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A Level H2 Chemistry Practice Paper 1

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TuitionGoWhere Practice Paper - Chemistry H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Chemistry Level: A-Level H2 Paper: Practice Paper — Acids, Bases & Salts Duration: 1 hour 30 minutes Total Marks: 60 Name: ___________________________ Class: ___________________________ Date: ___________________________

Instructions

Answer all questions in the spaces provided.
The number of marks for each question or part-question is shown in brackets [ ].
All numerical answers should be given to an appropriate number of significant figures unless otherwise stated.
The use of an approved scientific calculator is expected, where appropriate.
The Data Booklet is relevant to this paper.
You may lose marks if you do not show your working, if you do not use appropriate units, or if you do not give your final answer to a suitable degree of precision.

Section A: Multiple Choice [10 marks]

Questions 1–10 are multiple choice questions. Each question carries 1 mark. Write the letter corresponding to the correct answer in the space provided.

1. Which of the following is the conjugate base of $HSO_4^-$ ?

A. $H_2SO_4$ B. $SO_4^{2-}$ C. $H_3O^+$ D. $H_2SO_3$

Answer: ___________

2. A solution has a pH of 3.40. What is the concentration of $OH^-$ ions in this solution at 25 °C?

A. $2.51 \times 10^{-4}$ mol dm $^{-3}$ B. $3.98 \times 10^{-4}$ mol dm $^{-3}$ C. $2.51 \times 10^{-11}$ mol dm $^{-3}$ D. $3.98 \times 10^{-11}$ mol dm $^{-3}$

Answer: ___________

3. Which of the following salts will produce an aqueous solution with pH > 7?

A. $NH_4Cl$ B. $NaNO_3$ C. $K_2CO_3$ D. $AlCl_3$

Answer: ___________

4. The $K_a$ of a weak acid $HA$ is $4.0 \times 10^{-6}$ mol dm $^{-3}$ . What is the pH of a 0.050 mol dm $^{-3}$ solution of $HA$ ?

A. 2.70 B. 3.35 C. 3.85 D. 5.35

Answer: ___________

5. A buffer solution is prepared by mixing 50.0 cm $^3$ of 0.200 mol dm $^{-3}$ $CH_3COOH$ with 50.0 cm $^3$ of 0.200 mol dm $^{-3}$ $NaOH$ . Which statement about this mixture is correct?

A. The solution is a buffer because it contains a weak acid and its conjugate base. B. The solution is not a buffer because all the $CH_3COOH$ has been neutralised. C. The solution is a buffer because excess $CH_3COOH$ remains. D. The solution is not a buffer because it contains only $CH_3COONa$ and water.

Answer: ___________

6. In the titration of 25.0 cm $^3$ of 0.100 mol dm $^{-3}$ $NaOH$ with 0.100 mol dm $^{-3}$ $HCl$ , what is the pH at the equivalence point?

A. 1.0 B. 5.0 C. 7.0 D. 13.0

Answer: ___________

7. The solubility product, $K_{sp}$ , of $PbI_2$ is $8.7 \times 10^{-9}$ mol $^3$ dm $^{-9}$ at 25 °C. What is the solubility of $PbI_2$ in water at this temperature?

A. $1.3 \times 10^{-3}$ mol dm $^{-3}$ B. $2.6 \times 10^{-3}$ mol dm $^{-3}$ C. $1.7 \times 10^{-3}$ mol dm $^{-3}$ D. $8.7 \times 10^{-9}$ mol dm $^{-3}$

Answer: ___________

8. Which indicator is most suitable for a titration between a weak acid and a strong base where the equivalence point pH is approximately 8.5?

Indicator	pH range
Methyl orange	3.1 – 4.4
Bromothymol blue	6.0 – 7.6
Phenolphthalein	8.2 – 10.0
Thymol blue	1.2 – 2.8

A. Methyl orange B. Bromothymol blue C. Phenolphthalein D. Thymol blue

Answer: ___________

9. Which of the following best explains why the pH of a 0.100 mol dm $^{-3}$ solution of $CH_3COOH$ is higher than that of a 0.100 mol dm $^{-3}$ solution of $HCl$ ?

A. $CH_3COOH$ has a higher molecular mass than $HCl$ . B. $CH_3COOH$ is only partially dissociated in aqueous solution. C. $CH_3COOH$ is more soluble in water than $HCl$ . D. $CH_3COOH$ molecules are larger than $HCl$ molecules.

Answer: ___________

10. A solution contains 0.020 mol of $NH_3$ and 0.010 mol of $NH_4Cl$ in 1.00 dm $^3$ of solution. Given that $K_b$ for $NH_3$ is $1.8 \times 10^{-5}$ mol dm $^{-3}$ , what is the pH of this buffer solution?

A. 8.95 B. 9.25 C. 9.56 D. 10.25

Answer: ___________

Section B: Structured Questions [30 marks]

Answer all questions. Show your working clearly.

11. [6 marks]

(a) Define the term Brønsted–Lowry acid. [1]

(b) For the following reaction, identify the two conjugate acid–base pairs and label each species as an acid or a base:

$H_2PO_4^-(aq) + CO_3^{2-}(aq) \rightleftharpoons HPO_4^{2-}(aq) + HCO_3^-(aq)$

[3]

Pair 1: Acid ___________ / Base ___________

Pair 2: Acid ___________ / Base ___________

12. [5 marks]

A student titrates 25.0 cm $^3$ of 0.150 mol dm $^{-3}$ $NaOH$ solution with 0.100 mol dm $^{-3}$ $H_2SO_4$ .

(a) Write the balanced equation for this reaction. [1]

(b) Calculate the volume of $H_2SO_4$ required to reach the equivalence point. [2]

13. [6 marks]

A buffer solution is prepared by mixing 100 cm $^3$ of 0.300 mol dm $^{-3}$ methanoic acid ( $HCOOH$ ) with 100 cm $^3$ of 0.150 mol dm $^{-3}$ $NaOH$ .

$K_a$ for $HCOOH = 1.6 \times 10^{-4}$ mol dm $^{-3}$

(a) Calculate the number of moles of $HCOOH$ and $HCOO^-$ present in the buffer solution. [2]

(b) Calculate the pH of the buffer solution. [2]

(c) Explain, with reference to the equilibrium involved, how this buffer solution resists a change in pH when a small amount of dilute $HCl$ is added. [2]

14. [6 marks]

The solubility product, $K_{sp}$ , of silver chromate( $Ag_2CrO_4$ ) is $1.1 \times 10^{-12}$ mol $^3$ dm $^{-9}$ at 25 °C.

(a) Write an expression for $K_{sp}$ of $Ag_2CrO_4$ . [1]

(b) Calculate the solubility of $Ag_2CrO_4$ in water at 25 °C, in mol dm $^{-3}$ . [3]

(c) Predict and explain whether the solubility of $Ag_2CrO_4$ would increase, decrease, or remain the same if solid $AgNO_3$ is added to the saturated solution. [2]

15. [7 marks]

A student performs a titration to determine the concentration of a solution of ethanoic acid ( $CH_3COOH$ ) using 0.100 mol dm $^{-3}$ $NaOH$ . The following titration results were obtained:

Titration	Rough	1	2	3
Final reading / cm $^3$	24.80	24.30	32.10	24.40
Initial reading / cm $^3$	0.00	0.00	8.00	0.00
Volume used / cm $^3$	24.80	24.30	24.10	24.40

Volume of $CH_3COOH$ used = 25.0 cm $^3$

(a) Identify the anomalous result and explain why it should be excluded. [1]

(b) Calculate the mean titre to be used in your calculation. [1]

(d) The student used phenolphthalein as the indicator. State the colour change observed at the endpoint. [1]

(e) Explain why phenolphthalein is a suitable indicator for this titration. [1]

Section C: Data-Based & Extended Response [20 marks]

Answer all questions.

16. [8 marks]

The graph below shows the pH change when 0.100 mol dm $^{-3}$ $NaOH$ is added to 25.0 cm $^3$ of 0.100 mol dm $^{-3}$ of a monoprotic acid $HX$ .

<image_placeholder> id: Q16-fig1 type: graph linked_question: Q16 description: A pH titration curve showing pH (y-axis, range 0-14) versus volume of NaOH added in cm³ (x-axis, range 0-50 cm³). The curve starts at approximately pH 2.0 at 0 cm³, rises gradually, then has a steep vertical rise between 24 and 26 cm³, passing through pH 7 at 25.0 cm³, and levels off at approximately pH 12.0 by 40 cm³. The equivalence point is at 25.0 cm³. A half-equivalence point is visible at approximately 12.5 cm³ where pH ≈ 4.6. labels: y-axis: pH (0 to 14); x-axis: Volume of NaOH added / cm³ (0 to 50); equivalence point marked at (25.0, 7); half-equivalence point at approximately (12.5, 4.6); initial pH ≈ 2.0; final pH ≈ 12.0 values: Initial pH = 2.0; Equivalence point volume = 25.0 cm³; Equivalence point pH = 7; Half-equivalence point volume = 12.5 cm³; Half-equivalence point pH = 4.6; Final pH ≈ 12.0 must_show: The full S-shaped titration curve, clearly labelled axes with units, equivalence point at 25.0 cm³, initial pH around 2, steep rise region, and the half-equivalence point region where the curve is relatively flat (buffer region).

Use the graph to answer the following questions.

(a) From the graph, determine the initial concentration of the acid $HX$ if the acid was the limiting reagent. Explain your reasoning. [2]

(b) The pH at the half-equivalence point is 4.6. Calculate the $K_a$ of acid $HX$ . [2]

(d) Estimate the pH of the solution after 5.0 cm $^3$ of $NaOH$ has been added. Explain whether this region of the curve acts as a buffer. [2]

17. [6 marks]

Read the following passage and answer the questions that follow.

Blood is a buffered solution that maintains a pH between 7.35 and 7.45. The principal buffer system in blood involves carbonic acid ( $H_2CO_3$ ) and hydrogen carbonate ions ( $HCO_3^-$ ):

$H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)$

When excess acid enters the blood, the equilibrium shifts to the left, removing $H^+$ ions. When excess base enters the blood, the equilibrium shifts to the right, replenishing $H^+$ ions. The lungs and kidneys also help regulate blood pH by controlling the concentrations of $CO_2$ and $HCO_3^-$ respectively.

(a) Write an expression for $K_a$ for the above equilibrium. [1]

(b) Explain, using Le Chatelier's principle, how the carbonic acid–hydrogen carbonate buffer maintains blood pH when a small amount of acid is added. [2]

(c) A patient's blood pH drops to 7.10. State whether this condition is called acidosis or alkalosis, and suggest one way the body might attempt to correct this. [2]

(d) Explain why a very dilute solution of $HCl$ with pH 6.0 would not be suitable as a substitute for blood, even though it is close to blood pH. [1]

18. [6 marks]

A student wishes to determine the percentage by mass of calcium carbonate in a sample of impure chalk. The student weighs out 2.50 g of the chalk and adds it to 50.0 cm $^3$ of 1.00 mol dm $^{-3}$ $HCl$ (an excess). The reaction is:

$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$

After the reaction is complete, the excess $HCl$ is titrated with 0.500 mol dm $^{-3}$ $NaOH$ . The titration requires 32.60 cm $^3$ of $NaOH$ to neutralise the excess acid.

(a) Calculate the number of moles of $HCl$ initially added. [1]

(b) Calculate the number of moles of $NaOH$ used in the back-titration, and hence the number of moles of excess $HCl$ . [2]

(c) Calculate the number of moles of $HCl$ that reacted with $CaCO_3$ , and hence the percentage by mass of $CaCO_3$ in the chalk sample. [3]

Answers

TuitionGoWhere Practice Paper — Chemistry H2 A-Level

Answer Key: Practice Paper — Acids, Bases & Salts

Section A: Multiple Choice [10 marks]

1. B [1]

Explanation: A conjugate base is formed when a Brønsted–Lowry acid donates a proton ( $H^+$ ). $HSO_4^-$ loses one $H^+$ to become $SO_4^{2-}$ . Option A ( $H_2SO_4$ ) is the conjugate acid of $HSO_4^-$ , not the conjugate base. Option C ( $H_3O^+$ ) is unrelated to this conjugate pair. Option D ( $H_2SO_3$ ) is a different acid entirely.

2. C [1]

Explanation:

$pH = 3.40$ , so $[H^+] = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm $^{-3}$
At 25 °C, $K_w = [H^+][OH^-] = 1.00 \times 10^{-14}$ mol $^2$ dm $^{-6}$
$[OH^-] = \frac{K_w}{[H^+]} = \frac{1.00 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11}$ mol dm $^{-3}$

Common mistake: Students often select option A, which is $[H^+]$ rather than $[OH^-]$ . Always check what the question is asking for.

3. C [1]

Explanation: $K_2CO_3$ is a salt formed from a strong base ( $KOH$ ) and a weak acid ( $H_2CO_3$ ). The $CO_3^{2-}$ ion hydrolyses in water to produce $OH^-$ ions, making the solution alkaline (pH > 7).

$NH_4Cl$ : salt of weak base + strong acid → acidic solution
$NaNO_3$ : salt of strong base + strong acid → neutral solution
$K_2CO_3$ : salt of strong base + weak acid → alkaline solution ✓
$AlCl_3$ : salt of weak base + strong acid → acidic solution (cation hydrolysis)

4. B [1]

Explanation:

For a weak acid: $[H^+] = \sqrt{K_a \times c} = \sqrt{4.0 \times 10^{-6} \times 0.050} = \sqrt{2.0 \times 10^{-7}} = 4.47 \times 10^{-4}$ mol dm $^{-3}$
$pH = -\log(4.47 \times 10^{-4}) = 3.35$

Common mistake: Using $[H^+] = K_a \times c$ instead of $\sqrt{K_a \times c}$ , or forgetting to take the square root.

5. D [1]

Explanation:

Moles of $CH_3COOH$ = $0.0500 \times 0.200 = 0.0100$ mol
Moles of $NaOH$ = $0.0500 \times 0.200 = 0.0100$ mol
The $NaOH$ completely neutralises all the $CH_3COOH$ to form $CH_3COONa$ (0.0100 mol).
The resulting solution contains only $CH_3COONa$ (a salt) and water. There is no significant amount of $CH_3COOH$ or $CH_3COO^-$ remaining as a conjugate pair in comparable amounts, so it is not a buffer.

A buffer requires significant amounts of both a weak acid and its conjugate base. Here, the acid has been completely neutralised.

6. C [1]

Explanation: $NaOH$ is a strong base and $HCl$ is a strong acid. The salt formed ( $NaCl$ ) is derived from a strong acid and strong base, so it does not hydrolyse. The solution at the equivalence point is neutral: pH = 7.0.

Common mistake: Students who choose D (pH 13.0) are confusing the equivalence point with the starting pH of the $NaOH$ solution.

7. A [1]

Explanation:

$PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$
Let solubility = $s$ mol dm $^{-3}$ . Then $[Pb^{2+}] = s$ and $[I^-] = 2s$ .
$K_{sp} = [Pb^{2+}][I^-]^2 = s(2s)^2 = 4s^3$
$4s^3 = 8.7 \times 10^{-9}$
$s^3 = 2.175 \times 10^{-9}$
$s = \sqrt[3]{2.175 \times 10^{-9}} = 1.3 \times 10^{-3}$ mol dm $^{-3}$

Common mistake: Forgetting the stoichiometric coefficient of $I^-$ when writing the $K_{sp}$ expression. Since 2 moles of $I^-$ are produced per mole of $PbI_2$ , $[I^-] = 2s$ , not $s$ .

8. C [1]

Explanation: In a weak acid–strong base titration, the equivalence point occurs at pH > 7 (typically around 8–10) because the salt formed hydrolyses to produce a basic solution. The indicator must change colour within the steep portion of the titration curve near the equivalence point. Phenolphthalein (pH range 8.2–10.0) is the most suitable as its colour change range encompasses the equivalence point pH of ~8.5.

9. B [1]

Explanation: $HCl$ is a strong acid and dissociates completely in aqueous solution, so $[H^+] = 0.100$ mol dm $^{-3}$ and pH = 1.0. $CH_3COOH$ is a weak acid and only partially dissociates, so $[H^+] < 0.100$ mol dm $^{-3}$ and pH > 1.0. The key difference is the degree of dissociation, not molecular mass, solubility, or molecular size.

10. C [1]

Explanation:

For a basic buffer: $pOH = pK_b + \log\left(\frac{[\text{conjugate acid}]}{[\text{base}]}\right)$
$pK_b = -\log(1.8 \times 10^{-5}) = 4.74$
$pOH = 4.74 + \log\left(\frac{0.010}{0.020}\right) = 4.74 + \log(0.50) = 4.74 + (-0.30) = 4.44$
$pH = 14.00 - 4.44 = 9.56$

Common mistake: Using the $K_a$ expression instead of $K_b$ for a buffer involving a weak base. Since $NH_3$ is a weak base, use the $K_b$ form of the Henderson–Hasselbalch equation.

Section B: Structured Questions [30 marks]

11. [6 marks]

(a) [1]

A Brønsted–Lowry acid is a proton ( $H^+$ ) donor.

Mark: 1 mark for "proton donor" or " $H^+$ donor".

(b) [3]

Pair 1:

Acid: $H_2PO_4^-$ (donates $H^+$ to become $HPO_4^{2-}$ )
Base: $CO_3^{2-}$ (accepts $H^+$ to become $HCO_3^-$ )

Pair 2:

Acid: $HCO_3^-$ (conjugate acid of $CO_3^{2-}$ )
Base: $HPO_4^{2-}$ (conjugate base of $H_2PO_4^-$ )

Marking: 1 mark for each correctly identified pair (acid and base correctly labelled).

Teaching note: In any acid–base reaction, identify which species loses a proton (acid) and which gains a proton (base). The conjugate base is what remains after the acid donates $H^+$ ; the conjugate acid is formed when the base accepts $H^+$ .

(c) [2]

$H_2PO_4^-$ is amphiprotic — it can act as both an acid and a base.

As an acid (donates $H^+$ ): $H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}$

As a base (accepts $H^+$ ): $H_2PO_4^- + H^+ \rightleftharpoons H_3PO_4$

Marking: 1 mark for stating it is amphiprotic / can donate or accept $H^+$ ; 1 mark for two correct equations showing both behaviours.

12. [5 marks]

(a) [1]

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)$

Mark: 1 mark for correct balanced equation with state symbols accepted but not required.

(b) [2]

Moles of $NaOH$ = $\frac{25.0}{1000} \times 0.150 = 3.75 \times 10^{-3}$ mol
From the equation: 1 mol $H_2SO_4$ reacts with 2 mol $NaOH$
Moles of $H_2SO_4$ needed = $\frac{3.75 \times 10^{-3}}{2} = 1.875 \times 10^{-3}$ mol
Volume of $H_2SO_4$ = $\frac{1.875 \times 10^{-3}}{0.100} = 1.875 \times 10^{-2}$ dm $^3$ = 18.8 cm $^3$ (to 3 s.f.)

Marking: 1 mark for correct moles of NaOH and stoichiometric ratio; 1 mark for correct final answer with unit.

(c) [2]

Moles of $H_2SO_4$ added = $\frac{10.0}{1000} \times 0.100 = 1.00 \times 10^{-3}$ mol
Moles of $H^+$ from $H_2SO_4$ = $2 \times 1.00 \times 10^{-3} = 2.00 \times 10^{-3}$ mol
Moles of $NaOH$ initially = $3.75 \times 10^{-3}$ mol
Moles of $OH^-$ remaining = $3.75 \times 10^{-3} - 2.00 \times 10^{-3} = 1.75 \times 10^{-3}$ mol
Total volume = $25.0 + 10.0 = 35.0$ cm $^3$ = $0.0350$ dm $^3$
$[OH^-] = \frac{1.75 \times 10^{-3}}{0.0350} = 0.0500$ mol dm $^{-3}$
$pOH = -\log(0.0500) = 1.30$
$pH = 14.00 - 1.30 = \mathbf{12.70}$

Marking: 1 mark for correct excess moles of OH⁻ and total volume; 1 mark for correct pH.

13. [6 marks]

(a) [2]

Initial moles of $HCOOH$ = $\frac{100}{1000} \times 0.300 = 0.0300$ mol
Moles of $NaOH$ added = $\frac{100}{1000} \times 0.150 = 0.0150$ mol

The reaction: $HCOOH + NaOH \rightarrow HCOONa + H_2O$

Moles of $HCOOH$ remaining = $0.0300 - 0.0150 = 0.0150$ mol
Moles of $HCOO^-$ (from $HCOONa$ ) formed = $0.0150$ mol

Marking: 1 mark for correct moles of HCOOH remaining; 1 mark for correct moles of HCOO⁻ formed.

(b) [2]

Using the Henderson–Hasselbalch equation:

$pH = pK_a + \log\left(\frac{[HCOO^-]}{[HCOOH]}\right)$

$pK_a = -\log(1.6 \times 10^{-4}) = 3.80$
Total volume = $100 + 100 = 200$ cm $^3$ = $0.200$ dm $^3$
$[HCOOH] = \frac{0.0150}{0.200} = 0.0750$ mol dm $^{-3}$
$[HCOO^-] = \frac{0.0150}{0.200} = 0.0750$ mol dm $^{-3}$

$pH = 3.80 + \log\left(\frac{0.0750}{0.0750}\right) = 3.80 + \log(1) = 3.80 + 0 = \mathbf{3.80}$

Marking: 1 mark for correct pKa and concentrations; 1 mark for correct pH.

Teaching note: When $[HCOOH] = [HCOO^-]$ , the ratio is 1 and $\log(1) = 0$ , so $pH = pK_a$ . This occurs at the half-equivalence point of a titration.

(c) [2]

When a small amount of dilute $HCl$ is added, the $H^+$ ions from $HCl$ react with the $HCOO^-$ ions in the buffer:

$HCOO^-(aq) + H^+(aq) \rightarrow HCOOH(aq)$

By Le Chatelier's principle, the equilibrium:

$HCOOH(aq) \rightleftharpoons H^+(aq) + HCOO^-(aq)$

shifts to the left to partially remove the added $H^+$ . The added $H^+$ is consumed by the conjugate base ( $HCOO^-$ ), converting it into the weak acid ( $HCOOH$ ). Since the weak acid is only partially dissociated, the increase in $[H^+]$ is much smaller than it would be in an unbuffered solution, so the pH remains relatively constant.

Marking: 1 mark for the equation showing H⁺ reacting with HCOO⁻; 1 mark for explanation referencing Le Chatelier's principle / equilibrium shifting left / H⁺ being consumed.

14. [6 marks]

(a) [1]

$K_{sp} = [Ag^+]^2[CrO_4^{2-}]$

Mark: 1 mark for correct expression. Note the exponent of 2 on $[Ag^+]$ due to the stoichiometry of $Ag_2CrO_4$ .

(b) [3]

$Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$

Let the solubility = $s$ mol dm $^{-3}$

$[Ag^+] = 2s$
$[CrO_4^{2-}] = s$

$K_{sp} = (2s)^2 \times s = 4s^3$

$4s^3 = 1.1 \times 10^{-12}$

$s^3 = 2.75 \times 10^{-13}$

$s = \sqrt[3]{2.75 \times 10^{-13}} = \mathbf{6.50 \times 10^{-5}} \text{ mol dm}^{-3} \text{ (to 3 s.f.)}$

Marking: 1 mark for correct Ksp expression in terms of s; 1 mark for correct substitution; 1 mark for correct final answer.

(c) [2]

The solubility of $Ag_2CrO_4$ would decrease.

When $AgNO_3$ is added, it dissociates to provide $Ag^+$ ions, increasing $[Ag^+]$ in solution. By Le Chatelier's principle, the equilibrium:

$Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$

shifts to the left to partially remove the excess $Ag^+$ . This causes more $Ag_2CrO_4$ to precipitate, reducing its solubility. This is an example of the common ion effect.

Marking: 1 mark for "decrease"; 1 mark for explanation referencing Le Chatelier's principle / common ion effect.

15. [7 marks]

(a) [1]

Titration 2 (or the value 32.10 cm $^3$ / 24.10 cm $^3$ used) is anomalous. It deviates significantly from the other concordant values (24.30 and 24.40 cm $^3$ ) and is not within 0.10 cm $^3$ of the other consistent results.

Mark: 1 mark for correctly identifying the anomalous result and a valid reason.

(b) [1]

Mean titre = $\frac{24.30 + 24.40}{2} = \mathbf{24.35}$ cm $^3$

(Titrations 1 and 3 are concordant; the rough titration and titration 2 are excluded.)

Mark: 1 mark for correct mean.

(c) [3]

$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

Moles of $NaOH$ used = $\frac{24.35}{1000} \times 0.100 = 2.435 \times 10^{-3}$ mol
From the equation, mole ratio $CH_3COOH$ : $NaOH$ = 1 : 1
Moles of $CH_3COOH$ = $2.435 \times 10^{-3}$ mol
Concentration of $CH_3COOH$ = $\frac{2.435 \times 10^{-3}}{\frac{25.0}{1000}} = \frac{2.435 \times 10^{-3}}{0.0250} = \mathbf{0.0974}$ mol dm $^{-3}$ (to 3 s.f.)

Marking: 1 mark for correct moles of NaOH; 1 mark for correct use of 1:1 ratio; 1 mark for correct concentration.

(d) [1]

Colourless to pink (or colourless to pale pink).

Mark: 1 mark. In acidic/neutral solution, phenolphthalein is colourless; at the equivalence point of a weak acid–strong base titration (pH > 7), it turns pink.

(e) [1]

Phenolphthalein changes colour in the pH range 8.2–10.0, which falls within the steep vertical section of the titration curve for a weak acid–strong base titration. The equivalence point for such a titration occurs at approximately pH 8–9, which is within the indicator's range.

Mark: 1 mark for stating that the indicator's pH range coincides with the steep portion of the titration curve / equivalence point pH.

Section C: Data-Based & Extended Response [20 marks]

16. [8 marks]

(a) [2]

At the equivalence point, 25.0 cm $^3$ of 0.100 mol dm $^{-3}$ $NaOH$ is required to neutralise 25.0 cm $^3$ of acid $HX$ .

Moles of $NaOH$ = $\frac{25.0}{1000} \times 0.100 = 2.50 \times 10^{-3}$ mol
Since $HX$ is monoprotic, the mole ratio $HX$ : $NaOH$ = 1 : 1
Moles of $HX$ = $2.50 \times 10^{-3}$ mol
Concentration of $HX$ = $\frac{2.50 \times 10^{-3}}{0.0250} = \mathbf{0.100}$ mol dm $^{-3}$

Reasoning: At the equivalence point, moles of acid = moles of base (for a monoprotic acid). The volume of NaOH at the equivalence point (25.0 cm³) is read directly from the graph.

Marking: 1 mark for reading the equivalence point volume from the graph; 1 mark for correct calculation of concentration.

(b) [2]

At the half-equivalence point, $[HX] = [X^-]$ , so $pH = pK_a$ .

From the graph, at the half-equivalence point, $pH = 4.6$ .

Therefore, $pK_a = 4.6$ .

$K_a = 10^{-4.6} = \mathbf{2.5 \times 10^{-5}} \text{ mol dm}^{-3} \text{ (to 2 s.f.)}$

Marking: 1 mark for stating that at half-equivalence point, pH = pKa; 1 mark for correct Ka value.

(c) [2]

$HX$ is a weak acid.

Evidence from the graph:

The initial pH is approximately 2.0, not 1.0 (which would be expected for a 0.100 mol dm $^{-3}$ strong acid). This indicates incomplete dissociation.
The titration curve shows a buffer region (a relatively flat portion before the steep rise), which is characteristic of a weak acid–strong base titration.
The equivalence point pH is 7.0 (as stated), but more importantly, the curve shape with a gradual initial rise and a buffer region confirms weak acid behaviour.

Marking: 1 mark for stating "weak acid"; 1 mark for a valid explanation referencing the graph (initial pH > 1, or presence of buffer region).

(d) [2]

After adding 5.0 cm $^3$ of $NaOH$ , the pH would be approximately 3–4 (reading from the graph, approximately pH 3.5).

This region of the curve does act as a buffer. Before the equivalence point, the solution contains a mixture of unreacted $HX$ (weak acid) and its conjugate base $X^-$ (formed by neutralisation). This $HX$ / $X^-$ mixture constitutes a buffer system. The curve is relatively flat in this region, indicating that the pH changes only gradually as more $NaOH$ is added — this is the characteristic behaviour of a buffer.

Marking: 1 mark for estimated pH value (any value between 3.0 and 4.5 accepted); 1 mark for explaining that the region contains a mixture of HX and X⁻ which acts as a buffer.

17. [6 marks]

(a) [1]

$K_a = \frac{[H^+][HCO_3^-]}{[H_2CO_3]}$

Mark: 1 mark for correct expression. Water is omitted as it is a pure liquid.

(b) [2]

When a small amount of acid is added to the blood, the concentration of $H^+$ ions increases. By Le Chatelier's principle, the equilibrium:

$H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)$

shifts to the left, favouring the reverse reaction. The excess $H^+$ ions react with $HCO_3^-$ ions to form more $H_2CO_3$ . This removes the added $H^+$ from solution, minimising the change in pH. The buffer is effective because there are significant concentrations of both $H_2CO_3$ and $HCO_3^-$ present.

Marking: 1 mark for stating the equilibrium shifts left; 1 mark for explaining that H⁺ reacts with HCO₃⁻ to form H₂CO₃, minimising pH change.

(c) [2]

The condition is called acidosis (blood pH < 7.35).

The body might attempt to correct this by:

Increasing the rate and depth of breathing to expel more $CO_2$ , which shifts the equilibrium to the left, reducing $[H^+]$ and raising pH back towards normal.
Alternatively: The kidneys excrete more $H^+$ ions and reabsorb more $HCO_3^-$ ions.

Marking: 1 mark for "acidosis"; 1 mark for a valid corrective mechanism.

(d) [1]

A very dilute solution of $HCl$ has no buffering capacity. While its pH (6.0) is close to blood pH, it contains no conjugate acid–base pair to resist pH changes. If even a small amount of acid or base is added, the pH would change dramatically. Blood, on the other hand, contains the $H_2CO_3$ / $HCO_3^-$ buffer system that actively resists pH changes, maintaining pH within the narrow range of 7.35–7.45.

Marking: 1 mark for explaining that the HCl solution lacks buffering capacity / has no conjugate acid–base pair.

18. [6 marks]

(a) [1]

Moles of $HCl$ initially added = $\frac{50.0}{1000} \times 1.00 = \mathbf{0.0500}$ mol

Mark: 1 mark for correct answer.

(b) [2]

Moles of $NaOH$ used = $\frac{32.60}{1000} \times 0.500 = 0.0163$ mol
The reaction: $HCl + NaOH \rightarrow NaCl + H_2O$ (1:1 ratio)
Moles of excess $HCl$ = moles of $NaOH$ used = 0.0163 mol

Marking: 1 mark for correct moles of NaOH; 1 mark for correct moles of excess HCl (1:1 ratio).

(c) [3]

Moles of $HCl$ that reacted with $CaCO_3$ = $0.0500 - 0.0163 = 0.0337$ mol
From the equation: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$
Mole ratio $CaCO_3$ : $HCl$ = 1 : 2
Moles of $CaCO_3$ = $\frac{0.0337}{2} = 0.01685$ mol
Molar mass of $CaCO_3$ = $40.1 + 12.0 + 3(16.0) = 100.1$ g mol $^{-1}$
Mass of $CaCO_3$ = $0.01685 \times 100.1 = 1.687$ g
Percentage by mass of $CaCO_3$ = $\frac{1.687}{2.50} \times 100 = \mathbf{67.5\%}$ (to 3 s.f.)

Marking: 1 mark for correct moles of HCl reacted; 1 mark for correct moles and mass of CaCO₃; 1 mark for correct percentage.

Total: 60 marks