Grade 11Physical Chemistry
% dissociation of 0.024M solution of weak acid HA(ka = 2 × 10-3) is :
% dissociation of 0.024M solution of weak acid HA
(ka = 2 × 10-3) is :
1 Answer
Vikas TU
HA ============> H+ + A-
t=0 0.024 0 0
t=t 0.024(1-a) 0.024a 0.024a
Now from ionic eq. equal conc. method,
[H+] = [A-]
[H+]^2 = Ka*[HA] => 0.024*2*10^-3 => 48 x 10^-6
[H+] = root48) * 10^-3 M
Now,
Ca = root(48)*10^-3
a = root(48)/24 => 0.288*100 => 28.88%
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