pH is one of the most widely used numbers in science and one of the most consistently misunderstood, because it looks like a simple scale from 0 to 14 and is actually a logarithmic one. The definition is a single line, and almost every practical problem follows from it:
pH = −log₁₀[H⁺]
where [H⁺] is the hydrogen ion concentration in moles per litre. The rearrangement is the one used in calculations:
[H⁺] = 10^(−pH)
That exponent is the whole story. Because the scale is logarithmic, each unit of pH is a tenfold change in acidity, and that single fact explains why pH is the right scale for chemistry and the wrong scale for most other measurement.
The logarithmic scale, concretely
A solution at pH 3 has [H⁺] = 10⁻³ = 1 × 10⁻³ mol/L. A solution at pH 4 has [H⁺] = 10⁻⁴ mol/L. So the pH 3 solution is ten times more acidic. Going further, pH 2 is 100 times more acidic than pH 4, and pH 1 is 1,000 times more acidic than pH 4.
Two consequences that are easy to miss. First, the equal steps are multiplicative: the jump from pH 2 to pH 3 is the same factor as pH 7 to pH 8, even though the hydrogen ion concentration changes from 10⁻² to 10⁻³ in the first case and from 10⁻⁷ to 10⁻⁸ in the second. Second, the full range is enormous. pH 7 is neutral, so pH 2 is 10⁵ times more acidic than neutral water and pH 12 is 10⁵ times more alkaline. The 0–14 span is only the comfortable part of the scale: concentrated acids push the reading below 0 and concentrated alkalis push it above 14, so the scale is formally open-ended rather than bounded at those two numbers.
The pH calculator handles this inversion in both directions, which is the operation needed for nearly every problem that follows.
pOH and the 14
Water self-ionises: two H₂O molecules become H₃O⁺ and OH⁻. At 25 °C the product of the two concentrations is a constant, the ionic product of water:
Kw = [H⁺] × [OH⁻] = 1.0 × 10⁻¹⁴
Taking the negative logarithm of both sides gives the counterpart relation, valid at 25 °C:
pH + pOH = 14
So a solution at pH 4 has pOH 10, and a solution at pH 11 has pOH 3. The 14 is not universal — it depends on temperature, since Kw varies — and at 100 °C the sum is about 13.6, which matters in high-temperature industrial chemistry but not in ordinary laboratory work. Worked examples: 0.1 M NaOH fully dissociates, so [OH⁻] = 0.1 M, pOH = 1, pH = 13. Saturated calcium hydroxide gives [OH⁻] ≈ 0.020 M, pOH = 1.70, pH = 12.30.
Strong acids versus weak acids
A strong acid — HCl, HNO₃, H₂SO₄ — ionises completely in water. If you make 0.1 M HCl, essentially every molecule dissociates, [H⁺] = 0.1 mol/L, and:
pH = −log₁₀(0.1) = 1.00
A weak acid — acetic acid, lactic acid, carbonic acid — only partially ionises, because the conjugate base is not strong enough to pull the proton off. The extent is set by the acid dissociation constant Ka = [H⁺][A⁻]/[HA]. For acetic acid, Ka = 1.8 × 10⁻⁵ at 25 °C.
Where the acid is weak, its hydrogen is not equal to the formal concentration, and the whole of chemistry lab work follows from getting that distinction right.
Worked example 1: a solution at pH 4
Given pH = 4, invert the definition:
[H⁺] = 10^(−4) = 1.00 × 10⁻⁴ mol/L
Check: pH = −log₁₀(1 × 10⁻⁴) = 4.00 exactly. If this were a strong acid, that is 1 × 10⁻⁴ mol per litre of HCl. Note how much smaller that is than 0.1 M — the difference between a 1 in 10,000 solution and a 10% solution is the difference between pH 4 and pH 1, which is a factor of 1,000 in concentration.
Worked example 2: 0.1 M acetic acid
This is the example that shows why weak acids are weak. Formal concentration C = 0.10 mol/L, Ka = 1.8 × 10⁻⁵. For a weak acid, [H⁺] ≈ √(Ka × C), which gives:
[H⁺] ≈ √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol/L
pH = −log₁₀(1.34 × 10⁻³) = 2.87
The shortcut formula √(Ka × C) is an approximation that assumes the undissociated concentration stays close to C. It is valid when dissociation is under about 5%, and here only 1.34% of the acid dissociates, so the approximation holds well. Solving the quadratic exactly, x = (−Ka + √(Ka² + 4KaC)) ÷ 2, gives 1.33 × 10⁻³ mol/L and pH 2.88 — a difference of 0.01, well within the precision anyone cares about here.
The point to hold onto: a solution of 0.1 M acetic acid is pH 2.87, whereas 0.1 M hydrochloric acid is pH 1.00. A nearly hundredfold difference in hydrogen ion concentration from two solutions with identical formal concentrations, because one ionises completely and the other barely does at all. Anyone diluting from concentrated acid relies on getting this right.
Worked example 3: a buffer with equal concentrations
Now the payoff. Mix 0.10 M acetic acid with 0.10 M sodium acetate, the sodium salt supplying the conjugate base A⁻. The equilibrium now has both species present in known amounts, so the dissociation is suppressed and the pH is predictable by the Henderson–Hasselbalch equation, which is just Ka with the algebra rearranged:
pH = pKa + log([A⁻] ÷ [HA])
With [A⁻] = [HA] = 0.10 M, the ratio is exactly 1, log(1) = 0, so:
pH = pKa = −log₁₀(1.8 × 10⁻⁵) = 4.74
This is a result worth recognising on sight: a buffer made of equal concentrations of acid and conjugate base has a pH exactly equal to pKa, independent of how concentrated it is or how much water you added. Dilute a pH 4.74 buffer tenfold with water and it is still pH 4.74.
The logarithm is the whole mechanism. Because the pH depends on the ratio rather than the amounts, a buffer resists change in proportion to how far it already is from that ratio. Two more cases from the same equation:
- [A⁻]/[HA] = 10 → pH = 4.74 + 1 = 5.74
- [A⁻]/[HA] = 1/10 → pH = 4.74 − 1 = 3.74
So a tenfold change in one component moves the pH by exactly one unit — the same log-scale behaviour as the pH scale itself, now appearing in the buffer. The molarity calculator supplies the concentrations in mol/L, and the dilution calculator handles the case where you make the buffer from a stock.
Worked example 4: 0.010 M acetic acid
Same acid, ten times more dilute. Using the approximation:
[H⁺] ≈ √(1.8 × 10⁻⁵ × 0.010) = √(1.8 × 10⁻⁷) = 4.24 × 10⁻⁴ mol/L
pH = −log₁₀(4.24 × 10⁻⁴) = 3.37
The exact quadratic gives 4.15 × 10⁻⁴ mol/L and pH 3.38, so the approximation is starting to show. The reason is visible in the dissociation fraction: 4.24 × 10⁻⁴ ÷ 0.010 = 4.2% dissociated, approaching the 5% limit where the shortcut stops being reliable. Going one step further, 0.0010 M acetic acid dissociates 12.5% of the time, the approximation fails outright, and you need the full solution.
The useful pattern across the three concentrations: dilution of a weak acid by ten raises the pH by roughly half a unit, not a full unit, because the number of protons released depends on the equilibrium rather than the formal concentration. Diluting 0.1 M to 0.010 M moved pH from 2.87 to 3.37 — a shift of 0.50, because pH = ½(pKa − log C) for a weak acid. That half-unit is the signature of a weak acid and a quick way to check whether an answer is plausible.
Buffer capacity: when a buffer stops working
Resistance to added strong acid or base is not unlimited, and the limit is set by how much of each component is available. Formally:
β = 2.303 × C × Ka[H⁺] ÷ (Ka + [H⁺])²
where C is the total buffer concentration. The useful structural facts are simpler than the formula: capacity is maximal when pH = pKa, where the two terms in the numerator are equal and β = 2.303 × C ÷ 4; it scales linearly with C; and it falls off steeply on either side, because the derivative of the expression above is symmetric and drops as the ratio departs from 1.
Worked demonstration. Start with 1.00 L of buffer containing 0.100 M acetic acid and 0.100 M sodium acetate, so pH = 4.74. Add 0.0010 mol of solid NaOH — it converts 0.0010 mol of HA into A⁻:
pH = 4.7447 + log(0.1010 ÷ 0.0990) = 4.7534
Only 0.009 pH units. Add 0.010 mol and the same calculation gives 4.7447 + log(0.110 ÷ 0.090) = 4.83 — still under 0.1 units. The buffer is not weakening; it is still absorbing base efficiently, because 90% of the acid component is still available to neutralise it. Push to 0.050 mol and the acid is half consumed, the pH moves to 4.7447 + log(0.150 ÷ 0.050) = 5.22, and the useful range is clearly closing.
The practical rule: a buffer is effective over roughly pKa ± 1, and by the time 90% of one component is exhausted the resistance has largely collapsed. This is why buffer recipes specify a concentration and a ratio rather than just a pH, and why biological systems maintain buffers continuously instead of filling a vessel once. Concentrations are handled the same way as in the molar mass guide, and serial preparation in the dilution guide is how the ratios get made.
Where this shows up: blood, soil, and a soft drink
Blood pH is held between 7.35 and 7.45, and the 0.10-unit window is not arbitrary — the body dies outside roughly 6.8 to 7.8. What makes this achievable is a bicarbonate buffer governed by a variant of the same equation:
pH = 6.1 + log([HCO₃⁻] ÷ (0.03 × PCO2))
The numerator is the metabolic component and the denominator the respiratory one, which is what lets the body regulate pH two independent ways. A normal reading: pH = 6.1 + log(24 ÷ (0.03 × 40)) = 7.40. If HCO₃⁻ falls to 15 with PCO2 = 30, the equation gives 7.32 — a metabolic acidosis, where the buffer base is depleted. If HCO₃⁻ stays at 24 while PCO2 rises to 60, it gives 7.22 — a respiratory acidosis, where CO₂ has accumulated because it is not being exhaled fast enough. The two causes require different treatments, and this is precisely how they are told apart.
Soil pH matters because nutrient availability is not linear in it. Most crops do well between 6.5 and 7.5, but preferences diverge sharply: blueberries need 4.5–5.5 and will suffer iron deficiency above about 6, while brassicas and alfalfa tolerate 6.0–7.5. This is the log scale in a practical setting — a change of one pH unit in soil is a tenfold change in hydrogen ion concentration, and it shifts the balance between available and locked-up forms of iron, phosphorus and aluminium.
Cola has a pH of about 2.5, giving [H⁺] = 3.16 × 10⁻³ mol/L. That is a difference of 4.9 pH units from blood's 7.40, which on a logarithmic scale means roughly 80,000 times the hydrogen ion concentration. Yet drinking it does no acid damage to the stomach, for three reasons: the phosphoric and carbonic acids are weak, so comparatively few protons are actually released; the carbon dioxide is exhaled rather than absorbed, so the acid is effectively removed; and the stomach lining is protected by a layer of alkaline mucus with a pH near 8, maintained by bicarbonate secretion, which neutralises acid before it reaches the tissue.
Compare with gastric juice itself at pH 1.5, ten times more acidic than the cola — the stomach is the more extreme environment, and it is protected by that same mechanism plus the fact that the acid is a secretion rather than an intake. This is also why the buffer concept is not merely academic: a pH of 7.4 is maintained against a 6-unit chemical gradient, and the same log arithmetic above is what describes the margin.
Frequently asked questions
What does a change of one pH unit actually mean?
A factor of ten in hydrogen ion concentration. pH 3 means [H⁺] = 10⁻³ mol/L and pH 4 means 10⁻⁴ mol/L, so the pH 3 solution is ten times more acidic. Because the scale is logarithmic, the difference between pH 2 and pH 3 is the same magnitude of change as between pH 7 and pH 8, even though the absolute concentrations differ enormously.
Why is 0.1 M acetic acid pH 2.9 and not pH 1?
Because acetic acid is weak. Only about 1.3% of it dissociates at that concentration, since the equilibrium constant Ka = 1.8 × 10⁻⁵ is small. A 0.1 M solution of a strong acid like HCl fully ionises and does give pH 1, but acetic acid holds most of its hydrogen bound in undissociated molecules, so only a small fraction is free.
How does the Henderson-Hasselbalch equation work?
It comes from writing the acid dissociation equilibrium as a ratio: pH = pKa + log([A⁻]/[HA]). The pH is fixed primarily by the pKa, with the base-to-acid ratio providing a logarithmic correction. When the two concentrations are equal the ratio is 1, the logarithm is zero, and the pH equals the pKa exactly — which is why equal-concentration buffers resist added acid or base best.
What is buffer capacity and how do I increase it?
Buffer capacity is the amount of strong acid or base a solution can absorb before its pH changes appreciably, and it peaks when pH equals pKa. It scales with the total concentration of the buffer pair, so raising both concentrations increases capacity proportionally. Capacity also collapses as one component is exhausted, which is why a buffer stops working once roughly 90% of one species has been consumed.