A question about retention and organic solvent strength

[soiree]

Hi, I have a question that I cannot solve by myself, and has nobody nearby to ask, so decided to make topic here.

The question is about a famous equation about retention factor and solvent strength.

log k = log kw - sφ

And when some books describe this equation like this;
'For small molecules, the value of S is approximately 3,
and this predicts that a 10% change in organic modifier concentration will produce about a three-fold change in retention as measured by k.'
(quoted from http://books.google.co.kr/books?id=jn0e ... A6&f=false)
My local language textbook also said so(though they say S = 4, but I think that makes no much difference).

But I can't understand why 10% decrease in B(organic) makes three-fold retention increase when S is about 3-4.

As I thought, if there is concentration A and concentration B that is 10% increase to A, so is B = (A+0.1), the equation about two concentrations are

log Ka = log Kw - sA
log Kb = log Kw - sB = log Kw - sA - 0.1s
(Ka is retention factor when φ is A, and Kb also)

and to substract each side, the equation can change like

log(Ka/Kb) = 0.1s

If 10% increase of organic solvent make three-fold increase of k, the left side of equation will be approximately log 3, 0.4771. But the right is 0.3 when s = 3.
And I think 0.3 is closer to log 2 than log 3. If s = 4, still 0.4 ≃ log 2.5.

I am not sure about my thoughts for now, and need someone to check my bias. Please would anyone give me a enlightenment?


And another trivia question; is 'that' logarithm really common, not natural?

[tom jupille]

To answer in reverse order: it's done in log10 because most people compute that way.

On the main point, the observation that log(k') is a linear function of the fraction strong solvent is only an approximation. It's fairly good over a fairly wide range of situations, but it *is* still an approximation. That said, the S value can be interpreted as representing the number of strong solvent molecules required to displace a single analyte molecule from the stationary phase. In that sense it's analogous to the Z value in ion exchange (the slope of log(k') vs log (ionic strength), which turns out to be the ratio of the effective charge on the analyte to the effective charge on the driving (diplacing) ion.

Note that this is a really crude approximation. Stating that S values are between 3-4 is, IMHO, unjustifiably precise. I usually assume that for small molecules S ≈ 5 (i.e., somewhere between 2 and 10); for peptides, S ≈ 10 (i.e, somewhere between 5 and 20) and for proteins, S ≈ 25 (i.e., somewhere between about 10 and 50).

Now, if S ≈ 5, then log(Ka/Kb) ≈ 0.1 * 5 ≈ 0.5
Get rid of the logs, and you have 10^0.5 ≈ 3
So, a 10% change in aqueous/organic ratio should change k' values by about a factor of 3 (the "Rule of Three").

I can't understand why 10% decrease in B(organic) makes three-fold retention increase when S is about 3-4.

You're right, it doesn't. The only problem is that you are imputing more precision to the S-value estimates than they will support! If you have an S value of 3, then that 10% change will change k' by a factor of 2. If you have an S value of 4, k' will change by a factor of 2.5.

[soiree]

First of all, I was relieved that, at least, my calculations about that equation are not that wrong. But I recognized myself as a zealot of theory too. So I tested %B vs k experiment today, and got a real data. In my case, 10% B(It was ACN : THF = 1:1, as my lab uses frequently) change makes 4-fold retention change, and from that data, I solved S is about 6.3 in this case.

Also I just thought S as a unique constant related to molecular weight(some books also describes S as a mw-related constant, like S = 0.48*M^0.44), and did not noticed that S can be dealt with strength between analyte-stationary phase.

It is certain worth asking here. Appreciate for your answer.