Answer:
Kb = 2.23 x 10⁻⁵
Explanation:
Benz-CO-NH₂ + H₂O => Benz-CO-NH₃⁺OH⁻ ⇄ Benz-CO-NH₃⁺ + OH⁻
given pOH = 2.91 => [OH⁻] = 10E-2.91 = 1.23 x 10⁻³M
Benz-CO-NH₃⁺OH⁻ ⇄ Benz-CO-NH₃⁺ + OH⁻
At Equilibrium 0.68M 1.23 x 10⁻³M 1.23 x 10⁻³M
Kb = [Benz-CO-NH₃⁺][OH⁻]/[ Benz-CO-NH₃⁺OH⁻]
= (1.23 x 10⁻³)² M²/ 0.068M = 2.23 x 10⁻⁵
