The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10-10. what is the ph of an aqueous solution of 0.080 m sodium cyanide (nacn) ?

According to the reaction equation:

and by using ICE table:

CN - + H2O ↔ HCN + OH-

initial 0.08 0 0

change - X + X + X

Equ (0.08-X) X X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

= (1 x 10^-14) / (4.9 x 10^-10)

= 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X = 0.0013

∴ [OH] = X = 0.0013

∴ POH = - ㏒[OH]

= - ㏒0.0013

= 2.886

∴ PH = 14 - POH

= 14 - 2.886 = 11.11