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A molecule has an ionizable group with a pKa value of 8. At pH equal to...

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Chemistry

A molecule has an ionizable group with a pKavalue of 8. At pH equal to 7.0, estimate roughly how much of thegroup is deprotonated (expressed as %)? Use 2 decimal points. Donot use scientific notation to enter your answer.

Answer & Explanation Solved by verified expert
3.9 Ratings (392 Votes)

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
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