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pH and pOH practice problems
pH uses a logarithmic scale: a one-unit decrease corresponds to a tenfold increase in hydrogen-ion activity. Introductory dilute-solution problems usually approximate activity with molar concentration.
By ExamCat · Original study lesson · Published September 24, 2026
Understand the method
In that approximation, pH = −log₁₀[H⁺]. At 25°C, pH + pOH ≈ 14 for aqueous solutions. Keep the temperature assumption visible; neutral pH is not universally seven at every temperature.
Worked example, step by step
A dilute solution at 25°C has [H⁺] = 1.0 × 10⁻³ M. Find pH and pOH.
- Take the base-ten logarithm: log₁₀(10⁻³) = −3.
- Apply the leading minus sign: pH = 3.00.
- Use the 25°C relationship: pOH = 14.00 − 3.00 = 11.00.
- Interpret: the solution has greater hydrogen-ion concentration than a pH 4 solution, by a factor of ten.
A mistake to watch for
Do not read pH 3 as three times as acidic as pH 1. Also do not equate the concentration of every weak acid with [H⁺]; weak-acid dissociation requires additional reasoning.
Try it yourself: three questions
Write an answer before opening the explanation. For calculations, keep your units and intermediate steps; for concepts, state the reason for your choice.
1. What is the approximate pH for [H⁺] = 10⁻⁵ M?
Show answer and explanation
5. The negative sign converts a negative logarithm to a positive pH value.
2. How does [H⁺] compare between pH 2 and pH 4?
Show answer and explanation
The pH 2 solution has 100 times the hydrogen-ion concentration under the concentration approximation. There are two tenfold steps.
3. At 25°C, a solution has pOH = 3. What is pH?
Show answer and explanation
Approximately 11. The solution is basic under these conditions because its hydrogen-ion concentration is below the neutral value.
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Further reading
OpenStax Chemistry 2e provides textbook background. The worked scenario and practice questions on this page were written independently for ExamCat.
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