A 20 ml of 0.27M of solution of the salt NaA has a pH of 8.70. Calculate the pKa value of the acid HA.

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

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.