The pH of a 1 .00 x 10-2 M solution of cyanic acid(HOCN) is 2.77 at 25 degrees. Calculate Ka and pKa for HOCN from this result.

Answer 1

...........HOCN ==> H^+ + OCN^-
I..........0.01.....0.......0
C..........-x.......x.......x
E........0.01-x.....x.......x

The problem tells you pH = 2.77; therefore, 2.77 = -log(H^+); therefore, (H^+) = 0.00170 = x in he above./
Evaluate H^+< OCN^-, and HOCN and calculate Ka.
pKa = -log K.