution:

    Reaction: NaOH(aq) OH-(aq) + Na+(aq)
    Initial concentration:
    (mol L-1)
    0.10   0   0
    Change in concentration:
    (mol L-1)
    -0.10   +0.10   +0.10
    Equilibrium concentration:
    (mol L-1)
    0.10 - 0.10
    = 0
      0.10   0.10

    [OH-] = 0.10 mol L-1

  3. Calculate pOH : pOH = -log10[OH-]
    pOH = -log10[0.10] = 1
  4. Calculate %dissociation:
    %dissociation = [OH-]/[base initial] × 100

    [OH-] = 0.1 mol L-1

    [NaOH initial] = 0.1 mol L-1

    %dissociation = 0.1/0.1 × 100 = 100%

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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 × 10-5 at 25°C)

  1. Write the base dissociation equation:
    NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
  2. Write the equilibrium expression for the base dissociation:

    Kb = [NH4+][OH-]

    [NH3]
    1.8 × 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-

    Set up a R.I.C.E. Table as shown below to find the equilibrium concentration of each species in solution:

    Reaction: NH3 + H2 O NH4+ + OH-
    Initial concentrations:
    (mol L-1
    0.40   0   0
    Change in concentrations:
    (mol L-1
    -x   +x   +x
    Equilibrium concentrations:
    (mol L-1
    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] ≈ 0.40 mol L-1

  4. Substitute the concentration values into the expression for the base dissociation:

    Kb = [NH4+][OH-]

    [NH3]
    1.8 × 10-5 = [x][x]

    [0.40]

    1.8 × 10-5 = x2 ÷ 0.40

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

    x = √7.2 × 10-6 = 2.7 × 10-3

    [OH-] = x = 2.7 × 10-3 mol L-1

  5. Calculate pOH:
    pOH = -log10[OH-]

    pOH = -log[2.7 × 10-3] = 2.6

  6. Calculate %dissociation:
    %dissociation = [OH-]/[base initial] × 100

    [OH-] = 2.7 × 10-3

    [NH3 initial] = 0.40 mol L-1

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

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

Calculate the [H+] and pH for a 0.62 mol L-1 aqueous ammonia solution.
Kb = 1.8 × 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 × 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-

    Set up a R.I.C.E. Table as shown below to find the equilibrium concentration of each species in solution:

    Reaction: NH3 + H2O NH4+ + OH-
    Initial Concentrations:
    (mol L-1)
    0.62   0   0
    Change in Concentrations:
    (mol L-1)
    -x   +x   +x
    Equilibrium Concentrations:
    (mol L-1)
    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] ≈ 0.62 mol L-1

  4. Substitute the concentration values into the expression for the base dissociation:
    Kb = [NH4+][OH-]

    [NH3]
    1.8 × 10-5 = [x][x]

    [0.62]

    1.8 × 10-5 = x2 ÷ 0.62

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

    x = √1.1 × 10-5 = 3.3 × 10-3

    [OH-] = x = 3.3 × 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 × 10-3 = 3 × 10-12 mol L-1

  6. Calculate pH:
    pH = -log10[H+]

    pH = -log[3 × 10-12] = 11.5

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