The equilibrium expression for the hydrolysis of a weak base, A™, in water is: OK=[A]/[OH-][H3O+] K₁ = [OH-][HA]/[A¯][H3O+] OK = [OH-][A]/[HA] O Kb = [OH-][H3O*]/[A-] K₂ = [OH-][HA]/[A]
The equilibrium expression for the hydrolysis of a weak base, A™, in water is: OK=[A]/[OH-][H3O+] K₁ = [OH-][HA]/[A¯][H3O+] OK = [OH-][A]/[HA] O Kb = [OH-][H3O*]/[A-] K₂ = [OH-][HA]/[A]
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![The equilibrium expression for the hydrolysis of a weak base, A™, in water is:
OK=[A]/[OH-][H3O+]
K₁ = [OH-][HA]/[A-][H30*]
OK₂= [OH-][A-]/[HA]
O K₂ = [OH-][H3O*]/[A-]
K₁ = [OH-][HA]/[A-]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F82b39212-1ee3-4db2-a8fb-dc9be7b70f60%2F34e3c15a-50fb-4cb9-97eb-eebe421f817d%2F11i1f7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The equilibrium expression for the hydrolysis of a weak base, A™, in water is:
OK=[A]/[OH-][H3O+]
K₁ = [OH-][HA]/[A-][H30*]
OK₂= [OH-][A-]/[HA]
O K₂ = [OH-][H3O*]/[A-]
K₁ = [OH-][HA]/[A-]
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