Ka and the pH of Weak Acids

This lesson covers: 

1. The acid dissociation constant (K_a) for weak acids
2. Assumptions made when finding K_a for weak acids
3. Using K_a to calculate the pH of weak acids
4. Determining acid concentration or K_a from pH
5. The relationship between pK_a and K_a

K_a  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:

Ka ​=[HA][H+][A−]​

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.
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These assumptions simplify the K_a formula to:

Ka ​=[HA][H+]2​

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

There are certain situations where the K_a assumptions become inaccurate:

1. For stronger weak acids with K_a > 10^-2 mol dm^-3, the concentration of H⁺ ions becomes significant, and there is a notable difference between [HA]_equilibrium and [HA]_initial.
2. For very weak acids or very dilute solutions, the concentration of H⁺ ions from the dissociation of water becomes significant, and there is a notable difference between [H⁺]_equilibrium and [A^-]_equilibrium.

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 1.75×10−5 mol dm^-3 at 298 K.

Step 1: K_a equation

Ka ​=[CH3​COOH][H+]2​

Step 2: Rearrange K_a equation

[H+]=√ Ka ​× [CH3​COOH]​

Step 3: Substitution and correct evaluation

[H+]=√(1.75×10−5)×0.0100​=4.18×10−4 mol dm−3

Step 4: Calculate pH

pH =−log10​[H+]=−log10​(4.18×10−4)=3.38

Thus, the pH of the 0.01 mol dm^-3 ethanoic acid solution is 3.88.

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 1.34×10−5 mol dm^-3 at 298 K, calculate the acid's concentration.

Step 1: Calculate [H⁺]

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

Step 2: K_a equation

Ka ​=[CH3​CH2​COOH][H+]2​

Step 3: Rearrange K_a equation

[CH3​CH2​COOH]=Ka​[H+]2​

Step 4: Substitution and correct evaluation

[CH3​CH2​COOH]=(1.34×10−5)(1.29×10−3)2​=0.124 mol dm−3

Hence, the concentration of the propanoic acid solution is 0.124 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.

Step 1: Calculate [H⁺]

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

Step 2: K_a equation

Ka ​=[HF][H+]2​

Step 3: Substition and correct evaluation

Ka ​=0.100(7.24×10−4)2​=5.25×10−6 mol dm−3

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

The relationship between pK_a and K_a

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 pK_a from K_a

Given the K_a of carbonic acid is 4.5×10−7 mol dm−3, calculate its pK_a.

Step 1: pK_a equation

pK_a = -log₁₀(K_a)

Step 2: Substitution and correct evaluation

pKa ​=−log10​(4.5×10−7)=6.3

Thus, the pK_a of carbonic acid is 6.3.

Worked example 5 - Calculating K_a from pK_a

Given the pK_a of formic acid is 3.7, calculate its K_a value.

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

Therefore, formic acid’s Ka​ value is 2.0×10−4 mol dm−3

Using pK_a 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.

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

Ka ​=[C6​H5​COOH][H+]2​

Step 3: Rearrange K_a equation

[H+]=√ Ka ​× [C6​H5​COOH]​

Step 4: Substitution and correct evaluation

[H+]=√(6.31×10−5)×0.0500​=1.78×10−3 mol dm−3

Step 5: Calculate pH

pH =−log10​[H+]=−log10​(1.78×10−3)=2.75

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