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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.

CHEMISTRY, M5 EQ-Bank 26

The diagrams represent equipment used in an investigation to determine the chloride ion concentration in a water sample.
 

Describe how the chloride ion concentration in a water sample can be determined using the equipment in the diagrams. Include a relevant chemical equation.   (3 marks)

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  • When the silver nitrate solution is added, chloride ions present in the water sample will precipitate with the added silver ions described by the following equation:
  •    \(\ce{Ag+(aq) +Cl^-(aq) \rightleftharpoons AgCl(s)}\)
  • Adding excess silver nitrate ensures all chloride ions precipitate out.
  • The filtering apparatus is used to filter solid silver chloride. This solid is then dried to constant mass and weighed.
  • Using the molar mass of silver chloride, the number of moles of solid silver chloride produced is calculated. Silver chloride contains silver ions and chloride ions in a 1:1 molar ratio and using this ratio, the moles of chloride present is calculated.
  • The result represents the same number of moles of chloride in the original water sample. Using the volume of the water sample, its chloride ion concentration is calculated.
Show Worked Solution
  • When the silver nitrate solution is added, chloride ions present in the water sample will precipitate with the added silver ions described by the following equation:
  •    \(\ce{Ag+(aq) +Cl^-(aq) \rightleftharpoons AgCl(s)}\)
  • Adding excess silver nitrate ensures all chloride ions precipitate out.
  • The filtering apparatus is used to filter solid silver chloride. This solid is then dried to constant mass and weighed.
  • Using the molar mass of silver chloride, the number of moles of solid silver chloride produced is calculated. Silver chloride contains silver ions and chloride ions in a 1:1 molar ratio and using this ratio, the moles of chloride present is calculated.
  • The result represents the same number of moles of chloride in the original water sample. Using the volume of the water sample, its chloride ion concentration is calculated.

CHEMISTRY, M5 EQ-Bank 15 MC

What will happen when sulfuric acid is added to a saturated solution of sparingly soluble calcium sulfate?

  1. The concentration of calcium and sulfate ions will increase over time due to the presence of \(\ce{H^{+}}\) ions.
  2. The concentration of calcium and sulfate ions will decrease over time due to the presence of \(\ce{H^{+}}\) ions.
  3. The concentration of calcium and sulfate ions will increase over time due to the presence of \(\ce{SO4^{2-}}\) ions.
  4. The concentration of calcium and sulfate ions will decrease over time due to the presence of \(\ce{SO4^{2-}}\) ions.
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`D`

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  • The saturated solution of calcium sulfate is originally at equilibrium
  •    \(\ce{CaSO4(s) \rightleftharpoons Ca^2+(aq) + SO4^2-(aq)}\)
  • The addition of sulfuric acid increases the concentration of \(\ce{SO4^{2-}}\) (sulfate) ions in solution.
  • By Le Chatelier’s principle, the above equilibrium will shift left to counteract this, decreasing the concentration of calcium and sulfate ions over time.

`=>D`

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)`

Show Worked Solution

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

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

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

`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?

  1. `text{Ba(OH)}_(2) \ text{and} \ text{MgCl}_(2)`
  2. `text{Ba(OH)}_(2) \ text{and} \ text{MgSO}_(4)`
  3. `text{Ba(OH)}_(2) \ text{and} \ text{NaCl}`
  4. `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`