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.
Open the pH / pOH Calculator →Ionic equilibrium is where most Class 11–12 students lose marks. ABC Chemistry teaches it properly at the Gurugram coaching centre and through online classes across India, with home tuition available in Delhi-NCR — see abcchemistry.in.