Question

The ka for hydrocyanic acid (hcn) is 4.9 × 10−10. what is [oh−] in a 0.075 m solution of sodium cyanide (nacn)?

Answer 1

  by using ICE table:
 
            CN- +  H2O  ↔ HCN  + OH-

initial  0.075                    0          0

change -X                       +X         +X

Equ   (0.075-X)                   X          X


when Kb = Kw/Ka 

∴Kb = 1 x 10^-14 / 4.9 x 10^-10

         =  2 x 10^-5

and when Kb = [HCN][OH]/[CN-]

by substitution:

2 x 10^-5 = X^2 / (0.075 - X)      by solving for X

∴ x = 0.0012 M

∴[OH] = 0.0012 M