pH = -log [H+]
3.5 = -log [H+]
[H+] = 3.16*10^-4 M
let the acid be written as HA
HA <--------> H+ + A-
C 0 0 (initial)
C-x x x (at equilibrium)
x = [H+] = 3.16*10^-4 M
x = 78% of C
3.16*10^-4 = 0.78*C
C = 4.05*10^-4 M
Ka = x*x / (C-x)
= (3.16*10^-4 )*(3.16*10^-4 ) / (4.05*10^-4 - 3.16*10^-4 )
= 1.12*10^-3
pKa = -log Ka
= -log (1.12*10^-3)
= 2.95
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