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CHEMISTRY, M2 EQ-Bank 10

  1. A student is asked to prepare 500.0 mL of a 0.150 mol L\(^{-1}\) standard solution of oxalic acid \(\ce{(C2H2O4.2H2O)}\), and then to perform a dilution to produce 250.0 mL of a 0.0300 mol L\(^{-1}\) solution. Outline and explain each step in this process, including the calculations involved and choice of equipment.   (5 marks)

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  1. Justify the procedure in part (a.) by explaining two measures taken to ensure the accuracy of the standard solution and diluted solution produced.   (2 marks)

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Show Answers Only

a.    Calculate the mass of oxalic acid:

\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)

\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)

\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)

  • Measure 9.455 g of oxalic acid using an electronic balance.

Prepare the standard solution:

  • Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.
  • Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.
  • Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

\(V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}\)

  • Use a pipette to transfer 50.0 mL of the 0.150 mol L\(^{-1}\) solution into a 250.0 mL volumetric flask.
  • Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L\(^{-1}\).

b.    Answers could include two of the following:

  • A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.
  • The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimising any error in the amount of solute added to the solution. 
  • The pipette provides precise measurements crucial for accurate dilutions.
  • Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.
Show Worked Solution

a.    Calculate the mass of oxalic acid:

\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)

\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)

\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)

  • Measure 9.455 g of oxalic acid using an electronic balance.

Prepare the standard solution:

  • Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.
  • Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.
  • Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

\(V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}\)

  • Use a pipette to transfer 50.0 mL of the 0.150 mol L\(^{-1}\) solution into a 250.0 mL volumetric flask.
  • Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L\(^{-1}\).

b.    Answers could include two of the following:

  • A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.
  • The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimising any error in the amount of solute added to the solution. 
  • The pipette provides precise measurements crucial for accurate dilutions.
  • Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.

Filed Under: Concentration and Molarity Tagged With: Band 5, Band 6, smc-4261-20-Concentration Calculations, smc-4261-60-Dilutions, smc-4261-70-Standard Solutions, smc-4261-90-Practical Investigation

CHEMISTRY, M2 EQ-Bank 9 MC

Which of the following statements best describes a primary standard solution?

  1. It is a solution that must be prepared from a substance with a low molar mass to increase accuracy.
  2. It is a solution prepared using a substance with a precisely known and stable concentration, suitable for use in standardising other solutions.
  3. It is a solution prepared from any chemical, as long as the concentration is measured with a volumetric flask.
  4. It is a solution that can vary in concentration over time and requires frequent standardisation.
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\(B\)

Show Worked Solution
  • Primary standards are used to prepare primary solutions with known and accurate concentrations.
  • A primary standard is a substance that is pure, stable, has a high molar mass (to minimise weighing errors), and is not affected by the atmosphere (such as moisture absorption or \(\ce{CO2}\)​ interaction).

\(\Rightarrow B\)

Filed Under: Concentration and Molarity Tagged With: Band 4, smc-4261-70-Standard Solutions

CHEMISTRY, M2 EQ-Bank 6 MC

A chemist is given 200.0 mL of a 1.2 M solution of sodium sulfate and is asked to dilute it to form 50.0 mL of a 0.30 M sodium sulfate solution.

Which of the following options regarding the dilution is correct?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\ & \\
\rule[-1ex]{0pt}{0pt} \ & \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Volume of 1.2 M solution needed } & \textbf{Glassware to make} \\
\textbf{to dilute solution (mL)} \rule[-1ex]{0pt}{0pt}& \textbf{accurately known solution} \\
\hline
\rule{0pt}{2.5ex} 12.5 \rule[-1ex]{0pt}{0pt}& \text{beaker, measuring cylinder} \\
\hline
\rule{0pt}{2.5ex} 75 \rule[-1ex]{0pt}{0pt}& \text{beaker, measuring cylinder} \\
\hline
\rule{0pt}{2.5ex} 12.5 \rule[-1ex]{0pt}{0pt}& \text{volumetric flask, pipette} \\
\hline
\rule{0pt}{2.5ex} 75 \rule[-1ex]{0pt}{0pt}& \text{volumetric flask, pipette} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution
  • Using the dilution formula, \(c_1V_1=c_2V_2\)
  •    \(V_1 = \dfrac{c_2V_2}{c_1} = \dfrac{0.3 \times 50}{1.2} = 12.5\ \text{mL}\)
  • To prepare an accurately diluted solution, a volumetric flask and a pipette should be used, as these provide precise measurements. 

\(\Rightarrow C\)

Filed Under: Concentration and Molarity Tagged With: Band 4, smc-4261-60-Dilutions, smc-4261-70-Standard Solutions, smc-4261-90-Practical Investigation

CHEMISTRY, M2 EQ-Bank 5 MC

A standard solution is best described as:

  1. A solution prepared to an approximate concentration for general use.
  2. A solution with a precisely known concentration, used in quantitative chemical analysis.
  3. A solution containing only one type of solute molecule.
  4. A solution prepared by dissolving a solid solute in a small volume of solvent.
Show Answers Only

\(B\)

Show Worked Solution
  • A standard solution is a solution with a precise and known concentration, allowing accurate calculations in quantitative chemical analysis.
  • It is prepared by dissolving an exact amount of a primary standard in a specific volume of solvent.

\(\Rightarrow B\)

Filed Under: Concentration and Molarity Tagged With: Band 3, smc-4261-70-Standard Solutions

CHEMISTRY, M2 2011 HSC 18* MC

A household cleaning agent contains a weak base with the formula \( \ce{NaX}\). 1.00 g of this compound was dissolved in water to give 100.0 mL of solution. A 20.0 mL sample of the solution was mixed with 0.100 mol L\(^{-1}\) hydrochloric acid, and required 24.4 mL of the acid for neutralisation.

\(\ce{NaX + HCl -> NaCl + HX}\)

What is the molar mass of the weak base?

  1. 82.0 g mol\(^{-1}\)
  2. 84.0 g mol\(^{-1}\)
  3. 122 g mol\(^{-1}\)
  4. 410 g mol\(^{-1}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\ce{n(HCl)} = 0.100 \times 24.4 \times 10^{-3} = 2.44 \times 10^{-3}\ \text{mol}\)

\(\ce{n(NaX)}=2.44 \times 10^{-3}\ \text{mol in 20 mL sample.}\)

\(\ce{c(NaX)}=\dfrac{2.44 \times 10^{-3}}{20 \times 10^{-3}} =0.122\ \text{mol L}^{-1}\)

\(\ce{n(NaX)}=0.122 \times 0.1=0.0122\ \text{mol (in 100 mL sample)}\)

\(\ce{MM(NaX)}=\dfrac{1}{0.0122}=82\ \text{g mol}^{-1}\)

\(\Rightarrow A\)

Filed Under: Concentration and Molarity Tagged With: Band 6, smc-4261-20-Concentration Calculations, smc-4261-60-Dilutions, smc-4261-70-Standard Solutions

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