Fe³⁺ = 0.0950 M Fe³⁺ + H₂O → Fe(OH)²⁺ + H⁺ 0.0950 0 0 0.0950-x x x Ka = x² / (0.0950-x) 3.00 x 10⁻³ = x² / (0.0950-x) x² + (3.00 x 10⁻³) x - (2.85 x 10⁻⁴) = 0 By solving the value of x = 0.0154 [H⁺] = 0.0154 M pH = - log [H⁺] pH = - log (0.0154) pH = 1.81
