smc-3672-10-Mixed ionic solutions

CHEMISTRY, M5 2025 HSC 32

The following three solids were added together to 1 litre of water:

$\ce{0.006\ \text{mol}\ Mg(NO3)2}$
$\ce{0.010\ \text{mol}\ NaOH}$
$\ce{0.002\ \text{mol}\ Na2CO3}$.

Which precipitate(s), if any, will form? Justify your answer with appropriate calculations. (5 marks)

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All sodium and nitrate salts are soluble $\Rightarrow$ possible precipitates are $\ce{Mg(OH)2}$ and $\ce{MgCO3}$.

$\ce{\left[Mg^{2+}\right]=6 \times 10^{-3} \quad\left[ OH^{-}\right]=1 \times 10^{-2} \quad\left[ CO3^{2-}\right]=2 \times 10^{-3}}$

$\ce{\left[Mg^{2+}\right]\left[ OH^{-}\right]^2=6 \times 10^{-7} \quad \quad \quad \ \ \ K_{\textit{sp}}=5.61 \times 10^{-12}}$

$\ce{\left[ Mg^{2+}\right]\left[ CO3^{2-}\right]=1.2 \times 10^{-5} \quad \quad K_{\textit{sp}}=6.82 \times 10^{-6}}$

$\ce{Mg(OH)2}$ is a lot less soluble than $\ce{MgCO3}$ and will precipitate preferentially.

$\ce{\left[Mg^{2+}\right]\left[OH^{-}\right]^2 > K_{\textit{sp}}}$

$\Rightarrow \ce{Mg(OH)2}$ will precipitate.

$\ce{Mg^{2+}(aq) + 2OH-(aq) \rightarrow Mg(OH)2(s)}$

Since $K_{\textit{sp}}$ is small, assume reaction goes to completion.

$\ce{n(Mg^{2+})=0.006\ mol, n(OH^{-})=0.010\ mol}$

Using stoichiometric ratio $(1:2)$

$0.006 > \dfrac{0.010}{2}=0.005\ \ \Rightarrow \ce{Mg^{2+}}$ is in excess.

Concentration drops to: $6 \times 10^{-3}-5 \times 10^{-3}=1 \times 10^{-3}$.

Check if $\ce{MgCO3}$ will precipitate:

$\ce{\left[Mg^{2+}\right]\left[CO3^{2-}\right]}$ becomes $2 \times 10^{-6} <K_{\textit{sp}}$.

$\ce{\Rightarrow\ MgCO3}$ won’t precipitate.

Show Worked Solution

All sodium and nitrate salts are soluble $\Rightarrow$ possible precipitates are $\ce{Mg(OH)2}$ and $\ce{MgCO3}$.

$\ce{\left[Mg^{2+}\right]=6 \times 10^{-3} \quad\left[ OH^{-}\right]=1 \times 10^{-2} \quad\left[ CO3^{2-}\right]=2 \times 10^{-3}}$

$\ce{\left[Mg^{2+}\right]\left[ OH^{-}\right]^2=6 \times 10^{-7} \quad \quad \quad \ \ \ K_{\textit{sp}}=5.61 \times 10^{-12}}$

$\ce{\left[ Mg^{2+}\right]\left[ CO3^{2-}\right]=1.2 \times 10^{-5} \quad \quad K_{\textit{sp}}=6.82 \times 10^{-6}}$

♦♦ Mean mark 40%.

$\ce{Mg(OH)2}$ is a lot less soluble than $\ce{MgCO3}$ and will precipitate preferentially.

$\ce{\left[Mg^{2+}\right]\left[OH^{-}\right]^2 > K_{\textit{sp}}}$

$\Rightarrow \ce{Mg(OH)2}$ will precipitate.

$\ce{Mg^{2+}(aq) + 2OH-(aq) \rightarrow Mg(OH)2(s)}$

Since $K_{\textit{sp}}$ is small, assume reaction goes to completion.

$\ce{n(Mg^{2+})=0.006\ mol, n(OH^{-})=0.010\ mol}$

Using stoichiometric ratio $(1:2)$

$0.006 > \dfrac{0.010}{2}=0.005\ \ \Rightarrow \ce{Mg^{2+}}$ is in excess.

Concentration drops to: $6 \times 10^{-3}-5 \times 10^{-3}=1 \times 10^{-3}$.

Check if $\ce{MgCO3}$ will precipitate:

$\ce{\left[Mg^{2+}\right]\left[CO3^{2-}\right]}$ becomes $2 \times 10^{-6} <K_{\textit{sp}}$.

$\ce{\Rightarrow\ MgCO3}$ won’t precipitate.

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 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`

Show Worked Solution

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{)} \ 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 xx 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.	\(\ce{HF(aq)}\)	\(\ce{CF3OO-(aq)}\)