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*Revision Materials and Past Papers* 1 Atomic Structure 2 Amounts of Substance 3 Bonding 4 Energetics 5 Kinetics 6 Chemical Equilibria & Kc 7 Redox Equations 8 Thermodynamics 9 Rate Equations 10 Kp (Equilibrium Constant) 11 Electrode Potentials & Cells 12 Acids and Bases 13 Periodicity 14 Group 2: Alkaline Earth Metals 15 Group 7: The Halogens 16 Period 3 Elements & Oxides 17 Transition Metals 18 Reactions of Ions in Aqueous Solution 19 Intro to Organic Chemistry 20 Alkanes 21 Halogenoalkanes 22 Alkenes 23 Alcohols 24 Organic Analysis 25 Optical Isomerism 26 Aldehydes & Ketones 27 Carboxylic Acids & Derivatives 28 Aromatic Chemistry 29 Amines 30 Polymers 31 Amino Acids, Proteins & DNA 32 Organic Synthesis 33 NMR Spectroscopy 34 Chromatography RP1–RP12 Required Practicals

1.12 Acids and Bases (A-level only)

1.12.1 Bronsted-Lowry Acid-Base Equilibria 1.12.2 Definition and Determination of pH 1.12.3 The Ionic Product of Water, Kw 1.12.4 Weak Acids and Bases Ka for Weak Acids 1.12.5 pH Curves, Titrations and Indicators 1.12.6 Buffer Action

Weak Acids and Bases – Ka for Weak Acids

Specification Reference Physical Chemistry, Acids and bases 3.1.12.4

Quick Notes

  • Weak acids partially dissociate in solution:
  • AQA A-Level Chemistry diagram showing a weak acid HA partially dissociating into H+ and A−
  • Acid dissociation constant (Ka):
  • AQA A-Level Chemistry expression for Ka for a weak acid
  • pKa is related to Ka:
  • AQA A-Level Chemistry definition of pKa in terms of Ka
  • Stronger weak acids have a larger Ka and smaller pKa.
  • To calculate pH of a weak acid:
    • Use Ka = ([H+]2) / [HA]
    • Find [H+] = √(Ka × [HA])
    • Use pH = -log[H+]

Full Notes

Ka has been covered in more detail here.
This page is just what you need to know for AQA A-level Chemistry :)

Weak Acids and Their Partial Dissociation

A weak acid (HA) only partially dissociates in water:

AQA A-Level Chemistry diagram showing a weak acid HA partially dissociating into H+ and A−

Unlike strong acids (which fully dissociate), weak acids establish an equilibrium in solution between HA, H+ and A-.

The position of this equilibrium is based on the strength of the acid - the stronger the acid, the more the position lies to the right (more dissociation), meaning a higher concentration of H+ and A ions in the mixture.

Example ethanoic acid (CH3COOH):

AQA A-Level Chemistry example showing ethanoic acid CH3COOH establishing equilibrium with H+ and CH3COO−

CH3COOH ⇌ H+ + CH3COO
where HA = CH3COOH and A = CH3COO

Acid Dissociation Constant (Ka)

Like with any equilibrium system, the position of equilibrium for the dissociation of a weak acid can be described using an equilibrium constant.

Ka is the equilibrium constant for acid dissociation.

Expression for Ka:

AQA A-Level Chemistry expression for Ka for a weak acid

Example For ethanoic acid (CH3COOH): CH3COOH ⇌ CH3COO + H+
Ka = 1.75 × 10−5 mol dm−3 (shows weak dissociation).

Relationship Between Ka and pKa

pKa is a logarithmic measure of Ka:

AQA A-Level Chemistry definition of pKa in terms of Ka

To convert pKa to Ka, use Ka = 10−pKa

AQA A-Level Chemistry comparison graphic linking larger Ka to smaller pKa and vice versa

Example
Ethanoic acid: Ka = 1.75 × 10−5, so pKa = 4.76.
Carbonic acid: Ka = 4.3 × 10−7, so pKa = 6.37.
Since pKa of ethanoic acid is lower, it is stronger than carbonic acid.

Calculating pH of a Weak Acid

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

AQA A-Level Chemistry relation showing Ka used to calculate [H+] for a weak acid

Steps to Calculate pH of a Weak Acid:

AQA A-Level Chemistry step-by-step flow to obtain pH from Ka and initial [HA]
Photo of Matt
Matt’s exam tip

Don’t forget that when using Ka to find the pH of a weak acid, we are assuming two things - that [H+] = [A-] and that the [HA] at the start is the same as the [HA] 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)
  3. [H+] = 1.32 × 10−3 mol dm−3
  4. pH = -log(1.32 × 10−3)
  5. pH = 2.88

Summary

Concept Explanation
Ka Definition Ka = ([H+][A]) / [HA]
pKa Calculation pKa = -log(Ka)
Stronger Weak Acids Large Ka, Small pKa
Weaker Weak Acids Small Ka, Large pKa
pH of Weak Acid Use [H+] = √(Ka × [HA]), then pH = -log[H+]