Buffers (OCR A Level Chemistry A): Revision Note
Exam code: H432
Buffer solutions
A buffer solution is a solution which resists changes in pH when small amounts of acids or alkalis are added
A buffer solution is used to keep the pH almost constant by maintaining an almost constant concentration of hydrogen and hydroxide ions in a solution
Acidic buffers
Acidic buffers are made from a weak acid and a soluble salt of the acid
A common example is an aqueous mixture of ethanoic acid and sodium ethanoate
In solution, ethanoic acid partially ionises:
CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)
ethanoic acid ⇌ ethanoate
high conc ⇌ low conc
Sodium ethanoate is a salt which fully dissociates in solution:
CH3COONa + aq → Na+ (aq) + CH3COO- (aq)
sodium ethanoate → ethanoate ion
low conc. → high conc.
This results in a buffer solution with high concentrations of CH3COOH and CH3COO-
The establishes the equilibrium with hydrogen and ethanoate ions:
CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)
high conc. high conc.
When H+ ions (acid) are added:
The extra H⁺ ions react with CH3COO- to form CH3COOH
The equilibrium shifts left to reduce [H+]
This limits any drop in pH
Since there's a large supply of CH3COO-, its concentration doesn’t change much
So, the pH remains almost constant

When OH- ions (base) are added:
The OH- ions react with H+ to form water:
OH- (aq) + H+ (aq) → H2O (l)
The [H+] decreases, shifting the equilibrium right
CH3COOH (aq) → H+ (aq) + CH3COO- (aq)
More CH3COOH dissociates to replace H+ ions
So, the pH stays nearly constant due to the buffer action
The buffer works because it contains a reservoir of both the weak acid and its conjugate base, which can react with added H⁺ or OH⁻ to restore balance.

Calculating the pH of a buffer solution
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 =
This can be rearranged to give:
[H+] = Ka x
To simplify the calculations, logarithms are used such that the expression becomes:
-log10[H+] = -log10Ka x -log10
Since -log10 [H+] = pH, the expression can also be rewritten as:
pH = pKa + log10
This is known as the Hendersen-Hasselbalch equation
Worked Example
Calculate the pH of a buffer solution containing 0.305 mol dm-3 of ethanoic acid and 0.520 mol dm-3 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 =
Step 2: Rearrange the equation to find [H+]
[H+] = Ka x
Step 3: Substitute the values into the expression
[H+] = 1.74 x
[H+] = 1.02 x 10-5 mol dm-3
Step 4: Calculate the pH
pH = - log [H+]
pH = -log 1.02 x 10-5
pH = 4.99
Uses of buffers
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 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 malfunctioning and eventually lead to 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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