5.3.3Physical Chemistry and Transition Elements
pH Calculations — OCR A Level Chemistry A H432 Revision Notes & Topical Questions
5 min read 5 parts Lesson 16 of 40
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 5Strong Acids and Bases
Full dissociation makes these calculations direct.
Strong monobasic acid
- [H⁺] equals the acid concentration, so pH = −log[acid]; 0.0100 mol dm⁻³ HCl gives pH 2.00.
Strong base
- Find [OH⁻], then calculate [H⁺] = Kw/[OH⁻] and take −log;
- for Ba(OH)₂ remember [OH⁻] is twice the base concentration.
Dilution
- Diluting a strong acid ten-fold raises the pH by exactly one unit, because [H⁺] falls by a factor of ten.
Mixing
- For an excess of acid or base after mixing, work out the leftover moles and divide by the total volume.
Section 2 of 5Weak Acids
A weak acid needs its Ka because it is only partially ionised.
1
Write Ka = [H⁺][A⁻]/[HA] for the acid and note that HA ⇌ H⁺ + A⁻ forms equal amounts of H⁺ and A⁻.
2
Assume [H⁺] = [A⁻], which ignores H⁺ from the ionisation of water because it is negligible for an acid.
3
Assume the equilibrium [HA] is approximately the initial concentration, because only a tiny fraction has dissociated.
4
Simplify to Ka = [H⁺]²/[HA], so [H⁺] = √(Ka × [HA]).
5
Calculate pH = −log[H⁺], and check that the assumptions are reasonable, since [H⁺] should be much smaller than [HA].
Section 3 of 5Working Backwards
Working Backwards
Finding Ka
- Convert the measured pH to [H⁺] using 10^(−pH), then use Ka = [H⁺]²/[HA].
Finding concentration
- Rearrange [HA] = [H⁺]²/Ka to find the concentration of a weak acid from its pH.
Percentage dissociation
- Divide [H⁺] by the original acid concentration and multiply by 100 to obtain the fraction ionised.
Limits of approximation
- The assumption fails for concentrated, stronger weak acids or very dilute solutions, where the dissociation is not small.
Section 4 of 5Worked Example
Question Calculate the pH of (a) 0.020 mol dm⁻³ Ba(OH)₂ and (b) 0.100 mol dm⁻³ ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³, at 298 K.
- (a) [OH⁻] = 2 × 0.020 = 0.040 mol dm⁻³.
- [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.040 = 2.5 × 10⁻¹³ mol dm⁻³, so pH = 12.60.
- (b) [H⁺]² = Ka × [HA] = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶.
- [H⁺] = 1.32 × 10⁻³ mol dm⁻³, so pH = 2.88.
Answer (a) pH = 12.60; (b) pH = 2.88.
Section 5 of 5Exam Checklist & Tips
5.3.3 – Key Points for the Exam
- pH −log₁₀[H⁺], where the concentration is in mol dm⁻³.
- Percentage dissociation The proportion of acid molecules that have ionised, shown as a percentage.
- Approximation Treating [HA] at equilibrium as equal to its initial concentration.
- Dibasic base A base such as Ba(OH)₂ that releases two hydroxide ions per formula unit.
Common Mistakes
- Forgetting to double [OH⁻] for Ba(OH)₂ or to double [H⁺] when a strong dibasic acid is assumed fully ionised.
- Forgetting the square root when calculating [H⁺] for a weak acid.
- Using Ka directly as [H⁺] for the weak acid.
- Reporting the pH of a base without first converting [OH⁻] to [H⁺] or pOH.
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