Base Dissociation Constants (K_b) Chemistry Tutorial

Key Concepts

• K_b, 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 K_b

• K_b provides a measure of the equilibrium position

  (i) if K_b is large, the products of the dissociation reaction are favoured

  (ii) if K_b is small, undissociated base is favoured.

• K_b provides a measure of the strength of a base

  (i) if K_b is large, the base is largely dissociated so the base is strong

  (ii) if K_b 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] × 100

  (i) If %dissociation ≈ 100%, the base is a strong base

  (ii) If %dissociation is small, the base is a weak base

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Some Base Dissociation Constants (25^oC)

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

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

K_b(ammonia) = 1.8 × 10^-5 (smaller)

K_b(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.

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%

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

Calculate the [OH^-], pOH and %dissociation in 0.40 mol L^-1 NH₃(aq).
(K_b = 1.8 × 10^-5 at 25°C)

1. Write the base dissociation equation:

   NH_3(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 (K_b 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%

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

Calculate the [H⁺] and pH for a 0.62 mol L^-1 aqueous ammonia solution.
K_b = 1.8 × 10^-5 at 25^oC.

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 (K_b 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 25^oC, K_w, 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