SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

CHEMISTRY, M5 2023 HSC 31

Copper(\(\text{II}\)) ions \( \ce{(Cu^{2+})} \) form a complex with lactic acid \( \ce{(C3H6O3)} \), as shown in the equation.

\( \ce{Cu^{2+}(aq)} + \ce{2C3H6O3(aq)} \rightleftharpoons \Bigl[\ce{Cu(C3H6O3)2\Bigr]^{2+}(aq)} \)

This complex can be detected by measuring its absorbance at 730 nm. A series of solutions containing known concentrations of \( \Bigl[\ce{Cu(C3H6O3)_2\Big]^{2+}} \) were prepared, and their absorbances measured.
 

\( Concentration \ of \Bigl[\ce{Cu(C3H6O3)_2\Bigr]^{2+}} \) \( \text{(mol L}^{-1}) \) \( Absorbance \)
0.000 0.00
0.010 0.13
0.020 0.28
0.030 0.43
0.040 0.57
0.050 0.72
 
Two solutions containing \( \ce{Cu^{2+}} \ \text{and} \ \ce{C3H6O3} \) were mixed. The initial concentrations of each in the resulting solution are shown in the table.
 
\( Species \) \( Initial \ Concentration\)
\( (\text{mol L}^{-1}) \)
\( \ce{Cu^{2+}} \) 0.056
\( \ce{C3H6O3} \) 0.111

 
When the solution reached equilibrium, its absorbance at 730 nm was 0.66.

You may assume that under the conditions of this experiment, the only species present in the solution are those present in the equation above, and that \( \Bigl[ \ce{Cu(C3H6O3)_2\Bigr]^{2+}} \) is the only species that absorbs at 730 nm.

With the support of a line graph, calculate the equilibrium constant for the reaction.   (7 marks)
 

Show Answers Only

\(K_{eq}=1.3 \times 10^4\)

Show Worked Solution

\(\text{From graph:}\)

\(\text{0.66 absorbance}\  \Rightarrow\ \ \Big[\bigl[\ce{Cu(C3H6O3)2\bigr]^{2+}\Big]} = 0.046\ \text{mol L}^{-1} \)

\begin{array} {|l|c|c|c|}
\hline  & \ce{Cu^{2+}} & \ce{2C3H6O3(aq)} & \ce{\big[Cu(C3H6O3)2\big]^{2+}(aq)} \\
\hline \text{Initial} & \ \ \ \ 0.056 & \ \ \ \ 0.111 & 0 \\
\hline \text{Change} & -0.046 & -0.092 & \ \ \ +0.046 \\
\hline \text{Equilibrium} & \ \ \ \ 0.010 & \ \ \ \ 0.019 & \ \ \ \ \ \ 0.046 \\
\hline \end{array}

\(K_{eq}\) \(=\dfrac{\ce{\Big[\big[Cu(C3H6O3)2\big]^{2+}\Big]}}{\ce{\big[Cu^{2+}\big]\big[C3H6O3\big]^2}}\)  
  \(=\dfrac{0.046}{0.010 \times 0.019^2}\)  
  \(=1.3 \times 10^4\)  

CHEMISTRY, M5 EQ-Bank 24

When a sample of solid silver chloride is added to a `1.00 xx10^(-2)` mol L−1 sodium chloride solution, only some of the silver chloride dissolves.

Calculate the equilibrium concentration of silver ions in the resulting solution, given that the `K_(sp)` of silver chloride is `1.8 xx10^(-10)`.   (3 marks)

Show Answers Only

\(\ce{[Ag^+] = 1.80 \times 10^{-8} mol L^{-1}}\)

Show Worked Solution

 \(\ce{AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)}\)

\begin{array} {|l|c|c|c|}
\hline  & \ce{[AgCl(s)]} & \ce{[Ag^+(aq)]} & \ce{[Cl^-(aq)]} \\
\hline \text{Initial} &  & 0 & 1.00 \times 10^{-2} \\
\hline \text{Change} &  & +x & +x \\
\hline \text{Equilibrium} &  & x & 1.00 \times 10^{-2} +x \\
\hline \end{array}

\(\ce{Let \ $x$ = [Ag^+]}\)

\[\ce{$K_{sp}$ = [Ag^+][Cl^-]}\]

\(\ce{$K_{sp} = x$(1.00 \times 10^{-2} + $x$) = 1.80 \times 10^{-10}}\)

\(\ce{Since $x$\ is small, 1.00 \times 10^{-2} + $x$ ≈ 1.00 \times 10^{-2}}\)

\begin{aligned}
\ce{$x$(1.00 \times 10^{-2})} & \ce{= 1.80 \times 10^{-10}}  \\
\ce{$x$} & \ce{= 1.80 \times 10^{-8}}  \\
\ce{\therefore [Ag^+]} & \ce{= 1.80 \times 10^{-8} mol L^{-1}}  \\
\end{aligned}

CHEMISTRY, M5 2019 HSC 31

The following reaction occurs in an aqueous solution.

   \(\ce{HgCl4^2-(aq) + Cu^2+(aq) \rightleftharpoons CuCl4^2-(aq) + Hg^2+\ \ \ \ \ \ $K_{eq}$ = 4.55 \times 10^{-11}}\)

A solution containing a mixture of \(\ce{HgCl4^2-(aq)}\) and \(\ce{Cu^2+(aq)}\) ions is prepared. The initial concentration of each ion is 0.100 mol L ¯1 and there are no other ions present.

Calculate the concentration of \(\ce{Hg^2+(aq)}\) ions once the system has reached equilibrium.   (4 marks)

Show Answers Only

\(\ce{[Hg^2+] = 6.75 \times 10^{-7} mol L^{-1}}\)

Show Worked Solution

\[\ce{$K_{eq}$ = \frac{[CuCl4^2-][Hg^2+]}{[HgCl4^2-][Cu^2+]}}\]

\begin{array} {|l|c|c|c|c|}
\hline  & \ce{[HgCl4^2-]} & \ce{[Cu^2+]} & \ce{[CuCl4^2-]} & \ce{[Hg^2+]} \\
\hline \text{Initial} & 0.100 & 0.100 & 0 & 0 \\
\hline \text{Change} & -x & -x & +x & +x \\
\hline \text{Equilibrium} & 0.100-x & 0.100-x & x & x \\
\hline \end{array}

Since `x` is small  `=> 0.100-x~~0.100`

`4.55 xx 10^{-11}` `=(x xx x)/((0.100-x)(0.100-x))`  
`4.55 xx 10^{-11}` `=x^2/(0.100)^2`  
`x^2` `=4.55 xx 10^{-11} xx (0.100)^2`  
`x` `=sqrt(4.55 xx 10^{-11} xx (0.100)^2)`  
  `=6.75 xx 10^{-7}\ text{mol L}^{-1}`  

 
\(\therefore \ce{[Hg^2+] = 6.75 \times 10^{-7} mol L^{-1}}\)

CHEMISTRY, M5 2020 HSC 35

In aqueous solution, iodide ions `(text{I}^-)`react rapidly with iodine `(text{I}_2)` to form triiodide ions `text{I}_(3)^(\ -)`, making the equilibrium system shown in the chemical equation:

`text{I}^(-)(aq)+ text{I}_(2)(aq) ⇌ text{I}_(3)^(\ -)(aq)`

The following relationships can be derived from the reaction mechanism:

`[text{I}^(-)]_(eq)=2[text{I}_(2)]_(eq)`

`[text{I}^(-)]_(i nitial)=4[text{I}_(2)]_(eq)+3[text{I}_(3)^(\ -)]_(eq)`

where 'initial' designates the initial concentration and 'eq' designates the equilibrium concentration.

The absorbance of the solution in the UV-Vis spectrum is given by:

`A=[text{I}_(3)^(\ -)]xx2.76 xx10^(4)`

Determine the value of the equilibrium constant, given that  `A = 0.745`  at equilibrium and  `[text{I}^(-)]_(i nitial )=7.00 xx10^(-4)\ text{mol L}^(-1)`.  (4 marks)

Show Answers Only

`K_text{eq}= 564`

Show Worked Solution
`text{A}` \(\ce{= [I3–]_{eq} \times 2.76 × 10^4}\)
`0.745` \(\ce{= [I3–]_{eq} \times 2.76 × 10^4}\)
\(\ce{[I3–]_{eq}}\) `= 2.70 \times 10^(−5)\ text{mol L}^(–1)`

 

\(\ce{[I– ]_{initial}}\) \(\ce{= 4[I2 ]_{eq} + 3 [I3– ]_{eq}}\)
`7.00 xx 10^(−4)` `= 4 [text{I}_2 ]_text{eq} + (3 xx 2.70 xx 10^(−5))`
\(\ce{[I2 ]_{eq}}\) `=(7.00 xx 10^(−4)-(3 xx 2.70 xx 10^(−5)))/4`
  `= 1.55 xx 10^(−4)\ text{mol L}^(–1)`

 

`[text{I}^– ]_text(eq) = 2 [text{I}_2 ]_text(eq) = 2 xx (1.55 xx 10^(−4)) = 3.10 xx 10^(−4)\ text{mol L}^(–1)`

 

`K_text{eq}` `=[text{I}_3\^(\ -)]_text(eq) /[[text{I}^–]_text(eq) xx [text{I}_2]_text(eq)]`
  `= [2.70 xx 10^(−5)] / [3.10 xx 10^(−4) xx 1.55 xx 10^(−4)]`
  `= 564`

Mean mark 55%.

CHEMISTRY, M6 2020 HSC 14 MC

The equation for the autoionisation of water is shown.

\( \ce{2H2O(l)  \rightleftharpoons  \ H3O+(aq) + OH-(aq)} \)

At 50°C the water ionisation constant, `K_(w)`, is  `5.5 xx10^(-14)`.

What is the pH of water at 50°C?

  1.  5.50
  2.  6.63
  3.  6.93
  4.  7.00
Show Answers Only

`B`

Show Worked Solution

\(\ce{$K_w$ = [H3O+][OH–]}\)

Since \(\ce{[H3O+] = [OH–]}\):

`[text{H}_3 text{O}^+]^2` `=5.5 xx 10^(−14)`  
`[text{H}_3 text{O}^+]` `= 2.3 xx 10^(−7}\  text{mol L}^(–1)`  

 
`text{pH = −log}_(10) (2.3 xx 10^(−7) ) = 6.63`

`=> B`