Question

A 0.300 M solution of a weak base has a pH of 9.10. What is the base hydrolysis constant, Kb, for the weak base?

Answer 1

pH + pOH = 14

pOH = 14 - 9.10

pOH = 4.90

-Log[OH-] = 4.90                                                                                   (pOH = - Log[OH-] )

[OH-] = 10^-4.90

[OH-] = 1.26 * 10^-6 M

And for weak bases,

[OH-] = (Kb*C)^1/2

(1.26*10^-6)² = Kb * 0.300

Kb = 5.29*10^-12