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SmokeyV4L SmokeyV4L
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11 years ago
If the acid dissociation constant, Ka, for an acid HA is 8 x 10?4 at 25?C, what percent of the acid is dissociated in a 0.50?molar solution of HA at 25?C?
A) 0.08%
B) 0.2%
C) 1%
(D) 2%
(E) 4%

can you explain please how to get the answer thanks
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wrote...
#1
11 years ago
HA <=> H+ + A-
start
0.50
change
-x. . .. +x. . +x
at equilibrium
0.50-x. .x .. x

Ka = [H+][A-]/ [HA]= 8 x 10^-4 = (x)(x) / 0.50-x

x = 0.02

% dissociation = 0.02 x 100/ 0.50=4 %
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