Reversed-Phase HPLC, adsorbent porosity.

RP-HPLC

Before we start discussing the effect of the porosity on the HPLC retention and the relationships between the pore size, pore volume and surface area let first discuss why reversed-phase HPLC is so popular. Almost 80% of all separations are done in RP mode.

Normal phase HPLC was discovered almost 100 years ago by M.Tswett, and it was the first in the extensive use in the 50s and early 60s, when reversed-phase mode was introduced and became very popular, why?

Let me give you an analogy. Why do we prefer fluorescence detection to UV if we can (I mean if component has fluorescent ability)? Fluorescent detector is more sensitive, you say. Yes, but why?

In UV detector the whole intense radiation of UV lamp passes through the cell and hit photodiode. We are trying to detect minor intensity deprecations on the very high energy level. Temperature fluctuations, lamp instability may significantly limit our ability to detect a small amounts of our sample. On the other hand in the fluorescence detector we ideally have zero background energy level, and even the single quantum of energy from the single molecule could be detected.

Same situation we have in NP and RP HPLC. NP explore polar interactions of our analyte with the stationary phase, they are on the level of 10-15 Kcal/mole, it is high energy background. Our ability to distinguish minor differences in the interactions of two analytes with the surface is limited. On the other hand in RP HPLC we are mainly utilizing dispersive interactions, and they are on the order of magnitude lover. So, we have much lover background energy, and we can effectively separate closely related components.

 

Adsorbents porosity

For RP HPLC we need hydrophobic surface, and to be able to separate closely related (similar) components we need a big surface. Chromatographic process is an amplification of the minor separation steps by exposing the analytes to the huge surface.

For that we are using porous material, like silica. It has surface about 200 to 400 m2/g, and average column contains 1,3 - 1.5 g of the adsorbent.

That big surface we only can get if we will have lots of tiny pores in each particle. Average pore diameter is 100 .

What do you think, how the single spherical porous silica particle looks like? If you ever seen the sponge ball of 10 cm diameter (kids toy, "nerv-ball") this is an exact model of silica particle. Let us do simple calculations now.

Pores in the silica particles are cylindrical. Let assume that there is no pore size distribution, we have only 100 pores, and our surface area is 400 m2/g

The surface of cylindrical pore could be calculated as S=pDL, where D is the pore diameter, and L is the pore length. One gram of the adsorbent has total volume slightly higher than 1.5 ml. Using the above equation we can calculate what will be the total length of all pores in one g of the adsorbent, approximately.

Take a moment before reading the answer and try to guess by yourself, how big it is (how long is the total length of all pores in one gram of the HPLC adsorbent)?

It is almost ten million kilometers, almost the distance from the Earth to the Sun. This actually does not mean that molecules are traveling all that distance.

The ratio of the particle diameter to the pore diameter is 500/1 this means that we have significant difference in the active flow inside and around the adsorbent particles. Imagine the water flow around spongy type rocks. The flow through the tiny channels inside will be much slower than around. If our analyte eventually penetrates inside the pore space in the particle it already will be retained

Now let us talk about reversed phase material. This is hydrophobic porous silica which surface has been modified with the hydrophobic ligands. These ligands may significantly alter the porous space inside the silica particles.

Slide 10