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2(b)
[6 Marks]
To prepare the $5 \mathrm{ppm}$ chloride standard solution you are required prepare a $100 \mathrm{~cm}^{3}$ stock solution of $1000 \mathrm{ppm}$ chloride from the salt sodium chloride in a volumetric flask. Following this, you should dillute the stock solution to prepare $50 \mathrm{~cm}^{3}$ of an intermediate chloride solution containing $100 \mathrm{ppm}$ of chloride. Finally you should prepare $50 \mathrm{~cm}^{3}$ of a $5 \mathrm{ppm}$ 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: $22.99 \mathrm{~g} / \mathrm{mol}$ )
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 $\mathrm{Cl}$ & $2.5 \mathrm{~min}$ \\
\hline $\mathrm{Br}$ & $4.8 \mathrm{~min}$ \\
\hline $\mathrm{SO}_{4}^{2}$ & $8.7 \mathrm{~min}$ \\
\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 6 months ago

Solution

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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 $C_1V_1 = C_2V_2$, we can calculate the volumes required to dilute a stock solution to a lower concentration.

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2(c) Answer

The elution order is $\mathrm{Cl}^- < \mathrm{Br}^- < \mathrm{SO}_4^{2-}$ 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

$\mathrm{SO}_4^{2-}$ is retained the longest due to its higher charge density, followed by $\mathrm{Br}^-$ due to its larger size compared to $\mathrm{Cl}^-$, which elutes first.

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