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