Worked Solutions
Module 8: Applied Chemistry — Worked Solutions (HSC Chemistry)
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Worked examples for HSC Chemistry Module 8: Applied Chemistry. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Identifying ions by precipitation
Question
A colourless solution is known to contain either chloride ions, $\text{Cl}^-{}_{(aq)}$, or sulfate ions, $\text{SO}_4^{2-}{}_{(aq)}$, but not both. Describe a chemical test that would distinguish between them, including the expected observations, and write a balanced ionic equation for the positive result of your test for chloride.
Solution
Use two selective precipitation tests.
Add acidified silver nitrate, $\text{AgNO}_3{}_{(aq)}$: chloride gives a white precipitate of $\text{AgCl}_{(s)}$; sulfate gives no precipitate with silver.
Add acidified barium nitrate, $\text{Ba(NO}_3)_2{}_{(aq)}$: sulfate gives a white precipitate of $\text{BaSO}_4{}_{(s)}$; chloride gives none.
Ionic equation for the positive chloride test:
$$\text{Ag}^+{}_{(aq)} + \text{Cl}^-{}_{(aq)} \rightarrow \text{AgCl}_{(s)}$$
Acidify with dilute nitric acid first to remove carbonate, which would also precipitate and confuse the result.
The trick with ion tests is to pick a reagent that reacts with one ion but not the other, so the two answers look different. Two classic reagents do this here.
First, add a few drops of acidified silver nitrate, $\text{AgNO}_3{}_{(aq)}$. Silver ions grab chloride to form an insoluble salt, so a white precipitate of $\text{AgCl}_{(s)}$ means chloride is present. Sulfate doesn't precipitate with silver, so a clear solution points to sulfate instead.
To confirm, add acidified barium nitrate, $\text{Ba(NO}_3)_2{}_{(aq)}$: barium and sulfate form insoluble $\text{BaSO}_4{}_{(s)}$, a white precipitate, while chloride stays in solution.
The balanced ionic equation for the positive chloride result is:
$$\text{Ag}^+{}_{(aq)} + \text{Cl}^-{}_{(aq)} \rightarrow \text{AgCl}_{(s)}$$
We acidify first (with dilute $\text{HNO}_3$) because ions like carbonate would also give a white precipitate, and we don't want a false positive.
Two selective precipitation tests:
- Silver nitrate (acidified): $\text{Cl}^-$ → white $\text{AgCl}_{(s)}$; $\text{SO}_4^{2-}$ → no ppt
- Barium nitrate (acidified): $\text{SO}_4^{2-}$ → white $\text{BaSO}_4{}_{(s)}$; $\text{Cl}^-$ → no ppt
Positive chloride test:
$$\text{Ag}^+{}_{(aq)} + \text{Cl}^-{}_{(aq)} \rightarrow \text{AgCl}_{(s)}$$
Acidify (dilute $\text{HNO}_3$) to rule out carbonate.
Where the marks go
- 1 mark: Names a suitable reagent for chloride (acidified silver nitrate)
- 1 mark: Names a suitable reagent for sulfate (acidified barium nitrate)
- 1 mark: States the distinguishing observation (white precipitate) for each test
- 1 mark: Correct balanced ionic equation $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}_{(s)}$
Key idea
Distinguish anions with selective precipitation — silver nitrate precipitates chloride, barium nitrate precipitates sulfate; acidify first to exclude carbonate.
Example 2 — Instrumental analysis (AAS)
Question
A series of copper standard solutions is analysed by atomic absorption spectroscopy (AAS), producing a straight calibration line through the origin described by $A = 0.250\,c$, where $A$ is absorbance and $c$ is the copper concentration in $\text{mg L}^{-1}$. A water sample gives an absorbance of $A = 0.180$. Calculate the copper concentration in the sample, and explain why AAS is suitable for this measurement.
Solution
Rearrange the calibration equation for concentration:
$$c = \frac{A}{0.250} = \frac{0.180}{0.250} = 0.720\ \text{mg L}^{-1}$$
So the copper concentration is $0.720\ \text{mg L}^{-1}$.
AAS is suitable because it's highly sensitive and selective — it can detect metal ions at trace (parts-per-million) levels by measuring absorption at a wavelength specific to copper, so other metals don't interfere.
A calibration line just links what we measure (absorbance) to what we want (concentration), so we put our sample's absorbance into the equation and solve for $c$. Rearranging $A = 0.250\,c$:
$$c = \frac{A}{0.250} = \frac{0.180}{0.250} = 0.720\ \text{mg L}^{-1}$$
That gives a copper concentration of $0.720\ \text{mg L}^{-1}$.
Why AAS? Each metal absorbs light at its own characteristic wavelength, so by tuning the instrument to copper's wavelength we measure only copper, even when it's present in tiny, trace amounts. That sensitivity and selectivity is exactly what you need for monitoring a heavy metal in water.
Solve $A = 0.250\,c$ for $c$:
$$c = \frac{0.180}{0.250} = 0.720\ \text{mg L}^{-1}$$
Why AAS:
- Very sensitive — detects trace (ppm) metal levels
- Selective — uses a wavelength specific to copper, so no interference
Where the marks go
- 1 mark: Rearranges the calibration equation correctly
- 1 mark: Correct concentration $c = 0.720\ \text{mg L}^{-1}$
- 1 mark: Explains AAS suitability (sensitivity/selectivity at trace levels)
Key idea
A calibration line converts a measured absorbance into concentration; AAS is suited to trace metal analysis because it is sensitive and element-selective.