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CHEMISTRY, M5 EQ-Bank 29

The information in the table shows how the solubility of lead chloride is affected by temperature.  
 

Using a graph, calculate the solubility product \((K_{sp})\) of the dissolution of lead chloride at 50°C. Include a fully labelled graph and a relevant chemical equation in your answer   (6 marks)
 

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\(\ce{K_{sp} = 6.4 \times 10^{-5}}\)

Show Worked Solution

 

\(\ce{PbCl2(s) \rightleftharpoons Pb^2+(aq) + 2Cl^-(aq)}\)

\(\ce{Using the graph:}\)

\(\ce{Solubility (50°) = 0.7 g/100 g water = 7 g/L}\)

\(\ce{Converting to mol L^{-1}:}\)

\[\ce{MM(PbCl2) = 207.2 + 2 \times 35.45 = 278.1}\]

\[\ce{n = \frac{m}{MM} = \frac{7}{278.1} = 0.0252 mol L^{-1}}\]

\(\ce{[Pb^2+(aq)] = 0.0252 mol L^{-1}}\)
 

\(\ce{Mole ratio \ Pb^2+ : Cl^- = 1:2}\)

\(\Rightarrow \ce{[Cl^-]  = 2 \times 0.0252 = 0.0504 mol L^{-1}}\)
 

\begin{aligned}
\ce{$K_{sp}$} & \ce{= [Pb^2+][Cl^-]^{2}} \\
 & \ce{=0.0252 \times (0.0504)^{2}}  \\
 & \ce{= 6.4 \times 10^{-5}}  \\
\end{aligned}

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)
     
         

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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, M5 2020 HSC 20 MC

The graph shows the concentration of silver and chromate ions which can exist in a saturated solution of silver chromate.
 

Based on the information provided, what is the `K_{sp}` for silver chromate?

  1. `1.1 xx10^(-8)`
  2. `2.2 xx10^(-8)`
  3. `1.1 xx10^(-12)`
  4. `4.4 xx10^(-12)`
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`C`

Show Worked Solution

When \(\ce{[Ag+]} = 1 \times 10^{-4}\ \text{mol L}^{-1} \),

\( \ce{CrO4^2–} = 11 \times\ 10^{-5}\ \text{mol L}^{-1} \) 

\begin{align}
K_{sp} &= \ce{[Ag+]^2}\ce{[CrO4^2–]}\\
& =(1 \times 10^{−4})^2(11×10^{−5}) \\
&=1.1×10^{−12}\ \text{mol L}^{–1} \\
\end{align}

`=> C`


♦ Mean mark 43%.