Question

For the hydrolysis of a salt of a weak acid and a weak base, show that

a b

w

K K

K

h

K 

(5)

Answer 1

Solution.

For weak acid:

$Ka = \frac{[H^+][A^-]}{[HA]}$

For weak base:

$Kb = \frac{[B^+][OH^-]}{[BOH]}$

Final reaction:

$A^- + B^+ + H2O = HA + BOH$

$Kh = \frac{[HA][BOH]}{[A^-][B^+]}$

So, we get:

$Kh = \frac{Kw}{Ka \times Kb}$

Because:

$Kw = [H^+][OH^-]$

Answer:

$Kh = \frac{Kw}{Ka \times Kb}$