A Level H2 Chemistry Practice Paper 2

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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: 50 Name: ___________________________ Class: ___________________________ Date: ___________________________

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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 a Data Booklet is permitted.
• A periodic table and relevant constants are provided on the last page.
• Electronic calculators may be used.

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Section A: Multiple Choice [10 marks]

Questions 1–10: Choose the one correct answer for each question. Write your answer in the space provided.

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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: ________ [1]

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

Answer: ________ [1]

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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: ________ [1]

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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: ________ [1]

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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 not a buffer because it contains only $CH_3COO^-$ and excess $NaOH$. D. The solution is a buffer because it contains $CH_3COOH$ and $CH_3COO^-$ in equal amounts.

Answer: ________ [1]

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6. During the titration of a weak acid with a strong base, at the half-equivalence point, which relationship holds?

A. $pH = pK_a$ B. $pH = \frac{1}{2} pK_a$ C. $pH = pK_w$ D. $pH = pK_a + 1$

Answer: ________ [1]

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7. The solubility product, $K_{sp}$, of $Mg(OH)_2$ is $5.6 \times 10^{-12}$ mol$^3$ dm$^{-9}$ at 25 °C. What is the solubility of $Mg(OH)_2$ in water at 25 °C?

A. $1.1 \times 10^{-4}$ mol dm$^{-3}$ B. $1.8 \times 10^{-4}$ mol dm$^{-3}$ C. $2.2 \times 10^{-4}$ mol dm$^{-3}$ D. $1.1 \times 10^{-6}$ mol dm$^{-3}$

Answer: ________ [1]

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8. Which indicator is most suitable for a titration of 0.10 mol dm$^{-3}$ $NH_3$ ($K_b = 1.8 \times 10^{-5}$) with 0.10 mol dm$^{-3}$ $HCl$?

Indicator	pH range of colour change
Methyl orange	3.1 – 4.4
Bromothymol blue	6.0 – 7.6
Phenolphthalein	8.2 – 10.0
Thymol blue	8.0 – 9.6

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

Answer: ________ [1]

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9. A solution contains 0.10 mol dm$^{-3}$ $HCOOH$ ($K_a = 1.8 \times 10^{-4}$) and 0.15 mol dm$^{-3}$ $HCOONa$. What is the pH of this solution?

A. 3.57 B. 3.92 C. 4.12 D. 4.74

Answer: ________ [1]

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10. Which of the following best explains why the pH at the equivalence point of a titration between $CH_3COOH$ and $NaOH$ is greater than 7?

A. Excess $NaOH$ is present at the equivalence point. B. $CH_3COO^-$ is the conjugate base of a weak acid and undergoes hydrolysis to produce $OH^-$. C. $Na^+$ ions react with water to produce $OH^-$. D. $CH_3COOH$ is only partially neutralised.

Answer: ________ [1]

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Section B: Structured Questions [25 marks]

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11. [4 marks]

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

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(b) In the following reaction, identify the two conjugate acid–base pairs:

$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$

Pair 1: acid ________ / conjugate base ________ [1] Pair 2: acid ________ / conjugate base ________ [1]

(c) Explain why $NH_3$ can act as a Brønsted–Lowry base. [1]

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12. [5 marks]

A student titrates 25.0 cm$^3$ of 0.120 mol dm$^{-3}$ $NaOH$ solution with 0.100 mol dm$^{-3}$ $H_2SO_4$.

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

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(b) Calculate the volume of $H_2SO_4$ required to reach the equivalence point. [3]

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(c) Suggest a suitable indicator for this titration and state the colour change observed. [1]

Indicator: _______________________ Colour change: _______________________

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13. [6 marks]

The following data were obtained from a titration of 25.0 cm$^3$ of a solution of a weak monoprotic acid $HA$ with 0.100 mol dm$^{-3}$ $NaOH$:

Titration	Rough	1	2	3
Final reading / cm$^3$	24.80	24.35	24.30	24.30
Initial reading / cm$^3$	0.00	0.00	0.00	0.00
Volume used / cm$^3$	24.80	24.35	24.30	24.30

(a) Calculate a suitable average titre to be used in your calculations. Show clearly how you obtained this value. [2]

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(b) Calculate the concentration of the acid $HA$ in mol dm$^{-3}$. [2]

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(c) The pH of the solution at the equivalence point was found to be 8.72. Explain why the pH is not 7. [2]

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14. [5 marks]

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

($K_a$ for $HCOOH = 1.8 \times 10^{-4}$ mol dm$^{-3}$)

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

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

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(c) State and explain what happens to the pH when a small amount of dilute $HCl$ is added to this buffer. [1]

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15. [5 marks]

The solubility product, $K_{sp}$, of $CaF_2$ is $3.9 \times 10^{-11}$ mol$^3$ dm$^{-9}$ at 25 °C.

(a) Write an expression for the solubility product, $K_{sp}$, of $CaF_2$. [1]

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(b) Calculate the solubility of $CaF_2$ in water at 25 °C, in mol dm$^{-3}$. [3]

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(c) Predict and explain whether the solubility of $CaF_2$ would increase, decrease, or remain the same if dissolved in 0.10 mol dm$^{-3}$ $NaF$ solution instead of pure water. [1]

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Section C: Free Response [15 marks]

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16. [8 marks]

A student performs a titration to determine the concentration of a solution of ethanoic acid, $CH_3COOH$, using 0.100 mol dm$^{-3}$ $NaOH$.

25.0 cm$^3$ of $CH_3COOH$ solution is titrated with 0.100 mol dm$^{-3}$ $NaOH$. The following pH curve was obtained:

Generated graph for Q16.

(a) From the graph, determine the volume of $NaOH$ added at the equivalence point. [1]

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(b) Calculate the concentration of the $CH_3COOH$ solution. [2]

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(c) Using the graph, estimate the $K_a$ of $CH_3COOH$. Show your reasoning. [2]

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(d) Explain why phenolphthalein is a suitable indicator for this titration. Refer to the graph in your answer. [2]

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(e) The student repeats the titration using methyl orange as the indicator. Explain whether this would give an accurate result. [1]

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17. [7 marks]

A solution is prepared by dissolving ammonium chloride, $NH_4Cl$, in water.

($K_b$ for $NH_3 = 1.8 \times 10^{-5}$ mol dm$^{-3}$; $K_w = 1.0 \times 10^{-14}$ mol$^2$ dm$^{-6}$ at 25 °C)

(a) Write an equation to show the hydrolysis of the ammonium ion in water. [1]

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(b) Derive an expression for the acid dissociation constant, $K_a$, of $NH_4^+$ in terms of $K_w$ and $K_b$. [2]

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(c) Calculate the pH of a 0.200 mol dm$^{-3}$ solution of $NH_4Cl$. [3]

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(d) A student claims that adding solid $NH_4Cl$ to a solution of $NH_3$ would decrease the pH of the solution. Evaluate this claim. [1]

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Data and Constants

Constant	Value
$K_w$ at 25 °C	$1.0 \times 10^{-14}$ mol$^2$ dm$^{-6}$
Avogadro constant, $L$	$6.02 \times 10^{23}$ mol$^{-1}$

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End of Paper

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Marking note: Section A = 10 marks, Section B = 25 marks, Section C = 15 marks. Total = 50 marks.

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TuitionGoWhere Practice Paper — Chemistry H2 A-Level

Answer Key: Acids, Bases & Salts (Version 2)

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Section A: Multiple Choice

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 (gains a proton), not the conjugate base. Option C is the conjugate acid of water. Option D is unrelated.
• Common mistake: Students confuse conjugate acid with conjugate base. Remember: acid loses $H^+$ → conjugate base; base gains $H^+$ → conjugate acid.

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2. B [1]

• Explanation: pH = 3.40, so $[H^+] = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm$^{-3}$. Using $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$: $[OH^-] = \frac{1.0 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11} \text{ mol dm}^{-3}$
• Common mistake: Students may select A, which is $[H^+]$ not $[OH^-]$, or C, which is a calculation error.

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3. C [1]

• Explanation: $K_2CO_3$ is a salt of a strong base ($KOH$) and a weak acid ($H_2CO_3$). The $CO_3^{2-}$ ion hydrolyses in water to produce $OH^-$, making the solution alkaline (pH > 7). $NH_4Cl$ and $AlCl_3$ produce acidic solutions (cations of weak bases). $NaNO_3$ is a salt of a strong acid and strong base, giving pH ≈ 7.
• Teaching note: Salts of strong acid + weak base → acidic; weak acid + strong base → alkaline; strong + strong → neutral.

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4. C [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$^{-3}$. $pH = -\log(4.47 \times 10^{-4}) = 3.35$
• Wait — recalculating: $\sqrt{4.0 \times 10^{-6} \times 0.050} = \sqrt{2.0 \times 10^{-7}} = 4.472 \times 10^{-4}$. $pH = -\log(4.472 \times 10^{-4}) = 3.35$. Answer is B.
• Correction: The answer is B (3.35).
• Common mistake: Students may forget to take the square root or use the wrong formula.

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5. B [1]

• Explanation: Moles of $CH_3COOH$ = $0.050 \times 0.200 = 0.010$ mol. Moles of $NaOH$ = $0.050 \times 0.200 = 0.010$ mol. The acid and base react in a 1:1 ratio, so all the $CH_3COOH$ is completely neutralised to form $CH_3COONa$. The resulting solution contains only the salt $CH_3COONa$ (and water), with no remaining weak acid to form a buffer pair. A buffer requires significant amounts of both a weak acid and its conjugate base.
• Common mistake: Students assume any mixture of acid and base forms a buffer. A buffer requires excess weak acid remaining alongside its conjugate base.

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6. A [1]

• Explanation: At the half-equivalence point, exactly half the weak acid has been neutralised, so $[HA] = [A^-]$. Substituting into the Henderson–Hasselbalch equation: $pH = pK_a + log\frac{[A^-]}{[HA]} = pK_a + log(1) = pK_a + 0 = pK_a$
• Teaching note: This is a key concept — at half-equivalence, pH = $pK_a$, which allows experimental determination of $K_a$ from a titration curve.

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7. A [1]

• Explanation: Let the solubility of $Mg(OH)_2$ = $s$ mol dm$^{-3}$. $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ $K_{sp} = [Mg^{2+}][OH^-]^2 = s \times (2s)^2 = 4s^3$ $4s^3 = 5.6 \times 10^{-12}$ $s^3 = 1.4 \times 10^{-12}$ $s = \sqrt[3]{1.4 \times 10^{-12}} = 1.12 \times 10^{-4} \approx 1.1 \times 10^{-4} \text{ mol dm}^{-3}$
• Common mistake: Forgetting the coefficient 2 for $[OH^-]$ when setting up the $K_{sp}$ expression, leading to $s^3$ instead of $4s^3$.

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8. A [1]

• Explanation: This is a weak base ($NH_3$) titrated with a strong acid ($HCl$). The equivalence point pH is acidic (below 7) because the salt $NH_4Cl$ hydrolyses to produce an acidic solution. The equivalence point pH is approximately: $[NH_4^+] \approx 0.050 \text{ mol dm}^{-3} \quad (\text{after dilution})$ $K_a(NH_4^+) = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$ $[H^+] = \sqrt{K_a \times c} = \sqrt{5.56 \times 10^{-10} \times 0.050} = 5.27 \times 10^{-6}$ $pH \approx 5.28$ Methyl orange (pH range 3.1–4.4) is the closest match. The steep portion of the curve passes through the methyl orange range.
• Teaching note: For weak base–strong acid titrations, the equivalence point is acidic, so an indicator with an acidic pH range is needed.

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9. B [1]

• Explanation: Using the Henderson–Hasselbalch equation: $pH = pK_a + log\frac{[A^-]}{[HA]}$ $pK_a = -\log(1.8 \times 10^{-4}) = 3.74$ $pH = 3.74 + log\frac{0.15}{0.10} = 3.74 + log(1.5) = 3.74 + 0.18 = 3.92$
• Common mistake: Swapping $[A^-]$ and $[HA]$ in the log term, giving pH = 3.57 (option A).

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10. B [1]

• Explanation: At the equivalence point, all $CH_3COOH$ has been converted to $CH_3COONa$. The $CH_3COO^-$ ion is the conjugate base of a weak acid and undergoes hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$ This produces $OH^-$ ions, making the solution slightly alkaline (pH > 7).
• Common mistake: Choosing A (excess NaOH) — at the equivalence point, there is no excess reagent.

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Section B: Structured Questions

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11. [4 marks]

(a) [1] A Brønsted–Lowry acid is a proton ($H^+$) donor.

• Marking: Must mention "proton" and "donor" (or "donates").

(b) [2]

• Pair 1: acid = $H_2O$ / conjugate base = $OH^-$ [1]
• Pair 2: acid = $NH_4^+$ / conjugate base = $NH_3$ [1]
• Explanation: $H_2O$ donates $H^+$ to become $OH^-$. $NH_3$ accepts $H^+$ to become $NH_4^+$. The conjugate acid–base pairs are: $H_2O$/$OH^-$ and $NH_4^+$/$NH_3$.

(c) [1] $NH_3$ can act as a Brønsted–Lowry base because it has a lone pair of electrons on the nitrogen atom that can accept a proton ($H^+$).

• Marking: Must mention lone pair and ability to accept $H^+$.

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12. [5 marks]

(a) [1] $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$

• Marking: Correct formula of all species and balanced equation. Award 1 mark.

(b) [3]

• Moles of $NaOH$ = $\frac{25.0}{1000} \times 0.120 = 3.00 \times 10^{-3}$ mol [1]
• From the equation, mole ratio $H_2SO_4 : NaOH = 1 : 2$
• Moles of $H_2SO_4$ = $\frac{3.00 \times 10^{-3}}{2} = 1.50 \times 10^{-3}$ mol [1]
• Volume of $H_2SO_4$ = $\frac{1.50 \times 10^{-3}}{0.100} = 1.50 \times 10^{-2}$ dm$^3$ = 15.0 cm$^3$ [1]
• Marking: 1 mark for moles of NaOH, 1 mark for correct mole ratio application, 1 mark for final answer with unit.

(c) [1] Indicator: phenolphthalein [½] Colour change: pink to colourless [½] (or colourless to pink, depending on direction — here acid in burette, alkali in flask, so pink → colourless)

• Note: Since $H_2SO_4$ (strong acid) is added to $NaOH$ (strong base), the equivalence point is at pH 7. Either phenolphthalein or methyl orange is acceptable. However, phenolphthalein is more commonly used for strong acid–strong base titrations in school labs.

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13. [6 marks]

(a) [2]

• Titrations 2 and 3 are concordant (within 0.10 cm$^3$ of each other). [1]
• Average titre = $\frac{24.30 + 24.30}{2} = 24.30$ cm$^3$ [1]
• Marking: 1 mark for identifying concordant values, 1 mark for correct average. Titration 1 (24.35) is also close but the two identical values (24.30, 24.30) should be used. If student uses all three concordant values: $\frac{24.35 + 24.30 + 24.30}{3} = 24.32$ cm$^3$, this is also acceptable.

(b) [2]

• Moles of $NaOH$ = $\frac{24.30}{1000} \times 0.100 = 2.43 \times 10^{-3}$ mol [1]
• Since $HA$ is monoprotic, mole ratio $HA : NaOH = 1 : 1$
• Concentration of $HA$ = $\frac{2.43 \times 10^{-3}}{25.0/1000} = \frac{2.43 \times 10^{-3}}{0.0250} = 0.0972$ mol dm$^{-3}$ [1]

(c) [2] The pH at the equivalence point is 8.72 (greater than 7) because the salt formed (sodium salt of the weak acid, $NaA$) undergoes hydrolysis. The conjugate base $A^-$ reacts with water: $A^- + H_2O \rightleftharpoons HA + OH^-$ This produces $OH^-$ ions, making the solution slightly alkaline. [2]

• Marking: 1 mark for identifying hydrolysis of the conjugate base, 1 mark for the explanation/equation showing production of $OH^-$.

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14. [5 marks]

(a) [2]

• Initial moles of $HCOOH$ = $\frac{100}{1000} \times 0.300 = 0.0300$ mol
• Moles of $NaOH$ = $\frac{100}{1000} \times 0.150 = 0.0150$ mol [1]
• $NaOH$ reacts with $HCOOH$ in 1:1 ratio:
  • Moles of $HCOOH$ remaining = $0.0300 - 0.0150 = 0.0150$ mol
  • Moles of $HCOO^-$ formed = $0.0150$ mol [1]
• Marking: 1 mark for moles of NaOH, 1 mark for both remaining HCOOH and HCOO⁻ moles.

(b) [2]

• 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$^{-3}$
• $pK_a = -\log(1.8 \times 10^{-4}) = 3.74$ [1]
• $pH = 3.74 + log\frac{0.0750}{0.0750} = 3.74 + 0 = 3.74$ [1]
• Alternative (using $K_a$ expression): $pH = pK_a$ when $[HA] = [A^-]$, so $pH = 3.74$.

(c) [1] The pH will decrease only very slightly (remain almost unchanged). The added $H^+$ ions react with the $HCOO^-$ in the buffer: $HCOO^- + H^+ \rightarrow HCOOH$ This removes the added $H^+$ ions, minimising the pH change. The buffer resists changes in pH.

• Marking: Award 1 mark for stating pH remains almost unchanged with correct explanation involving the buffer action.

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15. [5 marks]

(a) [1] $K_{sp} = [Ca^{2+}][F^-]^2$

• Marking: Must include the squared term for $F^-$. No mark for $K_{sp}$ without the square.

(b) [3]

• Let solubility of $CaF_2$ = $s$ mol dm$^{-3}$ $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$ $[Ca^{2+}] = s, \quad [F^-] = 2s$ $K_{sp} = s \times (2s)^2 = 4s^3$ [1] $4s^3 = 3.9 \times 10^{-11}$ $s^3 = 9.75 \times 10^{-12}$ [1] $s = \sqrt[3]{9.75 \times 10^{-12}} = 2.14 \times 10^{-4} \text{ mol dm}^{-3}$ [1]
• Answer: $2.1 \times 10^{-4}$ mol dm$^{-3}$ (to 2 s.f.)
• Marking: 1 mark for correct $K_{sp}$ expression in terms of $s$, 1 mark for correct substitution, 1 mark for final answer.

(c) [1] The solubility of $CaF_2$ would decrease. The $NaF$ solution provides $F^-$ ions (common ion effect). According to Le Chatelier's principle, the increased $[F^-]$ shifts the equilibrium to the left (towards the solid), reducing the solubility of $CaF_2$.

• Marking: 1 mark for "decrease" with correct explanation referencing common ion effect or Le Chatelier's principle.

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Section C: Free Response

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16. [8 marks]

(a) [1] Volume of $NaOH$ at equivalence point = 22.5 cm$^3$ (read from the graph at the steepest point of the curve).

• Marking: Accept 22.4–22.6 cm$^3$ depending on reading precision from the graph.

(b) [2]

• Moles of $NaOH$ at equivalence = $\frac{22.5}{1000} \times 0.100 = 2.25 \times 10^{-3}$ mol [1]
• Mole ratio $CH_3COOH : NaOH = 1 : 1$
• Concentration of $CH_3COOH$ = $\frac{2.25 \times 10^{-3}}{25.0/1000} = \frac{2.25 \times 10^{-3}}{0.0250} = 0.0900$ mol dm$^{-3}$ [1]

(c) [2]

• At the half-equivalence point, volume of $NaOH$ = $\frac{22.5}{2} = 11.25$ cm$^3$
• From the graph, at 11.25 cm$^3$, pH ≈ 4.74 [1]
• At half-equivalence: $pH = pK_a$, so $pK_a = 4.74$
• $K_a = 10^{-4.74} = 1.8 \times 10^{-5}$ mol dm$^{-3}$ [1]
• Marking: 1 mark for identifying half-equivalence point and reading pH, 1 mark for calculating $K_a$.

(d) [2] Phenolphthalein changes colour in the pH range 8.2–10.0. From the graph, the equivalence point occurs at pH ≈ 8.7, which falls within the phenolphthalein colour-change range. The steep portion of the titration curve passes through the phenolphthalein range, so a sharp colour change (colourless to pink) is observed at the equivalence point, giving an accurate endpoint. [2]

• Marking: 1 mark for stating the pH range of phenolphthalein, 1 mark for linking it to the equivalence point pH from the graph.

(e) [1] Methyl orange changes colour in the pH range 3.1–4.4, which is well below the equivalence point pH of 8.7. The colour change would occur before the true equivalence point (in the buffer region), leading to an underestimate of the volume of $NaOH$ required and therefore an underestimate of the concentration of $CH_3COOH$. This would not give an accurate result.

• Marking: 1 mark for explaining that methyl orange changes colour too early, leading to inaccurate results.

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17. [7 marks]

(a) [1] $NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$

• Accept: $NH_4^+ \rightleftharpoons NH_3 + H^+$ (simplified form)
• Marking: 1 mark for correct equation showing $NH_4^+$ acting as an acid.

(b) [2]

• For the conjugate acid–base pair: $K_a(NH_4^+) \times K_b(NH_3) = K_w$ [1]
• Therefore: $K_a(NH_4^+) = \frac{K_w}{K_b(NH_3)}$ [1]
• Explanation: $NH_4^+$ and $NH_3$ are a conjugate acid–base pair. The product of $K_a$ of the conjugate acid and $K_b$ of the base equals $K_w$.
• Marking: 1 mark for the relationship $K_a \times K_b = K_w$, 1 mark for rearranging to $K_a = K_w / K_b$.

(c) [3]

• $K_a(NH_4^+) = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$ mol dm$^{-3}$ [1]
• For the weak acid $NH_4^+$ at concentration 0.200 mol dm$^{-3}$: $[H^+] = \sqrt{K_a \times c} = \sqrt{5.56 \times 10^{-10} \times 0.200} = \sqrt{1.11 \times 10^{-10}} = 1.05 \times 10^{-5} \text{ mol dm}^{-3}$ [1] $pH = -\log(1.05 \times 10^{-5}) = 4.98$ [1]
• Answer: pH = 4.98 (or 5.0 to 2 s.f.)
• Marking: 1 mark for calculating $K_a$, 1 mark for correct $[H^+]$ calculation, 1 mark for pH.

(d) [1] The student's claim is correct. Adding $NH_4Cl$ to $NH_3$ solution increases the concentration of $NH_4^+$ (the conjugate acid). According to Le Chatelier's principle, this shifts the equilibrium: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$ to the left, reducing $[OH^-]$ and therefore decreasing the pH. This is consistent with the buffer equation: increasing $[NH_4^+]$ decreases pH.

• Marking: 1 mark for agreeing with the claim and providing a correct explanation using Le Chatelier's principle or the Henderson–Hasselbalch equation.

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Mark Summary

Section	Marks
A: Q1–10 (Multiple Choice)	10
B: Q11	4
B: Q12	5
B: Q13	6
B: Q14	5
B: Q15	5
C: Q16	8
C: Q17	7
Total	50