The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the stems are used to make rhubarb pie. if pka1 = 1.23 and pka2 = 4.19, what is the ph of a 0.0269 m solution of oxalic acid?

Chemical reaction (dissociation) 1: C₂O₄H₂(aq) ⇄ C₂O₄H⁻(aq) + H⁺(aq).
Chemical reaction (dissociation) 2: C₂O₄H⁻(aq) ⇄ C₂O₄²⁻(aq) + H⁺(aq).
c(C₂O₄H⁻) = c(H⁺) = x.
c(C₂O₄H₂) = 0.0269 M.
pKa₁ = 1.23.
Ka₁ = 10∧(-1.23) = 0.059.
Ka₁ = c(C₂O₄H⁻) · c(H⁺) / c(C₂O₄H₂).
0.059 = x² / (0.0269 M - x).
Solve quadratic eqaution: x = c(H⁺) = 0.02 M.
pH = -log(0.02 M) = 1.7.

Sorry i dont understand the question can you rephase it?