A solution containing$\mathbf{3}\mathbf{.}\mathbf{47}\mathbf{mMX}$(analyte) and$\mathbf{1}\mathbf{.}\mathbf{72}\mathbf{mMS}$(standard) gave peak areas of$\mathbf{3473}$and $\mathbf{10222}$,respectively, in a chromatographic analysis. Then$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{mL}$of $\mathbf{8}\mathbf{.}\mathbf{47}\mathbf{mMS}$was added to$\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{mL}$of unknown$\mathbf{X}$,and the mixture was diluted to$\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{mL}$. This solution gave peak areas of$\mathbf{5428}$and$\mathbf{4431}$for$\mathbf{X}$and$\mathbf{S}$, respectively.

(a) Calculate the response factor for the analyte.

(b) Find the concentration of $\mathbf{S}\mathbf{\left(}\mathbf{mM}\mathbf{\right)}\mathbf{}\mathbf{in}\mathbf{}\mathbf{the}\mathbf{}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{mL}\mathbf{}\mathbf{mixture}\mathbf{.}$

(c) Find the concentration of $\mathbf{X}\mathbf{\left(}\mathbf{mM}\mathbf{\right)}\mathbf{}\mathbf{in}\mathbf{}\mathbf{the}\mathbf{}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{mL}\mathbf{}\mathbf{mixture}\mathbf{.}$

(d) Find the concentration of$\mathbf{X}\mathbf{}\mathbf{in}\mathbf{}\mathbf{the}\mathbf{}\mathbf{orignal}\mathbf{}\mathbf{unknown}\mathbf{.}$

In this task, we have a solution that contains$3.47\mathrm{mM}\times \left(\mathrm{analyte}\right)$and$1.72\mathrm{mMS}$(standard)and yields chromatographic peak areas of $3473$and$10222$.$1\mathrm{mL}$of $8.47\mathrm{mMS}$was mixed with $5\mathrm{mL}$of ${\mathrm{X}}_{\mathrm{r}}$unknown and diluted to $ $10\mathrm{mL}$, yielding peak areas of $5428$and $4431$for $\mathrm{X}$and$\mathrm{S}$,respectively.

(a)

$$\begin{gathered} \mathrm{The}:\#\mathrm{c}{34632}^{\text{'}\text{'}}>\mathrm{response}\mathrm{factore}\mathrm{response}\mathrm{factor}\mathrm{for}\mathrm{the}\mathrm{analyte}\mathrm{will}\mathrm{be}\mathrm{calculated}\mathrm{here}. \\ \frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{{\mathrm{A}}_{\mathrm{S}}}{\mathrm{S}} \\ \frac{3.473}{3.47\mathrm{mM}}=\mathrm{F}\frac{10222}{1.72\mathrm{mM}} \\ \mathrm{F}=0.1684 \\ \mathrm{The}\mathrm{response}\mathrm{factor}\mathrm{for}\mathrm{the}\mathrm{analyte}\mathrm{is}0.1684 \end{gathered}$$

(b)

$$\begin{gathered} \mathrm{The}\mathrm{concentration}\mathrm{of}\mathrm{S}\left(\mathrm{mM}\right)\mathrm{in}\mathrm{the}10\mathrm{m}\mathrm{L}\mathrm{mixture}\mathrm{is}: \\ \left[\mathrm{S}\right]=8.47\mathrm{mM}\times \frac{1\mathrm{mL}}{10\mathrm{mL}}=0.847\mathrm{mM} \\ \mathrm{The}\mathrm{concentration}\mathrm{of}\mathrm{S}\left(\mathrm{mM}\right)\mathrm{is}0.847\mathrm{mM} \end{gathered}$$

(c)

$\mathrm{The}\mathrm{concentration}\mathrm{of}\mathrm{X}\left(\mathrm{mM}\right)\mathrm{in}\mathrm{the}10\mathrm{m}\mathrm{L}\mathrm{mixture}\mathrm{is}:$

$$\begin{gathered} \frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{{\mathrm{A}}_{\mathrm{S}}}{\left[\mathrm{S}\right]} \\ \frac{5428}{\left[\mathrm{X}\right]}=0.1684\frac{4431}{0.847\mathrm{mM}} \\ \left[\mathrm{X}\right]=6.16\mathrm{mM} \\ \mathrm{The}\mathrm{concentration}\mathrm{of}\mathrm{X}\left(\mathrm{mM}\right)\mathrm{is}6.16\mathrm{mM} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{d}\right) \\ \mathrm{The}\mathrm{concentration}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{original}\mathrm{unknown}\mathrm{is}: \\ \mathrm{The}\mathrm{original}\mathrm{concentration}=2\times \left(\mathrm{diluted}\mathrm{concentration}\right)=2.6.16\mathrm{mM}=12.32\mathrm{mM} \\ \mathrm{The}\mathrm{concentration}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{original}\mathrm{unknown}\mathrm{is}12.32\mathrm{mM} \end{gathered}$$