pH of a Mixture of Acids and Bases
Mixing two solutions is a very common ionic-equilibrium question, and it is one where a wrong shortcut — averaging the two pH values directly — almost always gives the wrong answer, because pH is a logarithmic scale. This guide covers the three cases you will actually meet: acid mixed with acid, base mixed with base, and acid mixed with base (strong-strong mixtures).
The method: work in moles, not pH, until the last step
Acid + base mixing: compare moles of H⁺ (Cₐ×Vₐ) with moles of OH⁻ (C_b×V_b); the excess, divided by total volume, gives the final [H⁺] or [OH⁻]
pH = −log[H⁺] · pOH = −log[OH⁻] · pH + pOH = 14 (at 25°C)
Always convert every volume to litres before multiplying by concentration, and always finish the mole arithmetic before taking a logarithm.
Worked example 1 — two strong acids mixed together
Mix 100 mL of 0.1 M HCl with 200 mL of 0.2 M HCl. Find the pH.
Moles H⁺ from solution 1 = 0.1 × 0.100 L = 0.01 mol
Moles H⁺ from solution 2 = 0.2 × 0.200 L = 0.04 mol
Total moles H⁺ = 0.05 mol; total volume = 0.100 + 0.200 = 0.300 L
[H⁺] = 0.05 ÷ 0.300 = 0.1667 M
pH = −log(0.1667) = 0.78
Note this is neither pH 1 nor pH "half of anything" — mixing two acids never causes neutralisation, only dilution/concentration averaging of the H⁺ already present.
Worked example 2 — two strong bases mixed together
Mix 150 mL of 0.05 M NaOH with 50 mL of 0.15 M NaOH. Find the pH.
Moles OH⁻ from solution 1 = 0.05 × 0.150 L = 0.0075 mol
Moles OH⁻ from solution 2 = 0.15 × 0.050 L = 0.0075 mol
Total moles OH⁻ = 0.015 mol; total volume = 0.150 + 0.050 = 0.200 L
[OH⁻] = 0.015 ÷ 0.200 = 0.075 M
pOH = −log(0.075) = 1.12
pH = 14 − 1.12 = 12.88
Worked example 3 — strong acid + strong base, excess acid remaining
Mix 100 mL of 0.2 M HCl with 100 mL of 0.1 M NaOH. Find the pH.
Moles H⁺ = 0.2 × 0.100 L = 0.02 mol; moles OH⁻ = 0.1 × 0.100 L = 0.01 mol.
Since moles H⁺ > moles OH⁻, the acid is in excess: excess H⁺ = 0.02 − 0.01 = 0.01 mol.
Total volume = 0.100 + 0.100 = 0.200 L
[H⁺] = 0.01 ÷ 0.200 = 0.05 M
pH = −log(0.05) = 1.30
Worked example 4 — strong acid + strong base, excess base remaining
Mix 50 mL of 0.1 M HCl with 150 mL of 0.1 M NaOH. Find the pH.
Moles H⁺ = 0.1 × 0.050 L = 0.005 mol; moles OH⁻ = 0.1 × 0.150 L = 0.015 mol.
Since moles OH⁻ > moles H⁺, the base is in excess: excess OH⁻ = 0.015 − 0.005 = 0.010 mol.
Total volume = 0.050 + 0.150 = 0.200 L
[OH⁻] = 0.010 ÷ 0.200 = 0.05 M
pOH = −log(0.05) = 1.30
pH = 14 − 1.30 = 12.70
Common mistakes that cost marks
- Averaging the two pH values directly — pH is logarithmic, so this is almost never correct, even for mixing equal volumes.
- Treating acid + acid (or base + base) as a neutralisation — there is no reaction between two acids; it is only a dilution/mixing calculation.
- Forgetting to check which species is in excess before choosing whether to compute pH or pOH first, when mixing an acid with a base.
- Using pH + pOH = 14 at a temperature other than 25°C without checking — this relationship depends on the ionic product of water, Kw, which changes with temperature.
Where this is tested
| Exam / topic | Typical use |
|---|---|
| CBSE Class 11 Chemistry — Equilibrium (Ionic Equilibrium) | Mixing-problem numericals on strong acids and bases |
| JEE Main/Advanced, NEET | pH-after-mixing and titration-curve questions |
| IIT-JAM / CSIR-NET analytical chemistry | Mixture and titration-curve calculations, extended to weak-acid/base and buffer systems |
Check the arithmetic instantly. The pH/pOH calculator lets you verify a mixing calculation like the ones above before you commit to it on paper.
Open the pH/pOH Calculator →Ionic equilibrium is one of the most numerical-heavy Class 11 chapters. ABC Chemistry's live online classes across India build this skill step by step — details at abcchemistry.in.