Answer:
The value of [tex]K_b[/tex] is [tex]2.04\times 10^{-5}[/tex].
Explanation:
[tex]HCN\rightleftharpoons H^++CN^-[/tex]
The dissociation constant of the HCN (acid)= [tex]K_a=4.9\times 10^{-10}[/tex]
Ionic product of water = [tex]K_w=1\times 10^{-14}[/tex]
[tex]K_b[/tex] for [tex]CN^-[/tex](base) =[tex]K_b=?[/tex]
[tex]K_w=K_a\times K_b[/tex]
[tex]K_b=\frac{1\times 10^{-14}}{4.9\times 10^{-10}}=2.04\times 10^{-5}[/tex]
The value of [tex]K_b[/tex] is [tex]2.04\times 10^{-5}[/tex].
