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

It was ernest rutherford