For hydrocyanic acid, hcn, ka = 4.9 × 10-10. calculate the ph of 0.20 m nacn. what is the concentration of hcn in the solution? for the ph i got 10 is that right?

By using the ICE table :

initial    0.2 M            0           0
change -X                 + X           +X
Equ     (0.2 -X)            X               X

when Ka = (X) (X) / (0.2-X)
so by substitution:

4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X 
∴X ≈ 10^-6
∴[HCN] = 10^-6

and PH = -㏒[H+]
             = -㏒ 10^-6
             = 6

Ibelieve y y even though i've only taken chemistry for one year