In a titration, 20 mL of 0.27 M NH3 was titrated with 0.20 M HCl. Calculate the pH of the resulting solution after the addition 0 mL of the acid? (give your answer to two decimal places) (Kb ammonia = 1.8 x 10-5)

The pH of 20 mL of 0.27 M NH₃ (ammonia) solution after 0 mL of acid is 11.34.What is the pH of the solution ammonia solution after 0 mL of acid has been added?The pH of 20 mL of 0.27 M NH₃ (ammonia) solution after 0 mL of acid has been added is equal to the pH of the solution when no acid has been added.The pH of the NH₃ solution is calculated from the dissociation equation and the formula below:Dissociation equation: NH₃ + H₂O ⇄ NH₄⁺ + OH⁻Kb = [NH₄⁺] * [OH⁻] / [NH₃]Kb of NH₃= 1.8 * 10⁻⁵From the equation of dissociation above, [NH₄⁺] = [OH⁻]  = xKb = x²/ [NH₃]x² = Kb * [NH₃]x = √( Kb * [NH₃])x = √(1.8 * 10⁻⁵ * 0.27)x = 0.0022 M [OH⁻] = 0.0022pOH = - log [OH⁻]pOH = -log (0.0022)pOH = 2.66pH = 14 - 2.66pH = 11.34Learn more about pH at: https://brainly.com/question/26424076#SPJ1