pH Calculator (from [H₃O⁺], [OH⁻] or pH)
pH and pOH from a hydronium or hydroxide concentration, or both concentrations from a pH, with the working.
Inputs
- Working
- pH = −log(1.20e-3) = 2.9208
- [H₃O⁺] (mol/L)
- 1.2000e-3
- pOH
- 11.0792
- [OH⁻] (mol/L)
- 8.3333e-12
- At 25 °C
- acidic — 1.20e+4× the [H₃O⁺] of pure water
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- Two lines of HTML, unlimited views
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What each pH input means
- I know the
- Start from the hydronium (H₃O⁺) concentration, the hydroxide (OH⁻) concentration, or a pH.
- [H₃O⁺] concentration
- Hydronium-ion concentration in mol/L. For a strong acid such as HCl, it equals the acid's molarity.
- [OH⁻] concentration
- Hydroxide-ion concentration in mol/L. For a strong base such as NaOH, it equals the base's molarity.
- pH
- A pH value, to find the concentrations it stands for.
pH formula
pH = −log₁₀[H₃O⁺] · pOH = −log₁₀[OH⁻] · pH + pOH = 14.00 at 25 °C · [H₃O⁺] = 10^(−pH)
pH compresses hydronium-ion concentrations that span fourteen powers of ten into a number from 0 to 14: it is the negative logarithm of [H₃O⁺] in mol/L, so a concentration of 10⁻³ M is pH 3 and each unit of pH is a tenfold change. pOH does the same for hydroxide, and because water's ion product is 10⁻¹⁴ at 25 °C the two always add to 14. Neutral water has [H₃O⁺] = [OH⁻] = 10⁻⁷ M, pH 7. The page works from whichever quantity you have — hydronium concentration, hydroxide concentration or the pH itself — and returns the other three, with the logarithm or power of ten written out.
pH worked example
OpenStax's stomach-acid example: [H₃O⁺] = 1.2 × 10⁻³ M. pH = −log(1.2 × 10⁻³) = −(−2.92) = 2.92, so pOH = 14 − 2.92 = 11.08 and [OH⁻] = 10⁻¹¹·⁰⁸ = 8.3 × 10⁻¹² M. That is 12,000 times the hydronium concentration of pure water. From a hydroxide concentration of 10⁻⁵ M the route is the reverse: pOH 5, pH 9, basic.
More pH examples
0.002 M sodium hydroxide gives [OH⁻] = 0.002: pOH = −log(0.002) = 2.699, so pH = 14 − 2.699 = 11.301 — basic, with 20,000 times less hydronium than pure water.
A pH of 8.3, about that of a baking-soda solution: [H₃O⁺] = 10⁻⁸·³ = 5.0 × 10⁻⁹ mol/L and [OH⁻] = 2.0 × 10⁻⁶. The headline repeats it as 8.3, with pOH 5.7.
Where the pH calculator can give the wrong answer
- The 14.00 relation holds at 25 °C. Water's ion product rises with temperature, so at 37 °C neutral pH is about 6.8, and pH + pOH is a little under 14; the page assumes 25 °C.
- These formulas take the concentration of H₃O⁺ actually present. For a weak acid, that is not the acid's concentration — acetic acid at 0.1 M is only about 1.3% ionised, pH 2.9 not 1 — and an equilibrium calculation with Kₐ comes first.
- Strictly pH uses ion activity rather than concentration; in dilute solutions the two are close, in concentrated or salty ones they are not, which is why a meter and this arithmetic can disagree by a few tenths.
- pH below 0 and above 14 exist (concentrated strong acids and bases) but the page's pH input is limited to 0–14, the range that concentrations up to 1 M cover.
- Enter concentrations in mol/L. A 1% or 1 ppm figure has to be converted to molarity first.
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pH reference table
| pH | [H₃O⁺] (mol/L) | [OH⁻] (mol/L) | pOH |
|---|---|---|---|
| 0 | 1 | 10⁻¹⁴ | 14 |
| 2 | 10⁻² | 10⁻¹² | 12 |
| 4 | 10⁻⁴ | 10⁻¹⁰ | 10 |
| 6 | 10⁻⁶ | 10⁻⁸ | 8 |
| 7 (neutral) | 10⁻⁷ | 10⁻⁷ | 7 |
| 8 | 10⁻⁸ | 10⁻⁶ | 6 |
| 10 | 10⁻¹⁰ | 10⁻⁴ | 4 |
| 12 | 10⁻¹² | 10⁻² | 2 |
| 14 | 10⁻¹⁴ | 1 | 0 |
pH FAQ
- Take the negative logarithm (base 10) of the hydronium concentration in mol/L. [H₃O⁺] = 0.001 M: −log(0.001) = 3, so pH 3. For 2.5 × 10⁻⁴ M, −log = 3.60. Enter the concentration and the page shows the log.
- Raise 10 to the negative pH: [H₃O⁺] = 10^(−pH). pH 4.5 gives 10⁻⁴·⁵ = 3.16 × 10⁻⁵ mol/L. Choose 'pH' as the known value and the page gives both [H₃O⁺] and [OH⁻].
- pOH is −log[OH⁻], the hydroxide equivalent of pH. At 25 °C, pH + pOH = 14, so a solution with pOH 3 has pH 11. Given either concentration the page reports both values.
- A thousand times: each pH unit is a factor of ten in hydronium concentration, and three units is 10³. The page's last line states the factor relative to pure water.
- 2: a strong acid such as HCl ionises completely, so [H₃O⁺] = 0.01 = 10⁻², and −log(10⁻²) = 2. Each tenfold dilution raises the pH by one, until it nears 7.
- Only at 25 °C, where water's ion product is 1.0 × 10⁻¹⁴. At body temperature, 37 °C, it is about 2.4 × 10⁻¹⁴, so the sum is about 13.6 and neutral water has a pH near 6.8.