The equilibrium expression for the ionization of a weak monoprotic acid, HA, in water is:
A) Ka = [HA]/[OH-][H3O+]
B) Ka = [OH-][H3O+]/[HA]
C) Ka = [H3O+][HA]/[A-]
D) Ka = [H3O+][A-]/[HA]
E) Ka = [HA]/[H3O+][A-]
Correct Answer:
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