Chloroform is an internal standard in the determination of the pesticide DDT in a polarographic analysis in which each compound is reduced at an electrode surface. A mixture containing $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{mM}$chloroform and$\mathbf{0}\mathbf{.}\mathbf{800}\mathbf{mM}$DDT gave signals of$\mathbf{15}\mathbf{.}\mathbf{3}\mathbf{\mu A}$for chloroform and$\mathbf{10}\mathbf{.}\mathbf{1}\mathbf{\mu A}$for DDT. An unknown solution$\left(10.0\mathrm{mL}\right)$containing DDT was placed in a$\mathbf{100}\mathbf{-}\mathbf{mL}$volumetric flask and$\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{\mu L}$of chloroform (FM 119.39, density$\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{484}\mathbf{g}\mathbf{/}\mathbf{mL}\mathbf{)}$were added. After dilution to the mark with solvent, polarographic signals of and$\mathbf{8}\mathbf{.}\mathbf{7}\mathbf{\mu A}$were observed for the chloroform and DDT, respectively. Find the concentration of DDT in the unknown.

To find the response factor, a single mixture is used. This "one-point calibration curve" would be adequate to provide an appropriate response factor if there was no experimental error. Because there is always some experimental variability, a multipoint calibration curve is preferred to average out part of it.

$$\begin{gathered} \frac{\mathrm{signal}\mathrm{from}\mathrm{analyte}}{\mathrm{signal}\mathrm{from}\mathrm{standard}}=\mathrm{F}\frac{\mathrm{concentration}\mathrm{of}\mathrm{analyte}}{\mathrm{concentration}\mathrm{of}\mathrm{standard}} \\ \frac{{\mathrm{A}}_{\mathrm{x}}}{{\mathrm{A}}_{\mathrm{s}}}=\mathrm{F}\left(\frac{\left[\mathrm{X}\right]}{\left[\mathrm{S}\right]}\right) \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{A_s}{\left[\mathrm{S}\right]}\frac{10.1\mathrm{\mu A}}{0.8\mathrm{mM}} \\ =\mathrm{F}\frac{15.3\mathrm{\mu A}}{0.5\mathrm{mM}} \\ \mathrm{F}=0.4126 \end{gathered}$$

Calculate the mass of the chloroform.

$$\begin{gathered} \mathrm{m}(\mathrm{chloroform})=\mathrm{V}\times \mathrm{p} \\ \mathrm{m}(\mathrm{chloroform})=\left(10.2\times {10}^{-6}\mathrm{L}\right)\left(1484\mathrm{g}/\mathrm{L}\right)\mathrm{m}\left(\mathrm{chloroform}\right) \\ =0.01514\mathrm{g} \end{gathered}$$

Now calculate moles $$\begin{gathered} \mathrm{n}(\mathrm{chloroform})=\frac{\mathrm{m}}{\mathrm{M}}=\frac{0.011514\mathrm{g}}{119.38/\mathrm{mol}} \\ =0.1268\mathrm{mmol} \end{gathered}$$

$$\begin{gathered} 0.1268\mathrm{mL}\mathrm{chloroform}\mathrm{in}100\mathrm{mL}\mathrm{gives}\mathrm{us}\left[\mathrm{S}\right]=1.268, \\ \mathrm{which}\mathrm{we}\mathrm{may}\mathrm{use}\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{concentration}\mathrm{of}\left[\mathrm{X}\right] \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{{\mathrm{A}}_{\mathrm{s}}}{\left[\mathrm{S}\right]}\frac{8.7\mathrm{\mu A}}{\left[\mathrm{X}\right]} \\ =0.4126\times \frac{29.4\mathrm{\mu A}}{1.268\mathrm{mM}}\left[\mathrm{X}\right] \\ =0.909\mathrm{mM} \end{gathered}$$

Thus, the concentration of DDT in unknown would be:

$\left[\mathrm{DTD}\right]=\left(0.909\mathrm{mM}\right)\frac{100\mathrm{mL}}{10\mathrm{mL}}=9.09\mathrm{mM}$