AQA and OCR A · Year 13
Acid–base calculations, worked examples and question generator
Start a pH, Ka, Kw, buffer, neutralisation or titration-stage question. Review the method below if you get stuck.
Choose how you want to practise
Choose a mode and work the chemistry. Guided practice works well for a specific weak area; mixed practice is useful once the individual routes are secure.
Guided practice by topic
Five questions from one area. Reveal and check each calculation step when needed.
Mixed practice
A balanced five-question set: one question from each major part of the topic. This is the best mode for finding weak areas.
Generate one full exam question
No route is announced in advance. Choose a board and area, or let the generator choose.
Five-question exam set
One question from each major area. Work through it on screen, with marks and a summary at the end.
Your practice
Guided set
Question1 / 5
Marks0 / 0
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Your results
Acids 0.3.1 · site 3.03.1
Before you choose an equation
Work out what is present after any reaction
I’d settle the chemistry before touching the logarithms: complete any fast reaction, identify what remains, then choose the expression that matches that mixture.
- 1Complete any fast reaction in moles.
Neutralisation and reactions with added acid or base happen before the equilibrium calculation.
- 2Identify what remains.
Look for excess strong acid, excess strong base, a weak acid, a buffer, or a particular titration stage.
- 3Use the expression that matches that mixture.
The question may require pH directly, Kw, Ka, or a buffer ratio.
- 4Check the final concentration and units.
After mixing solutions, remember the total volume.
Quick method check
What is left after the reaction?
Excess strong acid
Find the amount of H+ left, divide by the final volume, then use pH = −log10[H+].
Open the equation reference
- pH and hydrogen ions
- pH = −log10[H+]; [H+] = 10−pH
- Water
- Kw = [H+][OH−]
- Weak acid
- Ka = [H+][A−] / [HA]
- Acidic buffer
- [H+] = Ka[HA] / [A−]
Print a set
Create an acid–base mini paper
Generate three or five questions. The mark scheme prints after the questions on a new page.
Choose the board, topic and level, then generate a paper.
Worked examples
Five routes worth recognising
Example 1Strong Group 2 hydroxide
Calculate the pH of 0.0200 mol dm−3 Ba(OH)2 at 25 °C. Kw = 1.00 × 10−14 mol2 dm−6.
[OH−] = 2 × 0.0200 = 0.0400 mol dm−3
[H+] = Kw / [OH−] = 2.50 × 10−13 mol dm−3
pH = −log10(2.50 × 10−13) = 12.60
Common wrong route: using 0.0200 mol dm−3 as [OH−]. Each Ba(OH)2 supplies two hydroxide ions.
Example 2Weak acid from Ka
Calculate the pH of 0.100 mol dm−3 ethanoic acid. Ka = 1.74 × 10−5 mol dm−3.
For CH3COOH ⇌ H+ + CH3COO−, let [H+] = [CH3COO−] = x.
Ka ≈ x2 / 0.100, so x = √(1.74 × 10−6) = 1.32 × 10−3 mol dm−3.
pH = 2.88.
Check: only about 1.3% has dissociated, so treating the remaining acid concentration as 0.100 mol dm−3 is reasonable.
Example 3Buffer after strong acid is added
A buffer initially contains 0.0200 mol CH3COOH and 0.0300 mol CH3COO−. It receives 0.00500 mol HCl. Ka = 1.74 × 10−5 mol dm−3.
H+ + CH3COO− → CH3COOH
After reaction: acid = 0.0250 mol; salt anion = 0.0250 mol.
[H+] = Ka × 0.0250 / 0.0250 = 1.74 × 10−5 mol dm−3.
pH = 4.76.
Update both buffer amounts after the added acid reacts, then use those new values in the Ka expression.
Example 4Strong acid and strong base mixed
25.0 cm3 of 0.100 mol dm−3 HCl is mixed with 15.0 cm3 of 0.120 mol dm−3 NaOH.
HCl = 0.100 × 0.0250 = 0.00250 mol.
NaOH = 0.120 × 0.0150 = 0.00180 mol.
Excess H+ = 0.000700 mol in 0.0400 dm3, so [H+] = 0.0175 mol dm−3.
pH = 1.76.
Two common slips: subtracting concentrations directly before finding amounts, and dividing by only one of the original volumes.
Example 5Half-equivalence on a weak-acid titration curve
At half-equivalence, half of the weak acid has been converted into its conjugate base. The amounts of HA and A− are therefore equal.
Ka = [H+][A−] / [HA]
When [A−] = [HA], Ka = [H+].
Therefore pH = pKa.
At half-equivalence, pH = pKa follows directly from the Ka expression because [A−] = [HA].
When an answer goes wrong
Find the first decision that failed
I used the starting concentrations after mixing
Complete neutralisation in moles first. Concentrations only describe the final mixture after you divide what remains by the combined volume.
My answer is out by a factor of 1000
Check every volume conversion. Values in cm3 must be divided by 1000 before using n = cV with concentration in mol dm−3.
My strong-base pH is too low
Count the hydroxide ions per formula unit, then use Kw. Ca(OH)2 and Ba(OH)2 each supply two OH−.
My calculated buffer change looks too large
Write the reaction of the added acid or base first. One buffer component decreases while the other increases.
I assumed neutral water always has pH 7
Neutral means [H+] = [OH−]. The numerical pH depends on Kw, which changes with temperature.
Which titration route applies here?
Compare the amount of titrant added with the amount needed for equivalence. Before, halfway and after equivalence contain different species.