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AusGrader has 94 NESA Chemistry questions on Quantitative Analysis, all with instant AI grading and detailed marking feedback.

NESA (HSC) Chemistry — Quantitative Analysis Questions & Answers | AusGrader

Q19

2025

NESA

1 mark

Q19

1 mark

0.1 mol of solid sodium acetate is dissolved in 500 mL of 0.1 mol L $^{-1}$ HCl in a beaker. This solution has a pH of 4.8.

500 mL of distilled water is then added to the beaker.

What is the pH of the final solution?

A

2.4

B

4.5

C

4.8

D

5.1

Q16

2024

NESA

1 mark

Q16

1 mark

Which of the following is the overall reaction that takes place when a strong acid is added to a buffer containing equal amounts of acetic acid and acetate ions?

A

$\text{HCOO}^- + \text{H}_3\text{O}^+ \rightarrow \text{HCOOH} + \text{H}_2\text{O}$

B

$\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$

C

$\text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$

D

$\text{CH}_3\text{COOH} + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{C(OH)}_2^+ + \text{H}_2\text{O}$

Q39

2024

NESA

4 marks

Q39

4 marks

Water and octan-1-ol do not mix. When an aqueous solution of bromoacetic acid ( $\text{BrCH}_2\text{COOH}$ ) is shaken with octan-1-ol, an equilibrium system is established between bromoacetic acid dissolved in the octan-1-ol and in the water.

\text{BrCH}_2\text{COOH}(aq) \rightleftharpoons \text{BrCH}_2\text{COOH}(octan\text{-}1\text{-}ol)

The equilibrium constant expression for this system is

K_{eq} = \frac{[\text{BrCH}_2\text{COOH}(octan\text{-}1\text{-}ol)]}{[\text{BrCH}_2\text{COOH}(aq)]}.

An aqueous solution of bromoacetic acid with an initial concentration of $0.1000 \text{ mol L}^{-1}$ is shaken with an equal volume of octan-1-ol. Bromoacetic acid does not dissociate in octan-1-ol but does dissociate in water, with $K_a = 1.29 \times 10^{-3}$ . When the system has reached equilibrium, the $[\text{H}^+]$ is $9.18 \times 10^{-3} \text{ mol L}^{-1}$ .

Calculate the equilibrium concentration of aqueous bromoacetic acid and hence, or otherwise, calculate the $K_{eq}$ for the octan-1-ol and water system.

$K_a = \frac{[\text{H}^+][\text{BrCH}_2\text{COO}^-]}{[\text{BrCH}_2\text{COOH}]}$

\begin{align*} [\text{BrCH}_2\text{COOH}](aq) &= \frac{[\text{H}^+][\text{BrCH}_2\text{COO}^-]}{K_a}\\ &= \frac{[9.18 \times 10^{-3}][9.18 \times 10^{-3}]}{1.29 \times 10^{-3}}\\ &= 0.0654 \text{ mol L}^{-1} \end{align*}

$[\text{BrCH}_2\text{COOH}](\text{octan-1-ol}) = 0.1000 - 0.0654 - 9.18 \times 10^{-3} = 0.0254 \text{ mol L}^{-1}$

K_{eq} = \frac{[\text{BrCH}_2\text{COOH}(\text{octan-1-ol})]}{[\text{BrCH}_2\text{COOH}(aq)]} = \frac{0.0254 \text{ mol L}^{-1}}{0.0654 \text{ mol L}^{-1}} = 0.390

Marking Criteria

Descriptor	Marks
Calculates the $K_{eq}$	4
Provides most steps of the calculation	3
Provides some steps of the calculation	2
Provides some relevant information	1
None of the above	0

Q33

2025

NESA

7 marks

Q33

7 marks

Chalk is predominantly calcium carbonate. Different brands of chalk vary in their calcium carbonate composition.

The table shows the composition of three different brands of chalk.

	Brand X	Brand Y	Brand Z
$\text{CaCO}_3$ (%)	85.5	83.9	82.4

The following procedure was used to determine the calcium carbonate composition of a chalk sample.

A sample of chalk was crushed in a mortar and pestle.
A 3.00 g sample of the crushed chalk was placed in a conical flask.
100.0 mL of $0.550 \text{ mol L}^{-1} \text{ HCl}(aq)$ was added to the sample and left to react completely, resulting in a clear solution.
Four 20 mL aliquots of this mixture were then titrated with $0.10 \text{ mol L}^{-1} \text{ KOH}$ .

The results of the titrations are recorded.

Burette volume (mL)	Trial 1	Trial 2	Trial 3	Trial 4
Final	7.80	14.90	22.10	29.25
Initial	0.00	7.80	14.90	22.10
Total used	7.80	7.10	7.20	7.15

Determine the brand of the chalk sample. Include a relevant chemical equation in your answer.

Exclude the outlier.

Average volume KOH used excluding Trial 1 = $\frac{7.10 + 7.20 + 7.15}{3} = 0.00715$ L

HCl( $aq$ ) + KOH( $aq$ ) $\rightarrow$ KCl( $aq$ ) + H $_2$ O( $l$ )

moles KOH = $0.10 \times 0.00715 = 0.000715$ mol KOH

Ratio HCl : KOH = 1 : 1
$\therefore$ 0.000715 mol HCl for each sample
$\therefore$ $0.000715 \times 5 = 0.003575$ mol total in sampled solution

Initial $n$ HCl = $0.550 \times 0.1000 = 0.0550$ mol HCl

$n$ HCl that reacted with CaCO $_3$ = $0.0550 - 0.003575$ = 0.051425 mol HCl

2HCl( $aq$ ) + CaCO $_3$ ( $s$ ) $\rightarrow$ CaCl $_2$ ( $aq$ ) + H $_2$ O( $l$ ) + CO $_2$ ( $g$ )

Ratio HCl : CaCO $_3$ = 2 : 1
$\therefore$ $\frac{0.051425}{2} = 0.0257125$ mol CaCO $_3$

Mass CaCO $_3$ = $0.0257125 \times MM \text{ CaCO}_3 \, (100.09)$ = 2.5735641 g

% CaCO $_3$ = $\frac{2.5735641}{3.00} \times 100 = 85.7854\% \approx 85.8\%$

Chalk sample has to be Brand X.

Marking Criteria

Descriptor	Marks
Correctly identifies the brand AND Provides all correct calculations AND Provides a balanced chemical equation including states	7
Provides substantially correct calculations AND Provides a balanced chemical equation	6
Provides the main steps in the calculation AND Provides a balanced chemical equation	5
Provides the main steps in the calculation AND Provides a balanced chemical equation	4
Provides some steps in the calculation	3
Provides some steps in the calculation	2
Provides some relevant information	1
None of the above	0

Q3

2024

QCAA

Paper 2

9 marks

Q3

The concentration, pH and dissociation constant ( $K_a$ ) of aqueous solutions of ethanoic acid and two unknown monoprotic acids, I and II, are shown.

Acid	Concentration (M)	pH	$K_a$
CH $_3$ COOH(aq)	0.2		$1.8 \times 10^{-5}$
I	0.2	1.9	$6.6 \times 10^{-4}$
II	0.1	1.1	$1.3 \times 10^6$

Q3a

3 marks

Compare the relative strength of an aqueous solution of acid I and CH $_3$ COOH(aq).

Similarity: both weak acids

Difference: ethanoic acid ( $CH_3COOH$ ) is weaker than acid I

Significance: Ethanoic acid dissociates less to produce a lower $[H^+]$

Marking Criteria

Descriptor	Marks
Identifies that both acids are weak acids	1
Identifies that ethanoic acid is weaker than acid I	1
Explains that ethanoic acid dissociates less to produce a lower [H⁺]	1

Q3b

3 marks

Determine whether an aqueous solution of acid I or acid II would have a higher electrical conductivity. Explain your reasoning.

Acid II has a $K_a$ greater than 1 and is therefore a strong acid, while acid I is a weak acid due to small $K_a$ . Acid II dissociates to produce more ions in an aqueous solution than acid I.
Therefore, acid II has a higher electrical conductivity than acid I.

Marking Criteria

Descriptor	Marks
Identifies that acid II is stronger than acid I	1
Explains that acid II dissociates to produce more ions in solution	1
Determines that the electrical conductivity of acid II will be greater than acid I	1

Q3c

3 marks

Calculate the pH of 0.2 M CH $_3$ COOH(aq). Show your working.

$[H_3O^+] = [CH_3COO^-] = x$

$[CH_3COOH] = \frac{[H_3O^+][CH_3COO^-]}{K_a}$

$0.2 = \frac{(x)^2}{1.8\times10^{-5}}$

$[H_3O^+] = 1.9 \times 10^{-3} \text{ M}$

$pH = -\log_{10}[H_3O^+]$

$pH = 3$

Marking Criteria

Descriptor	Marks
Identifies $[H_3O^+] = [CH_3COO^-]$	1
Determines $[H_3O^+]$	1
Calculates pH	1

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