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A Level H2 Chemistry Practice Paper 2

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Questions

TuitionGoWhere Exam Practice (AI)

Subject: Chemistry H2
Level: A-Level
Paper: Practice Paper 2 (Version 2 of 5)
Topic: Acids, Bases and Salts
Duration: 1 hour 15 minutes
Total Marks: 60

Name: _______________________
Class: _______________________
Date: _______________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in this question paper.
The use of an approved scientific calculator is expected where appropriate.
A Data Booklet is provided for reference.
At the end of the examination, fasten all your work securely together.
The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Structured Questions

1 A student is tasked with determining the concentration of a solution of ethanoic acid, $\text{CH}_3\text{COOH}$ , by titration against a standard solution of sodium hydroxide, $\text{NaOH}$ .

The student performs a rough titration followed by three accurate titrations. The burette readings are recorded below.

Titration	Rough	1	2	3
Final reading / $\text{cm}^3$	24.50	23.80	47.10	23.90
Initial reading / $\text{cm}^3$	0.00	0.00	23.80	0.00
Titre / $\text{cm}^3$	24.50

(a) Complete the table above by calculating the titre for titrations 1, 2, and 3. [1]

(b) Select the appropriate titres to calculate the mean titre. Explain your choice. [2]

(d) The concentration of the $\text{NaOH}$ solution is $0.100 \text{ mol dm}^{-3}$ . Calculate the concentration of the ethanoic acid solution if $25.0 \text{ cm}^3$ of the acid was used in each titration. [2]

(e) Explain why phenolphthalein is a suitable indicator for this titration, whereas methyl orange is not. Refer to the pH at the equivalence point in your answer. [3]

Answer space

...... [10]

2 Buffer solutions are essential in maintaining pH stability in biological and chemical systems.

(a) Define a buffer solution. [2]

(b) A buffer solution is prepared by mixing $50.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ ethanoic acid ( $\text{CH}_3\text{COOH}$ ) with $50.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ sodium ethanoate ( $\text{CH}_3\text{COONa}$ ). The $K_a$ of ethanoic acid is $1.7 \times 10^{-5} \text{ mol dm}^{-3}$ .

(i) Calculate the pH of this buffer solution. [3]

(ii) Calculate the new pH of the solution after adding $1.0 \text{ cm}^3$ of $1.0 \text{ mol dm}^{-3}$ $\text{HCl}$ to the buffer mixture. Assume volumes are additive. [4]

(c) Explain, with the aid of an equation, how this buffer solution resists changes in pH when a small amount of strong base ( $\text{OH}^-$ ) is added. [2]

Answer space

<br...... [11]

3 Solubility products ( $K_{sp}$ ) govern the precipitation of sparingly soluble salts.

The $K_{sp}$ of magnesium hydroxide, $\text{Mg(OH)}_2$ , is $1.8 \times 10^{-11} \text{ mol}^3 \text{ dm}^{-9}$ at $298 \text{ K}$ .

(a) Write the expression for the solubility product, $K_{sp}$ , of $\text{Mg(OH)}_2$ . [1]

(b) Calculate the solubility of $\text{Mg(OH)}_2$ in pure water at $298 \text{ K}$ in $\text{mol dm}^{-3}$ . [3]

(c) Calculate the maximum concentration of $\text{Mg}^{2+}$ ions that can exist in a solution with a pH of 10.0 before precipitation of $\text{Mg(OH)}_2$ occurs. [3]

(d) A student mixes $100 \text{ cm}^3$ of $0.010 \text{ mol dm}^{-3}$ $\text{MgCl}_2$ with $100 \text{ cm}^3$ of $0.010 \text{ mol dm}^{-3}$ $\text{NaOH}$ . Determine, by calculation, whether a precipitate of $\text{Mg(OH)}_2$ will form. [4]

Answer space

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Answer space

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4 The pH of solutions of weak acids depends on their dissociation constant ( $K_a$ ) and concentration.

(a) Write the expression for the acid dissociation constant, $K_a$ , for a generic weak acid $\text{HA}$ . [1]

(b) A $0.10 \text{ mol dm}^{-3}$ solution of a weak acid $\text{HA}$ has a pH of 2.90.

(i) Calculate the concentration of $\text{H}^+$ ions in the solution. [1]

(ii) Calculate the value of $K_a$ for this acid. State any assumptions made. [3]

(c) Sketch the titration curve for the titration of $25.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ $\text{HA}$ with $0.10 \text{ mol dm}^{-3}$ $\text{NaOH}$ . Label the equivalence point and the region where the solution acts as a buffer. [3]

Graph space

...... [8]

5 Indicators are weak acids or bases that change color over a specific pH range.

(a) Explain the mechanism of action of an acid-base indicator $\text{HIn}$ using equilibrium principles. [3]

(b) The table below shows the pH ranges for three common indicators.

Indicator	pH Range	Color Change
Methyl Orange	3.1 – 4.4	Red to Yellow
Bromothymol Blue	6.0 – 7.6	Yellow to Blue
Phenolphthalein	8.3 – 10.0	Colorless to Pink

Select the most suitable indicator for the titration of $0.1 \text{ mol dm}^{-3}$ ethanoic acid with $0.1 \text{ mol dm}^{-3}$ sodium hydroxide. Justify your choice by referring to the pH at the equivalence point. [2]

Answer space

...... [6]

6 Practical skills are vital in chemistry.

(a) Describe how you would rinse a burette before filling it with a standard solution of $\text{NaOH}$ . Explain why this step is necessary. [2]

(b) During a titration, a student overshoots the endpoint. What should the student do? [1]

(c) Explain why distilled water can be added to the conical flask during a titration without affecting the final result. [2]

Answer space

...... [5]

End of Section A

Section B: Free Response Questions

7 Discuss the importance of buffer solutions in biological systems, specifically focusing on the control of blood pH. Include relevant chemical equations and explain the consequences of pH deviation from the normal range. [10]

Answer space

...... [10]

Total Marks for Section B: 10
Total Marks for Paper: 60

End of Paper

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Answers

TuitionGoWhere Exam Practice (AI) - Answer Key

Subject: Chemistry H2
Level: A-Level
Paper: Practice Paper 2 (Version 2 of 5)
Topic: Acids, Bases and Salts

Section A: Structured Questions

1
(a)
Titre 1: $23.80 - 0.00 = 23.80 \text{ cm}^3$
Titre 2: $47.10 - 23.80 = 23.30 \text{ cm}^3$
Titre 3: $23.90 - 0.00 = 23.90 \text{ cm}^3$
[1]

(b)
Concordant titres are 1 and 3 (23.80 and 23.90).
Titre 2 (23.30) is anomalous/not concordant (differs by more than 0.10 cm³ from the others).
Rough titre is ignored.
[2]

(d)
Moles of $\text{NaOH} = \frac{23.85}{1000} \times 0.100 = 0.002385 \text{ mol}$
Ratio $\text{CH}_3\text{COOH} : \text{NaOH} = 1:1$
Moles of $\text{CH}_3\text{COOH} = 0.002385 \text{ mol}$
Concentration = $\frac{0.002385}{25.0/1000} = 0.0954 \text{ mol dm}^{-3}$
[2]

(e)
Ethanoic acid is a weak acid and NaOH is a strong base.
The equivalence point is at pH > 7 (approx pH 8-9) due to the hydrolysis of the ethanoate ion ( $\text{CH}_3\text{COO}^-$ ).
Phenolphthalein changes color in the range 8.3–10.0, which includes the equivalence point.
Methyl orange changes color in the range 3.1–4.4, which is far below the equivalence point, leading to a large titration error.
[3]

2
(a)
A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added.
[2]

(b)
(i)
$[\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-] = 0.05 \text{ mol dm}^{-3}$ (diluted by half, but ratio is 1:1)
Alternatively, since volumes and initial concentrations are equal, the ratio of moles is 1:1.
$\text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\text{salt}]}{[\text{acid}]} \right)$
$\text{p}K_a = -\log_{10}(1.7 \times 10^{-5}) = 4.77$
$\text{pH} = 4.77 + \log_{10}(1) = 4.77$
[3]

(ii)
Moles $\text{H}^+$ added = $\frac{1.0}{1000} \times 1.0 = 0.001 \text{ mol}$
Initial moles $\text{CH}_3\text{COOH} = 0.05 \times 0.050 = 0.0025 \text{ mol}$
Initial moles $\text{CH}_3\text{COO}^- = 0.05 \times 0.050 = 0.0025 \text{ mol}$
After adding $\text{H}^+$ :
$\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$
Moles $\text{CH}_3\text{COO}^- = 0.0025 - 0.001 = 0.0015 \text{ mol}$
Moles $\text{CH}_3\text{COOH} = 0.0025 + 0.001 = 0.0035 \text{ mol}$
Total volume = $101 \text{ cm}^3 = 0.101 \text{ dm}^3$
New $[\text{CH}_3\text{COO}^-] = \frac{0.0015}{0.101}$
New $[\text{CH}_3\text{COOH}] = \frac{0.0035}{0.101}$
$\text{pH} = 4.77 + \log_{10} \left( \frac{0.0015}{0.0035} \right) = 4.77 + \log_{10}(0.4286) = 4.77 - 0.37 = 4.40$
[4]

(c)
$\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$
When $\text{OH}^-$ is added, it reacts with $\text{H}^+$ to form water.
The equilibrium shifts to the right to replenish $\text{H}^+$ , minimizing the change in pH.
Alternatively: $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$
[2]

3
(a)
$K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2$
[1]

(b)
Let solubility be $s \text{ mol dm}^{-3}$ .
$[\text{Mg}^{2+}] = s$ , $[\text{OH}^-] = 2s$
$K_{sp} = s(2s)^2 = 4s^3$
$1.8 \times 10^{-11} = 4s^3$
$s^3 = 4.5 \times 10^{-12}$
$s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4} \text{ mol dm}^{-3}$
[3]

(c)
$\text{pH} = 10.0 \Rightarrow \text{pOH} = 4.0 \Rightarrow [\text{OH}^-] = 1.0 \times 10^{-4} \text{ mol dm}^{-3}$
$K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2$
$1.8 \times 10^{-11} = [\text{Mg}^{2+}](1.0 \times 10^{-4})^2$
$[\text{Mg}^{2+}] = \frac{1.8 \times 10^{-11}}{1.0 \times 10^{-8}} = 1.8 \times 10^{-3} \text{ mol dm}^{-3}$
[3]

(d)
After mixing, volume doubles, so concentrations halve.
$[\text{Mg}^{2+}] = 0.005 \text{ mol dm}^{-3}$
$[\text{OH}^-] = 0.005 \text{ mol dm}^{-3}$
Ionic Product (IP) = $[\text{Mg}^{2+}][\text{OH}^-]^2 = (0.005)(0.005)^2 = 1.25 \times 10^{-7}$
Since IP ( $1.25 \times 10^{-7}$ ) > $K_{sp}$ ( $1.8 \times 10^{-11}$ ), a precipitate will form.
[4]

4
(a)
$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$
[1]

(b)
(i)
$[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.90} = 1.26 \times 10^{-3} \text{ mol dm}^{-3}$
[1]

(ii)
Assumption: $[\text{H}^+] = [\text{A}^-]$ and $[\text{HA}]_{\text{eq}} \approx [\text{HA}]_{\text{initial}}$
$K_a = \frac{(1.26 \times 10^{-3})^2}{0.10} = \frac{1.59 \times 10^{-6}}{0.10} = 1.59 \times 10^{-5} \text{ mol dm}^{-3}$
[3]

(c)
Curve starts at pH ~2.9.
Gradual rise (buffer region).
Steep vertical section around equivalence point (pH ~8-9).
Levels off at high pH (~13).
Equivalence point marked at volume 25.0 cm³.
Buffer region marked around half-equivalence (12.5 cm³).
[3]

5
(a)
$\text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^-$
$\text{HIn}$ and $\text{In}^-$ have different colors.
In acid, high $[\text{H}^+]$ shifts equilibrium left (color of $\text{HIn}$ ).
In base, low $[\text{H}^+]$ shifts equilibrium right (color of $\text{In}^-$ ).
[3]

(b)
Phenolphthalein.
The titration involves a weak acid and strong base, so the equivalence point is at pH > 7.
Phenolphthalein's range (8.3–10.0) overlaps with the steep part of the titration curve at the equivalence point.
[2]

(c)
Indicators are weak acids/bases themselves. Adding too much would consume a significant amount of titrant, causing a titration error.
[1]

6
(a)
Rinse with a small amount of the $\text{NaOH}$ solution to be used.
This ensures the concentration of the solution in the burette is not diluted by residual water.
[2]

(b)
Discard the result and repeat the titration.
[1]

(c)
Adding water changes the volume but not the number of moles of acid in the flask.
The endpoint depends on the moles of acid reacting with the moles of base added.
Therefore, the volume of base required remains unchanged.
[2]

Section B: Free Response Questions

7
Key Points:

Normal Blood pH: 7.35–7.45. Deviation leads to acidosis (<7.35) or alkalosis (>7.45), which can be fatal.
Buffer System: The carbonic acid-hydrogencarbonate buffer system is the primary buffer in blood.
$\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$
Mechanism:
- When acid ( $\text{H}^+$ ) is added (e.g., from metabolism), it reacts with $\text{HCO}_3^-$ to form $\text{H}_2\text{CO}_3$ , which decomposes to $\text{CO}_2$ and $\text{H}_2\text{O}$ . $\text{CO}_2$ is exhaled by lungs.
- When base ( $\text{OH}^-$ ) is added, it reacts with $\text{H}^+$ to form water. Equilibrium shifts right to replenish $\text{H}^+$ , consuming $\text{H}_2\text{CO}_3$ . Kidneys regulate $\text{HCO}_3^-$ levels.
Consequences:
- Acidosis: Depresses CNS, coma, death.
- Alkalosis: Overexcitability of nervous system, muscle spasms, tetany.
Role of Lungs and Kidneys: Lungs control $[\text{CO}_2]$ (short term), kidneys control $[\text{HCO}_3^-]$ (long term).

[10]