L3 - pH of Weak Bases

objectives

Understandings:

• The expression for the dissociation constant of a weak base (Kb).

• For a conjugate acid base pair, Ka × Kb = Kw.

• The relationship between Kb and pKb is (pKb = -log Kb).

Applications and skills:

• Solution of problems involving [H+ (aq)], [OH–(aq)], pH, pOH, Ka, pKa, Kb and pKb.

• Discussion of the relative strengths of acids and bases using values of Ka, pKa, Kb and pKb.

Guidance:

• Only examples involving the transfer of one proton will be assessed.

• Students should state when approximations are used in equilibrium calculations.

The base dissociation constant - K_b

When bases interact with water they do so by removing a hydrogen ion creating hydroxide ions in solution.

NH_{3 (aq)} + H₂O ⇋ NH₄⁺_(aq) + OH^-_(aq)

Once again, the water concentration is effectively constant and allows us to define a new constant, Kb:

K_b = [NH₄⁺] [OH^-]

[NH₃]

Kb is referred to as the base equilibrium constant and gives a measure of the extent of the equilibrium. Large values for Kb means strong base. However, as for acids, the values are usually very small and the -log(10) form of the information is used to generate convenient sized numbers.

pKb = - log(10) Kb

The lower the pKb value the stronger the base.

Using a conjugate system to find pH

With bases it may be more convenient to use the conjugate acid of the base to define a new Ka value, this time for the reverse reaction. In the ammonia base equilibrium above, the conjugate acid is the ammonium ion, NH₄⁺, whose dissociation proceeds according to the following equation:

NH₄⁺ + H₂O ⇋ NH₃ + H₃O⁺

For which the acid dissociation constant is:

K_a = [NH₃] [H⁺]

[NH₄⁺]

This allows us to have a pKa value for a specified base, although it is actually the pKa value of the conjugate acid of the base.

Relationship of Ka and Kb with Kw

Ka x Kb = Kw

This means (at Room Temperature) Ka x Kb = 1x10^-14

And if we know this then that means we ca use all of the following relationships to calculate anything we need about acids and bases:

• pKa + pKb = 14
• pH + pOH = 14
• pH = 14 - pOH
• pKa = 14 - pKb
• pKw = pKa + pKb
• Ka x Kb = Kw