Answer to Question #147837 in General Chemistry for Noah

Question

The pH of a 0.35 M solution of a weak base
is 10.58. What is the Kb of the base?

Answer 1

Let the weak base be represented by A^-

[A^-] = 0.35M

pH = 10.58

Now, let's find the pOH

pH + pOH = 14

pOH = 14 - 10.58

pOH = 3.42

Let's look for the concentration of OH^-

pOH = -log[OH^-]

-log[OH^-] = 3.42

[OH^-] = 10^-3.42

[OH^-] = 3.80 x 10^-4M

The ionization of the weak base is given below

A^- + H₂O :::::::::::: HA + OH^-

The equilibrium concentrations are

[OH^-] = 3.80 x 10^-4M

[HA] = 3.80 x 10^-4M

[A^-] = 0.35 - 3.80 x 10^-4

[A^-] = 0.349M

K_b = [HA][OH^-]/[A^-]

K_b = (3.80 x 10^-4)²/(0.349)

K_b = 4.14 x 10^-7M