4.5.4Unit 4. Rates, Equilibria and Further Organic Chemistry
pH Calculations of Acids — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 5 parts Lesson 36 of 56 Unit quiz
Curriculum Equation Key pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: pH + pOH = 14
Section 1 of 5Definition and Strong Acids
pH is a logarithmic measure of hydrogen ion concentration.
Definition
- pH = −log₁₀[H⁺], and [H⁺] = 10⁻ᵖᴴ, where [H⁺] is in mol dm⁻³.
Strong monobasic acid
- Complete dissociation means [H⁺] equals the acid concentration, so 0.0250 mol dm⁻³ HCl has pH 1.60.
Rounding
- pH values are normally given to two decimal places, which matches the precision of the data.
Sense check
- A concentrated strong acid has a low pH; each tenfold dilution raises the pH by one unit.
Section 2 of 5Weak Acids
A weak acid needs Ka because only a fraction of it dissociates.
1
Write the Ka expression and note that [H⁺] = [A⁻] because each dissociation gives one of each ion.
2
Assume the equilibrium concentration of HA is the same as its starting concentration, because dissociation is tiny.
3
Rearrange to give [H⁺]² = Ka × [HA], so [H⁺] = √(Ka × [HA]).
4
Take −log₁₀ of the result to find the pH and quote it to two decimal places.
Section 3 of 5Assumptions and Checks
Assumptions and Checks
Ignore water
- Hydrogen ions from the ionisation of water are negligible compared with those from the acid.
Small dissociation
- The approximation [HA] equilibrium = [HA] initial is valid only if the acid is weak and not extremely dilute.
Reverse problem
- Given a pH, find [H⁺] = 10⁻ᵖᴴ and then use Ka to find the acid concentration or Ka.
Dibasic acids
- Treat strong dibasic acids like H₂SO₄ carefully, since the second dissociation is not complete.
Section 4 of 5Worked Example
Question Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³.
- [H⁺]² = Ka × [HA] = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶.
- [H⁺] = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³.
- pH = −log₁₀(1.32 × 10⁻³) = 2.88.
Answer pH = 2.88.
Section 5 of 5Exam Checklist & Tips
4.5.4 – Key Points for the Exam
- pH −log₁₀ of the hydrogen ion concentration.
- Monobasic An acid that releases one H⁺ per molecule.
- Dissociation Splitting of a molecule into ions.
- Approximation Assumption used to simplify calculations.
Common Mistakes
- Using the acid concentration as [H⁺] for a weak acid.
- Forgetting the square root in the weak acid expression.
- Entering the pH into the calculator without the negative sign on the logarithm.
- Writing pH with too many significant figures.
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