Buffers (AQA A Level Chemistry): Revision Note

Exam code: 7405

Stewart Hird

Written by: Stewart Hird

Reviewed by: Caroline Carroll

Updated on

Acid & Basic Buffers

  • A buffer solution is a solution that resists changes in pH when small amounts of acids or alkalis are added

    • A buffer solution is used to keep the pH almost constant

    • A buffer can consist of a weak acid - conjugate base or a weak base - conjugate acid

Ethanoic acid and sodium ethanoate as a buffer

  • A common acidic buffer solution is an aqueous mixture of ethanoic acid and sodium ethanoate

  • Ethanoic acid is a weak acid and partially ionises in solution to form a relatively low concentration of ethanoate ions

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

  • Sodium ethanoate is a salt that fully ionises in solution

CH3COONa (s) → Na+ (aq) + CH3COO- (aq)

  • There are reserve supplies of the acid (CH3COOH) and its conjugate base (CH3COO-)

    • The buffer solution contains relatively high concentrations of CH3COOH (due to partial ionisation of ethanoic acid) and CH3COO- (due to full ionisation of sodium ethanoate)

  • In the buffer solution, the ethanoic acid is in equilibrium with hydrogen and ethanoate ions

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

  • When H+ ions are added:

  • The equilibrium position shifts to the left as H+ ions react with CH3COO- ions to form more CH3COOH until equilibrium is re-established

  • As there is a large reserve supply of CH3COO-, the concentration of CH3COO- in solution doesn’t change much as it reacts with the added H+ ions

  • As there is a large reserve supply of CH3COOH, the concentration of CH3COOH in solution doesn’t change much as CH3COOH is formed from the reaction of CH3COO- with H+

  • As a result, the pH remains reasonably constant

  • When OH- ions are added:

  • The OH- reacts with H+ to form water

OH- (aq) + H(aq) → H2O (l)

  • The H+ concentration decreases

  • The equilibrium position shifts to the right, and more CH3COOH molecules ionise to form more H+ and CH3COO- until equilibrium is re-established

CH3COOH (aq) → H+ (aq) + CH3COO- (aq)

  • As there is a large reserve supply of CH3COOH, the concentration of CH3COOH in solution doesn’t change much when CH3COOH dissociates to form more H+ ions

  • As there is a large reserve supply of CH3COO- the concentration of CH3COO- in solution doesn’t change much

  • As a result, the pH remains reasonably constant

Diagram of ethanoic acid equilibrium showing CH3COOH ⇌ CH3COO⁻ + H⁺, with captions explaining OH⁻ removes H⁺ so equilibrium shifts to replace lost hydrogen ions
When hydroxide ions are added to the solution, the hydrogen ions react with them to form water; The decrease in hydrogen ions would mean that the pH would increase however the equilibrium moves to the right to replace the removed hydrogen ions and keep the pH constant

Basic Buffers

  • A basic buffer can be made from ammonia (NH3) and ammonium chloride (NH4Cl)

  • Ammonia is a weak base. It reacts with water only partially:

NH3 (aq) + H2O(aq) NH4+ (aq) + OH-(aq)

  • Ammonium chloride dissolves completely in water to produce ammonium ions:

NH4Cl(s) NH4+ (aq) + OH-(aq)

  • The buffer therefore contains a significant amount of both NH3 and NH4+

  • When acid is added, the added hydrogen ions (H⁺) react with the ammonia:

NH3 (aq) + H+(aq) NH4+ (aq)

  • This removes most of the added H+ before it can significantly increase the hydrogen ion concentration. As a result, the pH decreases only slightly

  • When alkali is added, the added hydroxide ions (OH⁻) react with the ammonium ions:

NH4+ (aq) + OH-(aq) NH3 (aq) + H2O(aq)

  • This removes most of the added OH⁻ by converting it into water. Again, the pH changes only slightly

Examiner Tips and Tricks

Remember that buffer solutions cannot cope with excessive addition of acids or alkalis, as their pH will change significantly. The pH will remain relatively constant only if small amounts of acid or alkali are added.

Buffer Calculations

  • The pH of a buffer solution can be calculated using:

    • The Ka of the weak acid

    • The equilibrium concentration of the weak acid and its conjugate base (salt)

  • To determine the pH, the concentration of hydrogen ions is needed, which can be found using the equilibrium expression

Ka = [salt][H+][acid] which can be rearranged to [H+] = Ka x [acid][salt]

  • To simplify the calculations, logarithms are used such that the expression becomes:

-log10[H+] = -log10Ka + (-log10 [acid][salt])

  • Since -log10 [H+] = pH, the expression can also be rewritten as:

pH = pKa + log10 [salt][acid]

  • This is known as the Henderson-Hasselbalch equation

Dilution

  • On dilution, the ratio [salt]/[acid] is unchanged, so [H+] and hence pH stay the same

Related topics

Examiner Tips and Tricks

When discussing the action of a buffer, examiners like to see students stating that "the pH remains reasonably constant" is supported with reasoning. From the Henderson-Hasselbalch equation, the ratio of salt to acid stays almost constant when both are in high concentration, so there is very little change in pH.

Note that when you change the sign of the log, the terms become inverted, so the salt concentration is on the top - be careful, it is very easy to get this the wrong way around.

-log (A/B) = +log(B/A)

Worked Example

Calculating the pH of a buffer solution.

Calculate the pH of a buffer solution containing 0.305 mol dm-3 of ethanoic acid and 0.520 mol dm-3 of sodium ethanoate.The Ka of ethanoic acid  = 1.74 × 10-5 mol dm-3

Answer

Ethanoic acid is a weak acid that ionises as follows:

CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

Step 1: Write down the equilibrium expression to find Ka

Ka=[CH3COO][H+][CH3COOH]

Step 2: Rearrange the equation to find [H+]

[H+]=Ka×[CH3COOH][CH3COO]

Step 3: Substitute the values into the expression

[H+]=1.74×105×[0.305][0.520]

= 1.02 x 10-5 mol dm-3

Step 4: Calculate the pH

pH = - log [H+]

= -log 1.02 x 10-5

= 4.99

Worked Example

Calculate the pH of a solution made from a weak acid and a strong base.

25.0 cm3 of 0.200 mol dm-3 ethanoic acid (CH₃COOH) is mixed with 20.0 cm3 of 0.100 mol dm-3 sodium hydroxide (NaOH).

The pKa of ethanoic acid is 4.76. Calculate the pH of the resulting solution.

Answer

Step 1: Calculate the initial moles

Moles of ethanoic acid:

25.01000×0.200 = 0.00500 mol

Moles of sodium hydroxide:

20.01000×0.100=0.00200 mol

Step 2: Write the reaction

CH3COOH (aq) + OH- H+ (aq) + CH3COO- (aq) + H2O(l)

The reaction is 1 : 1.

OH- is the limiting reagent, so it reacts completely.

Step 3: Construct an ICE table

Substance

Initial (mol)

Change (mol)

Equilibrium (final) (mol)

CH₃COOH

0.00500

-0.00200

0.00300

NaOH

0.00200

-0.00200

0

CH₃COO-

-

+0.00200

0.00200

Step 4: Apply the Henderson–Hasselbalch equation

pH = pKa + log10 [salt][acid]

Substitute the values:

pH = pKa + log10 [0.00200][0.00300]

pH = 4.76 + log10 (0.667)

pH = 4.76 -0.176

pH = 4.58

Examiner Tips and Tricks

The final pH must be given to exactly two decimal places (examiners penalise 1 dp or more than 2.

Applications of Buffers

Uses of buffer solutions in controlling the pH of blood

  • In humans, HCO3- ions act as a buffer to keep the blood pH between 7.35 and 7.45

  • Body cells produce CO2 during aerobic respiration

  • This CO2 will combine with water in the blood to form a solution containing H+ ions

CO2 (g) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)

  • This equilibrium between CO2 and HCO3- is extremely important

  • If the concentration of H+ ions is not regulated, the blood pH would drop and cause ‘acidosis

    • Acidosis refers to a condition in which there is too much acid in the body fluids, such as blood

    • This could cause body malfunction and eventually lead to a coma

  • If there is an increase in H+ ions

  • The equilibrium position shifts to the left until equilibrium is restored

CO2 (g) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)

  • This reduces the concentration of H+ and keeps the pH of the blood constant

  • If there is a decrease in H+ ions

    • The equilibrium position shifts to the right until equilibrium is restored

CO2 (g) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)

  • This increases the concentration of H+ and keeps the pH of the blood constant

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Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Content Creator

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.