Acid-base equilibria

Q: What are Brønsted-Lowry acids, Brønsted-Lowry bases, and conjugate acid-base pairs?

A Brønsted-Lowry acid is a proton donor and a base is a proton acceptor; a conjugate acid-base pair consists of two species that differ by exactly one single \( \text{H}^+ \) ion.

Q: How is \( \text{pH} \) defined and calculated for strong monoprotic and diprotic acids?

\( \text{pH} = -\log_{10}[\text{H}^+] \). For a strong monoprotic acid, \( [\text{H}^+] = [\text{Acid}] \); for a strong diprotic acid, \( [\text{H}^+] = 2 \times [\text{Acid}] \).

Q: What is the ionic product of water (\( K_w \)) and how does temperature affect it?

\( K_w = [\text{H}^+][\text{OH}^-] \) (which equals \( 1.00 \times 10^{-14}\,\text{mol}^2\text{dm}^{-6} \) at \( 298\,\text{K} \)); because water autoionisation is endothermic, increasing temperature increases \( K_w \) and lowers the \( \text{pH} \) of neutral water.

Q: How is the \( \text{pH} \) of a strong base calculated?

Find \( [\text{OH}^-] \) from complete dissociation, determine \( [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} \), and calculate \( \text{pH} = -\log_{10}[\text{H}^+] \).

Q: What is the acid dissociation constant (\( K_a \)) and what approximations are used to find the \( \text{pH} \) of a weak acid?

\( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \); assuming \( [\text{H}^+] \approx [\text{A}^-] \) and \( [\text{HA}]_{\text{eq}} \approx [\text{HA}]_{\text{initial}} \), \( [\text{H}^+] = \sqrt{K_a \times [\text{HA}]} \).

Q: What is an acidic buffer solution and how does it resist changes in \( \text{pH} \)?

An acidic buffer contains large reservoirs of a weak acid and its conjugate base; added \( \text{H}^+ \) reacts with the conjugate base salt and added \( \text{OH}^- \) reacts with the weak acid.

Q: What are the key features and equivalence point \( \text{pH} \) values for different types of acid-base titration curves?

Strong acid–strong base has \( \text{pH} = 7 \) at equivalence; weak acid–strong base has \( \text{pH} > 7 \); strong acid–weak base has \( \text{pH}

Q: How do acid-base indicators work and how do you choose a suitable indicator for a titration?

An indicator is a weak acid whose \( \text{HIn} \) and \( \text{In}^- \) forms have different colours; it is suitable only if its entire \( \text{pH} \) transition range falls completely within the vertical section of the titration curve.

Chemistry 1110 · CCEA A Level · 11 min read · Updated: Aug 31, 2026

Introduction to Acid-Base Equilibria

Welcome to Acid-Base Equilibria! In your AS studies, you learned that acids produce hydrogen ions and bases neutralise them. At A2 level, we dive deeper into the mathematical and chemical principles that govern acid-base behaviour in solution. From the regulation of human blood \( \text{pH} \) to calculating the exact acidity of industrial mixtures, understanding these equilibria is a core skill for every chemist. Don't worry if the calculations look intimidating at first—we will break down every single concept into clear, step-by-step methods!

1. Brønsted-Lowry Theory & Conjugate Pairs

Proton Donors and Acceptors

According to the Brønsted-Lowry theory:
• An acid is a proton donor (it gives away an \( \text{H}^+ \) ion).
• A base is a proton acceptor (it takes in an \( \text{H}^+ \) ion).

Conjugate Acid-Base Pairs

When an acid loses a proton, the species formed can potentially regain that proton—meaning it acts as a base! This related pair is called a conjugate acid-base pair. A conjugate pair always differs by exactly one single \( \text{H}^+ \) ion.

Consider the reaction between ethanoic acid and water:
\( \text{CH}_3\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)} \)
• \( \text{CH}_3\text{COOH} \) is the acid; its conjugate base is \( \text{CH}_3\text{COO}^- \).
• \( \text{H}_2\text{O} \) acts as a base; its conjugate acid is \( \text{H}_3\text{O}^+ \) (the hydroxonium ion, often simplified as \( \text{H}^+ \)).

Memory Aid: Remember the acronym BAD: Bases Accept, Donors are Acids.

Key Takeaway: An acid loses \( \text{H}^+ \) to become its conjugate base; a base gains \( \text{H}^+ \) to become its conjugate acid.

2. The \( \text{pH} \) Scale and Strong Acids

Understanding \( \text{pH} \)

The hydrogen ion concentration, \( [\text{H}^+] \), in solutions can vary across huge orders of magnitude (from over \( 1\,\text{mol dm}^{-3} \) down to \( 10^{-14}\,\text{mol dm}^{-3} \)). To make these numbers easier to handle, the Danish chemist Søren Sørensen introduced the logarithmic \( \text{pH} \) scale.

The fundamental definitions are:
\( \text{pH} = -\log_{10}[\text{H}^+] \)
\( [\text{H}^+] = 10^{-\text{pH}} \)

Calculating the \( \text{pH} \) of Strong Monoprotic and Diprotic Acids

A strong acid is fully dissociated (completely ionised) in aqueous solution.

Example 1: Strong Monoprotic Acid (\( \text{HCl} \), \( \text{HNO}_3 \))
For \( 0.050\,\text{mol dm}^{-3} \) \( \text{HCl} \):
Because \( \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \), the concentration of \( [\text{H}^+] = 0.050\,\text{mol dm}^{-3} \).
\( \text{pH} = -\log_{10}(0.050) = 1.30 \)

Example 2: Strong Diprotic Acid (\( \text{H}_2\text{SO}_4 \))
For \( 0.025\,\text{mol dm}^{-3} \) \( \text{H}_2\text{SO}_4 \):
Assuming complete dissociation: \( \text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-} \)
\( [\text{H}^+] = 2 \times 0.025 = 0.050\,\text{mol dm}^{-3} \)
\( \text{pH} = -\log_{10}(0.050) = 1.30 \)

Common Mistake to Avoid: Always check if your acid is diprotic (like sulfuric acid). Forgetting to multiply the acid concentration by 2 for diprotic acids is an easy way to lose marks!

Key Takeaway: For strong monoprotic acids, \( [\text{H}^+] = [\text{Acid}] \). Always write \( \text{pH} \) values to 2 decimal places.

3. The Ionic Product of Water (\( K_w \)) & Strong Bases

The Autoionisation of Water

Water conducts electricity very slightly because it undergoes self-ionisation (autoionisation):
\( \text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \quad \Delta H > 0 \text{ (endothermic)} \)

The equilibrium constant expression for this process is:
\( K_c = \frac{[\text{H}^+][\text{OH}^-]}{[\text{H}_2\text{O}]} \)

Because the degree of ionisation is extremely small, \( [\text{H}_2\text{O}] \) is essentially constant. Multiplying \( K_c \) by \( [\text{H}_2\text{O}] \) gives a new constant, the ionic product of water (\( K_w \)):
\( K_w = [\text{H}^+][\text{OH}^-] \)

At \( 298\,\text{K} \) (\( 25\,^\circ\text{C} \)), the standard value is:
\( K_w = 1.00 \times 10^{-14}\,\text{mol}^2\text{dm}^{-6} \)

Effect of Temperature on \( K_w \) and Water Neutrality

Because the dissociation of water is endothermic (\( \Delta H \text{ is positive} \)), increasing temperature shifts the equilibrium to the right (by Le Chatelier's principle).
• Higher temperature \( \implies K_w \) increases \( \implies [\text{H}^+] \) increases \( \implies \text{pH} \) decreases.
• Did you know? Pure water at \( 50\,^\circ\text{C} \) has a \( \text{pH} \) of approximately \( 6.63 \), yet it is still completely neutral because \( [\text{H}^+] = [\text{OH}^-] \)! A solution is only neutral when \( [\text{H}^+] = [\text{OH}^-] \), regardless of whether \( \text{pH} = 7 \).

Calculating the \( \text{pH} \) of Strong Bases

Strong bases (such as \( \text{NaOH} \), \( \text{KOH} \), or \( \text{Ba(OH)}_2 \)) completely dissociate in water.

Step-by-Step Calculation:
Find the \( \text{pH} \) of \( 0.040\,\text{mol dm}^{-3} \) \( \text{NaOH} \) at \( 298\,\text{K} \):
1. Determine \( [\text{OH}^-] \): \( [\text{OH}^-] = 0.040\,\text{mol dm}^{-3} \)
2. Use \( K_w \) to calculate \( [\text{H}^+] \):
\( [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.00 \times 10^{-14}}{0.040} = 2.50 \times 10^{-13}\,\text{mol dm}^{-3} \)
3. Calculate \( \text{pH} \):
\( \text{pH} = -\log_{10}(2.50 \times 10^{-13}) = 12.60 \)

Key Takeaway: For strong bases, find \( [\text{OH}^-] \), use \( [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} \), and take \( -\log_{10}[\text{H}^+] \).

4. Weak Acids and the Acid Dissociation Constant (\( K_a \))

Weak Acids

A weak acid only partially dissociates in aqueous solution:
\( \text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{A}^-\text{(aq)} \)

The equilibrium expression is given by the acid dissociation constant (\( K_a \)):
\( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \quad (\text{units: }\text{mol dm}^{-3}) \)

We also use \( \text{p}K_a \), where:
\( \text{p}K_a = -\log_{10} K_a \quad \text{and} \quad K_a = 10^{-\text{p}K_a} \)
• Larger \( K_a \) (or smaller \( \text{p}K_a \)) means a stronger weak acid.

Approximations for Calculating the \( \text{pH} \) of a Weak Acid

To calculate the \( \text{pH} \) of a weak acid from its initial concentration \( c \), we make two key assumptions:
1. \( [\text{H}^+] \approx [\text{A}^-] \) (assuming dissociation of water provides negligible \( [\text{H}^+] \)).
2. \( [\text{HA}]_{\text{equilibrium}} \approx [\text{HA}]_{\text{initial}} \) (assuming the degree of dissociation is so small that the initial concentration remains virtually unchanged).

Using these approximations:
\( K_a = \frac{[\text{H}^+]^2}{[\text{HA}]} \implies [\text{H}^+] = \sqrt{K_a \times [\text{HA}]} \)

Worked Example:
Calculate the \( \text{pH} \) of a \( 0.100\,\text{mol dm}^{-3} \) solution of ethanoic acid (\( K_a = 1.74 \times 10^{-5}\,\text{mol dm}^{-3} \)).
1. \( [\text{H}^+] = \sqrt{1.74 \times 10^{-5} \times 0.100} = \sqrt{1.74 \times 10^{-6}} = 1.319 \times 10^{-3}\,\text{mol dm}^{-3} \)
2. \( \text{pH} = -\log_{10}(1.319 \times 10^{-3}) = 2.88 \)

Key Takeaway: For a single weak acid solution, use \( [\text{H}^+] = \sqrt{K_a \times [\text{HA}]} \) then calculate \( \text{pH} = -\log_{10}[\text{H}^+] \).

5. Buffer Solutions

What is a Buffer Solution?

A buffer solution is a system that resists changes in \( \text{pH} \) when small amounts of acid or base are added, or when diluted.

An acidic buffer is made from a weak acid and its conjugate base salt (e.g., ethanoic acid, \( \text{CH}_3\text{COOH} \), and sodium ethanoate, \( \text{CH}_3\text{COONa} \)).

How an Acidic Buffer Works

In solution, two main species are present in large quantities:
1. Undissociated weak acid: \( \text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)} \) (large reservoir of \( \text{CH}_3\text{COOH} \))
2. Ethanoate ions from the fully dissolved salt: \( \text{CH}_3\text{COONa(aq)} \rightarrow \text{CH}_3\text{COO}^-\text{(aq)} + \text{Na}^+\text{(aq)} \) (large reservoir of \( \text{CH}_3\text{COO}^- \))

• When small amounts of \( \text{H}^+ \) are added:
The added \( \text{H}^+ \) reacts with the reservoir of conjugate base:
\( \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{CH}_3\text{COOH(aq)} \)
The added \( \text{H}^+ \) ions are removed, keeping the \( \text{pH} \) virtually constant.

• When small amounts of \( \text{OH}^- \) are added:
The added \( \text{OH}^- \) reacts with the weak acid:
\( \text{CH}_3\text{COOH(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_2\text{O(l)} \)
The added \( \text{OH}^- \) ions are removed, keeping the \( \text{pH} \) virtually constant.

Calculating the \( \text{pH} \) of an Acidic Buffer

Rearranging the \( K_a \) expression gives:
\( [\text{H}^+] = K_a \times \frac{[\text{Acid}]}{[\text{Salt}]} \)

Worked Example:
Calculate the \( \text{pH} \) of a buffer containing \( 0.200\,\text{mol dm}^{-3} \) \( \text{CH}_3\text{COOH} \) and \( 0.100\,\text{mol dm}^{-3} \) \( \text{CH}_3\text{COONa} \) (\( K_a = 1.74 \times 10^{-5}\,\text{mol dm}^{-3} \)).
1. \( [\text{H}^+] = 1.74 \times 10^{-5} \times \frac{0.200}{0.100} = 3.48 \times 10^{-5}\,\text{mol dm}^{-3} \)
2. \( \text{pH} = -\log_{10}(3.48 \times 10^{-5}) = 4.46 \)

Special Case: Half-Neutralisation Point
When \( [\text{Acid}] = [\text{Salt}] \), \( \frac{[\text{Acid}]}{[\text{Salt}]} = 1 \).
Therefore, \( [\text{H}^+] = K_a \), which means:
\( \mathbf{\text{pH} = \text{p}K_a} \)

Key Takeaway: A buffer needs reservoirs of both weak acid and conjugate base. Use \( [\text{H}^+] = K_a \times \frac{[\text{Acid}]}{[\text{Salt}]} \) to calculate buffer \( \text{pH} \).

6. Titration Curves (\( \text{pH} \) Curves)

A titration curve plots the \( \text{pH} \) of the mixture against the volume of standard solution added from a burette. The shape of the curve depends entirely on the combination of strong and weak acids and bases.

Key Features of a \( \text{pH} \) Curve:

• Initial \( \text{pH} \): Starts low for strong acids (\(\sim 1\)), higher for weak acids (\(\sim 3\)).
• Buffering region (weak acid only): A gentle slope before the equivalence point where acid and conjugate salt coexist.
• Equivalence point: The exact point where chemically equivalent amounts of acid and base have reacted.
• Vertical section: A rapid change in \( \text{pH} \) around the equivalence point.
• Final \( \text{pH} \): Ends high for strong bases (\(\sim 13\)), lower for weak bases (\(\sim 10\)).

The Four Combinations:

1. Strong Acid – Strong Base (e.g., \( \text{HCl} \) and \( \text{NaOH} \))
• Initial \( \text{pH} \approx 1 \), final \( \text{pH} \approx 13 \)
• Long, steep vertical section from \( \text{pH} \approx 3 \text{ to } 10 \)
• Equivalence point at \( \text{pH} = 7 \)

2. Weak Acid – Strong Base (e.g., \( \text{CH}_3\text{COOH} \) and \( \text{NaOH} \))
• Initial \( \text{pH} \approx 3 \), final \( \text{pH} \approx 13 \)
• Vertical section is shorter, typically from \( \text{pH} \approx 7 \text{ to } 10 \)
• Equivalence point is alkaline (\( \text{pH} > 7 \)) because the conjugate base (\( \text{CH}_3\text{COO}^- \)) hydrolyses with water to form \( \text{OH}^- \).
• At half the equivalence volume: \( \text{pH} = \text{p}K_a \).

3. Strong Acid – Weak Base (e.g., \( \text{HCl} \) and \( \text{NH}_3 \))
• Initial \( \text{pH} \approx 1 \), final \( \text{pH} \approx 9 - 10 \)
• Vertical section from \( \text{pH} \approx 3 \text{ to } 7 \)
• Equivalence point is acidic (\( \text{pH} < 7 \)) due to \( \text{NH}_4^+ \) hydrolysis.

4. Weak Acid – Weak Base (e.g., \( \text{CH}_3\text{COOH} \) and \( \text{NH}_3 \))
• No vertical section at all! Only a point of inflection.
• Important: Traditional indicators cannot be used for this titration because there is no sudden \( \text{pH} \) jump.

Key Takeaway: The midpoint of the vertical section marks the equivalence point. Strong-strong neutralises at \( \text{pH } 7 \), weak acid-strong base at \( \text{pH} > 7 \), and strong acid-weak base at \( \text{pH} < 7 \).

7. Acid-Base Indicators

How Indicators Work

An indicator is itself a weak acid where the undissociated molecule (\( \text{HIn} \)) has a completely different colour from its conjugate base (\( \text{In}^- \)):
\( \text{HIn(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{In}^-\text{(aq)} \)
• In acidic conditions (high \( [\text{H}^+] \)): equilibrium shifts left \( \implies \) Colour of \( \text{HIn} \) dominates.
• In alkaline conditions (low \( [\text{H}^+] \)): equilibrium shifts right \( \implies \) Colour of \( \text{In}^- \) dominates.

Choosing the Correct Indicator

The indicator's colour change takes place over a range: \( \text{pH} = \text{p}K_{\text{In}} \pm 1 \).
Rule: An indicator is suitable for a titration only if its entire \( \text{pH} \) range falls completely within the vertical section of the titration curve.

• Phenolphthalein (\( \text{pH} \) range: \( 8.3 - 10.0 \), colourless in acid \( \rightarrow \) pink in alkali):
Suitable for Strong Acid – Strong Base and Weak Acid – Strong Base titrations.

• Methyl Orange (\( \text{pH} \) range: \( 3.1 - 4.4 \), red in acid \( \rightarrow \) yellow in alkali):
Suitable for Strong Acid – Strong Base and Strong Acid – Weak Base titrations.

Key Takeaway: Always match the indicator range to the steep vertical jump on the \( \text{pH} \) curve. Neither indicator works for weak acid–weak base titrations.

Summary of Core Mathematical Formulas

Keep this handy summary as a quick revision checklist before exams:
• \( \text{pH} = -\log_{10}[\text{H}^+] \)
• \( [\text{H}^+] = 10^{-\text{pH}} \)
• \( K_w = [\text{H}^+][\text{OH}^-] = 1.00 \times 10^{-14}\,\text{mol}^2\text{dm}^{-6} \) (at \( 298\,\text{K} \))
• \( [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} \) (for strong bases)
• \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \)
• \( [\text{H}^+] = \sqrt{K_a \times [\text{HA}]} \) (for weak acids)
• \( \text{p}K_a = -\log_{10} K_a \)
• \( [\text{H}^+] = K_a \times \frac{[\text{Acid}]}{[\text{Salt}]} \) (for buffer solutions)
• \( \text{pH} = \text{p}K_a \) (at half-neutralisation point)

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What are Brønsted-Lowry acids, Brønsted-Lowry bases, and conjugate acid-base pairs?

A Brønsted-Lowry acid is a proton donor and a base is a proton acceptor; a conjugate acid-base pair consists of two species that differ by exactly one single \( \text{H}^+ \) ion.

How is \( \text{pH} \) defined and calculated for strong monoprotic and diprotic acids?

\( \text{pH} = -\log_{10}[\text{H}^+] \). For a strong monoprotic acid, \( [\text{H}^+] = [\text{Acid}] \); for a strong diprotic acid, \( [\text{H}^+] = 2 \times [\text{Acid}] \).

What is the ionic product of water (\( K_w \)) and how does temperature affect it?

\( K_w = [\text{H}^+][\text{OH}^-] \) (which equals \( 1.00 \times 10^{-14}\,\text{mol}^2\text{dm}^{-6} \) at \( 298\,\text{K} \)); because water autoionisation is endothermic, increasing temperature increases \( K_w \) and lowers the \( \text{pH} \) of neutral water.

How is the \( \text{pH} \) of a strong base calculated?

Find \( [\text{OH}^-] \) from complete dissociation, determine \( [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} \), and calculate \( \text{pH} = -\log_{10}[\text{H}^+] \).

What is the acid dissociation constant (\( K_a \)) and what approximations are used to find the \( \text{pH} \) of a weak acid?

\( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \); assuming \( [\text{H}^+] \approx [\text{A}^-] \) and \( [\text{HA}]_{\text{eq}} \approx [\text{HA}]_{\text{initial}} \), \( [\text{H}^+] = \sqrt{K_a \times [\text{HA}]} \).

What is an acidic buffer solution and how does it resist changes in \( \text{pH} \)?

An acidic buffer contains large reservoirs of a weak acid and its conjugate base; added \( \text{H}^+ \) reacts with the conjugate base salt and added \( \text{OH}^- \) reacts with the weak acid.

How is the \( \text{pH} \) of an acidic buffer calculated, and what is true at the half-neutralisation point?

\( [\text{H}^+] = K_a \times \frac{[\text{Acid}]}{[\text{Salt}]} \); at the half-neutralisation point, \( [\text{Acid}] = [\text{Salt}] \), which simplifies to \( \text{pH} = \text{p}K_a \).

What are the key features and equivalence point \( \text{pH} \) values for different types of acid-base titration curves?

Strong acid–strong base has \( \text{pH} = 7 \) at equivalence; weak acid–strong base has \( \text{pH} > 7 \); strong acid–weak base has \( \text{pH} < 7 \); weak acid–weak base has no vertical section.

How do acid-base indicators work and how do you choose a suitable indicator for a titration?

An indicator is a weak acid whose \( \text{HIn} \) and \( \text{In}^- \) forms have different colours; it is suitable only if its entire \( \text{pH} \) transition range falls completely within the vertical section of the titration curve.

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