Question

Determine pH of a weak base that has Kb = 3.78 x10-18

Answer 1

Solution:

The equation representing the ionization of a weak base (B):

B(aq) + H₂O = BH⁺(aq) + OH^-(aq)

where C_o = C_o(B) - initial concentration of a weak base solution.

[OH^-] = [BH⁺] = x

[B] = C_o(B) - x

In the problem statement, the initial concentration of a base solution is not provided (although it should be), therefore:

Kb = [x] * [x] / [C_o(B) - x] = 3.78*10^-18

Knowing the values of C_o(B), we substitute it in the quadratic equation:

x² + x*3.78*10^-18 - C_o(B)*3.78*10^-18 = 0.

Solving the quadratic equation, we find the values of x.

Then,

x= [OH^-]

pOH = -log[OH^-] = -log(x).

pH = 14 - pOH

pH = 14 + log(x).