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

Answer:

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:

                        $CN^{-}+ H_{2}O----------->HCN + OH^{-}$

Initial                      0.265                                  0         0

Change                 -x                                           +x        +x

Equilibrium        0.265-x                                     x          x

$K=\frac{[HCN][OH^{-}}{[CN^{-}]}$

This K is Kb of HCN

Kb = Kw / Ka

$Kb=\frac{10^{-14}}{4.9X10^{-10}}=2.04X10^{-5}$

Putting values

$2.04X10^{-5}=\frac{x^{2} }{(0.265-x)}$

x can be ignored in denominator as Kb is very low

$2.04X10^{-5}=\frac{x^{2} }{(0.265)}$

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