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Acids and bases pH notes and lesson for AP Chemistry (free PDF download)

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Acids and bases pH notes and lesson for AP Chemistry (free PDF download)

This AP Chemistry lesson on acids and bases pH notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on acids and bases pH and check yourself with the quiz.

Quick answer: A strong acid ionises completely, so its pH comes straight from its concentration. A weak acid ionises only partly and is described by Ka: for HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA]. Solve for [H⁺] with an ICE table, then take pH = −log[H⁺]. Conjugate pairs obey Ka × Kb = Kw.

Level: AP Chemistry · Curriculum: AP Chem Unit 8

Acids and bases pH notes: the key ideas

What is pH and how do you calculate it?

Definition

pH = −log₁₀[H₃O⁺], where [H₃O⁺] is the hydronium concentration in mol/dm³. A Brønsted–Lowry acid donates a proton and a base accepts one. When an acid HA loses a proton, the leftover A⁻ is its conjugate base.

Water ionises slightly, 2H₂O ⇌ H₃O⁺ + OH⁻, with an equilibrium constant called the ion-product of water. At 25 °C it is fixed, so pH and pOH always add to 14.

Formulas

Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C; pH + pOH = 14.00; pH = −log₁₀[H₃O⁺]; [H₃O⁺] = 10⁻ᵖᴴ. Concentrations are in mol/dm³; Kw and Ka have no unit in AP work.

The pH scale from 0 to 14 coloured from red (acidic) to violet (basic), with lemon juice at pH 2, black coffee at 5, pure water at 7, baking soda solution at 8.3, milk of magnesia at 10.5 and household ammonia at 11.501234567891011121314pHlemon juiceblack coffeepure waterbaking sodamilk of magnesiahousehold ammoniaacidicneutral at 25 °Cbasic
The pH scale at 25 °C. Each step down the scale means ten times more H₃O⁺.

How do strong and weak acids differ?

A strong acid ionises completely, so [H₃O⁺] equals the acid concentration (for a monoprotic acid). A weak acid ionises only partly and reaches an equilibrium described by its acid dissociation constant Ka. A larger Ka means a stronger weak acid, and a smaller pKa (= −log₁₀Ka) means the same thing.

Two particle diagrams: a strong acid where every HA has split into H₃O⁺ and A⁻, and a weak acid where most HA stays whole with only one H₃O⁺ and one A⁻ pairstrong acidH₃O⁺A⁻H₃O⁺A⁻H₃O⁺A⁻H₃O⁺A⁻no HA leftweak acidHAHAH₃O⁺A⁻HAHAHAHAmost HA stays whole
Particle view of the same number of acid molecules in water.
TypeExamplesHow to get [H₃O⁺] or [OH⁻]
strong acidsHCl, HBr, HI, HNO₃, HClO₄, H₂SO₄ (first step)equal to the acid concentration
strong basesgroup 1 hydroxides, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂[OH⁻] = concentration × number of OH⁻
weak acidsCH₃COOH, HF, HCN, HCO₃⁻solve with Ka and an ICE table
weak basesNH₃, CH₃NH₂, CH₃COO⁻solve with Kb and an ICE table
Worked example 1

Find the pH of 0.0250 mol/dm³ HNO₃ and of 0.0040 mol/dm³ Ba(OH)₂. HNO₃ is strong, so [H₃O⁺] = 0.0250 and pH = −log₁₀(0.0250) = 1.60. Ba(OH)₂ releases two OH⁻, so [OH⁻] = 2 × 0.0040 = 0.0080 mol/dm³, pOH = 2.10 and pH = 14.00 − 2.10 = 11.90.

How do you find the pH of a weak acid?

Write the Ka expression, set up an ICE table (initial, change, equilibrium) and let x be the amount of HA that ionises. For HA + H₂O ⇌ H₃O⁺ + A⁻, Ka = x² / (C − x). When Ka is small compared with the concentration C, x is tiny and C − x ≈ C, so x = √(Ka × C).

Key rule

The approximation is acceptable when x is less than 5 % of C. Always check it after calculating x. If x/C exceeds 5 %, solve the quadratic x² + Ka·x − Ka·C = 0 instead.

Worked example 2

Find the pH of 0.150 mol/dm³ ethanoic acid, Ka = 1.8 × 10⁻⁵. Ka = x²/0.150, so x² = 2.7 × 10⁻⁶ and x = 1.64 × 10⁻³ mol/dm³ = [H₃O⁺]. The check: 1.1 % of 0.150, below 5 %, so the approximation holds. pH = −log₁₀(1.643 × 10⁻³) = 2.78.

Bar chart: percent ionisation of ethanoic acid rises from 0.4 percent at 1.0 mol per dm cubed to 12.5 percent at 0.0010 mol per dm cubed02468101214percent ionised (%)0.42 %1 mol/dm³1.3 %0.1 mol/dm³4.2 %0.01 mol/dm³12.5 %0.001 mol/dm³concentration of ethanoic acid
More dilute weak acid, larger percentage ionised (Ka = 1.8 × 10⁻⁵).

How do weak bases and conjugate pairs work?

A weak base such as ammonia reacts with water: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, with Kb = [NH₄⁺][OH⁻]/[NH₃]. The method matches the acid case, except that x is [OH⁻] and you convert through pOH. A conjugate acid–base pair obeys one neat link.

Formula

Ka × Kb = Kw, so pKa + pKb = 14.00 for a conjugate pair. Ka is the constant of the acid, Kb that of its conjugate base. Both are dimensionless in AP calculations.

Worked example 3

Find the pH of 0.20 mol/dm³ NH₃, Kb = 1.8 × 10⁻⁵. x = [OH⁻] = √(1.8 × 10⁻⁵ × 0.20) = 1.90 × 10⁻³ mol/dm³. pOH = 2.72, so pH = 14.00 − 2.72 = 11.28.

Common mistake

Students write pH = −log₁₀(0.020) for 0.020 mol/dm³ of a weak acid. A weak acid never gives [H₃O⁺] equal to its concentration. Equally, never forget that Ba(OH)₂ gives two OH⁻ per formula unit.

Lina’s tip

Before you start, ask one question: strong or weak? Strong means no ICE table. Weak means write Ka, call the unknown x, and check the 5 % rule at the end.

Key points to remember
  • pH = −log₁₀[H₃O⁺]; pH + pOH = 14.00 at 25 °C (Kw = 1.0 × 10⁻¹⁴).
  • Strong acid or base: ionises fully, so [H₃O⁺] or [OH⁻] follows from the concentration.
  • Weak acid: Ka = x²/(C − x), and x ≈ √(Ka × C) if x under 5 % of C.
  • Conjugate pair: Ka × Kb = Kw, so pKa + pKb = 14.00.
  • Weak acids ionise to a larger percentage as they are diluted.

Strong and weak acids, bases and pH: frequently asked questions

What is the difference between pH and pOH?

pH measures hydronium concentration and pOH measures hydroxide concentration. At 25 °C they always add to 14.00, so a solution with pOH 3.20 has pH 10.80.

When can you use the square-root shortcut for a weak acid?

Use x = √(Ka × C) when the answer x is less than 5 % of the starting concentration C. If it is bigger, solve the full quadratic instead.

Why is a 0.10 mol/dm³ weak acid less acidic than 0.10 mol/dm³ HCl?

The weak acid ionises only a small fraction of its molecules, so far fewer H₃O⁺ ions are present. The two solutions still neutralise the same amount of base because the unionised acid keeps supplying protons.

How do you get Kb from Ka?

For a conjugate pair, Kb = Kw/Ka. For example, F⁻ has Kb = 1.0 × 10⁻¹⁴ / 6.8 × 10⁻⁴ = 1.5 × 10⁻¹¹.

Is pure water always at pH 7?

Only at 25 °C. Kw increases with temperature, so pure water at 50 °C has pH about 6.63 yet is still neutral because [H₃O⁺] = [OH⁻].

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