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CHEMISTRY, M8 2025 HSC 7 MC

Copper ions can form coloured complexes with water molecules and with chloride ions in dilute aqueous solutions.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Complex ion} \rule[-1ex]{0pt}{0pt}& \quad \quad \text{Colour} \quad \quad\\
\hline
\rule{0pt}{2.5ex} \left[\ce{Cu}\left(\ce{H2O}\right)_6\right]^{2+} \rule[-1ex]{0pt}{0pt}& \text{Blue} \\
\hline
\rule{0pt}{2.5ex} \left[ \ce{CuCl4}\right]^{2-} \rule[-1ex]{0pt}{0pt}& \text{Green} \\
\hline
\end{array}

Which of the following analytical techniques would be most suitable to distinguish between these two complexes?

  1. Infrared spectrophotometry
  2. Carbon-13 NMR spectroscopy
  3. UV-visible spectrophotometry
  4. Atomic absorption spectroscopy
Show Answers Only

\(C\)

Show Worked Solution
  • The complexes \(\left[\ce{Cu}\left(\ce{H2O}\right)_6\right]^{2+}\) (blue) and \(\left[ \ce{CuCl4}\right]^{2-}\) (green) differ in colour because each absorbs light of different wavelengths in the visible region of the electromagnetic spectrum.
  • UV-visible spectrophotometry measures how much light a compound absorbs at each wavelength in the UV–visible range, so it can clearly distinguish between the two complexes by their distinct absorption spectra.

\(\Rightarrow C\)

CHEMISTRY, M8 2024 HSC 25

The concentration of phosphate ions in washing machine waste water can be determined using colourimetry.

A sample of washing machine waste water was collected and diluted by quantitatively transferring 1.00 mL of the solution to a volumetric flask and making up the volume to 1.000 L with distilled water.

Standard phosphate solutions were prepared and analysed with a colourimeter using an accepted method.

The standard calibration graph is shown.
 

The diluted sample solution was then analysed using the same method as the standard solutions. The absorbance of this solution was found to be 0.64 .

Determine the concentration of phosphate ions in the sample of washing machine waste water, in mol L\(^{-1}\).   (4 marks)

Show Answers Only

\(8.2 \times 10^{-3}\ \text{mol L}^{-1}\)

Show Worked Solution

  • Absorbance value of 0.64 = diluted concentration of 0.78 mg L\(^{-1}\).
  • As the sample was originally diluted by a factor of 1000
  •   Original concentration \( =0.78 \times 1000 = 780\ \text{mg L}^{-1} = 0.78\ \text{g L}^{-1}\)
  • \(\ce{MM(PO4^{3-})} = 30.97 + 4(16.00) = 94.97\ \text{g mol}^{-1}\)
  • \(\ce{[PO4^{3-}]} = \dfrac{0.78}{94.97} = 8.2 \times 10^{-3}\ \text{mol L}^{-1}\)

CHEMISTRY, M8 2012 VCE 6

The iron content in multivitamin tablets was determined using atomic absorption spectroscopy.

The absorbances of four standards were measured.

Three multivitamin tablets were selected. Each tablet was dissolved in 100.0 mL of water. The absorbance of each of the three solutions was then measured.

The following absorbances were obtained.

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \ \textbf{Solution} \rule[1ex]{0pt}{0pt} & \textbf{Concentration} & \textbf{Absorbance} \\
& \textbf{mg/L} & \\
\hline
\rule{0pt}{2.5ex} \text{Standard 1} \quad \quad & 0.00 & 0.06 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 2} & 100.0 & 0.16 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 3} & 200.0 & 0.25 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 4} & 300.0 & 0.36 \\
\hline
\rule{0pt}{2.5ex} \text{Standard 5} & 400.0 & 0.46 \\
\hline
\rule{0pt}{2.5ex} \text{Tablet 1} & - & 0.39 \\
\hline
\rule{0pt}{2.5ex} \text{Tablet 2} & - & 0.42 \\
\hline
\rule{0pt}{2.5ex} \text{Tablet 3} & - & 0.45 \\
\hline
\end{array}

  1.  i.  Use the grid below to construct a calibration graph of the absorbances of the standard solutions.  (2 marks)

  2. ii.  Determine the average iron content, in milligrams, of the multivitamin tablets.  (2 marks)

Spectroscopic techniques work on the principle that, under certain conditions, atoms, molecules or ions will interact with electromagnetic radiation. The type of interaction depends on the wavelength of the electromagnetic radiation.

  1. Name one spectroscopic technique that you have studied this year.
    1. Which part of the electromagnetic spectrum does this technique use?  (1 mark)

    2. How does this part of the electromagnetic spectrum interact with matter? What information does this spectroscopic technique provide?  (2 marks)

Show Answers Only

a.i.  

a.ii.  \(35.5\ \text{mg}\)
 

b.i. Answers could include:

  • AAS (visible light)
  • UV-Vis (UV or visible light)
  • IR (Infrared radiation)
  • NMR (radio waves) 

b.ii. Spectroscopic technique: AAS (one of many possible – see b.i.)

  • During AAS energy of a certain frequency is transferred to electrons within atoms to move them into higher energy levels.
  • The absorption of the light indicates the concentration of the targeted element within the sample.

Show Worked Solution

a.i.  

a.ii. Average absorbance (tablets) \(=\dfrac{0.39+0.42+0.45}{3}=0.42\)

Using the graph: absorbance value of \(0.42 → 355\ \text{mg L}^{-1}\)

\(\ce{m(Fe) (100\ \text{ml}) = 355 \times 0.1 =35.5\ \text{mg}}\)
 

b.i.  Answers could include:

  • AAS (visible light)
  • UV-Vis (UV or visible light)
  • IR (Infrared radiation)
  • NMR (radiowaves) 

b.ii. Spectroscopic technique: AAS (one of many possible – see b.i.)

  • During AAS energy of a certain frequency is transferred to electrons within atoms to move them into higher energy levels.
  • The absorption of the light indicates the concentration of the targeted element within the sample.
♦ Mean mark (b.ii) 42%.

CHEMISTRY, M8 2013 VCE 8* MC

A forensic chemist tests mud from a crime scene to determine whether the mud contains zinc. Which one of the following analytical techniques would be best suited to this task?

  1. infrared spectroscopy
  2. mass spectroscopy
  3. atomic absorption spectroscopy
  4. nuclear magnetic resonance spectroscopy
Show Answers Only

\(C\)

Show Worked Solution
  • AAS is the only analytical technique that is specific to identifying metals within a solution.

\(\Rightarrow C\)

CHEMISTRY, M8 2014 VCE 16-17 MC

An atomic absorption spectrometer can be used to determine the level of copper in soils. The calibration curve below plots the absorbance of four standard copper solutions against the concentration of copper ions in ppm.

The concentrations of copper ions in the standard solutions were 1.0, 2.0, 3.0 and 4.0 mg L\(^{-1}\). (1 mg L\(^{-1}\) = 1 ppm)
 

Question 16

The concentration of copper in a test solution can be determined most accurately from the calibration curve if it is between

  1. 0.0 ppm and 5.0 ppm.
  2. 0.0 ppm and 4.0 ppm.
  3. 1.0 ppm and 4.0 ppm.
  4. 1.0 ppm and 5.0 ppm.


Question 17

If the test solution gave an absorbance reading of 0.40, what would be the concentration of copper ions in the solution in mol L\(^{-1}\)?

  1. 2.5
  2. 3.9 × 10\(^{-2}\)
  3. 3.9 × 10\(^{-5}\)
  4. 2.5 × 10\(^{-6}\)
Show Answers Only

\(\text{Question 16:}\ C\)

\(\text{Question 17:}\ C\)

Show Worked Solution

Question 16

  • The concentration of copper ions with be most accurate on the calibration curve in between the limits of the concentration of standard solutions used to produce the calibration curve.
  • Thus it will be most accurate between 1.0 ppm and 4.0 ppm.

\(\Rightarrow C\)
 

Question 17

From the graph: Absorbance of 0.4 → 2.5 ppm (2.5 mg L\(^{-1}\))

\(\ce{M(copper ions) = 63.55\ \text{g mol}^{-1}}\)

\(\ce{c(copper ions) = 2.5 \times 10^{-3}\ \text{g L}^{-1} = \dfrac{2.5 \times 10^{-3}}{63.55}=3.9 \times 10^{-5}\ \text{mol L}^{-1}}\)

\(\Rightarrow C\)

♦ Mean mark (Q17) 49%.

CHEMISTRY, M8 2015 VCE 4

UV-visible spectroscopy was used to measure the spectra of two solutions, \(\text{A}\) and \(\text{B}\). Solution \(\text{A}\) was a pink colour, while Solution \(\text{B}\) was a green colour.

The analyst recorded the absorbance of each solution over a range of wavelengths on the same axes. The resultant absorbance spectrum is shown below.
 

  1. If 10.00 mL of Solution \(\text{A}\) was mixed with 10.00 mL of Solution \(\text{B}\), which wavelength should be used to measure the absorbance of Solution \(\text{B}\) in this mixture? Justify your answer.   (2 marks)

The analyst used two sets of standard solutions and blanks to determine the calibration curves for the two solutions. The absorbances were plotted on the same axes. The graph is shown below.
 

  1. The analyst found that, when it was measured at the appropriate wavelength, Solution \(\text{A}\) had an absorbance of 0.2
  2. If Solution \(\text{A}\) was cobalt\(\text{(II)}\) nitrate, \(\ce{Co(NO3)2}\), determine its concentration in mg L\(^{–1}\)   (2 marks)
  3.    \(\ce{ M(Co(NO3)2) = 182.9 g mol^{–1} \quad \quad 1 mM = 10^{–3} M}\)
  4. In another mixture, the pink compound in Solution \(\text{A}\) and the green compound in Solution \(\text{B}\) each have a concentration of approximately 1.5 × 10\(^{-2}\) M.
  5. Could the analyst reliably use both of the calibration curves to determine the concentrations for Solution \(\text{A}\) and Solution \(\text{B}\) by UV-visible spectroscopy? Justify your answer.   (2 marks)

Show Answers Only

a.    Wavelength: 625 nm (600 to 650 was accepted)

  • At this wavelength there is the maximum absorbance of solution \(\text{B}\) with no significant absorbance (interference) from solution \(\text{A}\).

b.    \(1.8 \times 10^3\ \text{mgL}^{-1}\)

c.   \(1.5 \times 10^{-2}\ \text{M} = 15\ \text{mM}\).

  • This value lies within the calibration curve for solution \(\text{A}\), so the calibration curve can be used to determine the concentration of solution \(\text{A}\).
  • This value lies beyond the calibration curve for solution \(\text{B}\), therefore the curve can’t be used to determine the concentration of solution \(\text{B}\). The analyst would need to extrapolate the curve but this would not be reliable.

Show Worked Solution

a.    Wavelength: 625 nm (600 to 650 was accepted)

  • At this wavelength there is the maximum absorbance of solution \(\text{B}\) with no significant absorbance (interference) from solution \(\text{A}\). 

b.    From the graph:

\(\text{Absorbance of 0.2} \Rightarrow 10 \times 10^{-3}\ \text{mol L}^{-1}\)

\(\ce{n(Co(NO3)2) = 10 \times 10^{-3} \times 182.9 = 1.829\ \text{g L}^{-1} = 1.8 \times 10^3\ \text{mg L}^{-1}}\) 
 

c.   \(1.5 \times 10^{-2}\ \text{M} = 15\ \text{mM}\).

  • This value lies within the calibration curve for solution \(\text{A}\), so the calibration curve can be used to determine the concentration of solution \(\text{A}\).
  • This value lies beyond the calibration curve for solution \(\text{B}\), therefore the curve can’t be used to determine the concentration of solution \(\text{B}\). The analyst would need to extrapolate the curve but this would not be reliable.
♦♦ Mean mark (c) 30%.

CHEMISTRY, M8 2016 VCE 6

Brass is an alloy of copper and zinc.

To determine the percentage of copper in a particular sample of brass, an analyst prepared a number of standard solutions of copper\(\text{(II)}\) ions and measured their absorbance using an atomic absorption spectrometer (AAS).

The calibration curve obtained is shown below.
 

  1. A 0.198 g sample of the brass was dissolved in acid and the solution was made up to 100.00 mL in a volumetric flask. The absorbance of this test solution was found to be 0.13
  2. Calculate the percentage by mass of copper in the brass sample.   (3 marks)

  3. If the analyst had made up the solution of the brass sample to 20.00 mL instead of 100.00 mL, would the result of the analysis have been equally reliable? Why?   (2 marks)

  4. Name another analytical technique that could be used to verify the result from part a.   (1 mark)

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a.    55.6%

b.    No, this would increase the concentration of the copper solution by a factor of 5.

  • The concentration of the solution and absorbance would be too high and outside the range of the calibration curve (it can’t be assumed that the calibration curve remains linear beyond the range of the known data).

c.    Answers could include:

  • UV-vis spectroscopy, colorimetry, volumetric analysis, gravimetric analysis.
Show Worked Solution

a.    Absorbance of 0.13 → \(\ce{Cu^2+}\) concentration of 1.1 gL\(^{-1}\)  (see graph)

\(\text{In 100 mL:}\)

\(\ce{m(Cu^2+)}=1.1 \times 0.1=0.11\ \text{g}\)

\(\Rightarrow \ce{\% Cu^2+}=\dfrac{0.11}{0.198} \times 100=55.6\%\)
 

b.    No, this would increase the concentration of the copper solution by a factor of 5.

  • The concentration of the solution and absorbance would be too high and outside the range of the calibration curve.
  • It can’t be assumed that the calibration curve remains linear beyond the range of the known data. 
♦♦♦ Mean mark (b) 25%.

c.    Answers could include:

  • UV-vis spectroscopy, colorimetry, volumetric analysis, gravimetric analysis.

CHEMISTRY, M8 2023 HSC 2 MC

The technique illustrated is used to analyse chemical substances in a sample.
 

What is the technique shown?

  1. Flame test
  2. Mass spectrometry
  3. Atomic absorption spectroscopy
  4. Ultraviolet-visible spectrophotometry
Show Answers Only

\(C\)

Show Worked Solution

By elimination:

  • Although a flame test uses a flame to analyse chemical samples, it does not require a detector, prism, lamp or lens. Hence it is not the analytical method being demonstrated in the above diagram (eliminate A).
  • Mass Spectroscopy is used for organic compounds and requires an electromagnet which is not present in the above diagram (eliminate B).
  • Atomic Absorption Spectroscopy (used for inorganic compounds) is based on the idea that atoms can absorb light at a specific unique wavelength. The above image demonstrates this.
  • Ultraviolet-visible Spectrophotometry: although this uses the same principle as AAS to detect sample concentration, Ultraviolet-visible Spectrophotometry uses a different wavelength of light and requires a different apparatus to the one in the diagram above (eliminate D).

\(\Rightarrow C\)

CHEMISTRY, M8 EQ-Bank 28

Limestone \(\ce{(CaCO_3)}\) contributes to the hardness of water by releasing \(\ce{Ca^2^+}\) ions. The following chemical equation represents this reaction.

\(\ce{CaCO3($s$) + H_2O($l$) + CO_2($g$) \rightleftharpoons Ca^2^+($aq$) + 2HCO3^-($aq$)}\)      \((\Delta H<0)\)

It has been suggested that heating water reduces its hardness.

Explain how this suggestion can be tested accurately, validly and reliably.   (9 marks)

Show Answers Only
  • Atomic absorption spectroscopy (AAS) can be used to test if heating reduces water hardness.
  • It does this by calculating the concentrations of metal ions in solutions. AAS can calculate the concentration of \(\ce{Ca^{2+}}\) in heated and non-heated samples of water and any difference in the relative concentrations of \(\ce{Ca^{2+}}\) can be used to verify the suggestion.
  • It should be noted that a reduced concentration of \(\ce{Ca^{2+}}\) indicates that the water hardness is reduced.

Methodology of testing

  • Prepare standard solutions with known concentrations of \(\ce{Ca^{2+}}\) and measure their absorbance. Plot the concentrations against the absorbance of the standard solutions and draw a calibration curve (i.e. a line of best fit).
  • Measure the absorbance of a water sample before heating and another after heating. Using the absorbance and the calibration curve, calculate the concentration of \(\ce{Ca^{2+}}\) in each sample and compare the concentrations between the heated and unheated samples.
  • The AAS should be calibrated, at which point the concentration of calcium ions can be calculated to an accuracy in the parts per million (ppm). To ensure accurate calibration of the AAS, the standard solutions need to be prepared precisely which will involve the accurate weighing of solids and the use of a pipette or a similar instrument to measure solution volumes.
  • Water used in the experiment should be de-ionised (normal drinking water has an abundance of \(\ce{Na+}\) and \(\ce{Ca^{2+}}\)).
  • The margin of experimental error decreases when sufficient calibration samples are used and the measurement of absorbance of these samples is repeated and averaged. 
  • The reliability of results increases when many samples of heated and non-heated water are used to confirm that the concentrations of \(\ce{Ca^{2+}}\) in the heated water samples are consistently lower than the concentrations of \(\ce{Ca^{2+}}\) in the unheated water samples.
  • AAS can also be used to test the validity of the results. A hollow cathode lamp for calcium can direct light through the solution. This light has a specific wavelength that will only be absorbed by calcium ions. In this way, accurate measurements are made which can then be compared against the results and provide evidence of the validity of the original suggestion.
Show Worked Solution
  • Atomic absorption spectroscopy (AAS) can be used to test if heating reduces water hardness.
  • It does this by calculating the concentrations of metal ions in solutions. AAS can calculate the concentration of \(\ce{Ca^{2+}}\) in heated and non-heated samples of water and any difference in the relative concentrations of \(\ce{Ca^{2+}}\) can be used to verify the suggestion.
  • It should be noted that a reduced concentration of \(\ce{Ca^{2+}}\) indicates that the water hardness is reduced.

Methodology of testing

  • Prepare standard solutions with known concentrations of \(\ce{Ca^{2+}}\) and measure their absorbance. Plot the concentrations against the absorbance of the standard solutions and draw a calibration curve (i.e. a line of best fit).
  • Measure the absorbance of a water sample before heating and another after heating. Using the absorbance and the calibration curve, calculate the concentration of \(\ce{Ca^{2+}}\) in each sample and compare the concentrations between the heated and unheated samples.
  • The AAS should be calibrated, at which point the concentration of calcium ions can be calculated to an accuracy in the parts per million (ppm). To ensure accurate calibration of the AAS, the standard solutions need to be prepared precisely which will involve the accurate weighing of solids and the use of a pipette or a similar instrument to measure solution volumes.
  • Water used in the experiment should be de-ionised (normal drinking water has an abundance of \(\ce{Na+}\) and \(\ce{Ca^{2+}}\)).
  • The margin of experimental error decreases when sufficient calibration samples are used and the measurement of absorbance of these samples is repeated and averaged. 
  • The reliability of results increases when many samples of heated and non-heated water are used to confirm that the concentrations of \(\ce{Ca^{2+}}\) in the heated water samples are consistently lower than the concentrations of \(\ce{Ca^{2+}}\) in the unheated water samples.
  • AAS can also be used to test the validity of the results. A hollow cathode lamp for calcium can direct light through the solution. This light has a specific wavelength that will only be absorbed by calcium ions. In this way, accurate measurements are made which can then be compared against the results and provide evidence of the validity of the original suggestion.

CHEMISTRY, M8 2018 HSC 4 MC

Which of the following greatly enhanced scientific understanding of the effects of trace elements?

  1. Improved filtration techniques
  2. The development of atomic absorption spectroscopy
  3. The creation of new elements in particle accelerators
  4. The work of Le Chatelier in describing chemical equilibrium
Show Answers Only

`B`

Show Worked Solution
  • AAS allows trace elements to be detected at much lower concentrations than previous techniques.

`=>B`

CHEMISTRY, M8 2017 HSC 22

Atomic absorption spectroscopy was used to determine the concentration of zinc in a water sample. The absorbance of a series of standard solutions of known concentration of zinc was measured. The results are shown in the table.
 

  1. Plot the data on the grid and draw a line of best fit.   (3 marks)
     

  1. In order for water to be considered safe for drinking, the concentration of zinc must be less than 2.80 ppm.
  2. The absorbance of the water sample was 0.58. Explain whether this water is safe for drinking.   (2 marks)

Show Answers Only

a.   

b.   Using the line of best fit above:

  • 0.58 absorbance (y-axis) ~ 3.6 ppm of zinc concentration
  • Since 3.6 ppm > 2.8 ppm (safe level), the water is not safe to drink.
Show Worked Solution

a.   
   

b.   Using the line of best fit above:

  • 0.58 absorbance (y-axis) ~ 3.6 ppm of zinc concentration
  • Since 3.6 ppm > 2.8 ppm (safe level), the water is not safe to drink.

CHEMISTRY, M8 2019 HSC 29

Stormwater from a mine site has been found to be contaminated with copper\(\text{(II)}\) and lead\(\text{(II)}\) ions. The required discharge limit is 1.0 mg L¯1 for each metal ion. Treatment of the stormwater with \(\ce{Ca(OH)2}\) solid to remove the metal ions is recommended.

  1. Explain the recommended treatment with reference to solubility. Include a relevant chemical equation.   (2 marks)

  2. Explain why atomic absorption spectroscopy can be used to determine the concentrations of \(\ce{Cu^2+}\) and \(\ce{Pb^2+}\) ions in a solution containing both species.   (2 marks)

  3. The data below were obtained after treatment of the stormwater.
     
         
     
    To what extent is the treatment effective in meeting the required discharge limit of 1.0 mg L¯1 for each metal ion? Support your conclusion with calibration curves and calculations.   (7 marks)
     
         

Show Answers Only

a.   Recommended Treatment:

  • Calcium hydroxide is a slightly soluble compound, while copper\(\text{(II)}\) hydroxide and lead\(\text{(II)}\) hydroxide are very insoluble in water.
  • When these compounds are added to water, the metal ions tend to precipitate out of solution.
  • For example, the addition of solid calcium hydroxide to water produces calcium ions \(\ce{Ca^2+}\) and hydroxide ions \(\ce{OH-}\), which can then react with lead\(\text{(II)}\) ions (\(\ce{Pb^2+})\) and copper\(\text{(II)}\) ions \(\ce{Cu^2+}\) to form precipitates of lead\(\text{(II}\) hydroxide and copper\(\text{(II)}\) hydroxide, respectively.
  • These reactions are represented by the equations:
  •    \(\ce{Pb^2+ + 2OH- -> Pb(OH)2, \ \ Cu^2+ + 2OH- -> Cu(OH)2}\) 

b.   Atomic absorption spectroscopy (AAS):

  • Can be used for determining the concentration of metal ions in a sample by measuring the absorbance of light at specific wavelengths that are characteristic of each metal.
  • AAS uses light wavelengths that correspond to atomic absorption by the element of interest, and since each element has unique wavelengths that are absorbed, the concentration of that element can be selectively measured in the presence of other species.
  • As a result, AAS can be used to independently measure the concentrations of different metal ions, such as lead\(\text{(II)}\) ions and copper\(\text{(II)}\) ions in a sample containing both types. 

c.   Concentrations of ions:

\begin{array} {|l|c|c|c|}
\hline  \text{Sample }& \ce{Cu^2+ \times 10^{-5} mol L^{-1}} & \ce{Pb^2+ \times 10^{-5} mol L^{-1}} \\
\hline \text{Water (pre-treatment)} & 5.95 & 4.75 \\
\hline \text{Water (post-treatment)} & 0.25 & 0.85 \\
\hline \end{array}

  • Concentrations of copper and lead have been significantly reduced.
  • Convert concentrations to compare with standard:

\begin{array} {ccc}
\ce{Cu^2+}: & 5.95 \times 10^{-5} \times 63.55 \times 1000 = 3.78\ \text{mg L}^{-1} \\
 & 0.25 \times 10^{-5} \times 63.55 \times 1000 = 0.16\ \text{mg L}^{-1} \\
& \\
\ce{Pb^2+}: & 4.75 \times 10^{-5} \times 207.2 \times 1000 = 9.84\ \text{mg L}^{-1} \\
& 0.85 \times 10^{-5} \times 207.2 \times 1000 = 1.76\ \text{mg L}^{-1} \end{array}

 

   

Conclusion:

  • The concentration of copper ions has been reduced to a level that is lower than the discharge limit (0.16 < 1.0) but the lead ion concentration has not (1.76 > 1.0).
  • The treatment has only been partially successful.
Show Worked Solution

a.   Recommended Treatment:

  • Calcium hydroxide is a slightly soluble compound, while copper\(\text{(II)}\) hydroxide and lead\(\text{(II)}\) hydroxide are very insoluble in water.
  • When these compounds are added to water, the metal ions tend to precipitate out of solution.
  • For example, the addition of solid calcium hydroxide to water produces calcium ions \(\ce{Ca^2+}\) and hydroxide ions \(\ce{OH-}\), which can then react with lead\(\text{(II)}\) ions (\(\ce{Pb^2+})\) and copper\(\text{(II)}\) ions \(\ce{Cu^2+}\) to form precipitates of lead\(\text{(II}\) hydroxide and copper\(\text{(II)}\) hydroxide, respectively.
  • These reactions are represented by the equations:
  •    \(\ce{Pb^2+ + 2OH- -> Pb(OH)2, \ \ Cu^2+ + 2OH- -> Cu(OH)2}\) 

♦ Mean mark (a) 46%.

b.   Atomic absorption spectroscopy (AAS):

  • Can be used for determining the concentration of metal ions in a sample by measuring the absorbance of light at specific wavelengths that are characteristic of each metal.
  • AAS uses light wavelengths that correspond to atomic absorption by the element of interest, and since each element has unique wavelengths that are absorbed, the concentration of that element can be selectively measured in the presence of other species.
  • As a result, AAS can be used to independently measure the concentrations of different metal ions, such as lead\(\text{(II)}\) ions and copper\(\text{(II)}\) ions in a sample containing both types. 

c.   Concentrations of ions:

\begin{array} {|l|c|c|c|}
\hline  \text{Sample }& \ce{Cu^2+ \times 10^{-5} mol L^{-1}} & \ce{Pb^2+ \times 10^{-5} mol L^{-1}} \\
\hline \text{Water (pre-treatment)} & 5.95 & 4.75 \\
\hline \text{Water (post-treatment)} & 0.25 & 0.85 \\
\hline \end{array}


♦♦ Mean mark (b) 32%.
  • Concentrations of copper and lead have been significantly reduced.
  • Convert concentrations to compare with standard:

\begin{array} {ccc}
\ce{Cu^2+}: & 5.95 \times 10^{-5} \times 63.55 \times 1000 = 3.78\ \text{mg L}^{-1} \\
 & 0.25 \times 10^{-5} \times 63.55 \times 1000 = 0.16\ \text{mg L}^{-1} \\
& \\
\ce{Pb^2+}: & 4.75 \times 10^{-5} \times 207.2 \times 1000 = 9.84\ \text{mg L}^{-1} \\
& 0.85 \times 10^{-5} \times 207.2 \times 1000 = 1.76\ \text{mg L}^{-1} \end{array}

   

Conclusion:

  • The concentration of copper ions has been reduced to a level that is lower than the discharge limit (0.16 < 1.0) but the lead ion concentration has not (1.76 > 1.0).
  • The treatment has only been partially successful.

♦ Mean mark (c) 53%.

CHEMISTRY, M8 2019 HSC 20 MC

The manganese content in a 12.0 gram sample of steel was determined by measuring the absorbance of permanganate \(\ce{(MnO4^-)} \) using the following process.

The steel sample was dissolved in nitric acid and the \(\ce{Mn^2+(aq)}\)  ions produced were oxidised to \(\ce{MnO4^-(aq)}\) by periodate ions, \(\ce{IO4^-(aq)}\) , according to the following equation.

\(\ce{2Mn^2+(aq) + 5IO4^-(aq) + 3H2O(l) → 2MnO4^-(aq) + 5IO3^-(aq) + 6H+(aq)} \)

The resulting solution was made up to a volume of 1.00 L, then 20.0 mL of this solution was diluted to 100.0 mL. The absorbance at 525 nm of the resulting solution was 0.50.

A calibration curve for \(\ce{MnO4^-(aq)}\) was constructed and is shown below.
 

What was the percentage by mass of manganese in the steel sample?

  1. 0.019%
  2. 0.096%
  3. 0.48%
  4. 1.0%
Show Answers Only

`C`

Show Worked Solution

From graph, 0.5 absorbance corresponds to 0.25 mg L ¯1.

\(\ce{[MnO4^-]_{dilute}}\) `=(25 xx 10^(-3))/(54.94+16 xx 4)=2.1019 xx 10^(-4)\ text{mol L}^(-1)`  
\(\ce{[MnO4^-]_{conc}}\) `=2.1019 xx 10^(-4) xx 1/2=1.05095 xx 10^(-3)\ text{mol L}^(-1)`  

 
\(\ce{n(MnO4^-) = c \times V = 1.05095 \times 10^{-3} \times 1 = 1.05095 \times 10^{-3} mol}\)

\(\ce{n(Mn^2) = n(MnO4^-) = 1.05095 \times 10^{-3} mol}\)

\(\ce{m(Mn^2+)}=\text{n} \times \text{MM}=1.05095 \times 10^{-3} \times 54.94=5.774 \times 10^{-3} \text{g} \)

 `:.\ text{Mn}^(2+)text{(%)} = (5.774 xx 10^(-3))/12.0 xx 100text(%) = 0.48text(%)`

`=>C`


♦♦ Mean mark 43%.

CHEMISTRY, M8 2022 HSC 16 MC

A blue solution of copper(`text{II}`) sulfate was investigated using colourimetry. Orange light (λ = 630 nm) was used and the pathlength was 1.00 cm.

Which change would result in a higher absorbance value?

  1. Diluting the solution
  2. Using a higher intensity lamp
  3. Using blue light (λ = 450 nm)
  4. Setting the pathlength to 2.00 cm
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`D`

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  • Consider Beer-Lambert law, `A = epsilonlc`.
  • As the pathlength (`l`) increases, the absorbance (`A`) increases.

`=> D`


♦ Mean mark 45%.

CHEMISTRY, M8 2022 HSC 6 MC

A UV-visible spectrometer was used to obtain the spectra of solutions of substances `P` and `Q`. The absorbance spectra are shown.
 

Which wavelength would be appropriate to determine the concentration of `Q` in a mixture of the two solutions?

  1. 410 nm
  2. 475 nm
  3. 550 nm
  4. 630 nm
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`D`

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  • The most appropriate wavelength would be a wavelength that is mostly absorbed by `Q` but minimally interfered with by `P`.
  • Thus, the best wavelength is 630 nm because is a significant absorbance by `Q`, and an insignificant absorbance by `P`.

`=> D`


♦ Mean mark 41%.

CHEMISTRY, M8 2022 HSC 4 MC

An analytical chemist was using atomic absorption spectroscopy (AAS) to determine the manganese concentration in a sample.

The following diagram shows the absorbance lines of manganese.
 

The diagrams below show the emission spectra of four AAS lamps.

Which lamp should be used to determine the manganese concentration in the sample?
 

 

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`C`

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Consider Option C:

  • The emission spectrum of this lamp emits wavelengths of light that correspond to the absorption spectrum of `text{Mn}`.
  • This lamp would be best used to determine the `text{Mn}` concentration because it will emit wavelengths of light that are only absorbed by `text{Mn}`.

`=> C`

CHEMISTRY, M8 2021 HSC 14 MC

A sample of nickel was dissolved in nitric acid to produce a solution with a volume of 50.00 mL. 10.00 mL of this solution was then diluted to 250.0 mL. This solution was subjected to colorimetric analysis. A calibration curve for this analysis is given.
 

The solution gave an absorbance value of 0.30.

What was the mass of the sample of nickel?

  1. 0.0021 g
  2. 0.031 g
  3. 0.053 g
  4. 0.15 g
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`D`

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Absorbance  = 0.30

From the calibration curve, an absorbance of 0.30 corresponds to a \(\ce{[Ni2+]}\) of \(\pu{0.0021 mol L-1}\).

\( \ce{n[(Ni^2+)]} = \pu{0.0021 mol L-1} \)

\(\ce{n(Ni^2+)}\ \text{diluted}=\text{c × V}= 0.0021 \times 0.250= 0.000525\ \text{mol} \)

This is the moles of \(\ce{Ni^2+}\) in the diluted sample, which contained only 10.0 mL of the initial 50.0 mL of the original solution.

Thus, in order to find the number of moles in the original sample, multiply the number of moles by 5.

\(\ce{n(Ni^2+)}\ \text{undiluted} = \pu{5 \times 0.000525 = 0.002625 mol}\)

\(\ce{n(Ni^2+)}\) `= text{m} / text{MM}`  
\(\ce{m(Ni^2+)}\) `=\ text{n × MM}`  
  `= 0.002625 xx 58.69`  
  `= 0.15406125… `  
  `= 0.15\ text{g (2 d.p.)}`  

  
`=> D`


♦♦ Mean mark 38%.