Question

Describe how the solvents worked as the mobile phase of the liquid chromatography experiment. Why was it necessary to use different concentrations of aqueous isopropanol in the step-gradient separation?

Answer 1

• In liquid chromatography a highly polar (or, highly non-polar) liquid is kept as stationary phase. The mixture of components are allowed to be dissolved in it. Now, when a less polar solvent (more polar for 2nd type of stationary phase), termed as mobile phase, is flowed, then the less polar (more polar for 2nd type) components of the mixture gets dissolved in it and goes with it. Less is the polarity of the component, more it dissolves in mobile phase with higher distribution coefficient (for 2nd case more the polarity more the component gets dissolved in mobile phase) and more it moves with the mobile phase. Since the polarity of the components are never same, so, different component move with mobile phase with different velocity. So, they get seperated. Increasing or, the polarity of the mobile phase allows more polar (less polar) components to move. Generally chromatography is based on the first type.
• As said in the last line of the previous answer, if we flow a specified fixed composition mixture of solvents as mobile phase then how can we get different components of analyte mixture ? Say there are three components in the analyte mixture - a (most polar), b (medium polarity) and c (least polar). Suppose we are using a highly polar stationary phase. So, a will be mostly sticked to stationary phase, less will be b and least will be c. So, if we use a very less polar solvent ( having very small concentration of isopropanol ), it'll dissolve more c and c will go with it. We can say b will also move but not extensively as c. So, then how can we get a and b ? Surely we have to increase polarity of the mobile phase after we get c coming out of the column. Then b will move faster and we'll get b after few time. If we do not increase the polarity then the retention time for b will be much lengthy and there is a potential chance of loss of analyte. Again after getting b we have to increase polarity again to quickly get a, otherwise it wont come out of the stationary phase in the column. This is why step-gradient is important.