ou add 0.232 g NH4Cl to a 500-mL volumetric flask and dilute to the mark. What is the pOH of the final solution at equilibrium? For NH3, Kb= 1.8×10-5.

Answer 1

NH₄⁺ + H₂O ==> NH₃ + H₃O⁺

NH4+ is acting as an acid and so we need to find the Ka. KaKb = 1x10^-14 and Ka = 1x10^-14/1.8x10^-5

Ka = 5.55x10-¹⁰

Ka = 5.55x10^-10 = [NH₃][H₃O⁺]/[NH₄⁺] and [NH₄⁺] = (x)(x) / [NH4+]

FInding [NH4+] we have 0.232 g NH4Cl x 1 mol/53.5 g = 0.00434 mol/0.5L = 0.00867 M

5.55x10^-10 = (x)(x) / [0.00867]

x2 = 4.81x10^-12

x = 2.19x10^-6 M = [H₃O⁺]

pH = -log 2.19x10^-6 M = 5.66

pOH = 14 - pH

pOH = 8.34