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

  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 

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.