How to use it
- Choose What to calculate. For a weak acid, type its concentration and either its Ka or its pKa, not both.
- For a strong acid or base, type the concentration only. For a buffer, type the pKa of the acid and the two concentrations.
- Read the pH and the pOH, then the ion concentrations [H⁺] and [OH⁻].
- For a weak acid or base, compare the exact pH with the square-root approximation, and follow the Working line by line.
How pH, pOH and the ion concentrations are linked
pH is the negative logarithm of the hydrogen ion concentration, and pOH is the negative logarithm of the hydroxide ion concentration. Water ionises a little, and at 25 °C the product of the two concentrations is 1.0 × 10⁻¹⁴, so pH and pOH always add up to 14.00. This is the value used on this page.
pH = −log₁₀[H⁺] pOH = −log₁₀[OH⁻] [H⁺] × [OH⁻] = 1.0 × 10⁻¹⁴ pH + pOH = 14.00Strong and weak acids and bases
A strong acid or base is treated as fully ionised, so its concentration gives the ion concentration directly. The page keeps the ionisation of water in the strong-acid formula, which matters only at very low concentrations.
A weak acid HA ionises only a little. With x as [H⁺], the acid constant Ka = x² ÷ (C − x) becomes the quadratic x² + Ka·x − Ka·C = 0, and the page solves it exactly. The square-root approximation, x ≈ √(Ka·C), is shown beside it. The quadratic uses one simplification: the hydrogen ion from water is neglected, which is a good assumption when x is well above 10⁻⁷ mol/L.
x = (−Ka + √(Ka² + 4·Ka·C)) ÷ 2 x ≈ √(Ka·C)Buffers and the Henderson–Hasselbalch equation
A buffer is a weak acid and its conjugate base together. Its pH follows the Henderson–Hasselbalch equation. The pH equals the pKa when the two concentrations are equal, and it moves by one unit for each tenfold change in their ratio.
pH = pKa + log₁₀([A⁻] ÷ [HA])Worked example: acetic acid, 0.10 M, Ka 1.8 × 10⁻⁵
With C = 0.10 M and Ka = 1.8 × 10⁻⁵, the quadratic is x² + 1.8 × 10⁻⁵·x − 1.8 × 10⁻⁶ = 0. Its positive root is x = 0.00133 mol/L, so the pH is 2.88. The square-root approximation gives pH 2.87, and 1.33 % of the acid dissociates.
| Quantity | Value |
|---|---|
| [H⁺] | 0.00133 mol/L |
| pH (quadratic) | 2.88 |
| pH (square-root approximation) | 2.87 |
| Share that reacts | 1.33 % |
Frequently asked questions
Why does the calculator solve the quadratic instead of using the square-root approximation?
The approximation assumes that only a small part of the acid dissociates. The quadratic makes no such assumption, so it stays right for stronger weak acids and for dilute solutions. The page shows both, and a note appears when they differ by more than 5 %.
How are pH and pOH related?
pH is the negative logarithm of [H⁺] and pOH that of [OH⁻]. At 25 °C they add up to 14.00, so a solution with a pH of 3 has a pOH of 11.
Does the calculator work at other temperatures?
No. The page uses the value of 14.00 for pH plus pOH, which holds at 25 °C. The ionisation of water changes with temperature, and the neutral pH moves with it.
Why is a very dilute strong acid not as acidic as its concentration suggests?
Water also gives hydrogen ions, which become important when the acid is very dilute. At 1.0 × 10⁻⁸ mol/L of HCl the water's contribution is large, so the pH is about 6.98, not 8, and the page includes it.
When can I use the Henderson–Hasselbalch equation?
It works well when both the acid and its base are present in amounts much larger than the hydrogen and hydroxide ions that water adds. Then the pH depends only on the pKa and the ratio of the two concentrations.
Does the calculator keep my values or send them anywhere?
No. The results are worked out in your browser, and nothing you type is stored or sent to our server.
Sources
Figures and text last checked on Tuesday, November 10, 2026.