Calculate the hydronium ion concentration in an aqueous solution of 9.19×10-2 M ascorbic acid, H2C6H6O6 (aq). [H3O+] = ?

H₂C₆H₆O₆ (ascorbic acid) is a diprotic acid.
Refer to: http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf
H₂C₆H₆O₆(aq) + H₂O(l) ⇌ HC₆H₆O₆⁻(aq) + H₃O⁺(aq) …… Kₐ₁ = 7.9 × 10⁻⁵
HC₆H₆O₆⁻(aq) + H₂O(l) ⇌ C₆H₆O₆²⁻(aq) + H₃O⁺(aq) …… Kₐ₂ = 1.6 × 10⁻¹²
(Ka values from different sources may be slightly different, and this may lead to slightly different in answer.)

Initial [H₂C₆H₆O] = 9.19 × 10⁻² M = 0.0919 M

As Kₐ₁ ≫ Kₐ₂, It can be assume that all the H₃O⁺ ions are produced in the first dissociation of H₂C₆H₆O₆.
_________ H₂C₆H₆O₆(aq) + H₂O(l) ⇌ HC₆H₆O₆⁻(aq) + H₃O⁺(aq) ____ Kₐ₁ = 7.9 × 10⁻⁵
Initial: ____ 0.0919 M _______________ 0 M _________ 0 M
Change: _____ -y M ________________ +y M _________ +y M
Eqm: ___ (0.0919 - y) M _____________ y M __________ y M

As Kₐ₁ is very small, the dissociation of H₂C₆H₆O₆ is to a very small extent.
We can assume that (0.0919 ≫ y
Hence, [H₂C₆H₆O₆] at eqm = (0.0919 - y) M ≈ 0.0919 M

At eqm: Ka = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]
7.9 × 10⁻⁵ = y² / 0.0919
y = √{(7.9 × 10⁻⁵) × 0.0919} = 2.7 × 10⁻³

[H₃O⁺] = 2.7 × 10⁻³ M