What is the Kb of the base PO−34 given that a 0.48M solution of the base has a pH of 12.70? The equation described by the Kb value is

Answer 1

I'm assuming you mean PO₄^3- (the phosphate anion).

PO₄^3- + H₂O ==> HPO₄^2- + OH^-

Kb = [HPO₄^2-][OH^-] / [PO₄^3-]

We can find the [OH-] from the pH of 12.70.

pH + pOH = 14

14.0 - 12.7 = 1.3 = pOH

[OH] = 1x10^-1.3

[OH-] = 5.0x10^-2 M

[HPO₄^2-] = 5.0x10^-2 M

Kb = (5.0x10^-2)² / 0.48 = 2.5x10^-3 / 0.48

Kb = 5.21x10^-3