Calculate the percent ionization of 0.140 m lactic acid in a solution containing 8.5×10−3 m sodium lactate.

Chemical reaction 1: C₃H₆O₃(aq) ⇄ C₃H₅O₃⁻(aq) + H⁺(aq).
Chemical reaction 2: C₃H₅O₃Na(aq) → C₃H₅O₃⁻(aq) + Na⁺(aq).
c(lactic acid - C₃H₆O₃) = 0,140 M.
c(sodim lactate - C₃H₅O₃Na) = 8,5·10⁻³ M.
Ka(C₃H₆O₃) = 1,4·10⁻⁴.
[H⁺] = x.
[C₃H₅O₃⁻] = 0,0085 M + x.
[C₃H₆O₃] = 0,140 M - x.
Ka = [C₃H₅O₃⁻] · [H⁺] / [C₃H₆O₃].
1,4·10⁻⁴ = (0,0085 M + x) · x / (0,140 M - x).
1,4·10⁻⁴ = 0,0085x + x² / (0,140 - x).
x = 0,00188 M.
percent ionization of lactic acid: 
α = 0,00188 M ÷ 0,140 M · 100% = 1,35%.

Dmight be the correct answer