How many NESA Chemistry questions cover Properties of Acids and Bases?

AusGrader has 52 NESA Chemistry questions on Properties of Acids and Bases, all with instant AI grading and detailed marking feedback.

NESA (HSC) Chemistry — Properties of Acids and Bases Questions & Answers | AusGrader

Q3

2024

NESA

1 mark

Q3

1 mark

Which of the following compounds can be correctly described as an Arrhenius base when dissolved in water?

A

Sodium nitrate

B

Sodium sulfate

C

Sodium chloride

D

Sodium hydroxide

Q14

2023

NESA

1 mark

Q14

1 mark

What volume of $0.540 \text{ mol L}^{-1}$ hydrochloric acid will react completely with $1.34 \text{ g}$ of sodium carbonate?

A

11.7 mL

B

23.4 mL

C

29.9 mL

D

46.8 mL

Correct Answer

Correct. $1.34 \text{ g}$ of $\text{Na}_2\text{CO}_3$ is $0.0126 \text{ mol}$ , which requires $0.0253 \text{ mol}$ of $\text{HCl}$ due to the 1:2 molar ratio. Dividing this by the concentration ( $0.540 \text{ mol L}^{-1}$ ) gives $0.0468 \text{ L}$ , or $46.8 \text{ mL}$ .

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

Q22

2023

NESA

4 marks

Q22

4 marks

Explain how the following substances would be classified under the Arrhenius and Brønsted–Lowry definitions of acids. Support your answer with relevant equations.

$\text{HCl}(aq)$
$\text{NH}_4\text{Cl}(aq)$

According to Arrhenius, acids are hydrogen-containing compounds that dissociate in water to give $\text{H}^+$ ions. $\text{HCl}(aq)$ would be considered an acid by Arrhenius as it produces $\text{H}^+$ ions in water.

\text{HCl}(aq) \rightarrow \text{H}^+(aq) + \text{Cl}^-(aq)

Arrhenius would not recognise the salt $\text{NH}_4\text{Cl}$ as an acid, as the predominant ions present in aqueous solution are ammonium and chloride.

In Brønsted–Lowry theory, acids are defined as proton donors. $\text{HCl}(aq)$ is a proton donor and therefore a Brnsted–Lowry acid.

Ammonium chloride ( $\text{NH}_4\text{Cl}$ ) is classified as a Brnsted–Lowry acid as the ammonium ion donates a proton to water and forms a hydronium ion.

\text{NH}_4^+(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_3(aq) + \text{H}_3\text{O}^+(aq)

Marking Criteria

Descriptor	Marks
Classifies both substances with respect to Arrhenius and Brønsted–Lowry theories Provides two relevant chemical equations	4
Classifies one substance with respect to Arrhenius and Brønsted–Lowry theories Provides a relevant chemical equation OR Classifies both substances with respect to Arrhenius OR Brønsted–Lowry theory Provides a relevant chemical equation OR Classifies both substances with respect to Arrhenius and Brønsted–Lowry theories	3
Classifies one substance with respect to Arrhenius and Brønsted–Lowry theories OR Classifies both substances with respect to Arrhenius OR Brønsted–Lowry theory	2
• Provides some relevant information	1
None of the above	0

Q37

2025

NESA

4 marks

Q37

4 marks

Compound A has the molecular formula $\text{C}_5\text{H}_{10}$ . The information below shows some chemical reactions beginning with this compound.

Compound A reacts with $\text{H}^+/\text{H}_2\text{O}$ to produce compounds B and C.
Compound B does not react with $\text{H}^+/\text{Cr}_2\text{O}_7^{2-}$ .
Compound C reacts with $\text{H}^+/\text{Cr}_2\text{O}_7^{2-}$ to produce compound D.
Compound D does not react with $\text{Na}_2\text{CO}_3(aq)$ .

Determine the structure of compound A. Justify your answer using all the data given.

Compounds B and C are alkanols.

Compound B cannot be oxidised – so B must be a tertiary alkanol, therefore there must be branching.

Compound C can be oxidised but not to a carboxylic acid (no reaction with carbonate) – so the oxidation product is a ketone, so compound C is a secondary alkanol.

To explain the alkanols produced above, the C=C in compound A is not in the terminal position.

Therefore compound A is 2-methylbut-2-ene (C $_5$ H $_{10}$ ).

Marking Criteria

Descriptor	Marks
Correctly determines the structure of compound A AND Correctly justifies the structure using all the data provided	4
Correctly determines the structure of compound A AND justifies the structure using some of the data provided OR Correctly interprets all of the data provided	3
Correctly interprets some of the data provided	2
Provides some relevant information	1
None of the above	0

NESA Chemistry Properties of Acids and Bases

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