II.
pH of weakacid = 1/2(pka-logC)
c = concentration of weakacid = 0.1 M
pka = ?
2.39 = 1/2(x-log0.1)
x = pka = 3.78
ka = 10^-pka
ka = 10^-(3.78)
dissociation constant ( ka) = 1.66*10^-4
III.
A.
HF(aq)
<===> H^+(aq) + F^-(aq)
initial 0.5 M 0 0
at equil 0.5-x x x
ka = [H^+][F^-]/[HF]
7.1*10^-4 = x^2/(0.5-x)
x = [H^+]=[F^-] = 0.0185 M
[HF] = 0.5-0.018 = 0.482 M
pH = -log(H^+)
= -log(0.0185)
= 1.732
ka = cx^2
7.1*10^-4 = 0.5x^2
x = degree of dissociation = 0.0377
% of dissociation = 0.0377*100 = 3.77%
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