Weak Acid pH Calculator (and Weak Base pH from Ka or Kb)
By Santaji GadepatilinUpdated October 30, 2026
A weak acid only partly dissociates, so its pH depends on both its concentration and its Ka. This weak acid pH calculator solves the equilibrium exactly with the quadratic formula — and works for weak bases from Kb — giving the pH, [H⁺] or [OH⁻] and the percent dissociated. A pH sensor measures the same quantity directly.
How to use the weak acid pH calculator
Choose weak acid or weak base, enter its Ka or Kb and the initial concentration in mol/L. The calculator returns the equilibrium [H⁺] (or [OH⁻] for a base), the pH and the percent dissociated. If you only know the pKa, convert it first with the pKa calculator.
Weak acid pH formula
Ka = x² ÷ (C − x) ⇒ x = (−Ka + √(Ka² + 4 Ka C)) ÷ 2
Acid: pH = −log₁₀ x · Base: pH = 14 + log₁₀ x · % dissociated = x ÷ C × 100
Water’s own ionisation is ignored, which is accurate unless the solution is extremely dilute (below about 10⁻⁶ mol/L) or the acid extremely weak.
Examples
0.1 mol/L acetic acid (Ka = 1.8 × 10⁻⁵):
- x = (−1.8 × 10⁻⁵ + √((1.8 × 10⁻⁵)² + 4 × 1.8 × 10⁻⁵ × 0.1)) ÷ 2 = 1.3327 × 10⁻³ mol/L
- pH = −log₁₀(1.3327 × 10⁻³) = 2.875
- Percent dissociated = 1.333%
0.1 mol/L ammonia (Kb = 1.8 × 10⁻⁵): [OH⁻] = 1.3327 × 10⁻³ mol/L, pH = 14 + log₁₀(x) = 11.125.
pH of acetic acid and ammonia vs concentration (K = 1.8 × 10⁻⁵)
| Concentration | Acetic acid pH | Acetic acid % dissociated | Ammonia pH |
|---|---|---|---|
| 1 mol/L | 2.373 | 0.42% | 11.627 |
| 0.5 mol/L | 2.524 | 0.60% | 11.476 |
| 0.1 mol/L | 2.875 | 1.33% | 11.125 |
| 0.05 mol/L | 3.027 | 1.88% | 10.973 |
| 0.01 mol/L | 3.382 | 4.15% | 10.618 |
| 0.001 mol/L | 3.901 | 12.55% | 10.099 |
Notes
- pH 14 + log x assumes 25 °C; at other temperatures use the local pKw.
- For a mixture of a weak acid and its salt (a buffer), use the Henderson–Hasselbalch calculator instead.
- Strong acids are fully dissociated — use the pH calculator with [H⁺] = C.
Frequently asked questions
How do you calculate the pH of a weak acid?
Solve Ka = x² ÷ (C − x) for x = [H⁺], which gives x = (−Ka + √(Ka² + 4 Ka C)) ÷ 2, then pH = −log₁₀ x. For 0.1 M acetic acid this gives pH 2.88.
Can I use the shortcut x = √(Ka C)?
Yes, when the acid is less than about 5% dissociated (C ÷ Ka above roughly 400). The calculator always uses the exact quadratic, so it stays correct for dilute or stronger weak acids.
How is weak base pH calculated?
The same way with Kb — x is then [OH⁻], pOH = −log₁₀ x and pH = 14 − pOH at 25 °C. 0.1 M ammonia (Kb = 1.8 × 10⁻⁵) has pH 11.13.
Why does dilution increase percent dissociation?
Le Chatelier's principle — diluting shifts the equilibrium towards more ions. 0.1 M acetic acid is 1.3% dissociated; 0.001 M is about 12.5%.