How many NESA Chemistry questions cover Analysis of Inorganic Substances?

AusGrader has 20 NESA Chemistry questions on Analysis of Inorganic Substances, all with instant AI grading and detailed marking feedback.

HSC Chemistry — Analysis of Inorganic Substances Questions & Answers

Q22

2025

NESA

2 marks

Q22

2 marks

Outline why quantitative and qualitative analyses are BOTH important in determining water quality.

Reveal Answer

Qualitative analysis allows the identification of a pollutant present in the water. Quantitative analysis allows for the amount of the pollutant present to be determined.

Marking Criteria

Descriptor	Marks
Outlines the need for both analysis types	2
Provides some relevant information	1
None of the above	0

Q8

2024

NESA

1 mark

Q8

1 mark

Which pair of ions produce different colours in a flame test?

A

$\text{Br}^-$ and $\text{Cl}^-$

B

$\text{Ag}^+$ and $\text{OH}^-$

C

$\text{Cu}^{2+}$ and $\text{Ca}^{2+}$

D

$\text{CH}_3\text{COO}^-$ and $\text{H}_2\text{PO}_4^-$

Reveal Answer

A

$\text{Br}^-$ and $\text{Cl}^-$

Bromide and chloride are anions, which generally do not produce characteristic colours in a standard flame test.

B

$\text{Ag}^+$ and $\text{OH}^-$

Silver ions ( $\text{Ag}^+$ ) do not produce a characteristic flame colour, and hydroxide ( $\text{OH}^-$ ) is an anion which also does not produce a colour.

C

$\text{Cu}^{2+}$ and $\text{Ca}^{2+}$

Correct Answer

Copper(II) ions ( $\text{Cu}^{2+}$ ) produce a characteristic blue-green flame, while calcium ions ( $\text{Ca}^{2+}$ ) produce a brick-red flame.

D

$\text{CH}_3\text{COO}^-$ and $\text{H}_2\text{PO}_4^-$

Acetate and dihydrogen phosphate are polyatomic anions and do not produce characteristic colours in a flame test.

Q23

2025

NESA

3 marks

Q23

3 marks

A student attempted to determine the % w/w of sulfate in a sample of solid fertiliser. They used the procedure described below.

Weigh a clean, dry beaker.
Add fertiliser to the beaker and weigh again.
Add 250 mL of distilled water and stir thoroughly.
Add 20 mL of 0.1 mol L $^{-1}$ BaCl $_2$ solution.
Filter out the BaSO $_4$ precipitate, using distilled water to ensure all of the solid is transferred from the beaker to the filter paper.
Put the filter paper and precipitate onto a weighed watch glass and leave them to dry for 20 minutes in the sun.
Weigh the watch glass, the filter paper and the precipitate.
Calculate the % w/w.

Justify TWO changes that can be made to the procedure to ensure more accurate results.

Reveal Answer

Filter out any insoluble components before step 4, or else these will contribute to the mass of the precipitate.

Weigh the filter paper, or else the mass of the precipitate will be inaccurate.

Marking Criteria

Descriptor	Marks
Justifies TWO appropriate changes with justifications of both related to accuracy	3
Justifies ONE appropriate change OR Identifies TWO appropriate changes	2
Provides some relevant information	1
None of the above	0

Q4

2025

NESA

1 mark

Q4

1 mark

A student is presented with two clear colourless solutions. One contains Pb $^{2+}$ and the other Na $^+$ ions.

Which ion can be added to the solutions to identify the solutions?

A

I $^−$

B

NH $_4^+$

C

NO $_3^−$

D

CH $_3$ COO $^−$

Reveal Answer

A

I $^−$

Correct Answer

Adding iodide ions ( $I^-$ ) will form a distinct yellow precipitate of lead(II) iodide ( $PbI_2$ ) with $Pb^{2+}$ , while sodium iodide ( $NaI$ ) remains soluble, allowing the solutions to be easily distinguished.

B

NH $_4^+$

Ammonium ( $NH_4^+$ ) is a cation and will not react or form a precipitate with either $Pb^{2+}$ or $Na^+$ cations.

C

NO $_3^−$

All nitrate ( $NO_3^-$ ) salts are soluble in water, meaning no precipitate will form with either $Pb^{2+}$ or $Na^+$ to help distinguish them.

D

CH $_3$ COO $^−$

Most acetate ( $CH_3COO^-$ ) salts, including lead(II) acetate and sodium acetate, are soluble in water, so no visible reaction or precipitate will occur.

Q39

2020

SCSA

12 marks

Q39

Fluorescent lights are glass tubes which are coated on the inside with rare earth metal phosphates (such as cerium, lanthanum and terbium phosphates) that provide light. Cerium, lanthanum and terbium are expensive, so are recovered once the fluorescent light is no longer functional.

The key steps in one method proposed for recovery of these rare earth metals are summarised below:

Step 1: Physical separation of the rare earth metal phosphates from the glass and any metallic components. This gives an impure powder consisting of cerium, lanthanum and terbium phosphates.
Step 2: Add excess solid sodium carbonate to the powder and heat, completely converting each rare earth metal phosphate to its corresponding oxide, as shown by the following balanced equations:
$2 \text{ LaPO}_4\text{(s)} + 3 \text{ Na}_2\text{CO}_3\text{(s)} \rightarrow \text{La}_2\text{O}_3\text{(s)} + 2 \text{ Na}_3\text{PO}_4\text{(s)} + 3 \text{ CO}_2\text{(g)}$
$4 \text{ CePO}_4\text{(s)} + 6 \text{ Na}_2\text{CO}_3\text{(s)} + \text{O}_2\text{(g)} \rightarrow 4 \text{ CeO}_2\text{(s)} + 4 \text{ Na}_3\text{PO}_4\text{(s)} + 6 \text{ CO}_2\text{(g)}$
$2 \text{ TbPO}_4\text{(s)} + 3 \text{ Na}_2\text{CO}_3\text{(s)} \rightarrow \text{Tb}_2\text{O}_3\text{(s)} + 2 \text{ Na}_3\text{PO}_4\text{(s)} + 3 \text{ CO}_2\text{(g)}$
Step 3: Wash the product from Step 2 with water.
Step 4: Add hydrochloric acid to the washed product from Step 3 to leach (dissolve) only the rare earth metal oxides.
Step 5: Use solvent extraction to separate the different rare earth metals from each other and create separate solutions of each of them.
Step 6: Add oxalic acid to the separated solutions to precipitate the rare earth metal ions as oxalate salts.
Step 7: Heat the oxalate salts to recover the rare earth metals as pure oxides, namely $\text{La}_2\text{O}_3$ , $\text{Tb}_4\text{O}_7$ and $\text{CeO}_2$ .

A chemist used the above procedure to determine the percentage by mass of lanthanum, terbium and cerium in some fluorescent lights and, after completing Step 1, had recovered 1.20 kg of the coating chemicals.

Q39b

The mass of the solid sent from Step 3 to Step 4 was 1.16 kg. This solid was leached with 6.00 mol L $^{-1}$ HCl at a solid to liquid ratio of 150 g per litre. Analysis of the solution at the end of leaching showed that it contained lanthanum, terbium and cerium, with its lanthanum concentration being $8.65 \times 10^{-3} \text{ mol L}^{-1}$ .

Q39a

3 marks

At the completion of Step 2, the mass of the mixture had decreased by 11.3 g. Calculate the mass of sodium carbonate that reacted with the rare earth metal phosphates.

Reveal Answer

n(CO₂) = 11.3/44.01 = 0.257 mol

n(CO₂) = n(C) = n(Na₂CO₃) = 0.257 mol

m(Na₂CO₃) = 0.257 × 105.99 = 27.2 g

Marking Criteria

Descriptor	Marks
n(CO₂) = 11.3/44.01 = 0.257 mol	1
n(CO₂) = n(C) = n(Na₂CO₃) = 0.257 mol	1
m(Na₂CO₃) = 0.257 × 105.99 = 27.2 g	1

Q39b

5 marks

Calculate the percentage, by mass, of lanthanum in the fluorescent light coating chemical, given that the leaching efficiency for lanthanum was 86%.

Note that the balanced equation for the leaching of lanthanum with hydrochloric acid is:
$\text{La}_2\text{O}_3\text{(s)} + 6 \text{ HCl(aq)} \rightarrow 2 \text{ LaCl}_3\text{(aq)} + 3 \text{ H}_2\text{O(l)}$

Reveal Answer

Volume of HCl used: 1160/150 = 7.73 L

n(La) = cV = 8.65 × 10⁻³ × 7.73 = 0.0669 mol La in solution

Taking into account the leaching efficiency,
n(La in the solid that was leached) = 0.0669/0.86 = 0.0778

m(La in the solid that was leached) = 0.0778 × 138.9 = 10.8 g

% La in the coating chemical = (10.8/1200) × 100 = 0.900%

Marking Criteria

Descriptor	Marks
Volume of HCl used: 1160/150 = 7.73 L	1
n(La) = cV = 8.65 × 10⁻³ × 7.73 = 0.0669 mol La in solution	1
Taking into account the leaching efficiency, n(La in the solid that was leached) = 0.0669/0.86 = 0.0778	1
m(La in the solid that was leached) = 0.0778 × 138.9 = 10.8 g	1
% La in the coating chemical = (10.8/1200) × 100 = 0.900%	1

Q39c

4 marks

Analysis of the cerium-containing solution produced in Step 5 showed that its cerium concentration was 0.146 mol L $^{-1}$ . This solution, which had a volume of 424 mL, was added to 110 mL of aqueous 1.15 mol L $^{-1}$ oxalic acid during Step 6, resulting in the precipitation of cerium oxalate, $\text{Ce(C}_2\text{O}_4\text{)}_2$ . The balanced equation for this reaction is:

$\text{CeCl}_4\text{(aq)} + 2 \text{ H}_2\text{C}_2\text{O}_4\text{(aq)} \rightarrow \text{Ce(C}_2\text{O}_4\text{)}_2\text{(s)} + 4 \text{ HCl(aq)}$

Did the chemist add enough oxalic acid solution to precipitate all of the cerium? Use calculations to support your answer.

Reveal Answer

n(cerium) = 0.146 × 0.424 = 0.0619 mol

n(oxalic acid needed to react with cerium) = 2 × 0.0619 = 0.124 mol

n(oxalic acid available) = 0.110 × 1.15 = 0.127 mol

comparison of the moles of oxalic acid shows that enough oxalic acid was added

Marking Criteria

Descriptor	Marks
n(cerium) = 0.146 × 0.424 = 0.0619 mol	1
n(oxalic acid needed to react with cerium) = 2 × 0.0619 = 0.124 mol	1
n(oxalic acid available) = 0.110 × 1.15 = 0.127 mol	1
comparison of the moles of oxalic acid shows that enough oxalic acid was added	1

Q30

2023

NESA

4 marks

Q30

4 marks

A water sample contains at least one of the following anions at concentrations of $1.0 \text{ mol L}^{-1}$ .

bromide (Br $^-$ )
carbonate (CO $_3^{2-}$ )

Outline a sequence of tests that could be performed in a school laboratory to confirm the identity of the anion or anions present. Include expected observations and TWO balanced chemical equations in your answer.

Reveal Answer

Add aqueous nitric acid – bubbles indicate carbonate present:
Acid removes carbonate for further testing of sample
$2\text{H}^+(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)$
Add silver nitrate solution – creamy precipitate indicates bromide present
$\text{Ag}^+(aq) + \text{Br}^-(aq) \rightarrow \text{AgBr}(s)$ .

Marking Criteria

Descriptor	Marks
Demonstrates a thorough understanding of anion testing in an appropriate sequence with expected observations Includes TWO balanced chemical equations including states	4
Demonstrates a sound understanding of anion testing with expected observation(s) and/or a correct chemical equation	3
Demonstrates some understanding of anion testing	2
Provides some relevant information	1
None of the above	0

Q27

2024

NESA

4 marks

Q27

4 marks

The following procedure is proposed to test for the presence of lead(II) and barium ions in water at concentrations of $0.1\text{ mol L}^{-1}$ .

Add excess $0.1\text{ mol L}^{-1}$ sodium sulfate solution. If a precipitate is produced, then barium ions are present.
Filter any precipitate produced.
Add excess $0.1\text{ mol L}^{-1}$ sodium bromide solution to the filtrate. If a precipitate is produced, then lead(II) ions are present.

Explain why this procedure gives correct results when only barium ions are present, but not when both barium and lead(II) ions are present. Include ONE balanced chemical equation in your answer.

Reveal Answer

If only barium ions are present, they will give a precipitate in step 1 of this procedure ( $\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$ ), so the conclusion that only barium ions are present will be correct.

If lead(II) ions are present, they will also precipitate at step 1 of this procedure. This will leave insufficient lead(II) ions in solution to give a precipitate in step 3. The conclusion reached will be that only barium ions are present, which is incorrect.

Marking Criteria

Descriptor	Marks
Demonstrates a thorough understanding of the procedure and qualitative ion testing Includes a balanced chemical equation, including states	4
Demonstrates a sound understanding of the procedure and qualitative ion testing	3
Demonstrates some understanding of the procedure and qualitative ion testing	2
Provides some relevant information	1
None of the above	0

NESA Chemistry Analysis of Inorganic Substances

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