pH Calculator
Calculate pH, pOH and ion concentrations for strong or weak acids and bases. Solves the exact equilibrium rather than the square-root shortcut.
Calculator
- pH Strongly acidic
- 2
- pOH
- 12
- [H⁺]
- 1 × 10⁻² M
- [OH⁻]
- 1 × 10⁻¹² M
Strongly acidic. pH 2 at 25 °C
Assumes complete dissociation and one proton (or hydroxide) per formula unit. For Ca(OH)₂, double the concentration first. For H₂SO₄, doubling gives only an upper limit, because its second proton is only partly released (pKa₂ about 2). The contribution from water itself is included, which is why a 10⁻⁸ M acid correctly comes out just below pH 7 rather than at pH 8.
Citing this tool
Last updated . Add the date you accessed it as well, which a citation of a page that can change asks for. If a specific result matters, cite the permalink from the tool’s share row instead of this page: it reproduces the exact parameters.
The equation
Sorensen (1909), definition of pH
What the pH scale is measuring
pH is a logarithmic report of hydrogen ion concentration,
pH = −log₁₀[H⁺]. The logarithm exists because the
concentrations involved span more than fourteen orders of magnitude, from
around 1 mol/L in strong acid to 10⁻¹⁴ mol/L in strong alkali. One pH unit
is therefore a tenfold change in concentration, and a solution at pH 3 is a
hundred times more acidic than one at pH 5.
The familiar relation pH + pOH = 14 follows from the
dissociation of water, but it holds only at 25 °C. The
ion product of water is temperature dependent, so at 100 °C pKw
falls to about 12.28 and genuinely neutral water sits near
pH 6.14. That water is not acidic; it is neutral, because
hydrogen and hydroxide ions remain equal in concentration.
Worked example
A strong acid dissociates completely, so 0.01 M HCl is straightforward:
[H⁺] = 0.01 mol/L, givingpH = −log₁₀(0.01) = 2.00
A weak acid needs the equilibrium solved. For 0.1 M acetic
acid with pKa 4.76, so Ka = 1.74 × 10⁻⁵:
- Set up the quadratic:
x² + Ka·x − Ka·C = 0 - Substitute:
x² + 1.74e-5·x − 1.74e-6 = 0 - Solve:
x = (−1.74e-5 + √6.9603e-6) / 2 = 1.31e-3 mol/L - Convert:
pH = −log₁₀(1.31e-3) = 2.88
Only 1.31e-3 / 0.1 = 1.3% of the acid has dissociated, which is
why the acetic acid solution is about 76 times less acidic than
hydrochloric acid at the same formal concentration.
Where the shortcuts break down
The approximation [H⁺] = √(Ka·C) assumes dissociation is small
enough to ignore in the mass balance. It reproduces the acetic acid answer
above, but for chloroacetic acid at 0.01 M it predicts pH 2.43 against a
correct 2.51, because roughly a third of the acid has dissociated. Treat
anything above about 5% dissociation as requiring the full quadratic.
Very dilute strong acids fail for the opposite reason. A
10⁻⁸ M strong acid does not have pH 8, which would make it
alkaline. Including the hydrogen ions from water gives
[H⁺] = 1.05e-7 and pH 6.98: slightly acidic,
as expected. Below about 10⁻⁶ M, water is no longer a negligible source.
A very dilute weak acid fails the same way when water is left out. The quadratic above puts 10⁻⁸ M acetic acid at pH 8.00, on the alkaline side, where counting water’s own ions gives 6.98. So this tool includes them for weak acids and bases as well.
Common mistakes
- Forgetting the second proton. Diprotic acids such as sulfuric can release two, so the hydrogen ion concentration from full dissociation would be double the formal concentration. Sulfuric acid’s second proton is only partly released (pKa₂ about 2), so the true figure lies between the formal concentration and double it.
- Applying pH + pOH = 14 at any temperature. The sum equals pKw, which is 14.00 only at 25 °C.
- Using −log C for very dilute acid. Ignoring water’s contribution produces the nonsensical result of an acid with pH above 7.
- Confusing pKa with pH. pKa is a property of the acid and does not change with dilution; pH describes the particular solution.
Converting units first? Use the concentration conversion table.
![pH Calculator: the equation pH = -log₁₀[H⁺].](/figures/ph-calculator.png)
Common questions
Why is the pH of a very dilute strong acid not simply −log C?
Because water itself supplies hydrogen ions. Taking pH = −log C for a 10⁻⁸ M acid gives pH 8, which would make an acid basic. Including the water equilibrium gives about pH 6.98 instead, correctly just on the acidic side of neutral. This calculator always includes that term.
When does the [H⁺] ≈ √(Ka·C) approximation fail?
It fails for stronger weak acids and for dilute solutions, because it assumes the amount dissociated is negligible next to the starting concentration. At 0.01 M with a pKa of 2, the shortcut overestimates [H⁺] by about 60 percent, so this tool solves the equilibrium exactly instead, with water’s own ions included.
Does pH + pOH always equal 14?
Only at 25 °C. The sum equals pKw, which is temperature dependent. At 100 °C pKw falls to about 12.3, so neutral water is pH 6.14 rather than 7. All results here are quoted at 25 °C.