Acids, Bases & pH

Which species is acting as a Brønsted–Lowry base in the forward reaction HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻?

• AH₂O
• BHCO₃⁻✓
• CH₂CO₃
• DOH⁻

Answer: A Brønsted–Lowry base is a proton (H⁺) acceptor. Here HCO₃⁻ accepts a proton from water to become H₂CO₃, so HCO₃⁻ is the base and water is the acid in this forward reaction.

Calculate the pH of 0.0500 mol dm⁻³ hydrochloric acid (a strong monoprotic acid). Use pH = −log₁₀[H⁺].

• ApH = 1.30✓
• BpH = 2.00
• CpH = 1.70
• DpH = 0.05

Answer: HCl is a strong acid that fully dissociates, so [H⁺] = 0.0500 mol dm⁻³. pH = −log₁₀(0.0500) = 1.30.

Calculate the pH of 0.0200 mol dm⁻³ sodium hydroxide at 298 K (Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶).

• ApH = 1.70
• BpH = 2.00
• CpH = 12.30✓
• DpH = 12.00

Answer: NaOH is a strong base, so [OH⁻] = 0.0200 mol dm⁻³. [H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴ / 0.0200 = 5.0 × 10⁻¹³ mol dm⁻³. pH = −log₁₀(5.0 × 10⁻¹³) = 12.30. (Equivalently pOH = 1.70, pH = 14 − 1.70.)

A weak monoprotic acid HA has Ka = 1.8 × 10⁻⁵ mol dm⁻³. Calculate the pH of a 0.100 mol dm⁻³ solution, assuming [H⁺] = √(Ka × c).

• ApH = 4.74
• BpH = 5.74
• CpH = 1.00
• DpH = 2.87✓

Answer: [H⁺] = √(Ka × c) = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³. pH = −log₁₀(1.34 × 10⁻³) = 2.87. The 4.74 distractor is the pKa, not the pH.

The acid dissociation constant of ethanoic acid is Ka = 1.8 × 10⁻⁵ mol dm⁻³. What is its pKa?

• A5.00
• B1.80
• C4.74✓
• D9.26

Answer: pKa = −log₁₀(Ka) = −log₁₀(1.8 × 10⁻⁵) = 4.74. A smaller Ka (weaker acid) corresponds to a larger pKa.

Which mixture would act as an effective acidic buffer solution?

• AHydrochloric acid and sodium chloride
• BSodium hydroxide and sodium chloride
• CEthanoic acid and sodium ethanoate✓
• DHydrochloric acid and sodium hydroxide in equal moles

Answer: An acidic buffer needs a weak acid and the salt of that weak acid (its conjugate base). Ethanoic acid (weak acid) with sodium ethanoate provides a reservoir of both CH₃COOH and CH₃COO⁻ to resist pH change. A strong acid plus its salt (A) cannot buffer.

A buffer is made from a weak acid HA (pKa = 4.76) and its sodium salt. The concentration of the salt A⁻ is 0.20 mol dm⁻³ and that of the acid HA is 0.10 mol dm⁻³. Using pH = pKa + log₁₀([A⁻]/[HA]), calculate the pH.

• A5.06✓
• B4.46
• C4.76
• D9.52

Answer: pH = pKa + log₁₀([A⁻]/[HA]) = 4.76 + log₁₀(0.20/0.10) = 4.76 + log₁₀(2) = 4.76 + 0.30 = 5.06. Because there is more conjugate base than acid, the pH is above the pKa.

What is a Brønsted-Lowry acid?

• AA proton donor✓
• BA proton acceptor
• CAn electron pair donor
• DAn electron pair acceptor

Answer: The Brønsted-Lowry definition is about protons: acids donate them, bases accept them. Electron pair donors and acceptors are Lewis bases and Lewis acids, a broader definition.

How does a strong acid differ from a concentrated acid?

• AStrong means many moles per litre; concentrated means fully dissociated
• BStrong means it reacts quickly; concentrated means it reacts slowly
• CStrong means fully dissociated; concentrated means many moles per litre✓
• DStrong means corrosive; concentrated means dilute in water

Answer: Strength describes how completely the acid ionises, concentration describes how much is dissolved. Ethanoic acid can be concentrated but is always weak, and hydrochloric acid can be dilute but is always strong.

What does Ka measure?

• AThe extent to which a strong acid dissociates
• BThe concentration of hydrogen ions in a solution
• CThe volume of base needed to neutralise the acid
• DThe extent to which a weak acid dissociates✓

Answer: Ka is the equilibrium constant for the dissociation of a weak acid, so a larger Ka means a stronger weak acid. pKa is its negative logarithm, so a smaller pKa means stronger.

What is the pH of a 0.1 mol dm⁻³ solution of a strong monoprotic acid?

• A0.1
• B13
• C7
• D1✓

Answer: A strong acid dissociates fully, so [H⁺] equals 0.1 mol dm⁻³ and pH = −log(0.1) = 1. A pH of 13 would be a strong alkali of the same concentration.

How does a buffer solution resist pH change when acid is added?

• AThe conjugate base present reacts with the added hydrogen ions✓
• BThe weak acid present reacts with all of the added hydrogen ions
• CThe water present simply dilutes the added hydrogen ions away
• DThe buffer neutralises the acid completely and permanently

Answer: An acidic buffer holds a large reservoir of both the weak acid and its conjugate base. Added H⁺ is mopped up by the conjugate base and added OH⁻ by the weak acid, so pH barely moves until a reservoir is exhausted.

What is Kw at 298 K?

• A1.0 × 10⁻⁷ mol² dm⁻⁶
• B1.0 × 10⁻¹⁴ mol dm⁻³
• C1.0 × 10⁷ mol² dm⁻⁶
• D1.0 × 10⁻¹⁴ mol² dm⁻⁶✓

Answer: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K, which is why neutral water has pH 7. Kw rises with temperature, so neutral water is slightly below pH 7 when warmer.

Why does the pH change sharply near the endpoint of a strong acid-strong base titration?

• AA small volume added changes hydrogen ion concentration greatly✓
• BA large volume added changes hydrogen ion concentration slightly
• CThe indicator itself reacts with the acid at that point
• DThe reaction becomes exothermic close to the endpoint

Answer: Near equivalence almost all the acid has reacted, so each further drop of base makes a proportionally enormous difference to [H⁺]. Because pH is logarithmic that shows as a near-vertical section of the curve.

Why is phenolphthalein unsuitable for a strong acid-weak base titration?

• AIts range lies exactly at the equivalence point
• BIt reacts chemically with weak bases in solution
• CIt is colourless in both acidic and alkaline conditions
• DIts range lies outside the steep part of the curve✓

Answer: The equivalence point for a strong acid and weak base is below pH 7, whereas phenolphthalein changes around pH 8.3–10. Methyl orange, changing around pH 3.1–4.4, sits in the steep section instead.

What is a conjugate acid-base pair?

• ATwo species differing by one proton✓
• BTwo species differing by one electron
• CTwo species with identical pH values
• DAn acid and a base that fully neutralise

Answer: Removing a proton from an acid leaves its conjugate base, as with HCl and Cl⁻. Every Brønsted-Lowry reaction involves two such pairs exchanging a proton.

25.0 cm³ of 0.150 mol dm⁻³ hydrochloric acid is mixed with 20.0 cm³ of 0.120 mol dm⁻³ sodium hydroxide.

What is the pH of the resulting solution?

• A1.27
• B1.52✓
• C0.82
• D2.87

Answer: HCl = 0.0250 × 0.150 = 3.75 × 10⁻³ mol; NaOH = 0.0200 × 0.120 = 2.40 × 10⁻³ mol. Excess H⁺ = 1.35 × 10⁻³ mol in a total of 45.0 cm³, so [H⁺] = 0.0300 mol dm⁻³ and pH = 1.52. Dividing by only the 25.0 cm³ of acid gives 1.27; not dividing by any volume gives 2.87.

35.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide is added to 20.0 cm³ of 0.100 mol dm⁻³ ethanoic acid at 298 K. (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶)

What is the pH of the final mixture?

• A12.63
• B12.44✓
• C1.56
• D13.00

Answer: NaOH = 3.50 × 10⁻³ mol; ethanoic acid = 2.00 × 10⁻³ mol, so 1.50 × 10⁻³ mol of OH⁻ is left over in 55.0 cm³: [OH⁻] = 0.0273 mol dm⁻³. [H⁺] = Kw ÷ [OH⁻] = 3.67 × 10⁻¹³, so pH = 12.44. Using 35.0 cm³ instead of the total volume gives 12.63, and 1.56 is the pOH, not the pH.

A buffer is 500 cm³ of solution containing 0.200 mol dm⁻³ methanoic acid and 0.150 mol dm⁻³ sodium methanoate. 0.0100 mol of solid sodium hydroxide is dissolved in it (assume no volume change). (Ka of methanoic acid = 1.80 × 10⁻⁴ mol dm⁻³)

What is the pH after the addition?

• A3.72✓
• B3.62
• C3.52
• D3.74

Answer: Before: HCOOH = 0.100 mol, HCOO⁻ = 0.0750 mol. OH⁻ converts acid to its salt: HCOOH = 0.0900 mol, HCOO⁻ = 0.0850 mol. [H⁺] = Ka × [HA] ÷ [A⁻] = 1.80 × 10⁻⁴ × 0.0900 ÷ 0.0850 = 1.91 × 10⁻⁴, so pH = 3.72. The original buffer has pH 3.62; moving 0.0100 mol the wrong way (salt to acid) gives 3.52, and 3.74 is the pKa.

A 25.0 cm³ sample of benzoic acid solution was titrated with sodium hydroxide. The equivalence point was at 25.00 cm³ of alkali, where the pH was 8.60. After 12.50 cm³ of alkali had been added, the pH was 4.20.

What is Ka for benzoic acid?

• A6.3 × 10⁻⁵ mol dm⁻³✓
• B2.5 × 10⁻⁹ mol dm⁻³
• C1.3 × 10⁻⁴ mol dm⁻³
• D3.2 × 10⁻⁵ mol dm⁻³

Answer: At half-neutralisation (12.50 cm³), half the acid has become its salt, so [HA] = [A⁻] and Ka = [H⁺]. Ka = 10⁻⁴·²⁰ = 6.3 × 10⁻⁵ mol dm⁻³. The pH at the equivalence point (8.60) reflects the salt solution, so using it gives 2.5 × 10⁻⁹, which is far too small.

Indicator pH ranges: bromophenol blue 3.0–4.6; methyl red 4.4–6.2; bromothymol blue 6.0–7.6; phenolphthalein 8.3–10.0.

Which indicator is best for titrating propanoic acid with potassium hydroxide?

• APhenolphthalein✓
• BMethyl red
• CBromothymol blue
• DBromophenol blue

Answer: This is a weak acid against a strong base, so the equivalence point is above 7 (about pH 9) and the steep part of the curve runs from roughly pH 7 to 11. Phenolphthalein changes colour within that range. Bromothymol blue changes colour below the steep section, so it would change too early.