Chemistry Help UKFree A-level chemistry practice

AQA and OCR A · Year 13

Acid–base calculations, worked examples and question generator

Practise pH, Ka, Kw, buffers, neutralisation and titration stages. Start with one topic, use mixed mode, or generate a full exam question.

Before you choose an equation

Work out what is present after any reaction

Acid–base calculations become much easier when the chemistry is settled before the logarithms begin.

  1. 1
    Complete any fast reaction in moles.

    Neutralisation and reactions with added acid or base happen before the equilibrium calculation.

  2. 2
    Identify what remains.

    Look for excess strong acid, excess strong base, a weak acid, a buffer, or a particular titration stage.

  3. 3
    Use the expression that matches that mixture.

    The question may require pH directly, Kw, Ka, or a buffer ratio.

  4. 4
    Check 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−]

Practice studio

Choose how you want to practise

Pick the route that matches what you need today. You can change the board, topic or level before generating another set.

Guided practice by topic

Five questions from one area. Reveal and check each calculation step when needed.

Your practice

Guided set

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AQA-style practice [5 marks]

Acids 0.3.1 · site 1.90.1

Printable practice

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.

Do not put the original buffer amounts into Ka. The added acid changes both components in opposite directions.

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 rather than 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.

This is a consequence of the Ka expression, not a separate formula that works at every point on the curve.

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 buffer barely changed on paper, but not in my calculation

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.

I cannot tell which titration route to use

Compare the amount of titrant added with the amount needed for equivalence. Before, halfway and after equivalence contain different species.

Try next

Kc and equilibrium

Equilibrium expressions

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