Calculate the value of dissociation constant of weak monoacidic base if it dissociates to 2 % in 0.1 M solution?

Step-by-step solution

For a weak monoacidic base with concentration C = 0.1 , M and percentage dissociation 2 % , the degree of dissociation is = 0.02 . The dissociation constant K_b is given by the expression K_b = [B^+][OH^-] [BOH] . Substituting equilibrium concentrations yields K_b = C ^2 1 - . Since 1 , we approximate 1 - 1 , simplifying to K_b C ^2 . Substituting values gives K_b = (0.1) (0.02)^2 = 0.1 4 10⁻⁴ = 4 10⁻⁵ . The dissociation constant is 4 10⁻⁵ .