A 0.068 M solution of benzamide has a pOH of 2.91. What is the value of Kb for
this compound?

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

i think the answer is terrae but not 100 percent sure. ▓▬▬ §