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15240672380's Question
Chemistry
Posted 2 months ago

2(b)
[6 Marks]
To prepare the 5ppm5 \mathrm{ppm} chloride standard solution you are required prepare a 100 cm3100 \mathrm{~cm}^{3} stock solution of 1000ppm1000 \mathrm{ppm} chloride from the salt sodium chloride in a volumetric flask. Following this, you should dillute the stock solution to prepare 50 cm350 \mathrm{~cm}^{3} of an intermediate chloride solution containing 100ppm100 \mathrm{ppm} of chloride. Finally you should prepare 50 cm350 \mathrm{~cm}^{3} of a 5ppm5 \mathrm{ppm} working standard of chloride which you will use for analysis.

Explain how you would prepare the three solutions required showing all the required calculations in full. (Note: atomic mass of sodium: 22.99 g/mol22.99 \mathrm{~g} / \mathrm{mol} )

Q 2(c)
[5 Marks]

A mixture of three anions eluted as follows using a strong base anion exchanger:
\begin{tabular}{|l|l|}
\hline Anion & Retention time \\
\hline Cl\mathrm{Cl} & 2.5 min2.5 \mathrm{~min} \\
\hline Br\mathrm{Br} & 4.8 min4.8 \mathrm{~min} \\
\hline SO42\mathrm{SO}_{4}^{2} & 8.7 min8.7 \mathrm{~min} \\
\hline
\end{tabular}

Explain why the mixture eluted as shown in the table above based on the type of stationary phase used
[End of Question 2]
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Answer from Sia
Posted 2 months ago
Solution
1
Calculate the mass of NaCl for 1000 ppm solution: To prepare a 1000 ppm solution, we need to calculate the mass of sodium chloride (NaCl) that would be present in 100 cm³ of solution. Since ppm is equivalent to mg/L, we need 1000 mg of NaCl in 1 L of solution. For 100 cm³, we need 100 mg of NaCl
2
Prepare the intermediate solution: To prepare 50 cm³ of a 100 ppm chloride solution from the 1000 ppm stock solution, we use the dilution formula C1V1=C2V2C_1V_1 = C_2V_2. Here, C1=1000C_1 = 1000 ppm, V1V_1 is the volume of stock solution needed, C2=100C_2 = 100 ppm, and V2=50V_2 = 50 cm³. Solve for V1V_1
3
Prepare the working standard solution: To prepare 50 cm³ of a 5 ppm chloride solution from the 100 ppm intermediate solution, again use the dilution formula with C1=100C_1 = 100 ppm, V1V_1 is the volume of intermediate solution needed, C2=5C_2 = 5 ppm, and V2=50V_2 = 50 cm³. Solve for V1V_1
2(b) Answer
To prepare the 1000 ppm solution, weigh out 100 mg of NaCl. To prepare the 100 ppm intermediate solution, take 5 cm³ of the 1000 ppm stock solution and dilute to 50 cm³. To prepare the 5 ppm working standard, take 2.5 cm³ of the 100 ppm intermediate solution and dilute to 50 cm³.
Key Concept
Dilution calculations are essential for preparing solutions of desired concentrations from stock solutions.
Explanation
By using the dilution formula C1V1=C2V2C_1V_1 = C_2V_2, we can calculate the volumes required to dilute a stock solution to a lower concentration.
1
Explain the elution order: The elution order in ion-exchange chromatography is influenced by the charge and size of the ions. Strong base anion exchangers preferentially bind anions with higher charge densities
2
Apply to given anions: SO42\mathrm{SO}_4^{2-} has a higher charge density than Br\mathrm{Br}^- and Cl\mathrm{Cl}^-, so it is retained the longest. Br\mathrm{Br}^- is larger than Cl\mathrm{Cl}^-, so it is retained longer than Cl\mathrm{Cl}^-, which is the smallest and elutes first
2(c) Answer
The elution order is Cl<Br<SO42\mathrm{Cl}^- < \mathrm{Br}^- < \mathrm{SO}_4^{2-} because the strong base anion exchanger binds anions with higher charge densities more strongly, and larger anions are retained longer than smaller ones.
Key Concept
The elution order in ion-exchange chromatography is determined by the charge density and size of the ions.
Explanation
SO42\mathrm{SO}_4^{2-} is retained the longest due to its higher charge density, followed by Br\mathrm{Br}^- due to its larger size compared to Cl\mathrm{Cl}^-, which elutes first.

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