Ka is the acid dissociation constant

Weak acids, such as ethanoic acid (CH₃COOH), only partially dissociate in aqueous solution. This means the concentration of H⁺ ions is less than the initial concentration of the acid.

To measure how much a weak acid dissociates, we use the acid dissociation constant, K_a.

For a generic weak acid HA, the dissociation equilibrium is represented as:

HA_(aq) ⇌ H⁺_(aq) + A^-_(aq)

The formula for K_a based on this equilibrium is:

${\text{K}}_{\text{a }}=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}$

Where:

• [HA] is the concentration of the acid that has not dissociated.
• [H⁺] is the concentration of hydrogen ions.
• [A^-] is the concentration of the conjugate base.

K_a has units of mol dm^-3.

The larger the K_a value, the stronger the weak acid.

Assumptions for weak acids

When calculating the pH of a weak acid (HA) based on its concentration and K_a value, we typically make two important assumptions:

1. [HA]_equilibrium ≈ [HA]_initial - This is because the ionisation of a weak acid is so small that the concentration of undissociated HA molecules present at equilibrium is approximately the same as the initial concentration of the acid.
2. [H⁺]_equilibrium ≈ [A^-]_equilibrium - This is because the ionisation of water is negligible, so the concentration of H⁺ ions produced by the ionisation of water molecules present in the solution is ignored.

These assumptions simplify the K_a formula to:

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

These assumptions are valid only for weak acids because stronger acids dissociate more, significantly affecting the initial and equilibrium concentrations of HA.

Calculating pH of weak acids using K_a

The K_a value for a weak acid is constant at a specific temperature and does not depend on the concentration. This property allows us to calculate the pH of a weak acid solution if we know the K_a value and the initial concentration of the acid.

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

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

Step 3: Substitution and correct evaluation

$[{\text{H}}^{+}]=\sqrt{(1.75\times {10}^{-5})\times 0.0100}=4.18\times {10}^{-4}{\text{ mol dm}}^{-3}$

Step 4: Calculate pH

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

Thus, the pH of the 0.01 mol dm^-3 ethanoic acid solution is 3.38.

Determining acid concentration or K_a from pH

We can use the same principles to find either the starting concentration of a weak acid or its K_a value if the pH is given.

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

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

Hence, the concentration of the propanoic acid solution is 0.124 mol dm^-3.

Worked example 3 - Calculating the Ka 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

${\text{K}}_{\text{a }}=\frac{{(7.24\times {10}^{-4})}^2}{0.100}=5.25\times {10}^{-6}{\text{ mol dm}}^{-3}$

Therefore, the K_a of hydrofluoric acid in this solution is 5.25 × 10^-6 mol dm^-3.

The relationship between pKa and Ka

pK_a offers another way to express the acid dissociation constant, defined as:

pK_a = $-$log₁₀(K_a)

Conversely, we can find K_a from pK_a through:

K_a = 10^-pKa

The smaller the pK_a value, the stronger the weak acid.

Worked example 4 - Calculating pKa from Ka

Given the K_a of carbonic acid is $4.5\times {10}^{-7}{\text{ mol dm}}^{-3}$, calculate its pK_a. Give your answer to 2 significant figures.

Step 1: pK_a equation

pK_a = $-$log₁₀(K_a)

Step 2: Substitution and correct evaluation

${\text{pK}}_{\text{a }}=-{\text{log}}_{10}(4.5\times {10}^{-7})=6.3$

Thus, the pK_a of carbonic acid is 6.3.

Worked example 5 - Calculating Ka from pKa

Given the pK_a of formic acid is 3.7, calculate its K_a value. Give your answer to 2 significant figures.

Step 1: K_a equation

K_a = 10^-pKa

Step 2: Substitution and correct evaluation

K_a = 10^-3.7 = 2.0 x 10^-4 mol dm^-3

${\text{Therefore, formic acid’s K}}_{\text{a}}\text{ value is }2.0\times {10}^{-4}{\text{ mol dm}}^{-3}$

Using pKa in calculations

If you're given a pK_a value instead of K_a for a problem, convert pK_a to K_a before applying the K_a formula.

Worked example 6 - 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

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

Step 4: Substitution and correct evaluation

$[{\text{H}}^{+}]=\sqrt{(6.31\times {10}^{-5})\times 0.0500}=1.78\times {10}^{-3}{\text{ mol dm}}^{-3}$

Step 5: Calculate pH

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

Therefore, the pH of the 0.0500 mol dm^-3 benzoic acid solution is 2.75.