The ph of 0.255 m hcn is 4.95. What is the value of ka for hydrocyanic acid?

Ka = 4.9 X 10^-9

Explanation:

when the pH  is 4.95

the [H+] = 10^-4.95

[H+] = 1.122 e-5

since the equilibrium equation is:

HCN <=> 1 mole H+  &  1 mole CN-

then

[H+] = [CN-]

when

HCN <=>  H+  &   CN-

Ka = [ H+]  [CN-]  / [HCN]

Ka = [ 1.122 e-5]  [ 1.122 e-5]  / [0.255]

Ka = 1.256 e-10 / 0.255

Ka = 4.94 e-9

which rounds off to 2 sig figs into

Ka = 4.9 X 10^-9

p.s. it was the pH pf 4.95 that limited the calculations to 2 sig figs