Calculate pH for strong acids and alkalis
Find hydrogen-ion concentrations for strong acids and strong alkalis, then calculate pH using Kw where needed.
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
A strong monoprotic acid is 0.0010 mol dm⁻³. What is [H⁺] after complete dissociation?
What you will learn
- Use complete dissociation for a strong acid at the stated concentration
- Relate strong-alkali concentration to [OH⁻]
- Use Kw to calculate [H⁺] and pH for a strong alkali
Strong acid concentration gives hydrogen-ion concentration
For a monoprotic strong acid such as HCl, assume complete dissociation at the stated concentration: [H⁺] ≈ c(HCl). Then pH = −log₁₀[H⁺]. Include stoichiometric coefficients if the supplied strong acid produces more than one hydrogen ion per formula unit under the question's stated assumption.
Use the concentration after any dilution or mixing. A strong acid is extensively dissociated; strength is not the same as concentration.
Use Kw for a strong alkali
For a strong alkali such as NaOH, complete dissociation gives [OH⁻] ≈ c(NaOH). At the stated temperature, use Kw = [H⁺][OH⁻] to find [H⁺] = Kw/[OH⁻], then pH = −log₁₀[H⁺].
At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. Check that the final [H⁺] and pH move in the expected direction: a more concentrated alkali has a smaller [H⁺] and a higher pH.
Pause and explain
How do you find [H⁺] from [OH⁻] in a strong alkali?
Worked example
Calculate the pH of 0.010 mol dm⁻³ HCl and 0.0010 mol dm⁻³ NaOH at 298 K.
Show the worked solution
- HCl is monoprotic and fully dissociated, so [H⁺] = 0.010 mol dm⁻³.
- For NaOH, [OH⁻] = 0.0010 mol dm⁻³.
- Use Kw/[OH⁻], then take −log₁₀[H⁺].
Answer pH(HCl) = 2.00. For NaOH, [H⁺] = 1.00 × 10⁻¹¹ mol dm⁻³, so pH = 11.00.
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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.
Calculate the pH of 0.0020 mol dm⁻³ HNO₃, treating it as a strong monoprotic acid.
Give me a hint
Use complete dissociation and then the pH definition.
Compare my reasoning
- Set [H⁺] = 0.0020 mol dm⁻³.
- Calculate −log₁₀(0.0020).
- Give pH to two decimal places.
pH = 2.70.
Look for these in your work
- Monoprotic stoichiometry applied
- Negative logarithm used
- pH stated
At 298 K, calculate the pH of 0.020 mol dm⁻³ KOH.
Give me a hint
Find [OH⁻], use Kw to obtain [H⁺], then calculate pH.
Compare my reasoning
- [OH⁻] = 0.020 mol dm⁻³.
- [H⁺] = (1.00 × 10⁻¹⁴)/0.020 = 5.00 × 10⁻¹³ mol dm⁻³.
- Take −log₁₀[H⁺].
pH = 12.30.
Look for these in your work
- Hydroxide concentration identified
- Kw rearranged correctly
- pH calculation consistent
A student says a 0.10 mol dm⁻³ weak acid must have pH 1.00 because a 0.10 mol dm⁻³ strong monoprotic acid does. Explain the error.
Give me a hint
Compare complete dissociation with partial dissociation.
Compare my reasoning
- A strong acid is treated as fully dissociated.
- A weak acid establishes an equilibrium and has [H⁺] below its initial concentration.
- Its pH needs Ka and equilibrium data, not concentration alone.
The calculation assumes complete dissociation. A weak acid is only partially dissociated, so its [H⁺] is lower than 0.10 mol dm⁻³ and its pH is higher than 1.00; use Ka to calculate it.
Look for these in your work
- Strength and concentration distinguished
- Equilibrium recognised
- Missing data identified
Common mistakes
- Confusing acid strength with concentration
- Using [OH⁻] directly in pH = −log[H⁺]
- Forgetting acid stoichiometry or dilution
- Using a Kw value for the wrong temperature
What can you explain now?
A strong monoprotic acid is 0.0010 mol dm⁻³. What is [H⁺] after complete dissociation?
Compare with the explanation
0.0010 mol dm⁻³
One H⁺ is produced per acid molecule at the stated concentration.
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Make it stick
How does complete dissociation give [H⁺] for a strong acid and how does Kw link pH to a strong alkali?
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.