Questions on titration curves and choosing an indicator

Question 1

Describe how the pH curve for the titration of 25.0 cm³ of ethanoic acid with sodium hydroxide is measured in the laboratory, and how the pKa of ethanoic acid is found from it.

Accepted Answer

Indicative content

calibrate the pH meter with buffer solutions of known pH, such as pH 7.00 and pH 4.00, rinsing the electrode with deionised water between them
pipette 25.0 cm³ of the acid into a beaker, stir, and record the initial pH with the electrode bulb fully immersed
add sodium hydroxide from a burette in measured portions, recording the volume and the steady pH after each
use smaller additions, 0.10 cm³ or dropwise, near the equivalence point, and continue past it to about twice the equivalence volume
plot pH against volume added and read the equivalence volume from the middle of the vertical section
read the pH at half the equivalence volume, which equals the pKa of ethanoic acid

Levels

Level 3 (5–6 marks) all stages of the practical are described in a workable order, from calibrating the meter through the small additions near equivalence and continuing past it, the equivalence volume is read from the vertical section, and the pKa is read as the pH at half the equivalence volume
Level 2 (3–4 marks) a pH meter and burette are used, the pH is recorded after each addition and the curve plotted, and the pKa is read at half equivalence; what is lost is the calibration, the smaller additions near the end point, or the readings beyond equivalence
Level 1 (1–2 marks) sodium hydroxide is added and the pH noted; an indicator is often used in place of the meter, or the pKa is read at the equivalence point instead of half-way to it

Question 2

Deduce the equivalence volume, the pKa and the Ka of a weak monoprotic acid from the pH curve, recorded as 0.100 mol dm⁻³ sodium hydroxide was added to 25.0 cm³ of the acid. Identify the acid from its Ka in the table, and calculate its concentration.

Accepted Answer

B1 the vertical section, from about pH 6 to 10.5, is centred on 20.0 cm³, so the equivalence volume is 20.0 cm³ B1 half-neutralisation is at 10.0 cm³, where the curve reads pH = pKa = 3.75, so Ka = 10^(−3.75) = 1.78 × 10⁻⁴ mol dm⁻³; accept a pKa read as 3.7 to 3.8 B1 the Ka matches methanoic acid; allow error carried forward from a misread pKa M1 moles of acid = moles of NaOH at equivalence = 0.100 × 0.0200 = 2.00 × 10⁻³ mol, in 0.0250 dm³ A1 concentration of the acid = 0.0800 mol dm⁻³

Question 3

Explain why the pH at the equivalence point is 7.00 when hydrochloric acid is titrated with sodium hydroxide, but below 7 when hydrochloric acid is titrated with ammonia.

Accepted Answer

B1 at equivalence the first solution holds sodium chloride, and neither Na⁺ nor Cl⁻ reacts with water B1 so [H⁺] = [OH⁻] and the pH at equivalence is 7.00 B1 with ammonia the salt is ammonium chloride, and NH₄⁺ is a weak acid that donates protons to water B1 NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) makes the solution acidic, about pH 5.3; an answer naming Cl⁻ as the cause is rejected

Question 4

Calculate the volume of sodium hydroxide needed to reach the equivalence point with the benzoic acid in the table, and the volume and pH at the half-neutralisation point. For benzoic acid Ka = 6.31 × 10⁻⁵ mol dm⁻³.

Accepted Answer

M1 moles of acid = 0.0900 × 0.0200 = 1.80 × 10⁻³ mol, and the reaction is 1 : 1, so 1.80 × 10⁻³ mol of NaOH is needed A1 volume = 1.80 × 10⁻³ ÷ 0.120 = 0.0150 dm³ = 15.0 cm³, so half-neutralisation is at 7.50 cm³ M1 at half-neutralisation [HA] = [A⁻], so pH = pKa = −log(6.31 × 10⁻⁵) A1 pH = 4.20, to two decimal places

Question 5

Calculate the concentration, the Ka and the starting pH of a weak monoprotic acid from the titration data in the table, in which the acid is titrated with sodium hydroxide.

Accepted Answer

M1 moles of acid = moles of NaOH = 0.125 × 0.0160 = 2.00 × 10⁻³ mol, so [HA] = 2.00 × 10⁻³ ÷ 0.0200 = 0.100 mol dm⁻³ M1 8.00 cm³ is half the equivalence volume, so pKa = 3.86 and Ka = 10^(−3.86) = 1.38 × 10⁻⁴ mol dm⁻³ M1 [H⁺] = √(1.38 × 10⁻⁴ × 0.100) = 3.72 × 10⁻³ mol dm⁻³ A1 the starting pH = 2.43, to two decimal places

Question 6

Explain, using the pH curve, why phenolphthalein is not a suitable indicator for the titration of hydrochloric acid with ammonia, and whether bromocresol green, which changes colour over pH 3.8 to 5.4, would be suitable.

Accepted Answer

B1 the vertical section of this curve runs from about 3.5 to 7, and the equivalence pH is about 5.3 B1 phenolphthalein changes over 8.3 to 10.0, outside the vertical section, so it changes colour gradually and well away from the equivalence point B1 bromocresol green's range, 3.8 to 5.4, lies within the vertical section B1 so a single drop takes it through its colour change at equivalence, and it is suitable

Question 7

Explain how an indicator such as methyl orange, HIn, changes colour during a titration in which alkali is added to acid, and why only two or three drops of it are used.

Accepted Answer

B1 the indicator is a weak acid in equilibrium: HIn ⇌ H⁺ + In⁻, and HIn and In⁻ have different colours B1 in acid the high [H⁺] keeps the equilibrium to the left, so the colour of HIn, red, is seen B1 as alkali removes H⁺ the equilibrium shifts to the right, and the colour of In⁻, yellow, takes over B1 the indicator is itself a weak acid, so a larger amount would react with the alkali and change the titre

Question 8

Compare the pH curve for 0.100 mol dm⁻³ hydrochloric acid titrated with 0.100 mol dm⁻³ sodium hydroxide with the curve for 0.100 mol dm⁻³ ethanoic acid titrated with the same alkali.

Accepted Answer

B1 the hydrochloric acid curve starts at pH 1.00 and the ethanoic acid curve at a higher pH, 2.88 B1 the ethanoic acid curve has a buffer region before equivalence, passing through pH 4.76 at half-neutralisation B1 the vertical section is shorter for ethanoic acid, about 7 to 10.5, against about 3.5 to 10.5 for hydrochloric acid B1 the equivalence pH is 7.00 for hydrochloric acid and about 8.7 for ethanoic acid, though both reach equivalence at the same volume and finish at the same pH

Question 9

Explain why, beyond the equivalence point, the pH curves for hydrochloric acid and for ethanoic acid titrated with the same sodium hydroxide solution follow the same line, and calculate the pH of both mixtures when 35.0 cm³ of 0.100 mol dm⁻³ NaOH has been added to 25.0 cm³ of 0.100 mol dm⁻³ acid.

Accepted Answer

B1 after equivalence all the acid has reacted, and the pH is set by the concentration of excess OH⁻ from the sodium hydroxide B1 the OH⁻ formed by ethanoate ions reacting with water is negligible beside that excess M1 excess OH⁻ = 1.00 × 10⁻³ mol in 60.0 cm³, so [OH⁻] = 0.0167 mol dm⁻³ and [H⁺] = 6.00 × 10⁻¹³ mol dm⁻³ A1 pH = 12.22 for both mixtures

Question 10

Compare the pH curve for 0.100 mol dm⁻³ hydrochloric acid titrated with 0.100 mol dm⁻³ ammonia with the curve for 0.100 mol dm⁻³ ethanoic acid titrated with 0.100 mol dm⁻³ sodium hydroxide.

Accepted Answer

B1 the hydrochloric acid curve starts lower, at pH 1.00, against 2.88 for ethanoic acid B1 the vertical section runs from about 3.5 to 7 with ammonia, but from about 7 to 10.5 with sodium hydroxide B1 the equivalence pH is about 5.3 with ammonia, below 7, and about 8.7 with sodium hydroxide, above 7 B1 the ammonia curve levels off near pH 9.25, far below the 12.5 reached with sodium hydroxide; both reach equivalence at the same volume

Question 11

Identify a suitable indicator, methyl orange or phenolphthalein, for each titration, or state that neither is suitable: (a) ethanoic acid with sodium hydroxide; (b) hydrochloric acid with ammonia; (c) ethanoic acid with ammonia.

Accepted Answer

B1 (a) phenolphthalein, whose range lies within the vertical section of about 7 to 10.5 B1 (b) methyl orange, whose range lies within the vertical section of about 3.5 to 7 B1 (c) neither, because a weak acid–weak base curve has no vertical section

Question 12

Calculate the pH at 298 K after the sodium hydroxide in the table has been added to the hydrochloric acid.

Accepted Answer

M1 OH⁻ in excess = (0.100 × 0.0300) − (0.100 × 0.0250) = 5.00 × 10⁻⁴ mol M1 [OH⁻] = 5.00 × 10⁻⁴ ÷ 0.0550 = 9.09 × 10⁻³ mol dm⁻³, so [H⁺] = 1.00 × 10⁻¹⁴ ÷ 9.09 × 10⁻³ = 1.10 × 10⁻¹² mol dm⁻³ A1 pH = 11.96, to two decimal places

Question 13

Explain why the pH at the equivalence point is greater than 7 when ethanoic acid is titrated with sodium hydroxide.

Accepted Answer

B1 at equivalence the solution contains sodium ethanoate B1 the ethanoate ion is the conjugate base of a weak acid and accepts protons from water: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ B1 this produces OH⁻ ions, so [OH⁻] > [H⁺] and the pH is above 7

Question 14

Explain why the pH at the half-neutralisation point of a weak acid titrated with a strong base is equal to the pKa of the acid.

Accepted Answer

B1 at half-neutralisation half the HA has been converted to A⁻, so [HA] = [A⁻] B1 in Ka = [H⁺][A⁻]/[HA] the two concentrations cancel, so Ka = [H⁺] B1 taking −log of both sides gives pKa = pH

Question 15

Suggest the most likely cause of this error, and how it should be put right: a student finds the pKa of benzoic acid by the half-neutralisation method and gets 4.42, against a data book value of 4.20, although the equivalence volume was correct.

Accepted Answer

B1 a correct equivalence volume means the volume measurements were sound, and the error is in the pH readings B1 the pH meter was most likely not calibrated, or calibrated incorrectly, giving readings about 0.2 too high throughout B1 calibrate the meter with two buffers of known pH at the temperature of the titration immediately before use, and repeat

Question 16

Evaluate the use of methyl orange (range 3.1 to 4.4) as the indicator for a titration of propanoic acid with sodium hydroxide, using the pH curve.

Accepted Answer

B1 the vertical section for a weak acid with a strong base runs from about 7 to 10.5 B1 methyl orange would change colour at pH 3.1 to 4.4, on the gently rising part of the curve well before equivalence B1 so the titre would be far too small; methyl orange is unsuitable and phenolphthalein should be used

Question 17

Explain why no indicator gives a sharp end point in a titration of ethanoic acid with ammonia, and how the equivalence volume can be found instead.

Accepted Answer

B1 both the acid and the base are weak, so the pH changes gradually through equivalence, with no vertical section B1 any indicator therefore changes colour over several cubic centimetres, so no single drop marks the end point B1 follow the titration with a pH meter and take the equivalence volume where the curve is steepest; allow a conductivity meter

Question 18

Calculate the ratio [In⁻]/[HIn] for methyl orange, pKIn = 3.70, in a solution of pH 4.40, and state the colour seen.

Accepted Answer

M1 [In⁻]/[HIn] = KIn ÷ [H⁺] = 10^(−3.70) ÷ 10^(−4.40) = 10^0.70 B1 In⁻ is five times HIn, so the yellow alkaline colour is seen; allow yellow-orange A1 [In⁻]/[HIn] = 5.01, to three significant figures

Question 19

Justify the choice of phenolphthalein, which changes colour over pH 8.3 to 10.0, rather than methyl orange, which changes over pH 3.1 to 4.4, for the titration of ethanoic acid with sodium hydroxide.

Accepted Answer

B1 the equivalence point lies above pH 7, at about pH 8 to 9, because the ethanoate ion formed is a weak base that accepts protons from water B1 the near-vertical part of the curve runs from about pH 7 to pH 11, and the whole range of phenolphthalein lies inside it, so its colour changes with one drop at the equivalence point B1 methyl orange changes colour at pH 3.1 to 4.4, in the buffer region where the pH is rising slowly, so it would change gradually and well before the equivalence volume; reject 'methyl orange is for strong acids' alone

Question 20

Suggest why, in the pH-curve practical, the base is added in 2.00 cm³ portions at first but in single drops close to the equivalence point.

Accepted Answer

B1 away from equivalence the pH changes slowly, so large additions still give enough points to fix the curve B1 near equivalence one drop can change the pH by several units, so small additions are needed to put enough points on the vertical section to locate its middle

Acid	Ka / mol dm⁻³
Methanoic acid	1.78 × 10⁻⁴
Benzoic acid	6.31 × 10⁻⁵
Ethanoic acid	1.74 × 10⁻⁵

Solution	Volume / cm³	Concentration / mol dm⁻³
Benzoic acid, in the flask	20.0	0.0900
Sodium hydroxide, in the burette	to be found	0.120

Quantity	Value
Volume of acid	20.0 cm³
Concentration of sodium hydroxide	0.125 mol dm⁻³
Volume of sodium hydroxide at the equivalence point	16.0 cm³
pH after 8.00 cm³ of sodium hydroxide	3.86

Solution	Volume / cm³	Concentration / mol dm⁻³
Hydrochloric acid	25.0	0.100
Sodium hydroxide added	30.0	0.100