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 Tip
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.