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]

Question

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]

Accepted Answer

∻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 $C_1V_1 = C_2V_2$. Here, $C_1 = 1000$ ppm, $V_1$ is the volume of stock solution needed, $C_2 = 100$ ppm, and $V_2 = 50$ cm³. Solve for $V_1$. ‖ ‖ 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 $C_1 = 100$ ppm, $V_1$ is the volume of intermediate solution needed, $C_2 = 5$ ppm, and $V_2 = 50$ cm³. Solve for $V_1$. ‖ ∻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 $C_1V_1 = C_2V_2$, 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**: $\mathrm{SO}_4^{2-}$ has a higher charge density than $\mathrm{Br}^-$ and $\mathrm{Cl}^-$, so it is retained the longest. $\mathrm{Br}^-$ is larger than $\mathrm{Cl}^-$, so it is retained longer than $\mathrm{Cl}^-$, which is the smallest and elutes first. ‖ ∻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. ⚹◊What is the chemical equation for the preparation of a 5 ppm chloride standard solution from a 1000 ppm stock solution?⍭ Generate me a similar question◊

Sia