Answer:
The [tex]K_b[/tex] expression for the weak base equilibrium is:
[tex]K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}[/tex]
Explanation:
[tex](CH_3)_3N(aq)+H_2O(l)\rightlefharpoons (CH_3)_3NH^++OH^-(aq)[/tex]
The expression of the equilibrium constant of base [tex]K_c[/tex] can be given as:
[tex]K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}[/tex]
][tex]K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}[/tex]
As we know, water is pure solvent, we can put [tex][H_2O]=1[/tex]
[tex]K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}[/tex]
So, the the [tex]K_b[/tex] expression for the weak base equilibrium is:
[tex]K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}[/tex]
