17) use the concentrations and pH value from solution 9 and the Kw value to approximate the Ka for acetic acid?

pH = 7.92
pOH = 14.00 - 7.92 = 6.08
[H₃O⁺] at eqm = 10⁻⁶˙⁰⁸ M

___________C₂H₃O₂⁻(aq) + H₂O(l) ⇌ HC₂H₃O₂(aq) + OH⁻(aq) …. Kb = Kw / Ka
Initial ______ 0.1 M _______________ 0 M _______ 0 M
Change ___ -10⁻⁶˙⁰⁸ M __________ +10⁻⁶˙⁰⁸ M __ +10⁻⁶˙⁰⁸ M
At eqm __ (0.1 - 10⁻⁶˙⁰⁸) M _______ 10⁻⁶˙⁰⁸ M ___ 10⁻⁶˙⁰⁸ M

Kb = [HC₂H₃O₂] [OH⁻] / [C₂H₃O₂⁻]
Kw / Ka = [HC₂H₃O₂] [OH⁻] / [C₂H₃O₂⁻]
(1.00 × 10⁻¹⁴) / Ka = (10⁻⁶˙⁰⁸)² / (0.1 - 10⁻⁶˙⁰⁸)

Ka = (1.00 × 10⁻¹⁴) × (0.1 - 10⁻⁶˙⁰⁸) / (10⁻⁶˙⁰⁸)² = 1.45 × 10⁻³