Answers
use:
pH = -log [H3O+]
5.31 = -log [H3O+]
[H3O+] = 4.898*10^-6 M
Now use:
% ionisation = [H3O+]*100 / [HA]
= (4.898*10^-6)*100 / 0.0415
= 0.0118 %
Answer: 0.0118 %
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