Let the molecule be HA.
[H+]= 10^-pH= 10^-7
                    Â
HA <---> H+ Â Â +Â Â A-
initial:Â Â Â Â Â Â Â Â Â Â Â Â
1Â Â Â Â Â Â Â Â Â Â Â Â
10^-7Â Â Â Â Â Â Â Â Â Â Â Â Â
0
final:Â Â Â Â Â Â Â Â Â
1-x           Â
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
