[dun13203171]

Hi guys, we covered this briefly in class but I've been looking at book which has various questions using the equations below and I don't understand when and why to use which equation.

The book states for an acidic drug

Ph-pKa=log(S-S0/S0)

Basic drug

Ph-pKa=log(S0/S-S0)


In this example below, they determine the compound is a base, I understand that, but it states to use the equation above but they rearrange it, I'm not sure why and how they determine which of the two equations to use?




In the example below, they calculate the isoelectric point and they state to use the basic equation but if you study the picture below, they plug the necessary figures into the equation and then rearrange it to the acidic equation and I don't know why.

It's states on the previous page to use the basic equation for ph's below the isoelectric point and the acidic equation for ph's above the isoelectric point.



All in all, I just don't know when to use which of the two forms of the equation. If someone could spend some time helping explain it would help me a great deal.

Thanks in advance

[AWK]

All in this text is clear. Read it slowly and attentively a few times (3 - 5). The best idea - rather the whole chapter.

1. From dependence of solubility at different pH you can decide which formula you should use for calculation of pI. In the case of slightly different values - calculate the mean value.
or
2. Having known pI and S₀ you can found solubility at different pH as sum of S and S₀ from rearrangement formula (below pI or above pI).
Note, your example with tryptophan gives rather crude solubilities since both points are laying too close of pK_a points in the range of buffering (± 0.8 pH unit aroud pK_a - value 0.8 is used for diluted solutions).

[mjc123]

They are simply doing an algebraic rearrangement of the equation to isolate the quantity they want to determine. I assume you are familiar with this - e.g. if you were told that y = 14/(1-x), could you rearrange that to get an equation for x?
In question 2, they are not rearranging the basic equation to the acidic equation. The acidic equation is not a rearrangement of the basic equation, but a different equation. Can you see that? They are, again, simply rearranging the basic equation in order to get S (the desired quantity) on top.