smc-4261-60-Dilutions

CHEMISTRY, M2 EQ-Bank 14

The concentration of a solution of hydrochloric acid $\ce{HCl}$ is 1.80% (w/v). What is the molar concentration produced by diluting 20.0 mL of this solution to a total volume of 200.0 mL with deionised water? (3 marks)

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$0.049\ \text{mol/L}$

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1.8% (w/v) means that there is 1.8 g of solute in 100 mL.
In 20 mL, there is $1.8\ \text{g} \times 0.2 = 0.36\ \text{g}$ of $\ce{HCl}$.
$n\ce{(HCl)} = \dfrac{0.36}{1.008 + 35.45} = 0.00987\ \text{mol}$
The final concentration after the dilution will be:
$c = \dfrac{n}{V} = \dfrac{0.00987}{0.2} = 0.049\ \text{mol/L}$

CHEMISTRY, M2 EQ-Bank 10

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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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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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.

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*}

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$C$

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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.

$\Rightarrow C$

CHEMISTRY, M2 EQ-Bank 3 MC

What volume of water needs to be added to dilute 15.0 mL of a 0.200 mol L$^{-1}$ solution to a 0.0500 mol L$^{-1}$ solution?

45.0 mL
60.0 mL
75.0 mL
90.0 mL

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$A$

Show Worked Solution

Use the dilution formula $c_1V_1=c_2V_2$, where:

$c_1 = 0.200\ \text{mol L}^{-1}$ (initial concentration)

$V_1 = 15\ \text{mL}$ (initial volume)

$c_2 = 0.0500\ \text{mol L}^{-1}$ (final concentration)

$V_2 = \dfrac{c_1V_1}{c_2} = \dfrac{0.2 \times 15}{0.05} = 60\ \text{mL}$

$\therefore$ Volume of water to be added $=60-15=45\ \text{mL}$

$\Rightarrow A$

CHEMISTRY, M2 EQ-Bank 5

The compound potassium nitrate has the formula $\ce{KNO3}$

A student makes a solution of this compound by dissolving 40.0 g in 250.0 mL of distilled water. Calculate the concentration of this solution in mol L$^{-1}$. (2 marks)

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The student now requires 500.0 mL of a 5.0% (w/v) solution. What volume of the solution in part (a) is required to make this? (3 marks)

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a. $1.58\ \text{mol L}^{-1}$

b. $0.156\ \text{L}$

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a. $n\ce{(KNO3)}= \dfrac{m}{MM} = \dfrac{40.0}{39.10 + 14.01 + 3(16.00)} =0.396\ \text{mol}$

Concentration of solution $=\dfrac{n}{V} = \dfrac{0.396}{0.25} = 1.58\ \text{mol L}^{-1}$

b. A 5% solution requires 5 grams of solute in 100 mL of solution.

Therefore a 500 mL solution requires 25 grams of solute.

$n\ce{(KNO3)} = \dfrac{25}{101.11} = 0.247\ \text{mol}$

$c\ce{(KNO3)} = \dfrac{n}{V} = \dfrac{0.247}{0.5} = 0.494\ \text{mol L}^{-1}$

$c_1V_1$	$=c_2V_2$
$1.58 \times V_1$	$=0.494 \times 0.5$
$V_1$	$=\dfrac{0.247}{1.58}$
$V_1$	$=0.156\ \text{L}$

CHEMISTRY, M2 EQ-Bank 4

A student is required to dilute 150.00 mL solution of 3.00 mol L$^{-1}$ hydrochloric acid to produce 250.00 mL of 0.54 mol L$^{-1}$ hydrochloric acid.

Explain how the student should perform this dilution in a school laboratory. Include relevant calculations in your answer AND explain how the student should prepare any equipment they would use. (4 marks)

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The student should wash a 250 mL volumetric flask with distilled water only, if any hydrochloric acid was in the flask it would increase the final concentration to an unknown value.
A 50 mL volumetric pipette should be rinsed with distilled water to remove impurities then rinsed with the 3.00 M HCl solution to be diluted, as if any water was left in the pipette it would dilute the acid by an unknown factor.
The initial and final concentrations and volumes are related by the dilution equation:
$C_1 V_1 = C_2 V_2 $, where
$ C_1 = 3.00 \, \text{M} $ (initial concentration)
$ V_1 $ = volume to be pipetted
$ C_2 = 0.54 \, \text{M} $ (final concentration)
$ V_2 = 250.00 \, \text{mL} $ (final volume)
Rearrange to solve for $ V_1 $:
$ V_1 = \dfrac{C_2 V_2}{C_1} = \dfrac{0.54 \times 250.00}{3.00}= \dfrac{125.00}{3.00} = 45 \, \text{mL} $
Thus, the student would pipette 45 mL of the 3.00 M acid and deliver it to the 250 mL volumetric flask, making it up to the 250 mL line with distilled water. This would produce a 0.50 M dilution.

Show Worked Solution

The student should wash a 250 mL volumetric flask with distilled water only, if any hydrochloric acid was in the flask it would increase the final concentration to an unknown value.
A 50 mL volumetric pipette should be rinsed with distilled water to remove impurities then rinsed with the 3.00 M HCl solution to be diluted, as if any water was left in the pipette it would dilute the acid by an unknown factor.
The initial and final concentrations and volumes are related by the dilution equation:
$C_1 V_1 = C_2 V_2 $, where
$ C_1 = 3.00 \, \text{M} $ (initial concentration)
$ V_1 $ = volume to be pipetted
$ C_2 = 0.54 \, \text{M} $ (final concentration)
$ V_2 = 250.00 \, \text{mL} $ (final volume)
Rearrange to solve for $ V_1 $:
$ V_1 = \dfrac{C_2 V_2}{C_1} = \dfrac{0.54 \times 250.00}{3.00}= \dfrac{125.00}{3.00} = 45 \, \text{mL} $
Thus, the student would pipette 45 mL of the 3.00 M acid and deliver it to the 250 mL volumetric flask, making it up to the 250 mL line with distilled water. This would produce a 0.50 M dilution.

CHEMISTRY, M2 EQ-Bank 3

A student is required to dilute 100.00 mL solution of 2.00 mol L$^{-1}$ hydrochloric acid to produce 200.00 mL of 0.20 mol L$^{-1}$ hydrochloric acid.

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The student should wash a 200 mL volumetric flask with distilled water only, as if any hydrochloric acid was in the flask it would increase the final concentration to an unknown value.
A 20 mL volumetric pipette should be rinsed with distilled water to remove impurities then rinsed with the 2.00 M HCl solution to be diluted, as if any water was left in the pipette it would dilute the acid by an unknown factor.
The initial and final concentrations and volumes are related by the dilution equation:
$ C_1 V_1 = C_2 V_2 $, where
$ C_1 = 2.00 \, \text{M} $ (initial concentration)
$ V_1 $ = volume to be pipetted
$ C_2 = 0.20 \, \text{M} $ (final concentration)
$ V_2 = 200.00 \, \text{mL} $ (final volume)
Solve for $ V_1 $:
$ V_1 = \dfrac{C_2 V_2}{C_1} = \dfrac{0.20 \times 200.00}{2.00} = \dfrac{40.00}{2.00} = 20.00 \ \text{mL} $
Thus, the student would pipette 20.00 mL of the 2.00 M acid and deliver it to the 200 mL volumetric flask, making it up to the 200 mL line with distilled water. This would produce a 0.200 M or $1:10$ dilution.

Show Worked Solution

The student should wash a 200 mL volumetric flask with distilled water only, as if any hydrochloric acid was in the flask it would increase the final concentration to an unknown value.
A 20 mL volumetric pipette should be rinsed with distilled water to remove impurities then rinsed with the 2.00 M HCl solution to be diluted, as if any water was left in the pipette it would dilute the acid by an unknown factor.
The initial and final concentrations and volumes are related by the dilution equation:
$ C_1 V_1 = C_2 V_2 $, where
$ C_1 = 2.00 \, \text{M} $ (initial concentration)
$ V_1 $ = volume to be pipetted
$ C_2 = 0.20 \, \text{M} $ (final concentration)
$ V_2 = 200.00 \, \text{mL} $ (final volume)
Solve for $ V_1 $:
$ V_1 = \dfrac{C_2 V_2}{C_1} = \dfrac{0.20 \times 200.00}{2.00} = \dfrac{40.00}{2.00} = 20.00 \ \text{mL} $
Thus, the student would pipette 20.00 mL of the 2.00 M acid and deliver it to the 200 mL volumetric flask, making it up to the 200 mL line with distilled water. This would produce a 0.200 M or $1:10$ dilution.

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?

82.0 g mol$^{-1}$
84.0 g mol$^{-1}$
122 g mol$^{-1}$
410 g mol$^{-1}$

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$A$

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

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

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

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

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

$\Rightarrow A$

\(c_1V_1\)	\(=c_2V_2\)
\(1.58 \times V_1\)	\(=0.494 \times 0.5\)
\(V_1\)	\(=\dfrac{0.247}{1.58}\)
\(V_1\)	\(=0.156\ \text{L}\)