Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

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

📚 IAL Edexcel Linear (Chemistry) Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
0%read

Part 1 of 5Definition and Strong Acids

pH is a logarithmic measure of hydrogen ion concentration.

Definition
  • Definition pH = −log₁₀[H⁺], and [H⁺] = 10⁻ᵖᴴ, where [H⁺] is in mol dm⁻³.
Strong monobasic acid
  • Strong monobasic acid Complete dissociation means [H⁺] equals the acid concentration, so 0.0250 mol dm⁻³ HCl has pH 1.60.
Rounding
  • Rounding pH values are normally given to two decimal places, which matches the precision of the data.
Sense check
  • Sense check A concentrated strong acid has a low pH; each tenfold dilution raises the pH by one unit.

Part 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.

Part 3 of 5Assumptions and Checks

Assumptions and Checks
Ignore water
  • Ignore water Hydrogen ions from the ionisation of water are negligible compared with those from the acid.
Small dissociation
  • Small dissociation The approximation [HA] equilibrium = [HA] initial is valid only if the acid is weak and not extremely dilute.
Reverse problem
  • Reverse problem Given a pH, find [H⁺] = 10⁻ᵖᴴ and then use Ka to find the acid concentration or Ka.
Dibasic acids
  • Dibasic acids Treat strong dibasic acids like H₂SO₄ carefully, since the second dissociation is not complete.

Part 4 of 5Worked Example

Question Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³.

  1. [H⁺]² = Ka × [HA] = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶.
  2. [H⁺] = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³.
  3. pH = −log₁₀(1.32 × 10⁻³) = 2.88.

Answer pH = 2.88.

Part 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.

Last updated:

\n