91Ó°ÊÓ

Response factor,

A response factor in chromatography is defined as the percentage between the concentration of a component being studied and the detector's reaction to that molecule.

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

where,

${\mathrm{A}}_{\mathrm{x}}=$signal from analyte

$\left[\mathrm{X}\right]$concentration of analyte

localid="1663563213253" $\mathrm{F}$=response factor

${\mathrm{A}}_{\mathrm{s}}$= signal from standard

$\left[\mathrm{S}\right]$= Concentration of standard

Concentration of $\left[\mathrm{S}\right]$:

$\left[\mathrm{S}\right]=\mathrm{initialconcentration}×\mathrm{dilutionfactor}$

Given,

${\mathrm{A}}_{\mathrm{x}}=$signal from analyte $=423$

$\left[\mathrm{X}\right]$=bconcentration of analyte

$\mathrm{F}=$response factor

${\mathrm{A}}_{\mathrm{s}}$=signal from standard =$447$

$\left[\mathrm{S}\right]$= Concentration of standard

The response factor for the standard mixture has to be calculated.

$\frac{\mathrm{Ax}}{\left[\mathrm{X}\right]}=\mathrm{F}\left(\frac{\mathrm{As}}{\left[\mathrm{S}\right]}\right)$

$$\begin{gathered} \frac{423}{0.0837}×\mathrm{F}\left(\frac{0.0666}{447}\right) \\ \mathrm{F}=\frac{423}{0.0837}×\left(\frac{0.0666}{447}\right) \\ =0.7529 \\ =0.753 \end{gathered}$$

The mixture of unknown plus standard used to find the concentration of .

$\left[\mathrm{S}\right]=initialconcentration×dilutionfactor$

$$\begin{gathered} \left[\mathrm{S}\right]=(0.146\mathrm{M})\left(\frac{10.0}{25.0}\right) \\ =0.0584\mathrm{M} \end{gathered}$$

For an unknown mixture,

role="math" localid="1663564419679" $$\begin{gathered} \frac{\mathrm{Ax}}{\left[\mathrm{X}\right]}=\mathrm{F}\left(\frac{\mathrm{As}}{\left[\mathrm{S}\right]}\right)\frac{423}{\left[\mathrm{X}\right]} \\ =0.753\left(\frac{447}{0.0584}\right) \\ \left[\mathrm{X}\right]=\frac{423}{0.753}×\frac{0.0584}{447} \\ \left[\mathrm{X}\right]=0.7339 \end{gathered}$$

Because $\mathrm{X}$was diluted from $10.0$to $25.0\mathrm{ml}$, the mixture was created using .

Because $\mathrm{X}$was diluted from $10.0$to $25.0\mathrm{mL}$and mixed with $S$, the original concentration of $\mathrm{X}$in the unknown was $\mathrm{X}$.

The original concentration of $\mathrm{X}$in the unknown was $\mathrm{X}$.

$$\begin{gathered} \left[\mathrm{X}\right]=\left(\frac{25.0}{100.0}\right)\left(0.7339\right) \\ =0.1839 \\ =0.184\mathrm{M} \end{gathered}$$

Hence, the concentration of was $\left[\mathrm{X}\right]$calculated as $0.184\mathrm{M}$.