Question

Question #56dce

Answer 1

You can't calculate the `"p"K_"a"` or `"p"K_"b"` of an unknown drug if you know nothing about it.

But the `"p"K_"a"` and `"p"K_"b"` values are important, because they determine how much of the drug is in the ionic and uncharged forms at a given pH.

You can use the Henderson-Hasselbalch Equation to calculate that

`"% ionized" = "100 %"/( 1 + 10^(q("pH" –"p"K_"a"))`

where the charge `q` = +1 for an acid and -1 for a base.

When `"pH"= "p"K_"a"`, 50% of the drug is ionized and 50% is uncharged.

• The ionized form is water-soluble and cannot cross a membrane.

• The uncharged form is lipid-soluble and cannot cross a membrane

When you design a new drug, you can't often predict its `"p"K_"a"`, but you can make a good guess about its solubility.

Chemists use the partition ratios to measure relative solubilities in two solvents.

The partition ratio `P` is the ratio of the concentrations of the unionized drug in two immiscible phases, such as water and octan-1-ol.

`P_"o/w" = ([S]_o)/([S]_w)`

where `P_"o/w"` is the partition ratio between octanol and water, and `[S]` represents the solubility of the drug in the two solvents.

This is usually expressed in logarithmic form:

`logP= log (([S]_o)/([S]_w))`

`logP` is a measure of the lipophilicity of the drug.

Chemists have devised computer programs from a database of up to 12 000 compounds and over 200 types of molecular "fragments" to predict `logP` values.

The programs can predict `logP` within ±0.5 units 80 % of the time.