Given that
pH = 3.75 = -log[H+]
[H+]= antilog (-3.75) = 1.78 *10-4
(a) Percent dissociation = ([H+] / [HA]initial)*100% = ((1.78 *10-4) / 0.80)*100% = 0.02%
(b)Ka = [H+][A-] / [HA]left = (1.78 *10-4)2 / (0.8 - (1.78 *10-4)) = 3.96 *10-8
Since Both Ka and percent ionization is small. So acid is a weak acid.
3) A 0.80 M solution of an unknown acid, HXog), has a pH of 3.75. a)...
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