Calculate the experimental Ka for benzoic acid? ?

Molar mass of HC₆H₅CO₂ = (1.0×6 + 12.0×7 + 16.0×2) g/mol = 122.0 g/mol
Initial concentration of HC₆H₅CO₂ = (2.12/122.0 mol) / (1.2 L) = 0.01448 M

pH = log[H₃O⁺] = 2.75
Hence, [H₃O⁺] at equilibrium = 10⁻²˙⁷⁵ M = 0.001778 M

Consider the dissociation of HC₆H₅CO₂:
_                      HC₆H₅CO₂(aq) + H₂O(ℓ) ⇌ C₆H₅CO₂⁻(aq) + H₃O⁺(aq)   Kₐ = ?
_Initial:              0.01448 M                            0 M                  0 M
_Change:              -y M                                +y M                +y M
_Equilibrium:  (0.01448 - y) M                        y M                  y M

At equilibrium:
[H₃O⁺] = y M = 0.001778 M
y = 0.001778

Kₐ for the dissociation of HC₆H₅CO₂
= [C₆H₅CO₂⁻] [H₃O⁺] / [HC₆H₅CO₂]
= y² / (0.01448 - y)
= 0.001778² / (0.01448 - 0.001778)
= 2.49 × 10⁻⁴