Question

A molecule has an ionizable group with a pK_a value of 8. At pH equal to 7.0, estimate roughly how much of the group is deprotonated (expressed as %)? Use 2 decimal points. Do not use scientific notation to enter your answer.

Answer 1

Let the molecule be HA.

[H+]= 10^-pH= 10^-7

                      HA <---> H+    +   A-

initial:             1             10^-7              0

final:          1-x             10^-7+x          x

pKa = 8

Ka= 10^-pKa = 10^-8

use:

Ka= [H+][A-] / [HA]

10^-8 = (10^-7+x)*x/(1-x)

since Ka is very small, x will be small and hence it can be ignored as compared to1.

Also 10^-7 will be smaller as compared to x as x will be roughly 10^-4

so,

10^-8 = x*x/ 1

x = 1*10^-4

deprotonated fraction = 1-x = 1-10^-4 which will be roughly 1

So molecule will hardly deprotonate.

Answer : 1