How to Calculate pH for Strong and Weak Acids and Bases: Formulas and Worked Examples

For strong acids (HCl, HBr, HI, HNO₃, HClO₄, H₂SO₄): pH = −log[H⁺], where [H⁺] equals the acid concentration because strong acids dissociate 100%. A 0.01 M HCl solution has [H⁺] = 0.01 M, so pH = −log(0.01) = 2.0. Done.

For weak acids (anything not on the strong acid list — acetic acid, HF, HCN, etc.): the acid does NOT fully dissociate. You need the acid dissociation constant Ka and an ICE table to find [H⁺]. Set up the equilibrium: HA ⇌ H⁺ + A⁻. Ka = [H⁺][A⁻] / [HA]. Solve for [H⁺], then pH = −log[H⁺].

For strong bases (NaOH, KOH, Ca(OH)₂, Ba(OH)₂): find [OH⁻] directly (equal to the base concentration for monoprotic bases, 2x for diprotic like Ca(OH)₂). Then pOH = −log[OH⁻], and pH = 14 − pOH.

For weak bases (NH₃, amines): use Kb and an ICE table to find [OH⁻], then pOH = −log[OH⁻], then pH = 14 − pOH.

Snap a photo of any pH problem and ChemistryIQ identifies whether the species is a strong or weak acid/base, selects the correct method, and solves step by step — including the ICE table for weak acid/base problems.

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Strong acids dissociate completely. The concentration of H⁺ equals the initial acid concentration.

Example 1: What is the pH of 0.0050 M HNO₃? HNO₃ is a strong acid → [H⁺] = 0.0050 M. pH = −log(0.0050) = 2.30.

Example 2: What is the pH of 0.10 M H₂SO₄? H₂SO₄ is a strong diprotic acid. The first dissociation is complete: H₂SO₄ → H⁺ + HSO₄⁻. So [H⁺] from the first dissociation = 0.10 M. The second dissociation (HSO₄⁻ → H⁺ + SO₄²⁻) is partial (Ka₂ = 0.012), but for most gen chem classes, you treat H₂SO₄ as producing 2 H⁺ per molecule at moderate concentrations. [H⁺] ≈ 0.20 M. pH = −log(0.20) = 0.70.

Strong bases: NaOH → Na⁺ + OH⁻ (one OH⁻ per formula unit). Ca(OH)₂ → Ca²⁺ + 2OH⁻ (two OH⁻ per formula unit).

Example 3: What is the pH of 0.025 M NaOH? [OH⁻] = 0.025 M. pOH = −log(0.025) = 1.60. pH = 14 − 1.60 = 12.40.

Example 4: What is the pH of 0.010 M Ca(OH)₂? [OH⁻] = 2 × 0.010 = 0.020 M (diprotic base). pOH = −log(0.020) = 1.70. pH = 14 − 1.70 = 12.30.

The mistake students make: forgetting that Ca(OH)₂ produces 2 OH⁻ per formula unit. If you use [OH⁻] = 0.010 instead of 0.020, your pH is off by 0.30 — enough to get the answer wrong.

ChemistryIQ identifies diprotic species automatically and adjusts the stoichiometry — so you never forget the 2x factor.

Weak acids partially dissociate, and the extent of dissociation is described by Ka. The smaller the Ka, the weaker the acid (less dissociation).

Example: What is the pH of 0.10 M acetic acid (CH₃COOH, Ka = 1.8 × 10⁻⁵)?

Step 1: Write the equilibrium. CH₃COOH ⇌ H⁺ + CH₃COO⁻

Step 2: Set up the ICE table. CH₃COOH H⁺ CH₃COO⁻ I: 0.10 0 0 C: −x +x +x E: 0.10−x x x

Step 3: Write the Ka expression. Ka = [H⁺][CH₃COO⁻] / [CH₃COOH] = x² / (0.10 − x) = 1.8 × 10⁻⁵

Step 4: Simplify (the 5% rule). If x is small compared to 0.10 (which it usually is for weak acids), approximate: 0.10 − x ≈ 0.10. Then: x² / 0.10 = 1.8 × 10⁻⁵. x² = 1.8 × 10⁻⁶. x = 1.34 × 10⁻³ M.

Step 5: Check the 5% rule. x / 0.10 = 1.34%. Since 1.34% < 5%, the approximation is valid.

Step 6: Calculate pH. [H⁺] = x = 1.34 × 10⁻³. pH = −log(1.34 × 10⁻³) = 2.87.

The 5% rule: if x / initial concentration > 5%, you cannot use the approximation and must solve the quadratic equation. This happens with very weak acids at low concentrations or acids with relatively large Ka values. ChemistryIQ checks the 5% rule automatically and switches to the quadratic formula when needed.

When the approximation fails: x² + Ka·x − Ka·C₀ = 0. Use the quadratic formula: x = [−Ka + √(Ka² + 4Ka·C₀)] / 2. Only the positive root is physically meaningful.

Weak base pH follows the same ICE table logic, but you solve for [OH⁻] first, then convert to pH.

Example: What is the pH of 0.15 M NH₃ (Kb = 1.8 × 10⁻⁵)?

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

ICE table: NH₃ NH₄⁺ OH⁻ I: 0.15 0 0 C: −x +x +x E: 0.15−x x x

Kb = x² / (0.15 − x) = 1.8 × 10⁻⁵. Approximate: x² / 0.15 = 1.8 × 10⁻⁵. x² = 2.7 × 10⁻⁶. x = 1.64 × 10⁻³.

[OH⁻] = 1.64 × 10⁻³. pOH = −log(1.64 × 10⁻³) = 2.79. pH = 14 − 2.79 = 11.21.

Common exam traps:

Trap 1: Using Ka when you need Kb (or vice versa). If the problem gives you Ka for the conjugate acid and asks for the pH of the base, convert: Kb = Kw / Ka, where Kw = 1.0 × 10⁻¹⁴. Example: Ka of NH₄⁺ = 5.6 × 10⁻¹⁰. Kb of NH₃ = 1.0 × 10⁻¹⁴ / 5.6 × 10⁻¹⁰ = 1.8 × 10⁻⁵.

Trap 2: Forgetting to convert pOH to pH for bases. Students calculate pOH correctly and report it as the answer. pH = 14 − pOH. A basic solution has pH > 7. If your answer for a base is pH < 7, you reported pOH by mistake.

Trap 3: pH of a salt solution. Sodium acetate (NaCH₃COO) dissolved in water produces a basic solution because the acetate ion is the conjugate base of a weak acid. You must identify the conjugate base, find its Kb (= Kw / Ka of the parent acid), and use the ICE table method for a weak base. This question type combines acid-base knowledge with the calculation method.

ChemistryIQ handles all of these — including the Ka/Kb conversion, the salt hydrolysis identification, and the pOH-to-pH conversion that students frequently forget.