explainer
How to Find the pH of a Weak Acid (the ICE-Table Method)
By the LazyTools team · Published 2026-07-12 · Updated 2026-07-12 · 4 min read
Drop a strong acid like HCl to 0.1 M and the pH is 1. Do the same with acetic acid and the pH is about 2.87 — nearly a hundred times less acidic. Same concentration, wildly different pH, because a weak acid barely comes apart in water. To find its pH you can’t just take a logarithm; you have to solve an equilibrium. The tool for that is the ICE table.
Why −log(C) fails
For a strong acid, every molecule donates its proton, so [H⁺] equals the acid concentration and pH = −log(C). Simple.
A weak acid doesn’t. Acetic acid in water sits mostly as intact CH₃COOH molecules, with only a small fraction broken into H⁺ and CH₃COO⁻ at any moment. That fraction — the percent ionization — is often just 1–2%. So the real [H⁺] is a fraction of the concentration, and the pH is much higher (less acidic) than −log(C) predicts.
How much ionizes is set by the acid’s Ka, its ionization constant. A bigger Ka means a stronger weak acid; acetic acid’s Ka is 1.8×10⁻⁵ (a pKa of 4.74).
Setting up the ICE table
ICE stands for Initial, Change, Equilibrium. For a weak acid HA ⇌ H⁺ + A⁻ starting at concentration C:
| HA | H⁺ | A⁻ | |
|---|---|---|---|
| Initial | C | 0 | 0 |
| Change | −x | +x | +x |
| Equilibrium | C − x | x | x |
Here x is the amount that ionizes, which is exactly [H⁺] at equilibrium. Plug the equilibrium row into the Ka expression:
Ka = [H⁺][A⁻] / [HA] = (x)(x) / (C − x) = x² / (C − x)
Solving for x
Rearranging gives a quadratic:
x² + Ka·x − Ka·C = 0
which solves (taking the positive root) to x = (−Ka + √(Ka² + 4·Ka·C)) / 2. For 0.1 M acetic acid:
x = [H⁺] ≈ 1.33×10⁻³ M → pH = −log(1.33×10⁻³) ≈ 2.87
And the percent ionization is 1.33×10⁻³ ÷ 0.1 × 100 ≈ 1.3% — confirming that only a sliver of the acid came apart.
The “x is small” shortcut — and when it breaks
Textbooks often skip the quadratic by assuming x is tiny compared to C, so C − x ≈ C and Ka ≈ x²/C, giving x ≈ √(Ka·C). For acetic acid that shortcut gives 1.34×10⁻³ — close enough.
But it fails when x isn’t small: for a stronger weak acid, or a dilute solution, x becomes a real chunk of C and dropping it introduces error. The safe move is to always solve the full quadratic, which is what the weak acid & base pH calculator does — no approximation, so it’s accurate across the board.
The same idea, everywhere in equilibrium
Once the ICE table clicks, the rest of equilibrium follows the same pattern. A Ksp / solubility calculation is an ICE table for a dissolving salt; a general Kc equilibrium problem is an ICE table solved for the extent of reaction. Weak bases are identical with Kb and pOH. Learn it once and it powers the whole topic.
Just solve it
Enter your Ka (or Kb) and concentration into the weak acid & base pH calculator and it returns the pH, pOH, ion concentration and percent ionization — solving the exact quadratic, with quick presets for common acids and bases. Like every LazyTools tool, it runs entirely in your browser, nothing uploaded.
This uses the standard weak-acid equilibrium: Ka = x²/(C − x) solved exactly for x = [H⁺], with pH = −log x, for a monoprotic acid at 25 °C. Ka values are standard reference figures; enter the one from your data. Source: standard general-chemistry treatment of acid–base equilibria (e.g. IUPAC conventions, Zumdahl/Brown LeMay).
Frequently asked questions
How do I find the pH of a weak acid?
Set up an ICE table for HA ⇌ H⁺ + A⁻, which gives Ka = x²/(C − x) with x = [H⁺]. Solve that quadratic for x, then pH = −log x. For 0.1 M acetic acid (Ka = 1.8×10⁻⁵), x ≈ 1.33×10⁻³ and pH ≈ 2.87.
Why isn't the pH of a weak acid just −log of the concentration?
That formula only works for strong acids, which ionize 100%. A weak acid ionizes only partially — often a percent or two — so the actual [H⁺] is far smaller than the acid concentration, giving a much higher pH than −log(C) would.
What is an ICE table?
A table of Initial, Change, and Equilibrium concentrations. For a weak acid: Initial is C, 0, 0; Change is −x, +x, +x; Equilibrium is C−x, x, x. The equilibrium row goes into the Ka expression to solve for x.
Do I need the quadratic formula, or can I assume x is small?
The 'x is small' shortcut (Ka ≈ x²/C) works when the acid is weak and reasonably concentrated, but it fails for stronger weak acids or dilute solutions. Solving the full quadratic x² + Ka·x − Ka·C = 0 is always correct — which is what the calculator does.
What is percent ionization?
The fraction of acid that actually dissociates: [H⁺] ÷ C × 100. For 0.1 M acetic acid it's about 1.3%. Percent ionization increases as the solution gets more dilute, even though the pH rises.
How do I find the pH of a weak base?
The same way with Kb: solve for x = [OH⁻] from Kb = x²/(C − x), take pOH = −log x, then pH = 14 − pOH at 25 °C. The calculator has a weak-base mode that does this.