Brønsted-Lowry (SL)
Which of the following species can act as a Brønsted-Lowry base?
A. HCl
B. NH₃
C. H₂SO₄
D. CH₄
Answer: B. NH₃ (can accept a proton to form NH₄⁺)
According to Lewis theory, a base is:
A. A proton donor
B. A proton acceptor
C. An electron pair donor
D. An electron pair acceptor
C. An electron pair donor
What is the pH of a 0.01 mol dm⁻³ solution of HCl?
A. 1
B. 2
C. 12
D. 13
B. 2 (pH = -log[H⁺] = -log(0.01) = 2)
A buffer solution contains:
A. A strong acid and its conjugate base
B. A weak acid and its conjugate base
C. Two strong acids
D. A weak acid and a strong base
B. A weak acid and its conjugate base
Which indicator would be most suitable for a strong acid-strong base titration?
A. Methyl orange (pH range 3.1-4.4)
B. Bromothymol blue (pH range 6.0-7.6)
C. Phenolphthalein (pH range 8.3-10.0)
D. All of the above
D. All of the above (the equivalence point is at pH 7, and all these indicators change color near the steep vertical region of the curve)
Write the equation for the reaction between ethanoic acid (CH₃COOH) and water. Identify the conjugate acid-base pairs.
CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)
Conjugate pairs:
CH₃COOH/CH₃COO⁻ (acid/conjugate base)
H₃O⁺/H₂O (acid/conjugate base)
Identify the Lewis acid and Lewis base in the following reaction:
BF₃ + NH₃ → F₃B-NH₃
Lewis acid: BF₃ (accepts electron pair - has empty orbital on boron)
Lewis base: NH₃ (donates lone pair of electrons from nitrogen)
Calculate the pH of a 0.05 mol dm⁻³ solution of sodium hydroxide (NaOH) at 25°C.
[OH⁻] = 0.05 mol dm⁻³
pOH = -log(0.05) = 1.30
pH = 14 - 1.30 = 12.70
Explain how a buffer solution containing ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COONa) resists changes in pH when a small amount of acid is added.
When acid (H⁺) is added to the buffer, the ethanoate ions (CH₃COO⁻) from the sodium ethanoate react with the added H⁺ ions:
CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq)
This removes most of the added H⁺ ions, preventing a significant decrease in pH. The large reservoir of CH₃COO⁻ ions means the buffer can neutralize added acid without a major change in the ratio of acid to conjugate base.
25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution is titrated with 0.0500 mol dm⁻³ hydrochloric acid. Calculate the volume of acid required to reach the equivalence point.
Moles of NaOH = 0.100 × (25.0/1000) = 2.50 × 10⁻³ mol
Mole ratio HCl:NaOH = 1:1
Moles of HCl needed = 2.50 × 10⁻³ mol
Volume of HCl = (2.50 × 10⁻³)/0.0500 = 0.0500 dm³ = 50.0 cm³
Which species is amphiprotic?
A. Cl⁻
B. HCO₃⁻
C. NH₄⁺
D. SO₄²⁻
B. HCO₃⁻ (can donate a proton to form CO₃²⁻ or accept a proton to form H₂CO₃)
Which of the following can act as a Lewis acid but NOT as a Brønsted-Lowry acid?
A. H₃O⁺
B. AlCl₃
C. HNO₃
D. NH₄⁺
B. AlCl₃ (can accept electron pairs but has no proton to donate)
At 25°C, the ionic product constant for water (Kw) is 1.0 × 10⁻¹⁴. What is the concentration of H⁺ ions in pure water?
A. 1.0 × 10⁻¹⁴ mol dm⁻³
B. 1.0 × 10⁻⁷ mol dm⁻³
C. 1.0 × 10⁻⁶ mol dm⁻³
D. 1.0 mol dm⁻³
B. 1.0 × 10⁻⁷ mol dm⁻³ (in pure water, [H⁺] = [OH⁻] = √Kw)
The acid dissociation constant (Ka) for methanoic acid is 1.8 × 10⁻⁴. What is the pKa?
A. 2.74
B. 4.74
C. 3.74
D. 5.74
C. 3.74 (pKa = -log(1.8 × 10⁻⁴) = 3.74)
In a titration of a weak acid with a strong base, at the equivalence point:
A. pH = 7
B. pH < 7
C. pH > 7
D. pH cannot be determined
C. pH > 7 (the conjugate base of the weak acid hydrolyzes water, producing OH⁻ ions)
Hydrogen sulfate ion (HSO₄⁻) can act as both a Brønsted-Lowry acid and base. Write equations showing HSO₄⁻ acting as:
(i) An acid with water
(ii) A base with hydrochloric acid
(i) HSO₄⁻(aq) + H₂O(l) ⇌ SO₄²⁻(aq) + H₃O⁺(aq)
(ii) HSO₄⁻(aq) + HCl(aq) → H₂SO₄(aq) + Cl⁻(aq)
The formation of the complex ion [Cu(H₂O)₆]²⁺ can be described using Lewis acid-base theory.
(i) Identify the Lewis acid and Lewis base in this reaction.
(ii) Describe the type of bond formed between copper and water.
(iii) Explain why Cu²⁺ can act as a Lewis acid.
(i) Lewis acid: Cu²⁺; Lewis base: H₂O
(ii) Coordinate covalent bond (dative bond) - both electrons in the bond come from the oxygen atom in water
(iii) Cu²⁺ has empty d orbitals that can accept electron pairs from donor species. The positive charge also makes it electron-deficient and able to attract electron pairs.
A solution has a hydrogen ion concentration of 3.5 × 10⁻⁴ mol dm⁻³ at 25°C.
(i) Calculate the pH of the solution.
(ii) Calculate the hydroxide ion concentration.
(iii) State whether the solution is acidic, basic, or neutral.
(i) pH = -log(3.5 × 10⁻⁴) = 3.46
(ii) [OH⁻] = Kw/[H⁺] = (1.0 × 10⁻¹⁴)/(3.5 × 10⁻⁴) = 2.86 × 10⁻¹¹ mol dm⁻³
(iii) Acidic (pH < 7)
A buffer solution is made by mixing 50.0 cm³ of 0.200 mol dm⁻³ ammonia (NH₃) with 50.0 cm³ of 0.200 mol dm⁻³ ammonium chloride (NH₄Cl). The Kb for ammonia is 1.8 × 10⁻⁵.
(i) Calculate the pOH of the buffer solution using the Henderson-Hasselbalch equation (or equivalent).
(ii) Calculate the pH of the buffer solution.
(i) Since equal volumes and concentrations are mixed, [NH₃] = [NH₄⁺]
pOH = pKb + log([NH₄⁺]/[NH₃])
pKb = -log(1.8 × 10⁻⁵) = 4.74
pOH = 4.74 + log(1) = 4.74
(ii) pH = 14 - 4.74 = 9.26
Sketch and label a titration curve for the titration of 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (a weak acid) with 0.100 mol dm⁻³ sodium hydroxide. On your sketch:
(i) Label the equivalence point and estimate its pH
(ii) Label the half-equivalence point and explain its significance
(iii) Identify the buffer region
(iv) Suggest a suitable indicator
(i) Equivalence point occurs at 25.0 cm³ of NaOH added; pH ≈ 8.7-9.0 (basic due to hydrolysis of ethanoate ions)
(ii) Half-equivalence point at 12.5 cm³ of NaOH added; at this point pH = pKa because [CH₃COOH] = [CH₃COO⁻]
(iii) Buffer region: from start of titration until near equivalence point (approximately 0-20 cm³ added), where both weak acid and conjugate base are present in significant amounts
(iv) Phenolphthalein (changes color in pH range 8.3-10.0, which matches the steep rise at equivalence point)
Ammonia reacts with hydrogen chloride gas to form ammonium chloride. Using Brønsted-Lowry theory, explain this reaction in terms of proton transfer. Include the equation and identify which species is the acid and which is the base. Explain why this reaction goes to completion rather than establishing equilibrium.
NH₃(g) + HCl(g) → NH₄Cl(s) or NH₄⁺Cl⁻(s)
HCl is the Brønsted-Lowry acid (proton donor) and NH₃ is the Brønsted-Lowry base (proton acceptor). The proton is transferred from HCl to NH₃ forming NH₄⁺ and Cl⁻. This reaction goes to completion because HCl is a very strong acid and NH₃ is a relatively strong base, making the reverse reaction (NH₄⁺ donating a proton back to Cl⁻) extremely unfavorable. The products are much more stable than the reactants.
Consider the reaction: CO₂(g) + H₂O(l) → H₂CO₃(aq)
(i) Identify the Lewis acid and Lewis base, explaining your reasoning with reference to electron movement.
(ii) Draw a Lewis structure showing the coordinate bond formation.
(iii) Explain why this reaction can be described by Lewis theory but not by Brønsted-Lowry theory.
(i) Lewis acid: CO₂ (the carbon atom accepts an electron pair into its empty orbital); Lewis base: H₂O (oxygen donates a lone pair to carbon)
(ii) [Lewis structure should show the oxygen from water donating a lone pair to the carbon in CO₂, forming a coordinate bond]
(iii) This is a Lewis acid-base reaction because it involves electron pair donation/acceptance, but it is NOT a Brønsted-Lowry reaction because no proton is transferred in this step. Brønsted-Lowry theory only applies to proton transfer reactions.
25.0 cm³ of 0.100 mol dm⁻³ sulfuric acid (H₂SO₄) is diluted with water to make 500 cm³ of solution. Assume sulfuric acid is completely ionized and releases two protons per molecule.
(i) Calculate the concentration of H⁺ ions in the diluted solution.
(ii) Calculate the pH of the diluted solution.
(iii) Calculate the concentration of OH⁻ ions in the diluted solution at 25°C.
(i) Initial moles of H₂SO₄ = 0.100 × (25.0/1000) = 2.50 × 10⁻³ mol
Since H₂SO₄ releases 2 H⁺ per molecule: moles of H⁺ = 2 × 2.50 × 10⁻³ = 5.00 × 10⁻³ mol
[H⁺] = (5.00 × 10⁻³)/(500/1000) = 0.0100 mol dm⁻³
(ii) pH = -log(0.0100) = 2.00
(iii) [OH⁻] = Kw/[H⁺] = (1.0 × 10⁻¹⁴)/(0.0100) = 1.0 × 10⁻¹² mol dm⁻³
A buffer solution contains 0.150 mol dm⁻³ propanoic acid (C₂H₅COOH) and 0.100 mol dm⁻³ sodium propanoate (C₂H₅COONa). The Ka for propanoic acid is 1.3 × 10⁻⁵.
(i) Calculate the pH of the buffer solution.
(ii) Calculate the pH after adding 10.0 cm³ of 0.100 mol dm⁻³ HCl to 100 cm³ of this buffer solution.
(iii) Explain why the pH change is small.
(i) pH = pKa + log([A⁻]/[HA])
pKa = -log(1.3 × 10⁻⁵) = 4.89
pH = 4.89 + log(0.100/0.150) = 4.89 + (-0.176) = 4.71
(ii) Initial moles in 100 cm³:
C₂H₅COOH: 0.150 × 0.100 = 0.0150 mol
C₂H₅COO⁻: 0.100 × 0.100 = 0.0100 mol
Moles of HCl added: 0.100 × 0.010 = 0.00100 mol
After reaction with HCl:
C₂H₅COO⁻ + H⁺ → C₂H₅COOH
C₂H₅COOH: 0.0150 + 0.00100 = 0.0160 mol
C₂H₅COO⁻: 0.0100 - 0.00100 = 0.00900 mol
New concentrations in 110 cm³:
[C₂H₅COOH] = 0.0160/0.110 = 0.145 mol dm⁻³
[C₂H₅COO⁻] = 0.00900/0.110 = 0.0818 mol dm⁻³
pH = 4.89 + log(0.0818/0.145) = 4.89 + (-0.249) = 4.64
(iii) The pH only changes from 4.71 to 4.64 (a change of 0.07 pH units) because the buffer contains a large reservoir of the conjugate base (C₂H₅COO⁻) which reacts with the added H⁺ ions, converting them to the weak acid. The ratio of [A⁻]/[HA] changes only slightly, resulting in minimal pH change.
A student titrates 20.0 cm³ of 0.150 mol dm⁻³ propanoic acid (C₂H₅COOH, Ka = 1.3 × 10⁻⁵) with 0.100 mol dm⁻³ sodium hydroxide.
(i) Calculate the volume of NaOH required to reach the equivalence point.
(ii) Calculate the pH at the half-equivalence point.
(iii) Calculate the pH at the equivalence point. (Hint: At equivalence point, all acid is converted to its conjugate base; use Kb = Kw/Ka)
(iv) Explain why phenolphthalein is a better indicator choice than methyl orange for this titration.
(i) Moles of C₂H₅COOH = 0.150 × (20.0/1000) = 3.00 × 10⁻³ mol
Mole ratio = 1:1
Volume of NaOH = (3.00 × 10⁻³)/0.100 = 0.0300 dm³ = 30.0 cm³
(ii) At half-equivalence point, pH = pKa
pKa = -log(1.3 × 10⁻⁵) = 4.89
(iii) At equivalence point, total volume = 50.0 cm³
[C₂H₅COO⁻] = (3.00 × 10⁻³)/(50.0/1000) = 0.0600 mol dm⁻³
Kb = Kw/Ka = (1.0 × 10⁻¹⁴)/(1.3 × 10⁻⁵) = 7.69 × 10⁻¹⁰
C₂H₅COO⁻ + H₂O ⇌ C₂H₅COOH + OH⁻
[OH⁻] = √(Kb × [C₂H₅COO⁻]) = √(7.69 × 10⁻¹⁰ × 0.0600) = 6.79 × 10⁻⁶ mol dm⁻³
pOH = -log(6.79 × 10⁻⁶) = 5.17
pH = 14 - 5.17 = 8.83
(iv) Phenolphthalein (pH range 8.3-10.0) is better because the equivalence point is at pH 8.83, which falls within phenolphthalein's range. Methyl orange (pH range 3.1-4.4) would change colour too early in the titration, well before the equivalence point, giving inaccurate results.