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🧪pH Calculator - Weak Acids, Bases, Polyprotic Acids, Salts & Buffers

pH = −log[H⁺]. For a strong acid or base the pH follows directly from concentration, but weak acids and bases, polyprotic acids, salts and buffers require the dissociation equilibria and water autoionisation to be solved together. For example, 0.1 M acetic acid (pKa 4.75) gives [H⁺] ≈ √(c·Ka) ≈ 1.33×10⁻³ M and pH ≈ 2.88; 0.1 M ammonia has pH ≈ 11.12; 0.1 M phosphoric acid has pH ≈ 1.63; and 1×10⁻⁸ M HCl has pH 6.98, not 8, because the H⁺ from water itself can no longer be ignored. This calculator writes every component into the charge-balance equation and solves it by bisection on log[H⁺], giving an exact answer that includes every dissociation step and water. Alongside it shows the textbook simplest formula and the quadratic approximation, checks their validity conditions (c·Ka ≥ 20Kw, c/Ka ≥ 500) and reports how far they are from the exact value. Six modes cover strong acids and bases, weak monoprotic acids and bases (or your own Ka/Kb), polyprotic acids and each of their salts (carbonic, phosphoric, sulfuric, oxalic, citric and more, including amphiprotic NaHCO₃ and NaH₂PO₄), salt hydrolysis (sodium acetate, ammonium chloride, ammonium acetate), buffers (Henderson–Hasselbalch versus exact, buffer capacity and the pH change after adding strong acid or base) and a pH ↔ pOH ↔ [H⁺] ↔ [OH⁻] converter. Results include species fractions and concentrations plus an interactive alpha-versus-pH distribution chart, and Kw can be corrected for temperature.

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How to use pH Calculator

  1. 1Pick a mode or tap an example such as 0.1 M acetic acid, 0.1 M Na₂CO₃ or an acetate buffer
  2. 2Choose the substance and enter its concentration; custom weak acids and bases accept Ka or pKa
  3. 3Read pH, pOH, [H⁺] and [OH⁻], the step-by-step working and the species distribution chart

FAQ

When can I use [H⁺] = √(c·Ka) for a weak acid?
The simplest formula needs both c·Ka ≥ 20Kw (water's contribution is negligible) and c/Ka ≥ 500 (dissociation is small enough to use c for the equilibrium concentration). If only the first holds, solve the quadratic [H⁺]² + Ka[H⁺] − c·Ka = 0. The calculator shows all three results, flags whether each condition is met and reports the error.
Why is the pH of 1×10⁻⁸ M HCl not 8?
An acid solution cannot have pH above 7. At such low concentration water itself supplies more H⁺ (about 10⁻⁷ M) than the acid, so you must use [H⁺] = (c + √(c² + 4Kw))/2 ≈ 1.05×10⁻⁷ M, giving pH ≈ 6.98. The exact solver handles this automatically.
How do you calculate the pH of polyprotic acids and amphiprotic salts?
If Ka₁ is much larger than Ka₂, a polyprotic acid can be treated as monoprotic using Ka₁ (0.1 M H₃PO₄ gives pH ≈ 1.63). A normal salt such as Na₂CO₃ is treated as a base with Kb₁ = Kw/Ka₂. An acid salt like NaHCO₃ is amphiprotic, with pH ≈ ½(pKa₁ + pKa₂) ≈ 8.34. The tool also gives the exact solution and every species concentration.
How are buffer pH and buffer capacity calculated?
For a conjugate pair the Henderson–Hasselbalch equation gives pH = pKa + log(c_base/c_acid); an equimolar acetic acid/acetate buffer has pH ≈ 4.75 and buffers best within pKa ± 1. Buffer capacity is β = 2.303([H⁺] + [OH⁻] + c·α_acid·α_base). Enter an amount of added strong acid or base to see the buffer's pH shift by a few hundredths while the same amount in pure water shifts it by several units.