Answer
$(CH_3)_2NH + H_2O -> (CH_3)_2NH_2^+ + OH^- $
$Kb = \frac{[CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}$
Work Step by Step
1. Write and balance the reaction with water:
$(CH_3)_2NH + H_2O -> (CH_3)_2NH_2^+ + OH^- $
$1(CH_3)_2NH + 1H_2O -> 1(CH_3)_2NH_2^+ + 1OH^- $
2. Now, write the Kb formula:
$Kb = \frac{[Conjugate. acid]*[OH^-]}{[Base]}$
$Kb = \frac{[CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}$