pH and pOH Formulas — The Complete Relationship Explained
pH looks like a small topic and then quietly runs through half of your chemistry syllabus: ionic equilibrium, salt hydrolysis, buffers, indicators, titration curves, solubility, even biology and environmental science. Most of the marks lost here are not conceptual — they come from forgetting that pH + pOH = 14 only holds at 25 °C, or from treating a weak acid as if it were fully ionised. Here is the full relationship, with worked numbers.
The four formulas you need
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C
Taking −log of the third line: pH + pOH = 14 (at 25 °C)
Going backwards: [H⁺] = 10−pH and [OH⁻] = 10−pOH
What each term means
| Symbol | Meaning | Unit |
|---|---|---|
| [H⁺] | Molar concentration of hydrogen (hydronium) ions | mol L⁻¹ |
| [OH⁻] | Molar concentration of hydroxide ions | mol L⁻¹ |
| pH, pOH | Negative base-10 logarithm of those concentrations | no unit — it is a pure number |
| Kw | Ionic product of water; fixed at a given temperature | mol² L⁻² (usually quoted without units) |
The "p" simply means "take −log₁₀ of". That is also why pKa = −log Ka and pKw = −log Kw = 14 at 25 °C. Because the scale is logarithmic, a solution of pH 3 is ten times more acidic than one of pH 4, and a hundred times more acidic than one of pH 5.
Worked example 1 — Strong acid, easy numbers
Find pH, pOH and [OH⁻] for a solution with [H⁺] = 1.0 × 10⁻³ M.
pH = −log(1.0 × 10⁻³) = 3.00
pOH = 14 − 3.00 = 11.00
[OH⁻] = 10⁻¹¹ = 1.0 × 10⁻¹¹ M
Check with Kw: (1.0 × 10⁻³)(1.0 × 10⁻¹¹) = 1.0 × 10⁻¹⁴. ✔
Worked example 2 — Strong acid with a non-round concentration
Find the pH of 0.00200 M HCl.
HCl is a strong acid, so it ionises completely: [H⁺] = 2.00 × 10⁻³ M
pH = −log(2.00 × 10⁻³) = 3 − log 2.00 = 3 − 0.301 = 2.70
Useful shortcut for exam halls without a calculator: log 2 = 0.301, log 3 = 0.477, log 5 = 0.699, log 7 = 0.845. Almost every printed question uses one of these.
Worked example 3 — Strong base, and the trap in Ca(OH)₂
(a) Find the pH of 0.0100 M NaOH.
[OH⁻] = 1.00 × 10⁻² M → pOH = 2.00 → pH = 14 − 2.00 = 12.00
(b) Find the pH of 0.00500 M Ca(OH)₂.
One formula unit gives two hydroxide ions:
[OH⁻] = 2 × 0.00500 = 0.0100 M → pOH = 2.00 → pH = 12.00
Missing that factor of 2 gives pH 11.70 instead of 12.00 — a full mark gone.
Worked example 4 — Weak acid (this is the one that separates ranks)
A weak acid is only partly ionised, so [H⁺] is not equal to the acid concentration. For a weak monoprotic acid HA with dissociation constant Ka:
Find the pH of 0.100 M acetic acid, Ka = 1.8 × 10⁻⁵.
[H⁺] = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M
pH = −log(1.34 × 10⁻³) = 3 − log 1.34 = 3 − 0.128 = 2.87
Degree of ionisation = (1.34 × 10⁻³) ÷ 0.100 = 0.0134 = 1.34%, comfortably under 5%, so the approximation was justified. Note how far this is from pH 1.00, which is what you would wrongly get by treating acetic acid as strong.
Worked example 5 — Going backwards from pH
A soil sample has pH 4.40. Find [H⁺].
[H⁺] = 10−4.40 = 100.60 × 10−5 = 3.98 × 10⁻⁵
So [H⁺] ≈ 4.0 × 10⁻⁵ M — about forty times more acidic than neutral water.
The temperature limit nobody reads
Kw is a genuine equilibrium constant, and the self-ionisation of water is endothermic. Heat the water and Kw rises. At about 100 °C, Kw is roughly 5 × 10⁻¹³, so pKw ≈ 12.3 and the rule becomes pH + pOH ≈ 12.3. Neutral water at that temperature has pH ≈ 6.1 — and it is still neutral, because [H⁺] still equals [OH⁻].
Remember the real definition of neutrality: [H⁺] = [OH⁻], not "pH = 7". pH 7 is neutral only at 25 °C.
Common mistakes that cost marks
- Using pH + pOH = 14 at any temperature. It is a 25 °C result. If the question gives you a different Kw, use pKw instead of 14.
- Forgetting the number of ionisable groups. Ca(OH)₂ gives 2 OH⁻; H₂SO₄ gives 2 H⁺ (0.005 M H₂SO₄ has [H⁺] = 0.010 M, pH = 2.00).
- Treating a weak acid as fully ionised. For weak acids always go through Ka.
- Believing pH must lie between 0 and 14. 2 M HCl has a negative pH; 5 M NaOH has pH above 14. The 0–14 range is a convenience, not a law.
- Saying 10⁻⁸ M HCl has pH 8. An acid cannot be basic. At such extreme dilution the H⁺ from water itself dominates and the true pH is just under 7 — a favourite trap in JEE-level papers.
- Losing the minus sign. pH = −log[H⁺]. Dropping the minus gives a negative pH for every ordinary solution, which should immediately look wrong.
Where pH appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 10 | Acids, bases and salts; pH of everyday substances; indicators |
| CBSE/ICSE Class 11–12 | Ionic equilibrium, Ka/Kb, salt hydrolysis, buffers |
| JEE / NEET | Weak-acid pH, dilution effects, the 10⁻⁸ M trap, titration curves |
| IIT-JAM / CUET-PG | Buffer capacity, polyprotic acids, indicator selection |
| GATE / CSIR-NET | Analytical chemistry, biochemistry buffers, environmental chemistry |
Check your logs instantly. The pH / pOH calculator converts in every direction — [H⁺] to pH, pH to [OH⁻], pOH back to concentration — so you can verify a whole ionic-equilibrium question in seconds.
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