the ph of a 1.00*10^-2 M solution of cyanic acid (HOCN) is 2.77 at 25 C. Calculate Ka and pKa for HOCN from this result.

Answer 1

pH = 2.77 = -log(H^+)
(H^+) = 1.70E-3

.......HOCN ==> H^+ + OCN^-
I....0.0100.....0.......0
C......-x.......x.......x
E.....0.0100-x..x........x

Ka = (H^+)(OCN^-)/(HOCN)
Substitute into the Ka expression and solve for Ka. You know (H^+), (OCN^-) and (HOCN)
Then pKa = -log Ka.