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*Revision Materials and Past Papers* 1 Atomic Structure and the Periodic Table 2 Bonding and Structure 3 Redox I 4 Inorganic Chemistry and the Periodic Table 5 Formulae, Equations and Amounts of Substance 6 Organic Chemistry I 7 Modern Analytical Techniques I 8 Energetics I 9 Kinetics I 10 Equilibrium I 11 Equilibrium II 12 Acid-base Equilibria 13 Energetics II 14 Redox II 15 Transition Metals 16 Kinetics II 17 Organic Chemistry II 18 Organic Chemistry III 19 Modern Analytical Techniques II RP Required Practicals

12 Acid-base Equilibria

12.1 Acid–Base Theory and Core Definitions 12.2 pH, Ka, Kw, and pKa Calculations 12.3 Titration Curves and Indicators 12.4 Buffers and Their Action 12.5 Enthalpy Changes of Neutralisation

pH Calculations, Ka and Kw

Specification Reference Topic 12, points 4–6, 8–13, 15

Quick Notes

  • pH = –log[H⁺]
  • [H⁺] = 10–pH
  • Ka = [H⁺][A⁻] / [HA]
  • pKa = –log Ka
    • pKa is used to compare acid strength more conveniently
  • Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K
  • Strong acids: [H⁺] ≈ concentration of acid
  • Strong bases: [OH⁻] ≈ concentration of base
    • We can calculate [H⁺] in alkaline (basic) solutions using Kw
    • Edexcel A-Level Chemistry showing how Kw can be rearranged to find H+ and how H+ can be used to find pH

Full Notes

What is pH?

The pH scale measures the concentration of hydrogen ions (H⁺) in solution, on a logarithmic scale. This allows us to express a wide range of H⁺ concentrations more simply.

Edexcel A-Level Chemistry diagram showing the logarithmic expression for pH calculation.

A lower pH indicates a higher concentration of hydrogen ions (i.e. a more acidic solution).

To work backwards and calculate [H+] from pH:

Edexcel A-Level Chemistry diagram showing conversion between pH and hydrogen ion concentration.

This is useful when you’re given a pH value and need to calculate the hydrogen ion concentration, such as in weak acid problems.

Strong Acids and pH

Strong acids fully dissociate in solution.

This means the concentration of hydrogen ions equals the concentration of the acid (for monoprotic acids such as HCl).

For Example Strong acid pH

For 0.01 mol dm⁻³ HCl:
[H⁺] = 0.01 mol dm⁻³
pH = –log(0.01) = 2.00

Because there's no equilibrium to consider, these pH calculations are straightforward.

Weak Acids and Ka

Weak acids only partially dissociate in solution, setting up an equilibrium:

Edexcel A-Level Chemistry diagram showing dissociation equilibrium of a weak acid HA ⇌ H⁺ + A⁻.

The acid dissociation constant (Ka) is the equilibrium constant used for weak acid dissociation:

Edexcel A-Level Chemistry equation for the acid dissociation constant Ka.

Where:

The larger the Ka, the more the acid dissociates = stronger acid.
The smaller the Ka, the less it dissociates = weaker acid.

Introducing pKa

Ka values for weak acids are often very small, so it is sometimes easier to use pKa:

Edexcel A-Level Chemistry expression showing pKa calculation from Ka.

This is simply a more manageable number and a lower pKa = stronger acid.

You can also calculate Ka from pKa using:

Edexcel A-Level Chemistry equation showing conversion between Ka and pKa.

Calculating pH of a Weak Acid

You can calculate the pH of a weak acid using its Ka value and its concentration.

Since weak acids partially dissociate with H⁺ and A⁻ in a 1:1 ratio, we assume:

This means we can assume that [H⁺] × [A⁻] = [H⁺]².

Edexcel A-Level Chemistry expression showing derivation of [H⁺] from Ka and [HA].

Ka = ([H⁺]²) / [HA] and as a result [H⁺] = √(Ka × [HA]).

Steps to Calculate pH of a Weak Acid

  1. Use Ka expression:
    Ka = ([H⁺]²) / [HA]
  2. Rearrange to find [H⁺]:
    [H⁺] = √(Ka × [HA])
  3. Find pH:
    pH = –log[H⁺]
Edexcel A-Level Chemistry calculation steps for finding pH of a weak acid.
Photo of Matt
Matt’s exam tip

Don’t forget when using Ka to find the pH of a weak acid, we are assuming two things — one that [H⁺] = [A⁻] and that the [HA] at the start is the same as the concentration of weak acid at equilibrium. These are assumptions: for example the [HA] at equilibrium won’t be exactly the same as the initial [HA] as some will have dissociated, however the amount of dissociation is very small compared to the concentration meaning mathematically it isn’t significant enough for us to worry about.


Worked Example

Find the pH of 0.10 mol dm⁻³ ethanoic acid (Ka = 1.75 × 10⁻⁵).

  1. [H⁺] = √(Ka × [HA])
  2. [H⁺] = √(1.75 × 10⁻⁵ × 0.10) = 1.32 × 10⁻³ mol dm⁻³
  3. pH = –log(1.32 × 10⁻³) = 2.88

Using Experimental pH to Find Ka

You may be given the pH of a weak acid solution and asked to find Ka. Follow these steps:

  1. Convert pH to [H⁺]: [H⁺] = 10^(–pH)
  2. Assume [H⁺] = [A⁻]
  3. Use original acid concentration as [HA] (approximate)
  4. Plug into: Ka = [H⁺]² / [HA]

This is a common exam question.

Worked Example

A solution of a weak acid has a concentration of 0.100 mol dm⁻³ and a pH of 4.75. Calculate the Ka of the weak acid.

  1. Convert pH to [H⁺]: [H⁺] = 10^(–4.75) = 1.78 × 10⁻⁵ mol dm⁻³
  2. Assume [H⁺] = [A⁻] → [A⁻] = 1.78 × 10⁻⁵ mol dm⁻³
  3. Use original acid concentration as [HA] ≈ 0.100 mol dm⁻³
  4. Plug into Ka expression: (1.78 × 10⁻⁵)² / 0.100 = 3.17 × 10⁻⁹ mol dm⁻³

Final Answer: Ka = 3.17 × 10⁻⁹ mol dm⁻³

The Ionic Product of Water (Kw)

In water, a very small percentage of molecules ionise, releasing H⁺ and OH⁻ ions into solution.

Edexcel A-Level Chemistry diagram showing ionisation of water into H⁺ and OH⁻ ions.

This is a reversible process and an equilibrium is established.

The equilibrium constant for this is:

Edexcel A-Level Chemistry equation for ionic product of water Kw.

For a more detailed explanation on Kw and where it comes from, see here.

At 298 K: Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶
Kw is key for finding pH of bases, because strong bases increase [OH⁻]
.

Calculating pH of Strong Bases

To calculate the pH of a strong base like NaOH:

  1. [OH⁻] = concentration of the base
  2. Use Kw to find [H⁺]
  3. Use [H⁺] to find pH
Edexcel A-Level Chemistry worked example showing calculation of [H⁺] from [OH⁻] using Kw.
Worked Example

Find the pH of an NaOH solution with a concentration of 0.010 mol dm⁻³.

  1. [OH⁻] = 0.010
  2. [H⁺] = 1.0 × 10⁻¹⁴ / 0.010 = 1.0 × 10⁻¹²
  3. pH = –log(1.0 × 10⁻¹²) = 12.00

Using pKw

Just like Ka and pKa, we can define:

Edexcel A-Level Chemistry expression showing definition of pKw.

At 298 K, pKw = 14.

Summary