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Buffer Solutions

Learn and apply Buffer Solutions in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Buffer Solutions: Orientation

Buffers are an equilibrium-and-stoichiometry hybrid: neutralise the added strong acid/base first (using moles), then use the new [A⁻]/[HA] (or [B]/[BH⁺]) ratio to find the pH. This lesson teaches both the buffer action explanation and the calculation workflow.

If the acid-base foundations feel rusty, revisit Acids and Bases (Theories) and keep the Aqueous Equilibria hub open for linked methods.

Definitions (Must Know)

A. Buffer solution

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added.

B. Acidic buffer

An acidic buffer contains a weak acid and its conjugate base (e.g. CH₃COOH/CH₃COO⁻).

C. Basic buffer

A basic buffer contains a weak base and its conjugate acid (e.g. NH₃/NH₄ +).

Detailed Explanations

A. Buffer action: what happens when acid/base is added

For an acidic buffer HA/A⁻:

  • Adding acid: A-(aq) + H + (aq) → HA(aq)
  • Adding base: HA(aq) + OH-(aq) → A-(aq) + H₂O(l)

The buffer contains a “sink” for both H⁺ and OH⁻, so small additions mainly change the ratio [A⁻]/[HA] rather than causing a large change in [H⁺].

B. Where the Henderson–Hasselbalch form comes from (acidic buffer)

Start from:

Kₐ = [H⁺][A⁻]/[HA]

Rearrange:

[H⁺] = Kₐ([HA]/[A⁻])

Take - log ₁₀:

pH = pKₐ + log ₁₀([A⁻]/[HA])

C. Workflow: buffer pH after adding acid/base (exam method)

  1. Identify the conjugate pair (HA/A⁻ or BH + /B).
  2. Do the neutralisation first using moles (strong acid/base reacts completely).
  3. Convert updated moles to concentrations (or use the mole ratio if volume cancels).
  4. Use the Henderson–Hasselbalch form to find pH (or pOH for a basic buffer).

Stop before step 4 if the added strong acid or base consumes essentially all of one buffer component. For example, if 0.0100 mol A⁻ is treated with 0.0120 mol H⁺, excess strong acid remains and the mixture is no longer a buffer; calculate pH from that excess instead of inserting a zero or negative buffer amount into Henderson–Hasselbalch.

Mini example: If n(A⁻) drops from 0.0100 to 0.00900 after adding acid, the ratio [A⁻]/[HA] decreases, so the log term decreases and pH drops slightly.

D. Basic buffers (two equivalent ways)

For B/BH⁺, you can use:

  • pOH = pK_b + log ₁₀([BH⁺]/[B]), then pH = pK_w-pOH, or
  • pH = pKₐ + log ₁₀([B]/[BH⁺]) using pKₐ of BH⁺.

E. The ocean carbonate buffer and acidification

Dissolved carbon dioxide takes part in coupled equilibria:

CO₂(aq) + H₂O(l) ⇌ H₂CO₃(aq) H₂CO₃(aq) ⇌ H + (aq) + HCO₃-(aq) HCO₃-(aq) ⇌ H + (aq) + CO₃²⁻(aq)

The hydrogencarbonate/carbonate system can remove a small addition of acid because CO₃²⁻ accepts H⁺ to form HCO₃-. It can also respond to added base through proton donation by the acidic members of the system. This resistance is limited: a buffer reduces a pH change; it does not prevent one completely.

When more atmospheric CO₂ dissolves, the equilibria produce more H⁺. Ocean pH therefore falls, while [CO₃²⁻] falls as carbonate is converted to hydrogencarbonate. This is ocean acidification even though seawater remains alkaline: “acidification” means its pH is decreasing, not necessarily that pH has fallen below 7.

Worked Examples

Modelled example 1

Calculate an acidic buffer pH

Core

Problem

A buffer contains 0.200 mol dm⁻³ CH₃COOH and 0.100 mol dm⁻³ CH₃COO⁻. Given pKₐ = 4.76, find the pH.
Study the worked solution

  1. Choose the buffer relationship

    Method

    Use the acidic-buffer form of Henderson–Hasselbalch.

    Reason

    Both the weak acid and its conjugate base are present in appreciable concentrations.

    Working

    pH = pKₐ + log ₁₀([A⁻]/[HA]).

  2. Substitute the concentration ratio

    Method

    Place conjugate base over weak acid.

    Reason

    The formula uses [A⁻]/[HA], and the numerator is half the denominator.

    Working

    pH = 4.76 + log ₁₀(0.100/0.200) = 4.76 + log ₁₀(0.500).

  3. Evaluate and check

    Method

    Calculate the logarithm and compare with pKₐ.

    Reason

    A base-to-acid ratio below one must make pH lower than pKₐ.

    Working

    pH = 4.76-0.301 = 4.46, which is below 4.76 as expected.

Guided practice 2

Update a buffer after adding strong acid

About 8 min

Problem

100 cm³ of the buffer in Example 1 is treated with 1.00 × 10⁻³ mol of HCl. Find the new pH, assuming negligible volume change.

Try this before viewing the solution

Hints

Hint 1: react before equilibrium
Added H⁺ reacts completely with the conjugate-base member of the buffer.
Hint 2: update both amounts
Subtract 0.00100 mol from A⁻ and add the same amount to HA before using the buffer ratio.
View solution step by step

  1. Write the neutralisation

    Method

    React the added acid with ethanoate.

    Reason

    CH₃COO⁻ is the buffer component that removes added H⁺.

    Working

    CH₃COO⁻ + H⁺ → CH₃COOH.

  2. Find the initial amounts

    Method

    Convert the two concentrations in 0.100 dm³ to moles.

    Reason

    Strong-acid neutralisation must be completed using amounts before an equilibrium expression is applied.

    Working

    n(A⁻) = 0.0100 mol and n(HA) = 0.0200 mol.

  3. Update the conjugate pair

    Method

    Apply the one-to-one reaction change.

    Reason

    Every mole of added H⁺ consumes one mole of A⁻ and forms one mole of HA.

    Working

    n(A⁻) = 0.00900 mol and n(HA) = 0.0210 mol.

  4. Calculate the new pH

    Method

    Use the updated mole ratio; the common volume cancels.

    Reason

    Both components occupy the same final solution volume.

    Working

    pH = 4.76 + log ₁₀(0.00900/0.0210) = 4.39.

Common misconception 3

Update both buffer components after adding base

Find and correct the mistake

Learner claim

100 cm³ of the buffer in Example 1 receives 2.00 × 10⁻³ mol of NaOH. A learner reuses the initial 0.100/0.200 ratio and reports pH 4.46 because the volume change is negligible. Diagnose and correct the calculation.

Choose the first correction

Before applying Henderson–Hasselbalch

View solution step by step

  1. Locate the first error

    Method

    Reject the unchanged-ratio assumption.

    Reason

    Negligible volume change does not mean negligible chemical reaction.

    Working

    CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.

  2. Update the amounts

    Method

    Subtract hydroxide from HA and add the same amount to A⁻.

    Reason

    The neutralisation has one-to-one stoichiometry.

    Working

    n(HA) = 0.0200-0.00200 = 0.0180 mol; n(A⁻) = 0.0100 + 0.00200 = 0.0120 mol.

  3. Recalculate and check

    Method

    Use the new base-to-acid ratio.

    Reason

    Added base raises the ratio, so the pH must rise slightly.

    Working

    pH = 4.76 + log ₁₀(0.0120/0.0180) = 4.58.

Examiner practice 4

Calculate the pH of a basic buffer

4 marks

Problem

A buffer contains 0.250 mol dm⁻³ NH₃ and 0.150 mol dm⁻³ NH₄ +. Given pK_b(NH₃) = 4.74 and pK_w = 14.00 at 25°C, find the pH. [4 marks]

Try this before viewing the solution

View solution step by step

  1. State the basic-buffer form

    1 mark

    Method

    Use the base/conjugate-acid relationship.

    Reason

    The pair is NH₃/NH₄ + and the supplied constant is pK_b.

    Working

    pOH = pK_b + log ₁₀([BH⁺]/[B]).

  2. Substitute the ratio

    1 mark

    Reason

    The conjugate acid is in the numerator of this pOH form.

    Working

    pOH = 4.74 + log ₁₀(0.150/0.250).

  3. Find pOH

    1 mark

    Reason

    log ₁₀(0.600) = -0.222.

    Working

    pOH = 4.52.

  4. Convert to pH

    1 mark

    Method

    Subtract pOH from the stated pK_w.

    Reason

    At this temperature, pH + pOH = pK_w = 14.00.

    Working

    pH = 14.00-4.52 = 9.48.

Challenge 5

Form a buffer by partial neutralisation

Minimal support

Problem

200 cm³ of 0.100 mol dm⁻³ ethanoic acid is mixed with 50.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide. Given pKₐ = 4.76, calculate the final pH.

Try this before viewing the solution

Hints

Hint 1: identify the route
The mixture does not begin as a buffer; first decide whether the strong base partially or completely neutralises the weak acid.
Hint 2: build the pair
The hydroxide consumes 0.00500 mol of HA and forms the same amount of A⁻.
View solution step by step

  1. Calculate initial reacting amounts

    Method

    Convert both solutions to moles.

    Reason

    Neutralisation determines which species remain before equilibrium is considered.

    Working

    n(HA) = 0.200(0.100) = 0.0200 mol; n(OH⁻) = 0.0500(0.100) = 0.00500 mol.

  2. Form the conjugate pair

    Method

    Apply HA + OH⁻ → A⁻ + H₂O.

    Reason

    Hydroxide is limiting, so only part of the weak acid is converted.

    Working

    After reaction: n(HA) = 0.0150 mol and n(A⁻) = 0.00500 mol.

  3. Recognise the buffer

    Method

    Use Henderson–Hasselbalch with the mole ratio.

    Reason

    Both conjugate partners remain, and their common final volume cancels.

    Working

    [A⁻]/[HA] = 0.00500/0.0150 = 1/3.

  4. Calculate and check

    Method

    Evaluate the logarithmic ratio.

    Reason

    The ratio below one predicts a pH below pKₐ.

    Working

    pH = 4.76 + log ₁₀(1/3) = 4.28.

Challenge 6

Explain ocean acidification without saying the ocean becomes acidic

Minimal support

Problem

More atmospheric carbon dioxide dissolves in seawater. Explain why seawater pH falls, why carbonate-ion concentration falls, and why the phrase “ocean acidification” does not necessarily mean that seawater has pH below 7.

Try this before viewing the solution

Hints

Hint 1: first equilibrium
Start by showing how dissolved carbon dioxide can increase hydrogen-ion concentration.
Hint 2: carbonate response
Then ask what the extra hydrogen ions do to carbonate ions.
View solution step by step

  1. Link dissolved carbon dioxide to pH

    Method

    Show that more dissolved carbon dioxide shifts the carbonate system towards hydrogen ions and hydrogencarbonate ions.

    Reason

    The extra dissolved carbon dioxide supplies more weak acid to the equilibria.

    Working

    CO₂(aq) + H₂O(l) ⇌ H + (aq) + HCO₃-(aq); [H⁺] rises, so pH falls.

  2. Track the carbonate ions

    Method

    React the additional hydrogen ions with carbonate ions.

    Reason

    The carbonate/hydrogencarbonate pair helps resist the pH change by consuming some added H⁺.

    Working

    H + (aq) + CO₃²⁻(aq) ⇌ HCO₃-(aq); therefore [CO₃²⁻] falls.

  3. Interpret acidification accurately

    Method

    Describe a change in pH, not a compulsory crossing of pH 7.

    Reason

    “Acidification” means movement towards a lower pH. Seawater can remain alkaline while becoming less alkaline.

    Working

    Initial pH > 7 and lower final pH > 7 is still ocean acidification.

  4. Connect to carbonate availability

    Method

    State that less carbonate is available for organisms that form calcium carbonate structures.

    Reason

    The chemical effect follows from the measured fall in [CO₃²⁻], not simply from using the word “acid”.

    Working

    Ca²⁺(aq) + CO₃²⁻(aq) ⇌ CaCO₃(s).

Mind Stretchers

Mind stretcher 1Extension

You need a buffer at pH 5.00. You have three weak acids with pKₐ = 3.00, pKₐ = 4.76, and pKₐ = 6.10. Which acid is the best choice, and what ratio [A⁻]/[HA] is needed?

Show Answer

Mark scheme:

  • Choose pKₐ closest to the target pH → pick pKₐ = 4.76.
  • pH = pKₐ + log([A⁻]/[HA])
  • log([A⁻]/[HA]) = 5.00-4.76 = 0.24
  • [A⁻]/[HA] = 10^(0.24) = 1.74