Buffer Solutions

This lesson covers: 

  1. What buffers are
  2. The composition and mechanism of acidic buffers
  3. The composition and mechanism of basic buffers
  4. Applications of buffers
  5. How to calculate the pH of a buffer solution

Buffers resist changes in pH

A buffer is a solution that minimises alterations in pH when small quantities of acid or base are introduced.

  • Buffers do not completely prevent pH changes, but they significantly reduce them.
  • Buffers are effective only for limited amounts of added acid or base. When a buffer solution's capacity is exceeded, it loses its ability to resist pH changes, and the pH will change more dramatically with further additions of acid or base.
  • There are two types of buffers: acidic buffers and basic buffers.

Composition and mechanism of acidic buffers

Acidic buffers, characterised by a pH below 7, are created by mixing a weak acid with one of its salts. A common example is a solution of ethanoic acid (CH3COOH) and sodium ethanoate (CH3COONa).


In this solution, the weak acid (ethanoic acid) only slightly dissociates:

CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)


On the other hand, the salt (sodium ethanoate) fully dissociates into its ions when dissolved:

CH3COONa(aq) ➔ CH3COO-(aq) + Na+(aq)


As a result, the solution contains a large amount of undissociated ethanoic acid molecules and ethanoate ions from the salt.


The equilibrium position of this reaction shifts in response to changes in the concentration of H+ or OH- ions, following Le Chatelier's principle:

  1. When a small amount of acid is added, the H+ concentration increases. Most of the extra H+ ions combine with CH3COO- ions to form CH3COOH, shifting the equilibrium to the left and reducing the H+ concentration to a value close to its original level. Consequently, the pH remains relatively stable.
  2. When a small amount of base (e.g., NaOH) is added, the OH- concentration increases. Most of the extra OH- ions react with H+ ions to form water, removing H+ ions from the solution. This causes more CH3COOH to dissociate, forming H+ ions and shifting the equilibrium to the right. The H+ concentration increases until it is close to its original value, maintaining a stable pH.

Composition and mechanism of basic buffers

Basic buffers, characterized by a pH greater than 7, are created by mixing a weak base with one of its salts. A common example is a solution of ammonia (NH3, a weak base) and ammonium chloride (NH4Cl, a salt of ammonia).


In this solution, the salt fully dissociates:

NH4Cl(aq) ➔ NH4+(aq) + Cl-(aq)


Additionally, some of the ammonia molecules react with water:

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)


As a result, the solution contains a large amount of ammonium ions (NH4+) and ammonia molecules (NH3).


The equilibrium position of this reaction shifts in response to changes in pH:

  1. When a small amount of base is added, the OH- concentration increases. Most of the extra OH- ions react with NH4+ ions to form NH3 and H2O, shifting the equilibrium to the left and removing OH- ions from the solution. This minimises the change in pH.
  2. When a small amount of acid is added, the H+ concentration increases. Some of the H+ ions react with OH- ions to form H2O, causing the equilibrium to shift to the right to replace the consumed OH- ions. Additionally, some H+ ions react with NH3 molecules to form NH4+. These reactions remove most of the added H+ ions, minimising the change in pH.

Practical applications of buffers

Buffers have numerous practical applications in everyday life and various industries.


Some examples include:

  1. Shampoos - Most shampoos contain a pH 5.5 buffer to prevent hair from becoming rough when exposed to alkaline conditions, keeping it smooth and shiny.
  2. Biological washing powders - These products contain buffers to maintain the optimal pH for enzymes to function efficiently.
  3. Biological systems - Many biological buffer systems exist in the human body to ensure that tissues are maintained at the proper pH. For example, blood must remain at a pH close to 7.4, and it contains a buffer system to maintain this narrow range.

Calculating the pH of a buffer solution

To calculate the pH of an acidic buffer, you need to know the Ka of the weak acid and the concentrations of the weak acid and its salt.


The calculation requires the following assumptions:

  • The salt of the conjugate base is fully dissociated, so assume that the equilibrium concentration of A- is equal to the initial concentration of the salt.
  • HA is only slightly dissociated, so assume that its equilibrium concentration is equal to its initial concentration.


Here's an example of how to calculate the pH of a buffer solution.

Worked example 1 - Calculating the pH of a buffer solution

A buffer solution is made by adding 0.50 mol of ethanoic acid (CH3COOH) and 0.50 mol of sodium ethanoate (CH3COONa) to enough water to make 1.0 dm3 of solution. The Ka for ethanoic acid is 1.8 × 10-5 mol dm-3.

Calculate the pH of this buffer.


Step 1: Ka equation

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


Step 2: Rearrange Ka equation

[H+]= Ka × ([CH3COO] [CH3COOH])


Step 3: Calculate [H+]

Since [CH3COOH] = [CH3COO-], the Ka equation simplifies to:

[H+] = Ka = 1.8 x 10-5 mol dm-3


Step 4: Calculate pH

pH =log10[H+]=log10(1.8×10−5)=4.74

Therefore, the pH of the buffer solution is 4.74.

In addition to calculating the initial pH of a buffer solution, it is also important to understand how the pH changes when small amounts of acid or alkali are added. The following worked examples illustrate how to calculate the new pH of a buffer solution after such additions.

Worked example 2 - Calculating the pH of a buffer solution after adding acid

A buffer solution initially contains 0.50 mol of ethanoic acid (CH3COOH) and 0.50 mol of sodium ethanoate (CH3COONa) in 1.0 dm3 of solution. The Ka for ethanoic acid is 1.8×10−5 mol dm-3

Calculate the new pH of this buffer after adding 10 cm3 of 2.0 mol dm-3 hydrochloric acid (HCl).


Step 1: Conversion of cm3 into dm3

To convert from cm3 into dm3, divide by 1,000

10 cm3 = 0.010 dm3


Step 2: Calculate number of moles of HCl added

n = c × V =2.0×0.010=0.020 mol


Step 3: Calculate number of CH3COOH and CH3COO- moles after addition

The added H+ ions from HCl will react with CH3COO- to form CH3COOH:

 H+ + CH3COO- ➔ CH3COOH

Moles of CH3COOH (mol)Moles CH3COO- (mol)
Before addition of HCl0.500.50
Change+0.02-0.02
Moles after HCl addition0.520.48



Step 4: Calculate [CH3COOH] and [CH3COO-] after addition

Total volume after HCl addition = 1.010 dm3

[CH3COOH]=Vn=1.0100.52=0.51 mol dm−3

[CH3COO]=Vn=1.0100.48=0.48 mol dm−3


Step 5: Rearrange Ka equation

[H+]= Ka × ([CH3COO][CH3COOH])


Step 6: Substitution and correct evaluation

[H+]=(1.8×10−5)×(0.480.51)=1.95×10−5 mol dm−3


Step 7: Calculate pH

pH =log10[H+]=log10(1.95×10−5)=4.7

Therefore, the pH of the buffer solution after addition of HCl is 4.7.

Worked example 3 - Calculating the new pH of a buffer solution after adding alkali

Consider a buffer solution containing 0.40 mol dm-3 methanoic acid (HCOOH) and 0.60 mol dm-3 sodium methanoate (HCOONa) in 500 cm3 of solution. The Ka for methanoic acid is 1.6×10−4 mol dm-3.

Calculate the new pH of this buffer after adding 25 cm3 of 0.30 mol dm-3 sodium hydroxide (NaOH).


Step 1: Conversion of cm3 into dm3

To convert from cm3 into dm3, divide by 1,000

500 cm3 = 0.500 dm3

25 cm3 = 0.025 dm3


Step 2: Calculate number of HCOOH and HCOO- moles before addition

HCOOH:  n = c × V =0.40×0.500=0.20 mol

HCOO-:  n = c × V =0.60×0.500=0.30 mol$


Step 3: Calculate number of moles of NaOH added

n = c × V =0.30×0.025=7.5×10−3 mol


Step 4: Calculate number of HCOOH and HCOO- moles after addition

The added OH- ions from NaOH will react with HCOOH to form HCOO-:

OH- + HCOOH ➔ HCOO- + H2O

Moles of HCOOH (mol)Moles HCOO- (mol)
Before addition of NaOH0.200.30
Change-7.5 x 10-3+7.5 x 10-3
Moles after NaOH addition0.19250.3075


Step 5: Calculate [HCOOH] and [HCOO-] after addition

Total volume after NaOH addition = 0.525 dm3

[HCOOH]=Vn=0.5250.1925=0.37 mol dm−3

[HCOO]=Vn=0.5250.3075=0.59 mol dm−3


Step 6: Rearrange Ka equation

[H+]= Ka × ([HCOO][HCOOH])


Step 7: Substitution and correct evaluation

[H+]=(1.6×10−4)×(0.590.37)=1.00×10−4 mol dm−3


Step 8: Calculate pH

pH =log10[H+]=log10(1.00×10−4)=4.0

Therefore, the pH of the buffer solution after addition of NaOH is 4.0.