An alkaline buffer solution is the solution of a weak base and the salt of the weak base (that resists/opposes changes in pH upon addition of small amounts of acid or base).

0.20 mol NH₃ and 0.05 mol H₂SO₄ will produce an alkaline buffer, since 0.05 mol H₂SO₄ will neutralise half (0.10 mol) of the NH₃ (1:2 ratio), leaving unreacted weak base, NH₃, and the salt of the weak base, (NH₄)₂SO₄.

2NH₃ + H₂SO₄ → (NH₄)₂SO₄

0.20 mol NH₃ and 0.05 mol H₂SO₄ is therefore the correct answer.

Incorrect answers

0.20 mol NH₃ and 0.10 mol H₂SO₄ will result in all of the ammonia, NH₃, being neutralised (1:2 ratio) so this cannot form a buffer.

A combination of ethanoic acid, CH₃COOH, and sodium ethanoate, CH₃COONa, (in any reasonable quantities) will form a buffer solution, but that will be an acidic buffer.

A combination of an excess of ethanoic acid with sodium hydroxide will also form a buffer solution, since all of the sodium hydroxide will be neutralised and will produce sodium ethanoate, but an excess of NaOH will neutralise all the acid and the solution will not be a buffer (as is the case in this question).

A nucleophile is a substance which can donate a lone pair of electrons, which is an identical description to that of a Lewis base.

Likewise, an electrophile is equivalent to a Lewis acid as it can accept a lone pair of electrons.

Lewis base is therefore the correct answer.

CH₃COOH is an acid.

The other three options are all salts.

As a useful rule-of-thumb, salts formed from strong acids and strong bases (e.g. NaCl) and those from weak acids and weak bases give approximately neutral solutions of pH=7.

Salts formed from strong acids and weak bases (e.g. NH₄Cl) are acidic and those from weak acids and strong bases (e.g. CH₃COONa) are alkaline.

Salt hydrolysis, is due to the reaction of the conjugate bases/acids (salt ions) of weak acids/bases with water:

E.g. NH₄⁺ + H₂O ⇌ NH₄OH + H⁺

E.g. CH₃COO^– + H₂O ⇌ CH₃COOH + OH^–

CH₃CO₂Na is therefore the correct answer.

A Lewis acid is a substance that can accept a pair of electrons from a Lewis base, forming a co-ordinate (dative) bond.

A Bronsted-Lowry acid can donate an H⁺ ion.

Boron in BF₃ is electron deficient (it has an incomplete octet) and so it can accept a pair of electrons. BF₃ cannot act as a Bronsted-Lowry acid, as it has no hydrogens.

None of the other species can act as Lewis acids, as none of the atoms are electron deficient, and they all have complete octets (and cannot expand them); or two electrons in the case of hydrogen.

The correct answer is therefore BF₃.

Weak acids only partially dissociate: HA ⇌ H⁺ + A⁻

\(K_a = {{ [A^-][H^+] } \over [HA]}\)

Therefore the Ka for a stronger acid will be greater than for a weaker acid (more ions on top of fraction).

pKa = −log₁₀Ka

Therefore higher Ka leads to lower pKa (similarly to the relationship between [H⁺] and the pH scale).

Stronger acids will always have lower pH than weak acids of the same concentration because of the increased dissociation.

Without a calculator we need to remember that pKa is a logarithmic scale, so Ka = 1.4 × 10⁻³ will give a pKa of approx. 3 (the × 10⁻³ tells us this) and Ka = 1.0 × 10⁻¹⁰ will give a pKa of approx.10.

The strongest acid (with a pKa of approx. 3) is therefore CH₂ClCOOH, and so this will have the lowest pH.

Therefore CH₂ClCOOH (Ka = 1.4 × 10⁻³) is the correct answer.

The shapes of the four pH titration curves need to be learned.

The starting pH (∼3), high finishing pH and the steep inflection in the curve above pH 7 indicate that this is a weak acid, strong base pH curve. Therefore the curve could have been produced by titration of ethanoic acid with sodium hydroxide solution.

The equivalence point is the point at which the moles of acid = moles of base, and for a weak acid and strong base pH curve, this is typically around pH 9 (not pH 7).

At half the equivalence point (at 12.5cm³) half of the acid has been neutralised so the concentration of acid and salt is the same: [HA] = [A^–] which means this is a buffer solution, and so this part of the curve is known as the buffer region.

1 and 3 only is therefore the correct answer.

If the temperature is increased the equation will shift in the endothermic direction, which means the equilibrium will shift to the right.

Shifting the equilibrium to the right will increase the quantity of H⁺ ions and OH⁻ ions in the aqueous solution; [H⁺] and [OH⁻] will increase.

Kw = [H⁺] × [OH⁻]

Kw will therefore increase as the concentration of H⁺ and OH⁻ ions increases, and pH will decrease as H⁺ ion concentration increases, since pH=−log₁₀[H⁺].

Kw will increase and the pH will decrease is therefore the correct answer.

Indicators are weak acids that change colour when [HIn] = [In⁻] (In practice the colour changes across a pH range). At the point when [HIn] = [In⁻], K_a = [H⁺] and pH = pKa (see expression below; [HIn] and [In⁻] cancel each other out).

\(K_a = {{[H^+][In^-]} \over [HIn]}\)

Therefore indicators change colour across a range 'straddling' pH = pKa.

Therefore, for the indicator in this question, at pH 4.2, [Hin] = [In⁻] and since the protonated and deprotonated forms are yellow and blue, a 50:50 mixture will be green.

At pH 7.0 the indicator in this question will be blue (not yellow), since less H+ will push the equilibrium right and [Hin] << [In⁻]

2 and 3 only is therefore the correct answer.

The shapes of the four pH titration curves need to be learned.

This is a strong acid, weak base pH curve. Notice the low starting pH, finishing pH of ∼11, and the steep inflection in the curve below pH 7.

The equivalence point is at pH 5/6.

Hydrochloric acid and ammonia solution (strong acid and weak base) is therefore the correct answer.

A combination of ethanoic acid, CH₃COOH, and the salt of the weak acid e.g. sodium ethanoate, CH₃COONa, (in any reasonable quantities) will form a buffer solution, and that will be an acidic buffer (with a pH < 7).

A combination of an excess of ethanoic acid with sodium hydroxide will also form a buffer solution, since all of the sodium hydroxide will be neutralised and will produce sodium ethanoate (CH₃COOH + NaOH → CH₃COONa + H₂O), and this results in a mixture of unreacted ethanoic acid and sodium ethanoate.

12.5 cm³ of NaOH and pH = 4.8 is therefore the correct answer, since this volume of NaOH will neutralise half of the ethanoic acid, and the resulting buffer will be acidic (pH < 7).