Ph = -lg[H⁺]
CH3COOH ⇄ H⁺ + CH3COO⁻
initial: [CH3COOH] = 0,1mol/l; [H⁺] =0; [CH3COO⁻] = 0
la dizolvare : x x x
la echilibru: (0,1 - x) x x
Ka ={ [H⁺]·[CH3COO⁻]} /[CH3COOH]
[H⁺] = [CH3COO⁻
Ka = x² /(0,1-x) = 1,8·10⁻⁵ ptr. x f.f. mic 0,1 - x ≈ 0,1 [H⁺] =√(c·Ka)
⇒x² = 1,8·10⁻⁶ x = 1,34·10⁻³ mol / l = [H⁺] ⇒
⇒ Ph = - lg 1,34·10⁻³ = 3 - lg1,34 = 3-0,127 = 2,873
