pH Calculations of Acids (Edexcel International A Level (IAL) Chemistry): Revision Note
Exam code: YCH11
Acids - pH Calculations
Strong acids
Strong acids are completely ionised in solution
HA (aq) → H+ (aq) + A- (aq)
Therefore, the concentration of hydrogen ions, H+, is equal to the concentration of acid, HA
The number of hydrogen ions formed from the ionisation of water is very small relative to the [H+] due to ionisation of the strong acid and can therefore be neglected
The total [H+] is therefore the same as the [HA]
Worked Example
What is the pH of 0.01 mol dm-3 hydrochloric acid?
Answer
[HCl] = [H+] = 0.01 mol dm-3
pH = - log[H+]
pH = - log[0.01] = 2.00
The pH of dibasic acids
Dibasic or diprotic acids have two replaceable protons and will react in a 1:2 ratio with bases
Sulfuric acid is an example
H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)
You might think that being a strong acid it is fully ionised so the concentration of the hydrogen is double the concentration of the acid
This would mean that 0.1 mol dm-3 would be 0.2 mol dm-3 in [H+] and have a pH of 0.69
However, measurements of the pH of 0.1 mol dm-3 sulfuric acid show that it is actually about pH 0.98, which indicates it is not fully ionised
The ionisation of sulfuric acid occurs in two steps
H2SO4 → HSO4- + H+
HSO4- ⇌ SO42- + H+
Although the first step is thought to be fully ionised, the second step is suppressed by the abundance of hydrogen ions from the first step creating an equilibrium
The result is that the hydrogen ion concentration is less than double the acid concentration
Weak acids
The pH of weak acids can be calculated when the following is known: The concentration of the acid The Ka value of the acid From the Ka expression we can see that there are three variables:
However, the equilibrium concentration of [H+] and [A-] will be the same since one molecule of HA dissociates into one of each ion
This means you can simplify and re-arrange the expression to
Ka x [HA] = [H+]2
[H+]2 = Ka x [HA]
Taking the square roots of each side
[H+] = √(Ka x [HA])
Then take the negative logs
pH = -log[H+] = -log√(Ka x [HA])
Worked Example
pH calculations of weak acids
Calculate the pH of 0.100 mol dm-3 ethanoic acid at 298 K.
Ka = 1.74 × 10-5 mol dm-3
Answer
Ethanoic acid is a weak acid which ionises as follows:
CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)
Step 1: Write down the equilibrium expression to find Ka
Ka =
Step 2: Simplify the expression
The ratio of H+ to CH3COO- ions is 1:1
The concentration of H+ and CH3COO- ions are therefore the same
The expression can be simplified to:
Ka =
Step 3: Rearrange the expression to find [H+]
[H+] =
Step 4: Substitute the values into the expression to find [H+]
[H+] = = 1.32 x 10-3 mol dm-3
Step 5: Find the pH
pH = -log[H+]
pH = -log(1.32 x 10-3) = 2.88
Examiner Tips and Tricks
Assumptions
Sometimes exam questions will ask what assumptions are made in calculating the pH of a weak acid. There are two being made in these calculations.
We assume the concentration of the anion, [A-], is the same as [H+] when dissociation occurs, which is why we can simplify [H+][A-] to [H+]2
We assume the concentration of [HA]eqm is the same as [HA]initial, since the degree of dissociation is very small. In other words we say the dissociation of [HA] is negligible or so small it can be ignored.
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