The problems are on the acids and bases pH worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A strong acid ionises completely, so its pH comes straight from its concentration. These worked answers correct the 20 problems of the AP Chemistry worksheet on acids and bases pH, from pH of a strong acid to dilution and ionisation: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the equation, the substitution and the result, with the final value in bold. Compare your method as well as your number, because AP markers award credit for correct setup even when an arithmetic slip occurs.
Level: AP Chemistry · Curriculum: AP Chem Unit 8
Problem 1 pH of a strong acid
1. HCl is strong, so [H₃O⁺] = 0.0100 mol/dm³.2. pH = −log₁₀(0.0100) = 2.000.
Problem 2 pH of a strong base
1. pOH = −log₁₀(0.0030) = 2.52.2. pH = 14.00 − 2.52 = 11.48.
Problem 3 Hydronium from pH
1. [H₃O⁺] = 10⁻³·⁴⁰ = 3.98 × 10⁻⁴ mol/dm³.2. Neutral water has 1.0 × 10⁻⁷ mol/dm³, so the ratio is 10³·⁶ = about 3981 times larger.
Problem 4 Classify the species
HNO₃: strong acid. HF: weak acid. NH₃: weak base. KOH: strong base. CH₃COOH: weak acid. HBr: strong acid. Strong: HNO₃, KOH, HBr; weak: HF, NH₃, CH₃COOH.Problem 5 Calcium hydroxide
1. Ca(OH)₂ gives two OH⁻, so [OH⁻] = 2 × 0.050 = 0.100 mol/dm³.2. pOH = 1.000, so pH = 13.000.
Problem 6 Conjugate pairs
HCO₃⁻ donates a proton, so it is the acid and CO₃²⁻ is its conjugate base. NH₃ accepts a proton, so it is the base and NH₄⁺ is its conjugate acid. Pairs: HCO₃⁻/CO₃²⁻ and NH₄⁺/NH₃.Problem 7 Dilution of a strong acid
1. c₂ = 0.50 × 10.0 / 500 = 0.010 mol/dm³.2. pH = −log₁₀(0.010) = 2.00.
Problem 8 Neutral water at 50 °C
1. [H₃O⁺] = √Kw = √(5.5 × 10⁻¹⁴) = 2.35 × 10⁻⁷ mol/dm³.2. pH = 6.63.
3. Neutral means [H₃O⁺] = [OH⁻]. This holds at every temperature, but the value of each shifts as Kw grows.
Problem 9 Hydrocyanic acid
Ka = x²/0.10, so x = √(4.9 × 10⁻¹¹) = 7.00 × 10⁻⁶ mol/dm³. The ionised fraction is far below 5 %, so the shortcut holds. pH = 5.15.Problem 10 Fluoride as a base
1. pKa = −log₁₀(6.8 × 10⁻⁴) = 3.17.2. Kb = Kw/Ka = 1.0 × 10⁻¹⁴ / 6.8 × 10⁻⁴ = 1.5 × 10⁻¹¹.
3. pKb = 14.00 − 3.17 = 10.83.
Problem 11 HCl against ethanoic acid
1. HCl: pH = 1.00. Ethanoic acid: x = √(1.8 × 10⁻⁶) = 1.342 × 10⁻³ mol/dm³, pH = 2.87.2. Both contain 2.50 × 10⁻³ mol of acid. As OH⁻ removes H₃O⁺, the weak acid keeps ionising, so every molecule is eventually neutralised: equal volumes.
Problem 12 Ka from a pH reading
1. [H₃O⁺] = 10⁻³·⁰⁰ = 1.0 × 10⁻³ mol/dm³.2. [A⁻] = 1.0 × 10⁻³ and [HA] = 0.100 − 0.001 = 0.099 mol/dm³. Ka = (1.0 × 10⁻³)² / 0.099 = 1.0 × 10⁻⁵.
Problem 13 Percent ionisation of HF
1. x = √(6.8 × 10⁻⁴ × 0.500) = 0.0184 mol/dm³.2. pH = 1.73.
3. Percent = 0.01844/0.500 × 100 = 3.7 %, below 5 %, so valid.
Problem 14 Methylamine
x = [OH⁻] = √(4.4 × 10⁻⁴ × 0.40) = 0.0133 mol/dm³ (3.3 % of 0.40, valid). pOH = 1.88, so pH = 12.12.Problem 15 Ammonium chloride solution
1. Ka = Kw/Kb = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰.2. x = √(5.56 × 10⁻¹⁰ × 0.10) = 7.45 × 10⁻⁶ mol/dm³; pH = 5.13. The solution is acidic because NH₄⁺ donates protons.
Problem 16 Mixing acid and base
1. n(H⁺) = 0.0250 × 0.100 = 2.50 × 10⁻³ mol; n(OH⁻) = 0.0150 × 0.200 = 3.00 × 10⁻³ mol.2. OH⁻ is in excess by 5.0 × 10⁻⁴ mol.
3. Volume 40.0 cm³, so [OH⁻] = 0.0125 mol/dm³, pOH = 1.90, pH = 12.10.
Problem 17 When the shortcut fails
1. x = √(6.8 × 10⁻⁶) = 2.608 × 10⁻³ mol/dm³; that is 26 % of 0.0100, which fails the rule (pH would be 2.58).2. x² + 6.8 × 10⁻⁴x − 6.8 × 10⁻⁶ = 0 gives x = 2.290 × 10⁻³ mol/dm³, so pH = 2.64. The shortcut underestimates the pH by about 0.06.
Problem 18 Very dilute acid
1. A pH above 7 would mean a basic solution, but adding an acid cannot make water basic. Water itself supplies 1.0 × 10⁻⁷ mol/dm³ of H₃O⁺, which the student ignored.2. [H₃O⁺] = 1.0 × 10⁻⁸ + [OH⁻], and [OH⁻] = Kw/[H₃O⁺]. So x² − 1.0 × 10⁻⁸x − 1.0 × 10⁻¹⁴ = 0, giving x = 1.051 × 10⁻⁷ mol/dm³ and pH = 6.98.
Problem 19 Two weak acids
1. Ka(X) = 10⁻⁴·²⁰ = 6.31 × 10⁻⁵; Ka(Y) = 10⁻⁵·⁶⁰ = 2.51 × 10⁻⁶.2. x(X) = 3.552 × 10⁻³ mol/dm³, pH = 2.45; x(Y) = 7.088 × 10⁻⁴ mol/dm³, pH = 3.15.
3. Ratio = 5.01, which is about 5.0. It equals √(Ka ratio) = √(10¹·⁴) = 10⁰·⁷, because x depends on √Ka.
Problem 20 Dilution and ionisation
1. Before: x = 1.342 × 10⁻³ mol/dm³, pH = 2.87. After: x = 4.243 × 10⁻⁴ mol/dm³, pH = 3.37.2. Before: 1.34 %. After: 4.24 %.
3. Diluting lowers every concentration, so Q = [H₃O⁺][A⁻]/[HA] falls below Ka. The system shifts right to restore equilibrium, so a larger fraction ionises. The pH rises by only 0.5, not 1.
