Define pH, Ka, pKa and Kw
Use the mathematical definitions of pH, Ka, pKa and Kw and convert between concentration and pH.
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
If [H⁺] increases tenfold, how does pH change?
What you will learn
- Use pH = −log₁₀[H⁺] and [H⁺] = 10⁻ᵖᴴ
- Write Ka for a weak acid equilibrium
- Use pKa = −log₁₀Ka and Kw = [H⁺][OH⁻]
Definitions connect logarithms and concentrations
pH = −log₁₀[H⁺(aq)] and [H⁺(aq)] = 10⁻ᵖᴴ mol dm⁻³. A one-unit decrease in pH means a tenfold increase in hydrogen-ion concentration.
For HA(aq) ⇌ H⁺(aq) + A⁻(aq), Ka = [H⁺][A⁻]/[HA] at equilibrium. pKa = −log₁₀Ka, so a smaller pKa corresponds to a larger Ka and a stronger weak acid.
Water equilibrium relates hydrogen and hydroxide ions
Kw = [H⁺(aq)][OH⁻(aq)]. At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ in the syllabus data. Use the temperature and Kw value given when conditions differ.
When a question gives pH, use [H⁺] = 10⁻ᵖᴴ. Keep concentration units and logarithm base clear; pH itself has no concentration unit.
Pause and explain
Which expression defines pKa?
Worked example
Calculate pH when [H⁺] = 2.0 × 10⁻³ mol dm⁻³, then find [H⁺] at pH 4.00.
Show the worked solution
- Use pH = −log₁₀[H⁺] for the first solution.
- Substitute 2.0 × 10⁻³.
- Use [H⁺] = 10⁻ᵖᴴ for pH 4.00.
Answer pH = 2.70 (3 s.f.); at pH 4.00, [H⁺] = 1.00 × 10⁻⁴ mol dm⁻³.
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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 pH for [H⁺] = 5.0 × 10⁻⁵ mol dm⁻³.
Give me a hint
Apply the negative base-10 logarithm.
Compare my reasoning
- Write pH = −log₁₀[H⁺].
- Substitute 5.0 × 10⁻⁵.
- Round consistently with the concentration.
pH = 4.30 (2 d.p.).
Look for these in your work
- Correct definition used
- Logarithm sign correct
- Reasonable precision
Find [H⁺] in a solution of pH 2.50.
Give me a hint
Rearrange pH = −log₁₀[H⁺].
Compare my reasoning
- Use [H⁺] = 10⁻ᵖᴴ.
- Substitute 2.50.
- State mol dm⁻³.
[H⁺] = 10⁻²·⁵⁰ = 3.16 × 10⁻³ mol dm⁻³.
Look for these in your work
- Equation rearranged
- Power evaluated
- Concentration unit stated
An acid has Ka = 1.8 × 10⁻⁵. Calculate pKa and state how a larger Ka changes pKa.
Give me a hint
Take the negative base-10 logarithm of Ka.
Compare my reasoning
- pKa = −log₁₀(1.8 × 10⁻⁵).
- Evaluate to two decimal places.
- Compare the sign of the logarithmic relationship.
pKa = 4.74. A larger Ka gives a smaller pKa.
Look for these in your work
- Negative logarithm used
- Value calculated
- Inverse Ka–pKa relationship stated
Common mistakes
- Forgetting the negative sign in pH = −log[H⁺]
- Using natural logarithms instead of log₁₀
- Treating pKa as log₁₀Ka rather than −log₁₀Ka
- Using Kw at a different temperature without checking
What can you explain now?
If [H⁺] increases tenfold, how does pH change?
Compare with the explanation
It decreases by 1
The negative base-10 logarithm falls by one for a tenfold increase.
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Make it stick
Which logarithmic definitions connect pH to [H⁺], pKa to Ka, and Kw to both aqueous ions?
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.