smc-3672-20-Calcs given K(sp)

CHEMISTRY, M5 2023 HSC 17 MC

What mass of lead$\text{(II)}$ iodide (MM = 461 g mol$^{-1}$) will dissolve in 375 mL of water?

0.233 g
0.293 g
0.369 g
0.621 g

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$A$

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→ $\ce{PbI2 \rightleftharpoons Pb^2+ + 2I^-}\ \ \ \ \ \ \ K_{sp} = 9.8 \times 10^{-9}\ \text{(from data sheet)}$

$K_{sp}$	$= \ce{[Pb^2+][ 2I^-]^2}$
$9.8 \times 10^{-9}$	$= [x][ 2x]^2$
$x^3$	$=\dfrac{9.8 \times 10^{-9}}{4}$
$x$	$=0.001348\ \text{mol L}^{-1}$

$\text{Mass in 1 L}\ = 0.001348 \times 461 = 0.62147\ \text{g}$

$\text{Mass in 375 mL}\ = 0.62147 \times 0.375 = 0.233\ \text{g}$

$\Rightarrow A$

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, 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+]} $.

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

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

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.

Show Worked Solution

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%.

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

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

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%.

CHEMISTRY, M5 2022 HSC 17 MC

A 2.0 g sample of silver carbonate (MM = 275.81 g mol ¯¹) was added to 100.0 mL of water in a beaker. The solubility of silver carbonate at this temperature is `1.2 × 10^(-4)` mol L ¯¹. It was then diluted by adding another 100.0 mL of water.

What is the ratio of the concentration of silver ions in solution before and after dilution?

`1: 1`
`1: 2`
`2: 1`
`4: 1`

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`A`

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The maximum moles of `text{Ag}_2 text{CO}_3` that can be dissolved in 100.0 mL is:

`text{n(Ag}_2 text{CO}_3 text{)}_max`	`= 1.2 × 10^(−4) xx 0.1000`
	`= 1.2 xx 10^(−5) text{mol}`

The number of moles of silver carbonate added to the water is:

`text{n(Ag}_2 text{CO}_3)`	`= text{m} / text{MM}`
	`= 2.0 / 275.81`
	`= 7.2514 xx 10 ^(-3) text{mol}`

→ Thus, the moles of `text{Ag}_2 text{CO}_3` added is more than the maximum moles of `text{Ag}_2 text{CO}_3` able to be dissolved. i.e. the solution would be saturated before and after dilution.

→ As a result, the solution would have the same `text{Ag}_2 text{CO}_3` concentration before and after, thus, the ratio is 1:1.

`=> A`

♦♦♦ Mean mark 15%.

\(K_{sp}\)	\(= \ce{[Pb^2+][ 2I^-]^2}\)
\(9.8 \times 10^{-9}\)	\(= [x][ 2x]^2\)
\(x^3\)	\(=\dfrac{9.8 \times 10^{-9}}{4}\)
\(x\)	\(=0.001348\ \text{mol L}^{-1}\)