Answer:
pKa = 3.97
Explanation:
Let's state the equations:
NaA → A⁻ + Na⁺
A⁻ + H₂O ⇄ OH⁻ + HA Kb
We don't have the concentration of [OH⁻] but we do know pH, so we can determine, pOH and after that, [OH⁻].
14 - pH = pOH → 14 - 8.70 = 5.30
10^-pOH = [OH⁻] → 10⁻⁵'³⁰ = 5.01×10⁻⁶ M
Initially we can know, the moles of base, we had.
20 mL . 0.27M = 5.4 mmoles. So now, let's make the expression for Kb.
Kb = [OH⁻] . HA / [A⁻]
As the [OH⁻] is so low, we can assume, there where no loses and the salt is well concentrated. So:
Kb = (5.01×10⁻⁶)² / 0.27 → 9.29×10⁻¹¹
Kw = Ka . Kb, then Ka = Kw/ Kb
Ka = 1×10⁻¹⁴ /9.29×10⁻¹¹ = 1.08×10⁻⁴
pKa = - log Ka → - log 1.08×10⁻⁴ = 3.97
