It has hydrolysis:
A- + H2O = HA + OH-
From the expression of Kb we have:
Kb = [HA] * [OH-] / [A-] = X ^ 2 / 0.25 - X
It is assumed that - X is negligible and clears:
X = [OH-] = √Kb * 0.25 = √4.2x10 ^ -10 * 0.25 = 1.02x10 ^ -5 M
The pOH and the pH are calculated:
pOH = - log 1.02x10 ^ -5 = 5
pH = 14 - 5 = 9
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