smc-3672-25-Solubility rules

CHEMISTRY, M5 2023 HSC 16 MC

A solution contains potassium iodide and potassium chloride. It was analysed by performing a precipitation titration using silver nitrate. The titration curve for this reaction is shown, where \( \ce{pAg}=-\log _{10}\left[\ce{Ag}^{+}\right]\).

Why is this a valid and correct procedure for quantifying the amount of each anion present in the mixture?

\( \ce{AgCl} \) would precipitate out first, followed by \( \ce{AgI} \).
\( \ce{AgI}\) would precipitate out first, followed by \( \ce{AgCl} \).
Both \( \ce{AgI} \) and \( \ce{AgCl} \) precipitate out of the solution together.
Neither \( \ce{AgCl} \) nor \( \ce{AgI} \) would precipitate out of the solution.

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\(B\)

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This procedure is only correct and valid if the ions precipitated out at different times.
Therefore \( \ce{AgI}\) would precipitate out first as it is less soluble than \( \ce{AgCl} \).

\(\Rightarrow B\)

♦ Mean mark 41%.

CHEMISTRY, M5 2019 HSC 3 MC

Which of the following metal carbonates has the highest molar solubility?

Calcium carbonate
Copper`text{(II)}` carbonate
Iron`text{(II)}` carbonate
Lead`text{(II)}` carbonate

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`A`

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Calcium carbonate has the greatest solubility product among the listed metal carbonates.
Since all of these metal carbonates release the same amount of ions in solutions, their solubility products can be directly compared to determine which substance is most soluble.

`=>A`

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`

CHEMISTRY, M5 2022 HSC 31

Silver ions form the following complex with ammonia solution.

\( \ce{Ag+(aq) + 2NH3(aq) \rightleftharpoons [Ag(NH3)2]+(aq)}\)

The equilibrium constant is `1.6 × 10^(7)` at 25°C.

In order to determine the free \( \ce{Ag+}\) concentration in an aqueous ammonia solution, a student carried out a precipitation titration with \( \ce{NaI(aq)}\) as the titrant.
Evaluate the suitability of this method. (3 marks)

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If 0.010% of the total silver ions in solution are present as \( \ce{Ag+(aq)}\) at equilibrium, calculate the equilibrium concentration of aqueous ammonia in this solution. (4 marks)

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a. The method is not suitable.

Adding \( \ce{NaI}\) would cause the \( \ce{I-}\) ions to precipitate with the \( \ce{Ag+}\) ions to form \( \ce{AgI}\)
\( \ce{AgI(s) \rightleftharpoons Ag+(aq) + I- (aq)}\)
As a result, this would decrease \( \ce{[Ag+]}\), and disturb the equilibrium.
According to Le Chatelier’s Principle, the equilibrium will shift to the right in an attempt to counteract the change and increase \( \ce{[Ag+]}\).
\( \ce{Ag+(aq) + 2NH3(aq) \rightleftharpoons [Ag(NH3)2]+(aq)}\)
As a result, \( \ce{[Ag(NH3)2]+}\) would shift to the left and increase \( \ce{[Ag+]} \).

b. \({K_{eq}=\dfrac{\ce{[[Ag(NH3)2]+]}}{\ce{[Ag+][NH3]^2}}} \)

\[\ce {[[Ag(NH3)2]+] = \frac{99.99%}{0.010%} \times [Ag+] \ \ \ …\ (1)}\]

Substitute (1) into \(\ce {K_{eq}:}\)

`1.6 xx 10^7`	`= [(99.99%) xx [text{Ag}^+ ]] / [(0.010%) xx [text{Ag}^+ ][text{NH}_3 ]^2]= [99.99%] / [0.010% [text{NH}_3 ]^2]`
`[text{NH}_3 ]^2`	`=(99.99%)/(1.6 xx 10^7 xx 0.010%)`
`[text{NH}_3 ]`	`=sqrt((99.99%)/(1.6 xx 10^7 xx 0.010%))= 0.025 text{mol L}^-1`

Therefore, the concentration of \(\ce {NH3}\) at equilibrium is 0.025 mol L¯¹.

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a. The method is not suitable.

Adding \( \ce{NaI}\) would cause the \( \ce{I-}\) ions to precipitate with the \( \ce{Ag+}\) ions to form \( \ce{AgI}\)
\( \ce{AgI(s) \rightleftharpoons Ag+(aq) + I- (aq)}\)
As a result, this would decrease \( \ce{[Ag+]}\), and disturb the equilibrium.
According to Le Chatelier’s Principle, the equilibrium will shift to the right in an attempt to counteract the change and increase \( \ce{[Ag+]}\).
\( \ce{Ag+(aq) + 2NH3(aq) \rightleftharpoons [Ag(NH3)2]+(aq)}\)
As a result, \( \ce{[Ag(NH3)2]+}\) would shift to the left and increase \( \ce{[Ag+]} \).

♦ Mean mark (a) 40%.

b. \({K_{eq}=\dfrac{\ce{[[Ag(NH3)2]+]}}{\ce{[Ag+][NH3]^2}}} \)

\[\ce {[[Ag(NH3)2]+] = \frac{99.99%}{0.010%} \times [Ag+] \ \ \ …\ (1)}\]

Substitute (1) into \(\ce {K_{eq}:}\)

`1.6 xx 10^7`	`= [(99.99%) xx [text{Ag}^+ ]] / [(0.010%) xx [text{Ag}^+ ][text{NH}_3 ]^2]= [99.99%] / [0.010% [text{NH}_3 ]^2]`
`[text{NH}_3 ]^2`	`=(99.99%)/(1.6 xx 10^7 xx 0.010%)`
`[text{NH}_3 ]`	`=sqrt((99.99%)/(1.6 xx 10^7 xx 0.010%))= 0.025 text{mol L}^-1`

Therefore, the concentration of \(\ce {NH3}\) at equilibrium is 0.025 mol L¯¹.

♦ Mean mark (b) 40%.

CHEMISTRY, M5 2022 HSC 19 MC

What is the molar solubility of iron(`text{II}`) hydroxide?

`2.3 × 10^(-6)\ text{mol}\ text{L}^(-1)`
`2.9 × 10^(-6)\ text{mol}\ text{L}^(-1)`
`3.7 × 10^(-6)\ text{mol}\ text{L}^(-1)`
`4.9 × 10^(-9)\ text{mol}\ text{L}^(-1)`

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`A`

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`text{Fe(OH)}_2 (s) ⇌ text{Fe}^(2+) (aq) + 2 text{OH}^– (aq)`

Solids are not included in the `K_(sp)` expression

\begin{array} {|l|c|c|}
\hline & \text{Fe}^{2+} & \ \ \text{OH}^– \ \ \\
\hline \text{Initial} & 0 & 0 \\
\hline \text{Change} & + x & + 2x \\
\hline \text{Equilibrium} & x & 2x \\
\hline \end{array}

`text{K}_(sp)`	`= [text{Fe}^(2+)][text{OH}^–]^2`
`4.87 xx 10^(−17) `	`= x xx (2x)^2`
`4.87 xx 10^(−17)`	`= 4x^3`
`:.x`	`= root3((4.87 xx 10^(−17))/4)`
	`=2.30 xx 10^(−6) text{mol L}^(–1)`

`=> A`

♦ Mean mark 45%.