OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2019: Question 20
14 marks · Hard difficulty · Calculations
Calculate pH of weak acids and buffer solutions, determine titration curves, choose suitable indicators, and explain assumptions in acid dissociation.
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Question text
20 This question is about weak acids.
The Ka values of three weak acids are shown in Table 20.1.
Weak acid K / mol dm−3
a
iodic(V) acid, HIO (aq) 1.78 × 10−1
propanoic acid, C H COOH(aq) 1.35 × 10−5
hydrocyanic acid, HCN(aq) 6.17 × 10−10
Table 20.1
(a) Calculate the pH of 0.0800 mol dm−3 C H COOH(aq).
Give your answer to 2 decimal places.
pH = … [2]
(b) A student adds a total of 45.0 cm3 of 0.100 mol dm−3 NaOH(aq) to 25.0 cm3 of 0.0800 mol dm−3
C2H5COOH(aq) and monitors the pH throughout.
(i) Show by calculation that 20.0 cm3 of NaOH(aq) is required to reach the end point.
[1]
(ii) Calculate the pH of the final solution.
Give your answer to 2 decimal places.
pH = … [4]
(iii) On the axes below, sketch a pH curve for the pH changes during the addition of 45.0 cm3
of 0.100 mol dm−3 NaOH(aq) to 25.0 cm3 of 0.0800 mol dm−3 C H COOH(aq).
pH 7
0 10 20 30 40 50
volume NaOH(aq) added / cm3
[3]
(iv) The student considers using the four indicators in Table 20.2 for the titration.
Indicator pH range
Cresol red 0.2 – 1.8
Bromophenol blue 3.0 – 4.6
Cresol purple 7.6 – 9.2
Indigo carmine 11.6 – 14.0
Table 20.2
Explain which indicator would be most suitable for the titration.
… [1]
(v) The student repeats the experiment starting with 25.0 cm3 of 0.0800 mol dm−3 HCN(aq)
and adding a total of 45.0 cm3 of 0.100 mol dm−3 NaOH(aq).
Predict one similarity and one difference between the pH curve with C2H5COOH(aq)
and the pH curve with HCN(aq). Use the information in Table 20.1, and your answer to
(b)(iii).
Similarity …
Difference …
[2]
(c) The student calculates the pH of 0.0800 mol dm−3 HIO (aq). The student assumes that the
equilibrium concentration of HIO3(aq) is the same as the initial concentration of HIO3(aq).
The student measures the pH, and finds that the measured pH value is different from the
calculated pH value.
Explain why the measured pH is different from the calculated pH.
… [1]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
20 (a) FIRST CHECK THE ANSWER ON ANSWER LINE 2 2.2 ×2 ALLOW ECF throughout
If answer = 2.98 award 2 marks
-------------------------------------------------------------------------- ONLY ALLOW pH mark by ECF if Ka AND 0.080
[H+] = √(Ka × [C H COOH]) = 1.039 × 10–3 (mol dm–3) used and AND pH <7
pH = –log 1.039 × 10–3 = 2.98 (Must be to 2 DP) Common errors (Must be to 2 DP)
One mark for pH = 5.97 (No square root):
One mark for pH = 0.92 OR pH = 5.15 (Using
incorrect Ka values)
(b) (i) 25.0 1 2.5 ALLOW 0.02 dm3 if unit given
n(C2H5COOH) = (0.0800 × ) =) 0.002 (mol)
1000
AND Mark is for WORKING which could all be shown as
0.002 3 1 step
V(NaOH) = × 1000 = (= 20(.0) cm )
0.100
ALLOW method showing 20cm3 NaOH contains
the same moles as acid
n(C2H5COOH) = 0.08(00) x 0.025(0) = 0.002 (mol)
and
n(NaOH) = 0.02(00) x 0.1 = 0.002(00) (mol)
20 b (ii) FIRST CHECK THE ANSWER ON ANSWER LINE 4 ALLOW ECF throughout
H432/01 Mark Schemes June 2019
AO
Question Answer 20 Marks Guidance
element
If answer = 12.55 award 4 marks
-------------------------------------------------------------------------- For first mark ALLOW
(Excess volume of NaOH = 25(.0) cm3)
Excess mol of NaOH: – 25.0
n(OH )excess = 0.100 × = 0.0025 (mol)
1000
n(OH–) = n(OH–) – n(C H COOH)
excess 2 5
Common errors
45.0 25.0 1.2 ×1 If initial V(NaOH) = 45 cm3
= (0.100 × 1000) – (0.0800 × ) –
1000 [OH ] = 0.0643 (mol)
[H+] = 1.56 × 10–13 (mol dm–3)
= 0.0045 – 0.002 = 0.0025 (mol) 2.6 ×3 pH = 12.81 award three marks (no 1st mark)
Concentration of OH–: – +
If n(OH )excess is used in [H ] calculation
0.0025 n(OH–) = 0.0025 (mol)
[OH–] = ( ) = 0.0357 (mol dm–3) excess
70.0 × 10–3 –14
+ 1.00 × 10 –12 –3
[H ] = = 4.(00) × 10 (mol dm )
0.0025
Concentration of H+: nd
pH = 11.40 award three marks (no 2 mark)
1.00 × 10–14
[H+] = ( ) = 2.8 × 10–13 (mol dm–3)
0.0357
ALLOW pOH method for last two marks
pOH = – log[OH–] =1.447
Conversion to pH:
pH = (–log 2.8 × 10–13) = 12.55
pH = 14 – 1.447 = 12.55
ALLOW ECF for conversion from [H+] to pH
provided value calculated is above 7 and from
derived [H+]
20 b (iii) Shape 3 2.3 ×1 If pH curves wrong way round (i.e. adding acid to
H432/01 Mark Schemes June 2019
21 AO
element
Slight rise/flat, AND (near) vertical, AND then slight rise/flat alkali),
ONLY award mark for End point (~ 20 cm3)
2.4 ×2
pH
Vertical section within the extremes of pH 5 to 12 and a
minimum range of three pH units
AND middle of vertical section (equivalence point) needs to
be above pH 7
End point
Vertical section at ~ 20 cm3 NaOH
(iv) cresol purple 1 3.3 ALLOW pH range (of the indicator) matches
equivalence point
AND ALLOW end point/colour change matches
pH range matches vertical section/rapid pH change equivalence point
OR IGNORE colour change matches end point
end point/colour change matches vertical section/rapid pH Colour change is the same as end point
change
(v) similarity: end point / volume (20 cm3) of NaOH needed to 2 End point must not refer to same pH
3.2 ×2
neutralise
OR
final pH / shape of curve after end point
difference: HCN higher starting pH
OR ALLOW different equivalence point
HCN shorter vertical section IGNORE different starting pH
H432/01 Mark Schemes June 2019
AO
22 element
20 (c) HIO3 dissociation is not negligible / dissociates to a 1 3.3 ALLOW use of HA
significant extent Ignore [HIO3]equilibrium < [HIO3 ]initial/undissociated
OR
Large Ka and HIO3 is ‘stronger’ (weak) acid ALLOW
OR [HIO3]equilibrium ~ [HIO3]undissociated is no longer a valid
[HIO3 ]eqm is significantly lower than [HIO3 ]initial/undissociated assumption
ALLOW
[HIO3] has a larger Ka so the assumption that
[HIO3] at equilibrium = [HIO3] initially so
assumption is not valid
Total 15
How to answer it
Acids, Bases, pH Curves and Indicators
This question assesses core physical chemistry concepts regarding weak acids and strong bases. Key skills include calculating the pH of weak acids using Ka expressions, performing titration stoichiometry calculations, determining excess reagent pH after neutralisation, sketching and interpreting titration curves, selecting appropriate indicators based on pH ranges, comparing acid strengths, and evaluating the validity of standard weak acid approximations.
Part (a): Calculating pH of a Weak Acid
Calculate the pH of 0.0800 mol dm⁻³ C₂H₅COOH(aq). Give your answer to 2 decimal places.
📐 Step-by-Step Calculation
- State the Ka expression: Ka = [H⁺][C₂H₅COO⁻] / [C₂H₅COOH]
- Apply assumptions: Assume [H⁺] = [C₂H₅COO⁻] and [C₂H₅COOH]₀ is approximately equal to equilibrium concentration. Therefore, Ka = [H⁺]² / [HA].
- Rearrange for [H⁺]: [H⁺] = √(Ka × [HA]) = √(1.35 × 10⁻⁵ × 0.0800) = 1.039 × 10⁻³ mol dm⁻³
- Calculate pH: pH = -log(1.039 × 10⁻³) = 2.98
❌ Common Errors & Traps
- Forgetting the square root: Multiplying Ka by concentration without square-rooting gives pH = 5.97 (loses marks).
- Incorrect Ka value: Using Ka values from the wrong row in Table 20.1.
- Rounding too early: Always keep full calculator figures until the final logarithmic step.
Part (b)(i): Stoichiometry of Titration End Point
Show by calculation that 20.0 cm³ of NaOH(aq) is required to reach the end point.
✅ Correct Working
Moles of acid (C₂H₅COOH) = concentration × volume = 0.0800 × (25.0 / 1000) = 0.002 mol
Since the reaction is a 1:1 stoichiometry, moles of NaOH needed = 0.002 mol.
Volume of NaOH = moles / concentration = 0.002 / 0.100 = 0.0200 dm³ = 20.0 cm³
🧠 Exam Technique
This is a "show that" question. You must explicitly show every step of your calculation clearly. State the moles of acid first, use the 1:1 molar ratio, and finish by demonstrating the final volume calculation.
Part (b)(ii): pH of the Final Solution (Excess Base)
Calculate the pH of the final solution after adding 45.0 cm³ of 0.100 mol dm⁻³ NaOH to 25.0 cm³ of 0.0800 mol dm⁻³ C₂H₅COOH. Give your answer to 2 decimal places.
📐 Step-by-Step Calculation
- Moles of initial NaOH added: 0.100 × (45.0 / 1000) = 0.00450 mol
- Moles of acid neutralized: 0.0800 × (25.0 / 1000) = 0.00200 mol
- Calculate excess OH⁻ moles: 0.00450 - 0.00200 = 0.00250 mol
- Find total volume: 45.0 cm³ + 25.0 cm³ = 70.0 cm³ = 0.0700 dm³
- Calculate [OH⁻]: 0.00250 / 0.0700 = 0.03571 mol dm⁻³
- Calculate [H⁺] using Kw (1.00 × 10⁻¹⁴): [H⁺] = (1.00 × 10⁻¹⁴) / 0.03571 = 2.80 × 10⁻¹³ mol dm⁻³
- Convert to pH: pH = -log(2.80 × 10⁻¹³) = 12.55
❌ Common Errors
- Forgetting total volume: Dividing excess moles by initial volumes instead of the combined total volume (70.0 cm³).
- Using wrong initial volume for NaOH: Failing to subtract the reacted acid moles from the total added alkali.
Part (b)(iii): Sketching the pH Curve
Sketch a pH curve for the addition of 45.0 cm³ of 0.100 mol dm⁻³ NaOH(aq) to 25.0 cm³ of 0.0800 mol dm⁻³ C₂H₅COOH(aq).
💡 Key Features Required
- Initial pH: Starts around pH 3 (based on part a calculations).
- Buffer region: Gradual upward slope early on, showing the buffer region before the vertical inflection.
- Equivalence point: Steep vertical section located at 20.0 cm³, spanning at least 3 pH units, and centered above pH 7 (approx. pH 8–11) because the salt formed is alkaline.
- End/Final plateau: Curves off towards the final calculated pH value (~12.55) as excess strong base dominates.
🧠 Examiner Commentary
Examiners heavily penalize curves where the steep vertical inflection does not align with the calculated equivalence volume (20.0 cm³), or if the curve starts high like a strong acid-strong base titration.
Part (b)(iv): Indicator Selection
Explain which indicator would be most suitable for the titration from Table 20.2.
✅ Correct Answer
Cresol purple
💡 Explanation & Reasoning
An indicator is suitable if its pH range falls entirely within (or matches) the rapid vertical pH change / equivalence point of the titration curve. Cresol purple has a pH range of 7.6 – 9.2, which matches the steep vertical section of this weak acid-strong base titration.
Part (b)(v): Comparing Weak Acid Titration Curves
Predict one similarity and one difference between the pH curve with C₂H₅COOH and HCN. Use Table 20.1.
✅ Similarities & Differences
Similarity: The end point / volume of NaOH needed to neutralise remains identical at 20.0 cm³ (since concentration and volume of both acids are the same).
Difference: HCN is a much weaker acid than C₂H₅COOH (smaller Ka: 6.17 × 10⁻¹⁰ vs 1.35 × 10⁻⁵). Therefore, HCN has a higher starting pH and a shorter/less pronounced vertical inflection region.
❌ Common Errors
Students often incorrectly state that the equivalence point pH changes or that the final pH value will be completely different. Focus strictly on initial start points and stoichiometric equivalence volumes.
Part (c): Evaluating Weak Acid Assumptions
Explain why the measured pH is different from the calculated pH for 0.0800 mol dm⁻³ HIO₃(aq).
💡 Key Concept: Why Assumptions Fail
Iodic(V) acid (HIO₃) has a relatively large Ka value (1.78 × 10⁻¹). This means it dissociates to a significant extent in solution.
The standard weak acid calculation assumes that equilibrium concentration [HIO₃]eq is approximately equal to the initial concentration [HIO₃]initial because dissociation is minimal (< 5%). Because HIO₃ is a stronger weak acid, this assumption breaks down.
🧠 Top-Level Guidance
To score the mark, you must explicitly mention that dissociation is not negligible, or note that the large Ka value invalidates the standard approximation that [HA]eq = [HA]initial.
Topics
Module 5: Physical chemistry and transition elements · 5.1 Rates, equilibrium and pH
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.