smc-3672-10-Mixed ionic solutions

CHEMISTRY, M5 2023 HSC 34

When 125 mL of a magnesium nitrate solution is mixed with 175 mL of a 1.50 mol L$^{-1} $ sodium fluoride solution, 0.6231 g of magnesium fluoride (MM = 62.31 g mol$^{-1} $) precipitates. The $ K_{s p} $ of magnesium fluoride is 5.16 × 10$^{-11} $.

Calculate the equilibrium concentration of magnesium ions in this solution. (5 marks)

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$7.90 \times 10^{-11}\ \text{mol L}^{-1} $

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$\ce{MgF2(s) \rightleftharpoons Mg^{2+}(aq) + 2F-(aq)}$

$\ce{n(MgF2) = \dfrac{\text{m}}{\text{MM}} = \dfrac{0.6231}{62.31} = 1.000 \times 10^{-2} \text{mol}} $

$\ce{n(F^{-})_{init} = c \times V = 1.50 \times 0.175 = 0.263\ \text{mol}} $

$\ce{n(F^{-})_{after} = 0.263-2 \times 1.00 \times 10^{-2} = 0.243\ \text{mol}} $

$\ce{[F^{-}]_{after} = \dfrac{\text{n}}{\text{V}} = \dfrac{0.243}{0.300} = 0.808\ \text{mol L}^{-1}} $

♦ Mean mark 45%.

$K_{sp} = \ce{[Mg^{2+}][F^{-}]^2 }$

$\text{Since}\ K_{sp}\ \text{is small}\ \ \Rightarrow \text{assume}\ \ce{[F^{-}]_{eq} = 0.808\ \text{mol L}^{-1}} $

$\ce{[Mg^{2+}]_{eq} = \dfrac{5.16 \times 10^{-11}}{0.808^2} = 7.90 \times 10^{-11}\ \text{mol L}^{-1}} $

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)

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

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

\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, 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.

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

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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)

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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.

→ 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.

c. Concentrations of ions:

♦♦ Mean mark (b) 32%.

→ Concentrations of copper and lead have been significantly reduced.

→ Convert concentrations to compare with standard:

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 2019 HSC 18 MC

Consider the following equilibrium.

$\ce{HF(aq) + CF3COO-(aq) \rightleftharpoons F-(aq) + CF3COOH(aq) \ \ \ \ \ \ $K_{eq}$ = 3.80 \times 10^{-4}} $

Which row of the table correctly identifies the strongest acid and the strongest base in this system?

	$\text{Strongest acid}$	$\text{Strongest base}$
A.	$\ce{CF3OOH(aq)}$	$\ce{F-(aq)}$
B.	$\ce{CF3OOH(aq)}$	$\ce{CF3OO-(aq)}$
C.	$\ce{HF(aq)}$	$\ce{F-(aq)}$
D.	$\ce{HF(aq)}$	$\ce{CF3OO-(aq)}$

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`A`

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→ The small size of `K_(eq)` means that there is a higher concentration of reactants than products at equilibrium.

→ This shifts the equilibrium towards the reactants, which means that the reverse reaction is more likely to occur.

→ Because $\ce{F-}$ is more likely to accept a proton than $\ce{CF3COO-}$, it is a stronger base.

→ On the other hand, $\ce{CF3COO-}$ is more likely to donate a proton than $\ce{HF}$, making it the stronger acid.

`=>A`

♦♦♦ Mean mark 23%.

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)

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`K_text{eq}= 564`

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`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, M5 2020 HSC 17 MC

The following apparatus was set up in a temperature-controlled laboratory.

Excess solid sodium hydroxide is added to the beaker.

Which row of the table correctly identifies the change in the `text{CuSO}_(4)(s)` mass and the colour of the solution after several days?

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`D`

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$\ce{CuSO4(s) \rightleftharpoons Cu^2+ (aq) + SO4^2– (aq)}$

→ The addition of $\ce{NaOH}$ to the solution would result in a precipitate $\ce{Cu(OH)2}$ and thus decreases the amount of $\ce{Cu^2+}$ ions.

→ According to Le Chatlelier’s Principle, the system would shift right in an attempt to counteract the change and increase $\ce{[Cu^2+]}$, thereby decreasing the mass of the precipitate.

→ The blue colour will fade since the $\ce{[Cu^2+]}$ in the final solution is less.

`=>D`

♦♦ Mean mark 35%.

CHEMISTRY, M5 2022 HSC 35

A precipitate of strontium hydroxide `\text{Sr}(\text{OH})_2`, (`MM` = 121.63 g mol ¯¹) was produced when 80.0 mL of 1.50 mol L ¯¹ strontium nitrate solution was mixed with 80.0 mL of 0.855 mol L ¯¹ sodium hydroxide solution. The mass of the dried precipitate was 3.93 g.

What is the `K_{sp}` of strontium hydroxide? (5 marks)

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`3.06 × 10^(−4)`

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$ \ce{Sr(NO3)2(aq) + 2NaOH(aq) -> Sr(OH)2(s) + 2NaNO3(aq)} $

$ \ce{n(Sr(NO3)2) = c \times V = 1.50 \times 0.0800= 0.120\ mol}$

$ \ce{n(NaOH) = 0.855 \times 0.0800 = 0.0684 \text{mol} }$

`text{NaOH = limiting reagent}, \ \ \ text{Sr(NO}_3)_2 = text{excess reagent}`

`text{n(Sr(OH)}_2text{) produced}\ = 1 / 2 xx 0.0684= 0.0342 text{mol}`

`text{Thus, 0.0342 moles of Sr(OH)}_2 text{can be produced in solution.}`

`text{n(Sr(OH)}_2)\ text{precipitate}`	`= text{m} / text{MM}`
	`=3.93/121.63`
	`=0.0323111 text{mol}`
`text{n(Sr(OH)}_2)\ text{in solution}`	`=text{n(Sr(OH)}_2)\ text{produced} – text{n(Sr(OH)}_2)\ text{precipitate}`
	`=0.0342-0.0323111`
	`= 0.0018889 text{mol}`

$ \ce{Sr(OH)2(s) \rightleftharpoons Sr^2+(aq) + 2OH-(aq)} $

$ \ce{n(Sr(NO3)2)_{init} = 0.120 – \dfrac{1}{2} \times 0.0684 = 0.0858\ \text{mol}} $

\begin{array} {|l|c|c|}
\hline & \text{Sr}^{2+} & \text{OH}^– \\
\hline \text{Initial} & 0.0858 & 0 \\
\hline \text{Change} & +0.0018889 & +2 \times 0.0018889 \\
\hline \text{Equilibrium} & 0.0877 & 0.00378 \\
\hline \end{array}

`text{V (total)} = 0.08 + 0.08= 0.16\ text{L}`

`[text{Sr}^(2+)]= text{n} / text{V} = 0.0877 / 0.16 = 0.548 text{mol L}^(–1)`

`[text{OH}^–]=0.00378 / 0.16= 0.0236 text{mol L}^–1`

`K_(sp)`	`= [text{Sr}^(2+)][text{OH}^–]^2`
	`= 0.548 xx (0.02362)^2`
	`= 3.06 × 10^(−4)`

♦ Mean mark 43%

CHEMISTRY, M5 2020 HSC 11 MC

Equal volumes of two 0.04 mol L ¯¹ solutions were mixed together.

Which pair of solutions would give the greatest mass of precipitate?

`text{Ba(OH)}_(2) \ text{and} \ text{MgCl}_(2)`
`text{Ba(OH)}_(2) \ text{and} \ text{MgSO}_(4)`
`text{Ba(OH)}_(2) \ text{and} \ text{NaCl}`
`text{Ba(OH)}_(2) \ text{and} \ text{Na}_(2) text{SO}_(4)`

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`B`

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$\ce{Ba(OH)2(aq) + MgSO4(aq) → BaSO4(s) + Mg(OH)2(s)}$

→ Reaction B produces 2 molecules of precipitate

→ Reactions A and D produce 1 molecule of precipitate each

→ Reaction C does not produce a precipitate.

`=> B`

	\(\text{Strongest acid}\)	\(\text{Strongest base}\)
A.	\(\ce{CF3OOH(aq)}\)	\(\ce{F-(aq)}\)
B.	\(\ce{CF3OOH(aq)}\)	\(\ce{CF3OO-(aq)}\)
C.	\(\ce{HF(aq)}\)	\(\ce{F-(aq)}\)
D.	\(\ce{HF(aq)}\)	\(\ce{CF3OO-(aq)}\)