HPLC peak should generally not have an asymmetry factor, B/A in figure $\mathbf{23}\mathbf{-}\mathbf{14}$,outside the range$0.9-1.5$

1. Sketch the shape of a peak with an asymmetry of 1.8
2. What might you do to correct the asymmetry?

When the distance from the start of the peak to the centre and the distance from the centre to the end of the peak differs, the peak is asymmetric. It is denoted by the asymmetry factor:

$\mathrm{asymmetry}\mathrm{factor}=\frac{\mathrm{B}}{\mathrm{A}}$

Where A is the distance from the start of the peak to the centre and B is the distance from the centre to the end of the peak.

If the asymmetry factor is $1.8$that means that B is $1.8$times greater than A. We are measuring A and B at the $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$of the height of the peak

As you can see from the sketch the value of A is 2, and the value of B is 3.6 , so:

$\mathrm{asymmetry}\mathrm{factor}=\frac{3.6}{2}$

$\mathrm{asymmetry}\mathrm{factor}=1.8$

This sketch satisfies the given asymmetry.

b) We can use a split injection, reduce the injection volume.

increase the column thickness. A guard column or a pressure pulsed injection can also be used.

Hence,

$\mathrm{asymmetry}\mathrm{factor}=\frac{3.6}{2}$

$\mathrm{asymmetry}\mathrm{factor}=1.8$

b) We can use a split injection, reduce the injection volume. increase the column thickness. A guard column or a pressure pulsed injection can also be used