Question

What is the expression K_n for the neutralization reaction in terms of known K_b values and equilibrium constants? Then solve for K_n

CN^-_(aq) + H₃O⁺ <====> HCN_(aq) + H₂O_(l)

Answer 1

the reaction given is

CN- + H30+ ---> HCN + H20

the equilirbium expression is given by

Kn = [HCN ] / [CN-] [H30+]

now

consider hydrolysis of CN-

we get

CN- + H20 ---> HCN + OH-

Kb = [HCN ] [OH-] / [CN-]

also

for water

H20 + H20 ---> H30+ + OH-

we know that

Kw = [H30+] [OH-]

now

we got

Kw = [H30+] [OH-]

Kb = [HCN ] [OH-] / [CN-]

Kn = [HCN] / [H30+] [CN-]

we can see that

Kn = Kb / kw

now

we know that

Kw = 10-14

also

Kb for HCN = 1.62 x 10-5

so

using these values

we get

Kn = 1.62 x 10-5 / 10-14

Kn = 1.62 x 10^9

so

the value of Kn is 1.62 x 10^9