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18 Nov 2019
Ka values for strong acids and bases are so large that you will often see the reaction written with a single arrow ("completely dissociated in solution") as if there was no back reaction. I would prefer for you to think of it as an equilibrium with such a large equilibrium constant that the back reaction is negligible. When we want to write a Ka or a Kb expression, we react the acid or base directly with water and write out the equilibrium. To quantify the strength of the conjugate base C in aqueous solution, we look at the reaction of just the this ion in water by itself. The CI will be on the reactant side and is now defined as the base in the reaction. The base, Cl (proton acceptor) reacts with water by accepting the Ht ion to form the new products. Could you write this equation? Take a minute to write out this equation below Enter your answer in the order: base + acid yields conjugate acid + conjugate base. To save time, you can skip the phase labels this time. Since you are working with a base (CI), the equilibrium constant is now Kb. chemPad ⤠Help [Ci] Greek ⼠Look at what happens when we multiply the Ka for the acid times the Ko for its conjugate base. Everything cancels except the terms for [O] and [H3O+]. The product of [OH-] X [H30*) is the water constant or Kw (1.0 x 10.1 Ka X Kb [HCI] [ci] = [OH-][H3O+] = Kw = 1.0 X 10-14 Using the water constant, we can calculate the K, for any conjugate base if we know the Ka for the acid (and vice versa).
Ka values for strong acids and bases are so large that you will often see the reaction written with a single arrow ("completely dissociated in solution") as if there was no back reaction. I would prefer for you to think of it as an equilibrium with such a large equilibrium constant that the back reaction is negligible. When we want to write a Ka or a Kb expression, we react the acid or base directly with water and write out the equilibrium. To quantify the strength of the conjugate base C in aqueous solution, we look at the reaction of just the this ion in water by itself. The CI will be on the reactant side and is now defined as the base in the reaction. The base, Cl (proton acceptor) reacts with water by accepting the Ht ion to form the new products. Could you write this equation? Take a minute to write out this equation below Enter your answer in the order: base + acid yields conjugate acid + conjugate base. To save time, you can skip the phase labels this time. Since you are working with a base (CI), the equilibrium constant is now Kb. chemPad ⤠Help [Ci] Greek ⼠Look at what happens when we multiply the Ka for the acid times the Ko for its conjugate base. Everything cancels except the terms for [O] and [H3O+]. The product of [OH-] X [H30*) is the water constant or Kw (1.0 x 10.1 Ka X Kb [HCI] [ci] = [OH-][H3O+] = Kw = 1.0 X 10-14 Using the water constant, we can calculate the K, for any conjugate base if we know the Ka for the acid (and vice versa).
Jarrod RobelLv2
19 Feb 2019