Verified answer

There are several ways to do this. Since you are given Kb, the equilibrium between the weak base, water and its conjugate acid is:

B + H2O <--> BH+ + OH-

Kb = [BH+][OH-]/[B] = 1.0 X 10^-5

Substituting the concentrations of the base and its conjugate acid gives:

1.0 X 10^-5 = (0.410) [OH-] / (0.140)

[OH-] = 3.41 X 10^-6

From this, and the expression for Kw,

[H3O+] = 1.0 X 10^-14 / 3.41 X 10^-6 = 2.9X10^-9

and pH = - log 2.9X10^-9 = 8.53

You could also use Kb to calculate Ka and the pKa, and then use the Henderson-Haselbalch equation:

Ka X Kb = Kw

Ka = 1X10^-14 / 1X10^-5 = 1.0 X 10^-9

pKa = 9.00

Then, pH = pKa + log [base]/[acid]

pH = 9.00 + log 0.14 / 0.41 = 8.53

1st convert Kb to Ka= (Ka= Kw/Kb) so ( 1.0x10^-14 / 1.0x^-5) = 1x10^-9 Ka

2nd use the Henderson-Hasselblach equaiton which is (( pH= pKa + log (conjugate base/ conjugate acid))

3rd plug the numbers pH= 1x10^-9 + log (0.140/0.410) = 8.53