How to calculate pH
For a strong acid, it fully dissociates, so [H⁺] equals the concentration and pH = −log[H⁺]. For a strong base, [OH⁻] equals the concentration, pOH = −log[OH⁻], and pH = 14 − pOH. Weak acids only partly dissociate: set up Ka = x²/(C − x) where x = [H⁺] and Ka = 10^(−pKa), then solve, the shortcut x ≈ √(Ka·C) works when dissociation is small, and the exact quadratic works always. Weak bases work the same way with Kb and [OH⁻]. This tool solves the exact quadratic so it stays accurate even for concentrated or fairly strong weak acids where the approximation breaks down.
Assumes 25 °C (K_w = 1×10⁻¹⁴) and a monoprotic acid or base. pKa and pKb relate by pKa + pKb = 14 for a conjugate pair.
Working on buffer problems specifically? See the 3 MCAT buffer question types, explained.
Related tools: Buffer calculator · Acid-base titration curve · Biochem & MCAT equation sheet · Buffer pKa table · all biochem tools.
Worked example 1: 0.1 M strong acid (the default)
A strong acid fully dissociates, so this is the simplest case, no equilibrium to solve, just a direct log.
Worked example 2: 0.1 M acetic acid (weak acid, pKa 4.76)
Acetic acid is a classic weak acid; the tool preloads its textbook pKa of 4.76.
Here only about 1.3% of the acid dissociates (1.31×10⁻³ M out of 0.1 M total), so the quick approximation and the exact quadratic agree. This is the normal case for a dilute, genuinely weak acid.
Worked example 3: where the approximation actually breaks
A more concentrated, less-weak acid: 0.01 M with pKa = 2.0 (Ka = 0.01), deliberately chosen so Ka and C are comparable.
With 61.8% of the acid dissociated, "C − x ≈ C" is a bad assumption, and the two methods disagree by 0.21 pH units, enough to matter on an exam or in a real measurement. This is exactly why this tool always solves the exact quadratic rather than the shortcut.
FAQ
Why isn't a weak acid's pH just −log(concentration)?
That shortcut only works for strong acids, which dissociate essentially completely. A weak acid only partially dissociates, so [H⁺] is smaller than the total concentration. You have to solve the equilibrium expression to find the real [H⁺].
When does the √(Ka·C) approximation fail?
It assumes the dissociated fraction is small enough that C − x ≈ C, true when less than about 5% dissociates. It breaks down for more concentrated or less-weak acids, as shown in example 3 above.
How do you find the pH of a strong base?
It fully dissociates, so [OH⁻] equals its concentration. Compute pOH = −log[OH⁻], then pH = 14 − pOH (at 25°C, where pH + pOH = 14).
How are pKa and pKb related for a conjugate pair?
pKa + pKb = 14 at 25°C. A strong conjugate base (low pKb) pairs with a weak conjugate acid (high pKa), and vice versa. Use this to convert between the two when you're only given one.
Practice problems
1. Find the pH of a 0.05 M weak base with pKb = 4.75.
Show answer
2. Find the pH of a 0.02 M strong base solution.