The pH of a. 0727 M aqueous sodium cyanide, KCN, solution at 25 degrees C

Kb = [HA} [OH-] / [A-] where [A-] represents the concentration of CN - (.068M)

Kb = Kw / Ka = 1 x10-14 / 4.9 x 10-10 = 2 x 10-5

Since this is a salt solution which could be considered to have formed from the neutralization of a strong base (NaOH) and a weak acid (HCN), the Na + will have no effect on the pH of the solution while the CN - ion will undergo hydrolysis:

CN - + H2O - - > HCN + OH-

Based on this equation, the quantities of HCN and OH - produced must be the same and therefore [HCN]=[OH-]. We will set this equal to x.

Plugging into the original equation yields:

2 x 10-5 = x2 /.068 M

Solving for x yields 1.2 x 10-3 whidh is equal to the [OH-]

The pOH then is equal to - log (1.2x10-3) = 2.9

The pH of the solution would be 14 - 2.9 = 11.1