Sodium cyanide, nacn, is a salt formed from the neutralization of the weak acid hydrocyanic acid, hcn, with the strong base sodium hydroxide. given that the value of ka for hydrocyanic acid is 4.90×10−10, what is the ph of a 0.265 m solution of sodium cyanide at 25∘c?

pH = 11.37

Explanation:

Sodium cyanide will dissociate into sodium ion and cyanide ion. This cyanide ion will get hydrolyzed.The ICE table for hydrolysis of cyanide ion is:

Initial                      0.265                                  0         0

Change                 -x                                           +x        +x

Equilibrium        0.265-x                                     x          x

This K is Kb of HCN

Kb = Kw / Ka

Putting values

x can be ignored in denominator as Kb is very low

x= 2.33X10⁻³ M = [OH⁻]

pOH = -log[OH⁻]

pOH = -log(2.33X10⁻³ )

pOH = 2.63

pH = 14- pOH = 14-2.63 = 11.37  

the millers tale is the answer.