Question

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?

Answer 1

pH = 11.37

Step-by-step 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:

`CN^(-)+ H_(2)O----------->HCN + OH^(-)`

Initial 0.265 0 0

Change -x +x +x

Equilibrium 0.265-x x x

`K=([HCN][OH^(-))/([CN^(-)])`

This K is Kb of HCN

Kb = Kw / Ka

`Kb=(10^(-14))/(4.9X10^(-10))=2.04X10^(-5)`

Putting values

`2.04X10^(-5)=(x^(2) )/((0.265-x))`

x can be ignored in denominator as Kb is very low

`2.04X10^(-5)=(x^(2) )/((0.265))`

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