Chromatography Forum

Hi guys,

I'm having trouble answering this question after a practical i did on reversed phase HPLC and was wondering if perhaps any of you could help me.

Qs; "The retention in this form of chromatography can be approximated by the equation:
log k = log k(w) - S theta
where theta is the volume fraction of organic solvent in mobile phase
k = retention factor
k(w) = a constant related to the retention time of the substance if pure water was used as mobile phase
S is a constant for the compound on a given column with a given binary aqueous/organic mobile phase

Explain whether this eq holds for your results by plotting an appropriate graph for toluene? If so, use the graph to estimate the retention time of toluene in 40% methanol/ 60% water."

No values for the constant have been given and have one chromatogram of Toluene with mobile phase of 80% methanol/20% water and came out with a retention of 3.099.
If anyone can give me any pointers on how to tackle this I would be very grateful as I really have no idea at the moment.

Thanks very much

Can't be done with the data given. You are one parameter short.

You need an additional chromatogram for toluene (at a different %B) *or* you need to assume a value of S for toluene in that system (somewhere around 3 would be a good guess).

And, by the way, I've always seen that last parameter shown as "Φ" (phi, the Greek rough equivalent to the Latin "F" -- which makes sense as representing the "Fraction" of strong solvent), not "Θ" (theta).

Ahh sorry I also have a mixture which contained toluene, that I ran at different % of methanol, so have worked out;

at 80% methanol toluene had a retention of 3.099
at 90% - 2.406
at 70% - 4.526

Can I work it out with this data?

Thanks again

Yes -- with a caveat (see the last paragraph)

First, you have to convert the retention times to k' values. ("details are left to the student as an exercise")..
Then, calculate the logs of the k' values [ the log(k) values]
Now, your equation is in the form y = mx + b where y = log(k) and x = Φ. Which means that you can solve for m [ = S ) and b [ = log(kw) ].

Or, since your problem explicitly calls for a graph, simply put your two data points on a graph and connect the dots.

The catch is that the question is still misleading, because with only two points you can't actually evaluate whether the relationship holds, you can only assume that it does. You would need to run additional experiments to see whether the points all fall on the same straight line. In addition, using these data to estimate retention at 40% organic is running a *very* large extrapolation (as in, 10-15 fold increase in retention). At that point, it's more guesswork than estimation.

Graphically, you connect the dots, as mentioned. But you cannot determine k' without a value for retention time of an un-retained solute.

So you cover this in a course? May I ask the type, level of class?

yeah thanks,

am still quite confused because we haven't really covered this equation in lectures or anything

This is for an advanced analytical science module on level 3 of a forensic science degree.

Think i'll try speaking to my tutor about it but hes not very helpful! lol

Yes -- with a caveat (see the last paragraph)

First, you have to convert the retention times to k' values. ("details are left to the student as an exercise")..
Then, calculate the logs of the k' values [ the log(k) values]
Now, your equation is in the form y = mx + b where y = log(k) and x = Φ. Which means that you can solve for m [ = S ) and b [ = log(kw) ].

Or, since your problem explicitly calls for a graph, simply put your two data points on a graph and connect the dots.

The catch is that the question is still misleading, because with only two points you can't actually evaluate whether the relationship holds, you can only assume that it does. You would need to run additional experiments to see whether the points all fall on the same straight line. In addition, using these data to estimate retention at 40% organic is running a *very* large extrapolation (as in, 10-15 fold increase in retention). At that point, it's more guesswork than estimation.

I have calculated the k' values and the log k values for toluene at 70% and 90% methanol. I was now going to try plotting the values on the graph, but could you please tell me what would go on the X axis and what would go on the Y axis?

Volume fraction of organic on the x axis, in decimal form 0.7 for 70%..

But how can you calculate k' without knowledge of retention time of an unretained solute?

I know what the time for the unretained solute is now - 1.690 (uracil)

It was pretty thoroughly explained in the previous posts. You have retention times for toluene and for uracil (which is unretained). That means you can calculate the k' values for toluene at two different organic solvent compositions. Graph the log(k') vs the fraction strong solvent [that's the % expressed as a decimal fraction; e.g., 50% = 0.50 ]. You only have two points, so just connect the dots and extrapolate, and "Bob's yer uncle".