Questions on weak acids, Ka and pKa; Kw and strong bases

Question 1

Describe how a student could find Ka and pKa for benzoic acid from a single pH measurement on a solution of known concentration, and explain the assumptions in the calculation and when they would fail.

Accepted Answer

Indicative content

measure the pH of a benzoic acid solution of accurately known concentration with a pH meter calibrated against buffers
convert the reading to [H⁺] with [H⁺] = 10^(−pH), carrying the unrounded value into the next step of the calculation
assume [H⁺] = [A⁻], because the hydrogen ions from the dissociation of water are negligible beside those from the acid
assume [HA] at equilibrium equals the concentration made up, because only a small fraction of the weak acid dissociates
calculate Ka = [H⁺]² ÷ [HA] in mol dm⁻³, then pKa = −log Ka, giving pKa to two decimal places with no units
the second assumption fails if more than about 5% of the acid dissociates, as in very dilute solutions or for acids with a large Ka

Levels

Level 3 (5–6 marks) every stage is present in order: the calibrated pH measurement on a solution of known concentration, the conversion to [H⁺], both assumptions stated with the reason for each, Ka and pKa calculated correctly with units handled, and the condition under which the small-dissociation assumption fails
Level 2 (3–4 marks) the measurement, the conversion to [H⁺] and the Ka expression are right and one assumption is stated; why it holds, or when it fails, is left vague, or the pKa step is not finished
Level 1 (1–2 marks) the pH is measured and Ka written down; a frequent slip is to take [H⁺] as equal to the acid concentration, as for a strong acid, and no assumption is justified

Question 2

Calculate Ka and pKa for a weak monoprotic acid, given that a 0.0800 mol dm⁻³ solution of it has a pH of 3.10 at 298 K.

Accepted Answer

M1 [H⁺] = 10^(−3.10) = 7.94 × 10⁻⁴ mol dm⁻³ M1 Ka = [H⁺]² / [HA] = (7.94 × 10⁻⁴)² ÷ 0.0800 A1 Ka = 7.89 × 10⁻⁶ mol dm⁻³, to three significant figures; allow 7.88 to 7.90 A1 pKa = −log(7.89 × 10⁻⁶) = 5.10, to two decimal places; allow error carried forward from Ka

Question 3

Explain why the pH of pure water falls as its temperature rises, and why the water remains neutral.

Accepted Answer

B1 the dissociation H₂O ⇌ H⁺ + OH⁻ is endothermic B1 a rise in temperature shifts the equilibrium to the right, so Kw increases B1 [H⁺] increases, so pH = −log[H⁺] falls below 7 B1 [OH⁻] increases by the same amount, so [H⁺] = [OH⁻] and the water is still neutral; reject 'neutral because pH 7'

Question 4

Compare 0.100 mol dm⁻³ hydrochloric acid with 0.100 mol dm⁻³ ethanoic acid in their pH, their rate of reaction with magnesium ribbon, and the volume of 0.100 mol dm⁻³ sodium hydroxide needed to neutralise 25.0 cm³ of each.

Accepted Answer

B1 hydrochloric acid has the lower pH, 1.00 against 2.88, because it is fully dissociated B1 magnesium reacts faster with the hydrochloric acid, because its [H⁺] is higher B1 both need 25.0 cm³ of sodium hydroxide, the same volume B1 because as OH⁻ removes H⁺ from ethanoic acid the equilibrium shifts right until all the acid has reacted

Question 5

Calculate the mass of ethanoic acid, CH₃COOH, dissolved in the solution described in the table.

Accepted Answer

M1 [H⁺] = 10^(−3.10) = 7.94 × 10⁻⁴ mol dm⁻³ M1 [CH₃COOH] = [H⁺]² ÷ Ka = (7.94 × 10⁻⁴)² ÷ 1.74 × 10⁻⁵ = 0.0363 mol dm⁻³ M1 moles in 500 cm³ = 0.0363 × 0.500 = 0.0181 mol A1 mass = 0.0181 × 60.0 = 1.09 g, to three significant figures

Question 6

Give the expression for the acid dissociation constant, Ka, of propanoic acid, CH₃CH₂COOH, with its units, and give the relationship between pKa and Ka, saying whether a larger pKa means a stronger or a weaker acid.

Accepted Answer

B1 Ka = [H⁺][CH₃CH₂COO⁻] / [CH₃CH₂COOH] B1 the units of Ka are mol dm⁻³ B1 pKa = −log Ka, so a larger pKa means a smaller Ka and a weaker acid

Question 7

Calculate the pH of 0.0150 mol dm⁻³ barium hydroxide, Ba(OH)₂, at 298 K. Barium hydroxide is fully dissociated.

Accepted Answer

M1 each Ba(OH)₂ gives two OH⁻, so [OH⁻] = 2 × 0.0150 = 0.0300 mol dm⁻³ M1 [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0300 = 3.33 × 10⁻¹³ mol dm⁻³ A1 pH = 12.48, to two decimal places; 12.18 from [OH⁻] = 0.0150 is rejected

Question 8

Calculate the pH of 0.0500 mol dm⁻³ methanoic acid at 298 K. Ka = 1.78 × 10⁻⁴ mol dm⁻³.

Accepted Answer

M1 Ka = [H⁺]² / [HCOOH], taking [H⁺] = [HCOO⁻] and [HCOOH] = 0.0500 mol dm⁻³ M1 [H⁺] = √(1.78 × 10⁻⁴ × 0.0500) = 2.98 × 10⁻³ mol dm⁻³ A1 pH = 2.53, to two decimal places

Question 9

Calculate the concentration of ethanoic acid that has a pH of 3.00 at 298 K. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

Accepted Answer

M1 [H⁺] = 10^(−3.00) = 1.00 × 10⁻³ mol dm⁻³ M1 [CH₃COOH] = [H⁺]² ÷ Ka = (1.00 × 10⁻³)² ÷ 1.74 × 10⁻⁵ A1 [CH₃COOH] = 0.0575 mol dm⁻³, to three significant figures

Question 10

Calculate the pH of pure water and the pH of 0.100 mol dm⁻³ sodium hydroxide at 50 °C, where Kw = 5.48 × 10⁻¹⁴ mol² dm⁻⁶.

Accepted Answer

M1 in pure water [H⁺] = √(5.48 × 10⁻¹⁴) = 2.34 × 10⁻⁷ mol dm⁻³, giving pH 6.63 M1 in the NaOH, [H⁺] = 5.48 × 10⁻¹⁴ ÷ 0.100 = 5.48 × 10⁻¹³ mol dm⁻³ A1 pH of the sodium hydroxide = 12.26, to two decimal places

Question 11

Explain the two approximations made when the pH of a weak acid HA is calculated from its Ka and concentration.

Accepted Answer

B1 [H⁺] = [A⁻], because the H⁺ from the dissociation of water is negligible compared with that from the acid B1 [HA] at equilibrium equals the initial concentration of the acid B1 because the acid is only slightly dissociated, so the amount of HA lost is negligible

Question 12

Explain why diluting a weak acid tenfold raises its pH by about 0.5 units, not by 1 unit.

Accepted Answer

B1 for a weak acid [H⁺] = √(Ka × [HA]), so [H⁺] varies with the square root of the concentration B1 a tenfold dilution divides [H⁺] by √10, about 3.2, which is a change of log √10 = 0.5 in pH B1 dilution shifts HA ⇌ H⁺ + A⁻ to the right, so a greater fraction of the acid dissociates and [H⁺] falls by less than the concentration

Question 13

Suggest why the usual weak-acid approximations give a poor value for the pH of 0.00100 mol dm⁻³ dichloroethanoic acid, which has Ka = 5.0 × 10⁻² mol dm⁻³.

Accepted Answer

B1 the approximation assumes [HA] at equilibrium equals the initial concentration, 0.00100 mol dm⁻³ B1 with so large a Ka and so dilute a solution, most of the acid dissociates, about 98% B1 so [HA] is far below 0.00100 mol dm⁻³; the approximation gives pH 2.15, while the true pH is about 3.01

Question 14

Suggest why chloroethanoic acid, ClCH₂COOH, pKa 2.87, is a stronger acid than ethanoic acid, pKa 4.76.

Accepted Answer

B1 chlorine is more electronegative than hydrogen and draws electron density away from the carboxylate group B1 the negative charge on the ClCH₂COO⁻ ion is spread out, so the ion is more stable B1 the anion is therefore less ready to accept a proton back, so the equilibrium lies further to the right and Ka is larger

Question 15

Evaluate a student's method for the pH of 1.00 × 10⁻⁴ mol dm⁻³ ethanoic acid. The student uses [H⁺] = √(Ka × [HA]) and gets 4.38; the pH found without approximations is 4.47.

Accepted Answer

B1 the method assumes the acid is only slightly dissociated, so that [HA] equals 1.00 × 10⁻⁴ mol dm⁻³ B1 at this low concentration about a third of the acid is dissociated, so [HA] is well below the value used and [H⁺] is overestimated B1 the method is reliable at higher concentrations, below about 5% dissociation, but gives a pH 0.09 too low here

Question 16

Explain why 0.100 mol dm⁻³ ammonia solution has a pH of about 11.1, while 0.100 mol dm⁻³ sodium hydroxide has a pH of 13.00.

Accepted Answer

B1 sodium hydroxide is a strong base, fully dissociated, so [OH⁻] = 0.100 mol dm⁻³ B1 ammonia is a weak base, only partially protonated in NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, so [OH⁻] is much less than 0.100 mol dm⁻³ B1 a lower [OH⁻] means a higher [H⁺] = Kw ÷ [OH⁻], so the pH of the ammonia is lower

Question 17

Calculate the pH of pure water and of 0.0100 mol dm⁻³ sodium hydroxide at 0 °C, where pKw = 14.94.

Accepted Answer

M1 in pure water pH = pOH, so pH = ½pKw = 14.94 ÷ 2 A1 the pH of pure water at 0 °C is 7.47 A1 for the NaOH, pOH = −log(0.0100) = 2.00, so pH = 14.94 − 2.00 = 12.94

Question 18

Show that the pH of 0.100 mol dm⁻³ ethanoic acid at 298 K is 2.88. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

Accepted Answer

M1 Ka = [H⁺][CH₃COO⁻] ÷ [CH₃COOH], with [H⁺] = [CH₃COO⁻] and [CH₃COOH] taken as 0.100 mol dm⁻³, so [H⁺]² = Ka × 0.100 M1 [H⁺] = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³ A1 pH = −log₁₀(1.32 × 10⁻³) = 2.88, to two decimal places; the two assumptions, that very little acid dissociates and that water supplies negligible H⁺, should be stated

Question 19

State the expression for the ionic product of water, Kw, and give its value with units at 298 K.

Accepted Answer

B1 the ionic product of water is Kw = [H⁺][OH⁻] B1 Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K

Question 20

Calculate the pH of 0.0250 mol dm⁻³ potassium hydroxide at 298 K.

Accepted Answer

M1 [OH⁻] = 0.0250 mol dm⁻³, so [H⁺] = Kw ÷ [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 0.0250 = 4.00 × 10⁻¹³ mol dm⁻³ A1 pH = −log(4.00 × 10⁻¹³) = 12.40, to two decimal places

Question 21

Describe how the pH of a strong base is calculated from its concentration at 298 K.

Accepted Answer

B1 find [OH⁻] from the concentration of the base, multiplying by the number of OH⁻ per formula unit B1 calculate [H⁺] = Kw ÷ [OH⁻], then pH = −log[H⁺]

Question 22

Suggest why a value of Ka found from one pH reading of a weak acid of known concentration is more affected by a badly calibrated pH meter than a value found at the half-neutralisation point.

Accepted Answer

B1 from one reading Ka = [H⁺]² ÷ [HA], so pKa = 2pH + log[HA] and an error in pH is doubled in pKa B1 at half-neutralisation pKa equals the pH reading, so the same error appears in pKa only once

Quantity	Value
Volume of solution	500 cm³
pH at 298 K	3.10
Ka of ethanoic acid at 298 K	1.74 × 10⁻⁵ mol dm⁻³
Mr of CH₃COOH	60.0