Question

1. What is the percent ionization of a 1.38 mol/L weak base if its K_b = 2.7 x 10^-6?
2. What is Kb for N₂H₄ if the pH of a 0.158M solution of N₂H₅Cl is 4.5?

Answer 1

1. Kb=a^2xC

a=✓(Kb/C)=✓(2.7x10-6/1.38)= 0,00139875721. = 0.1398 %

2. [H+]=antipH=3.16227766017E-05

a=3.16227766017E-05/0.158 = 0,000200144156

Kb=a^2xC = 0,000200144156^2 x 0.158 = 6,32911395e-9.