Calculate the[OH-] of a solution of 0.1861M CH3CH2CH2NH3CL if Kb= 5.1*10-4 for CH3CH2CH2NH2

Answer 1

Let's call that long salt just BH3Cl and that ionizes into BH3^+ and Cl^-. It's the BH3^+ that is hydrolyzed.
BH3^+ + H2O ==> H3O^+ + BH2
Ka = (Kw/Kb) = (H3O^+)(BH2)/(BH3^+)
Set up an ICE chart and substitute into the Ka expression; solve for (H3O^+) and convert to (OH^-).