Calculate pH for a weak acid
Use Ka and equilibrium concentrations to calculate the hydrogen-ion concentration and pH of a weak acid.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 25. Use the equilibrium expressions and data stated in the question; written practice is self-checked and human subject-teacher review remains pending.
Before you begin
- Use AS equilibrium expressions, mole calculations, logarithms, and Brønsted–Lowry acid–base ideas.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For HA ⇌ H⁺ + A⁻, which concentrations appear in Ka?
What you will learn
- Write Ka for HA ⇌ H⁺ + A⁻
- Relate equilibrium concentrations to initial acid concentration
- Use a justified weak-dissociation approximation to calculate pH
Set up the acid dissociation equilibrium
For HA(aq) ⇌ H⁺(aq) + A⁻(aq), Ka = [H⁺][A⁻]/[HA] at equilibrium. If the initial acid concentration is c and x dissociates, then [H⁺] = x, [A⁻] = x and [HA] = c − x.
When x is small relative to c, use c − x ≈ c, giving x ≈ √(Ka c). Check the approximation by comparing x/c; if it is not small, solve the full equilibrium expression rather than forcing the approximation.
Convert concentration to pH at the end
After finding equilibrium [H⁺], calculate pH = −log₁₀[H⁺]. The Ka expression uses equilibrium concentrations, not the starting concentration for every species.
A small Ka means a small degree of dissociation at the same initial concentration. Do not use the strong-acid shortcut [H⁺] = c for a weak acid.
Pause and explain
When can c − x be approximated as c?
Worked example
Estimate the pH of 0.10 mol dm⁻³ ethanoic acid when Ka = 1.8 × 10⁻⁵ mol dm⁻³.
Show the worked solution
- Use [H⁺] ≈ √(Ka c) if dissociation is small.
- Calculate √(1.8 × 10⁻⁵ × 0.10).
- Take −log₁₀ of the resulting [H⁺].
Answer [H⁺] ≈ 1.34 × 10⁻³ mol dm⁻³; x/c ≈ 1.34%, so the approximation is reasonable. pH ≈ 2.87.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Estimate pH for 0.20 mol dm⁻³ HA with Ka = 2.0 × 10⁻⁶ mol dm⁻³.
Give me a hint
Use x ≈ √(Ka c), check x/c, then take the logarithm.
Compare my reasoning
- x ≈ √(2.0 × 10⁻⁶ × 0.20).
- Check x is small compared with 0.20 mol dm⁻³.
- Calculate pH = −log₁₀x.
[H⁺] ≈ 6.32 × 10⁻⁴ mol dm⁻³ and pH ≈ 3.20; x/c ≈ 0.32%, so the approximation is valid.
Look for these in your work
- Ka and c multiplied
- Approximation checked
- pH calculated from equilibrium [H⁺]
A weak acid has [H⁺] = 0.0015 mol dm⁻³ at equilibrium. Calculate its pH.
Give me a hint
Use the pH definition with the equilibrium hydrogen-ion concentration.
Compare my reasoning
- Write pH = −log₁₀[H⁺].
- Substitute 0.0015 mol dm⁻³.
- Round to two decimal places.
pH = 2.82.
Look for these in your work
- Equilibrium concentration used
- Logarithm sign correct
- Reasonable rounding
A learner calculates weak-acid pH as −log₁₀(0.10) for a 0.10 mol dm⁻³ acid. Explain what information is missing.
Give me a hint
The initial concentration is not generally the equilibrium [H⁺] for a weak acid.
Compare my reasoning
- Write the dissociation equation HA ⇌ H⁺ + A⁻.
- Use Ka to find the equilibrium dissociation amount.
- Then calculate pH from the resulting [H⁺].
The calculation treats the weak acid as fully dissociated. Ka is needed to determine equilibrium [H⁺]; only then can pH = −log₁₀[H⁺] be used.
Look for these in your work
- Weak-acid equilibrium identified
- Ka requirement stated
- pH step put in correct order
Common mistakes
- Using initial concentration in place of every equilibrium concentration
- Assuming complete dissociation
- Forgetting to check whether c − x ≈ c is justified
- Taking the logarithm before calculating [H⁺]
What can you explain now?
For HA ⇌ H⁺ + A⁻, which concentrations appear in Ka?
Compare with the explanation
Equilibrium [H⁺], [A⁻] and [HA]
Ka is defined using equilibrium concentrations.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How does Ka turn an initial weak-acid concentration into equilibrium [H⁺] and then pH?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.