Strong Acids and Strong Bases

Core Concept

A strong acid or strong base dissociates essentially 100% in water, so its equilibrium lies completely to the product side and the calculation collapses to simple stoichiometry: the concentration of H₃O⁺ (or OH⁻) equals the concentration of the acid (or base), multiplied by the number of protons/hydroxides per formula unit. There are only six strong acids to memorize — HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄ — and the strong bases are the group 1 hydroxides plus the heavier group 2 hydroxides.

Key Ideas

• Six strong acids: HCl, HBr, HI (hydrohalic), HNO₃ (nitric), H₂SO₄ (sulfuric, first proton), HClO₄ (perchloric).
• Strong bases: Group 1 hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) and Group 2 hydroxides Ca(OH)₂, Sr(OH)₂, Ba(OH)₂.
• "Strong" ≠ "concentrated": strong is about extent of ionization, not molarity; a dilute strong acid is still a strong acid.
• Complete ionization means [H₃O⁺] = C(acid) for monoprotic strong acids; [OH⁻] = C(base) for monohydroxide bases.
• Diprotic/dihydroxide stoichiometry: H₂SO₄ gives (mostly) 2 H₃O⁺ per molecule; Ca(OH)₂ gives 2 OH⁻ per formula unit.

Equations and Variables

• HCl(aq) + H₂O → H₃O⁺ + Cl⁻ (complete; single arrow)
• HNO₃ + H₂O → H₃O⁺ + NO₃⁻
• NaOH(s) → Na⁺ + OH⁻ (complete dissociation)
• For monoprotic strong acid: [H₃O⁺] = Cₐ, pH = −log Cₐ
• For monohydroxide strong base: [OH⁻] = C_b, pOH = −log C_b, pH = 14.00 − pOH
• For Ca(OH)₂: [OH⁻] = 2 × C_b (because each formula unit releases two OH⁻)

How It Works

1. Write the dissociation with a single arrow — for strong electrolytes there is no meaningful reverse reaction.
2. Read off the stoichiometry: one HCl gives one H₃O⁺; one Ca(OH)₂ gives two OH⁻.
3. Multiply the formal concentration by the number of H⁺ (or OH⁻) per formula unit to get the ion concentration.
4. Take the negative log to get pH (from [H₃O⁺]) or pOH (from [OH⁻]), then convert with pH + pOH = 14.00.
5. For very dilute strong acid (near 10⁻⁷ M), the acid's own H₃O⁺ becomes comparable to water's autoionization, and you must add the 1.0 × 10⁻⁷ M contribution from water — a special case beyond simple stoichiometry.

Worked Example

Calculate the pH of 0.025 M Ca(OH)₂ and of 0.0050 M HClO₄.

For Ca(OH)₂, each unit gives 2 OH⁻:

[OH⁻] = 2 × 0.025 M = 0.050 M

pOH = −log(0.050) = 1.30

pH = 14.00 − 1.30 = 12.70

For HClO₄ (monoprotic, complete):

[H₃O⁺] = 0.0050 M

pH = −log(0.0050) = 2.30

How it works

1. Write the dissociation with a single arrow — for strong electrolytes there is no meaningful reverse reaction.
2. Read off the stoichiometry: one HCl gives one H₃O⁺; one Ca(OH)₂ gives two OH⁻.
3. Multiply the formal concentration by the number of H⁺ (or OH⁻) per formula unit to get the ion concentration.
4. Take the negative log to get pH (from [H₃O⁺]) or pOH (from [OH⁻]), then convert with pH + pOH = 14.00.
5. For very dilute strong acid (near 10⁻⁷ M), the acid's own H₃O⁺ becomes comparable to water's autoionization, and you must add the 1.0 × 10⁻⁷ M contribution from water — a special case beyond simple stoichiometry.

Common confusions

• "Concentrated = strong." — Concentration is how much is present; strength is the fraction that ionizes. They are independent.
• "HF is a strong acid." — No; HF is weak (strong H–F bond, small Ka). HCl, HBr, HI are the strong hydrohalic acids.
• "H₂SO₄ donates both protons completely." — The first proton dissociates completely; the second (HSO₄⁻) is weak (Ka₂ ≈ 1.2 × 10⁻²).
• "Mg(OH)₂ is a strong base." — It is only sparingly soluble; even though it is a Group 2 hydroxide, its low solubility limits the OH⁻ it can deliver, so treat the strong Group 2 bases as Ca, Sr, Ba (and note solubility matters).
• "Forgetting the factor of 2." — Ca(OH)₂ produces 2 OH⁻ per unit; the most common pH error is using C_b instead of 2C_b.

Quick review

• Strong acids: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄ (complete ionization).
• Strong bases: Group 1 hydroxides + Ca(OH)₂, Sr(OH)₂, Ba(OH)₂.
• pH = −log[H₃O⁺] with [H₃O⁺] from stoichiometry; account for 2 H⁺/2 OH⁻.
• Strength ≠ concentration.
• Dilute-solution special case: add water's 1.0 × 10⁻⁷ M H₃O⁺ when acid concentration is near that value.