Ascorbic acid ( H2C6H6O2 ) has 𝐾a1=8.00×10−5 and 𝐾a2=1.60×10−12 . What's the pH of a 0.281 M ascorbic acid ( H2C6H6O2 ) solution?

In calculation of pH, we only consider Kₐ₁ because Kₐ₁ ≫ Kₐ₂.

           H₂C₆H₆O₂(aq) + H₂O(ℓ) ⇌ HC₆H₆O₂⁻(aq) + H₃O⁺(aq)     Kₐ₁ = 8.00 × 10⁻⁵
Initial:    0.281 M                              0 M                 0 M
Change:   -y M                               +y M               +y M
Eqm:   (0.281 - y) M                         y M                 y M
           ≈ 0.281 M

At equilibrium:
Kₐ₁ = [HC₆H₆O₂⁻] [H₃O⁺] / [H₂C₆H₆O₂]
8.00 × 10⁻⁵ = y² / 0.281
y = √(0.281 × 8.00 × 10⁻⁵) = 4.74 × 10⁻³
pH = -log[H₃O⁺] = -log(4.74 × 10⁻³) = 2.32

====
OR:
pH
= -(1/2)[log(Kₐ₁) + log[H₂C₆H₆O₂]ₒ]
= -(1/2)[log(8.00 × 10⁻⁵) + log(0.281)]
= 2.32

[H+] = sqrt( Ka *c)

= sqrt(8.00×10−5*0.281 = 0.00474M

pH = 2.32