Dissociation constants of weak acids and bases

The dissociation of a weak acid (HA) in aqueous solution can be represented by the following equilibrium:

HA_(aq) ⇌ H⁺_(aq) + A^-_(aq)

The equilibrium constant for this dissociation is called the acid dissociation constant (K_a):

$K_{a } = \frac{[H^{+}][A^{-}]}{[HA]}$

Similarly, the dissociation of a weak base (B) is represented by:

B_(aq) + H₂O_(l) ⇌ BH⁺_(aq) + OH^-_(aq)

The equilibrium constant is called the base dissociation constant (K_b):

$K_{b} = \frac{[BH^{+}][OH^{-}]}{[B]}$

Note that the concentration of water is not included in these expressions as it is effectively constant in dilute aqueous solutions.

Comparing strengths of acids and bases

The magnitudes of K_a and K_b indicate the extent of dissociation and therefore the relative strengths of acids and bases:

• Stronger acids have larger K_a values as they dissociate to a greater degree.
• Stronger bases have larger K_b values for the same reason.

Often K_a and K_b are expressed using p-notation, where:

$pK_{a } = -logK_{a}$ and $pK_{b } = -logK_{b}$.

This notation is useful because it allows for easier comparison of acid and base strengths:

• Smaller pK_a values correspond to larger K_a values and therefore stronger acids.
• Smaller pK_b values correspond to larger K_b values and therefore stronger bases.

For example, methanoic acid (HCOOH, pK_a = 3.75) is a stronger acid than ethanoic acid (CH₃COOH, pK_a = 4.76), as it has a smaller pK_a value.

The table below lists several acids in order of increasing strength:

Similarly, triethylamine ((CH₃)3_{N, pK}b = 4.20) is a stronger base than ammonia (NH₃, pK_b = 4.75), as it has a smaller pK_b value.

The table below lists several bases in order of increasing strength:

Calculating pH using dissociation constants

If the dissociation constant and concentration of a weak acid or base are known, the pH of its solution can be calculated using the following assumptions:

1. [H⁺] ≈ [A^-] for a weak acid HA - The dissociation of a weak acid produces equal amounts of H^+^ and A^-^ ions. Since the ionisation of water is negligible compared to the dissociation of the acid, we can assume that nearly all of the H^+^_(aq)_ ions in the solution are produced by the acid.
2. [OH^-] ≈ [BH⁺] for a weak base B - The dissociation of a weak base produces equal amounts of OH^-^ and BH^+^ ions. Since the ionisation of water is negligible compared to the reaction of the base, we can assume that nearly all of the OH^-^_(aq)_ ions in the solution are produced by the base.
3. [HA] and [B] at equilibrium are approximately equal to their initial concentrations - Only a small fraction of a weak acid or base dissociates in water, so the equilibrium concentrations of undissociated HA and B are approximately equal to their initial concentrations.

With these approximations, the K_a and K_b expressions can be simplified to:

$K_{a }\approx\frac{[H^{+}]^{2}}{[HA]}$ and $K_{b }\approx\frac{[OH^{-}]^{2}}{[B]}$

The following worked examples demonstrate how to use these equations in pH calculations.

Worked example 1 - Calculating the pH of a weak acid solution

Calculate the pH of a 0.0100 mol dm^-3 solution of ethanoic acid, given that its K_a is $1.75 \times 10^{-5}$ mol dm^-3 at 298 K. Give your answer to 2 decimal places.

Step 1: K_a equation

$\text{K}_{\text{a }} = \frac{[\text{H}^+]^2}{[\text{CH}_3\text{COOH}]}$

Step 2: Rearrange K_a equation

$[\text{H}^+] =\sqrt{\text{ K}_{\text{a }} \times\text{ }[\text{CH}_3\text{COOH}]}$

Step 3: Substitution and correct evaluation

$[\text{H}^+] = \sqrt{(1.75 \times 10^{-5}) \times0.0100}=4.18\times10^{-4}\text{ mol dm}^{-3}$

Step 4: Calculate pH

$\text{pH }= -\text{log}_{10} [\text{H}^+] = -\text{log}_{10}(4.18 \times 10^{-4})=3.38$

Worked example 2 - Calculating the concentration of propanoic acid from pH

Given a propanoic acid solution's pH is 2.89 and its K_a is $1.34 \times 10^{-5}$ mol dm^-3 at 298 K, calculate the acid's concentration. Give your answer to 3 significant figures.

Step 1: Calculate [H⁺]

[H⁺] = 10^-pH = 10^-2.89 = 1.29 x 10^-3 mol dm^-3

Step 2: K_a equation

$\text{K}_{\text{a }} = \frac{[\text{H}^+]^2}{[\text{CH}_3\text{CH}_2\text{COOH}]}$

Step 3: Rearrange K_a equation

$[\text{CH}_3\text{CH}_2\text{COOH}] = \frac{[\text{H}^+]^2}{\text{K}_{\text{a}}}$

Step 4: Substitution and correct evaluation

$[\text{CH}_3\text{CH}_2\text{COOH}] = \frac{(1.29 \times 10^{-3})^2}{(1.34 \times 10^{-5})}=0.124\text{ mol dm}^{-3}$

Worked example 3 - Calculating the K_a of hydrofluoric acid

Given a hydrofluoric acid (HF) solution's pH is 3.14 and its concentration is 0.100 mol dm^-3, calculate the K_a for hydrofluoric acid. Give your answer to 3 significant figures.

Step 1: Calculate [H⁺]

[H⁺] = 10^-pH = 10^-3.14 = 7.24 x 10^-4 mol dm^-3

Step 2: K_a equation

$\text{K}_{\text{a }} = \frac{[\text{H}^+]^2}{[\text{HF}]}$

Step 3: Substition and correct evaluation

$\text{K}_{\text{a }} = \frac{(7.24\times10^{-4})^2}{0.100}=5.25\times10^{-6}\text{ mol dm}^{-3}$

Worked example 4 - Calculating the pH of a benzoic acid solution

Calculate the pH of a 0.0500 mol dm^-3 solution of benzoic acid, given its pK_a is 4.20. Give your answer to 2 decimal places.

Step 1: Calculate K_a

K_a = 10^-pKa = 10^-4.20 = 6.31 x 10^-5 mol dm^-3

Step 2: K_a equation

$\text{K}_{\text{a }} = \frac{[\text{H}^+]^2}{[\text{C}_6\text{H}_5\text{COOH}]}$

Step 3: Rearrange K_a equation

$[\text{H}^+]=\sqrt{\text{ K}_{\text{a }} \times\text{ }[\text{C}_6\text{H}_5\text{COOH}]}$

Step 4: Substitution and correct evaluation

$[\text{H}^+] = \sqrt{(6.31 \times 10^{-5}) \times0.0500}=1.78\times10^{-3}\text{ mol dm}^{-3}$

Step 5: Calculate pH

$\text{pH }= -\text{log}_{10}[\text{H}^+] = -\text{log}_{10}(1.78 \times 10^{-3})=2.75$

The relationship between K_a, K_b and K_w

For any conjugate acid-base pair, HA and A^-, the following equilibria exist in aqueous solution:

HA_(aq) ⇌ H⁺_(aq) + A^-_(aq) $K_{a }=\frac{[H^{+}][A^{-}]}{[HA]}$

A^-_(aq) + H₂O_(l) ⇌ HA_(aq) + OH^-_(aq) $K_{b }=\frac{[HA][OH^{-}]}{[A^{-}]}$

Adding these two equations together gives the autoionization of water:

H₂O_(l) ⇌ H⁺_(aq) + OH^-_(aq) $K_{w } = [H^{+}][OH^{-}] = K_{a}\times{K_{b}}$

Taking logarithms of both sides:

pK_w = pK_a + pK_b

At 25°C, pK_w = 14.0, so for any conjugate acid-base pair at this temperature:

$pK_{a } = 14.0 - pK_{b}$ or $pK_{b } = 14.0 - pK_{a}$

These relationships show that the stronger the acid (lower pK_a), the weaker its conjugate base (higher pK_b), and vice versa. However, they apply only to conjugate pairs, not to unrelated acids and bases.

Worked example 5 - Calculating the pK_b of diethylamine

Calculate the pK_b of diethylamine (C₂H₅)₂NH, at 25°C, given a 0.0120 mol dm^-3 solution of diethylamine has a pH of 11.0. Give your answer to 3 significant figures.

Step 1: Calculate pOH

$pOH = 14.0 - pH = 14.0 - 11.0 = 3.00$

Step 2: Calculate [OH^-]

[OH^-] = 10^-pOH = 10^-3.00 = 1.00 × 10^-3 mol dm^-3

Step 3: K_b equation

$K_{b }= \frac{[OH^{-}]^{2}}{[(C_2H_5)_2NH]}$

Step 4: Substitution and correct evaluation

$K_{b }= \frac{(1.00\times10^{-3})^2}{0.0120}=8.33\times10^{-5}\text{ mol dm}^{-3}$

Step 5: Calculate pK_b

$pK_{b }= -log(8.33\times10^{-5})=4.08$

Worked example 6 - Calculating the pH of ammonia solution

Calculate the pH of a 0.200 mol dm^-3^ solution of ammonia, given that the pK_b_ of ammonia is 4.75. Give your answer to 2 decimal places.

Step 1: Calculate K_b_

K_b_ = 10^-pK_b_^ = 10^-4.75^ = 1.78 × 10^-5^ mol dm^-3^

Step 2: K_b equation

$K_b =\frac{[OH^{-}]^{2}}{[NH_{3}]}$

Step 3: Rearrange K_b equation

$[OH^{-}]=\sqrt{K_{b}\times{[NH_{3}]}}$

Step 4: Substitution and correct evaluation

$[OH^{-}]=\sqrt{(1.78\times10^{-5})\times0.200}=1.89\times10^{-3}\text{ mol dm}^{-3}$

Step 5: Calculate pOH

$pOH = -log(1.89\times10^{-3})=2.72$

Step 6: Calculate pH

$pH = 14.0 - pOH = 14.0 - 2.72=11.28$