 # Base Dissociation Constants (Kb) Chemistry Tutorial

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

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

• Kb provides a measure of the strength of a base
(i) if Kb is large, the base is largely dissociated so the base is strong

(ii) 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] × 100

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

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

No ads = no money for us = no free stuff for you!

## 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:

 base formula 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.

Do you know this?

Play the game now!

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

Do you understand this?

Take the test now!

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

Can you apply this?

Join AUS-e-TUTE!

Do the drill now!

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

Can you apply this?

Join AUS-e-TUTE!

Do the drill now!