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Determine the position of the equilibrium for the reaction of acetic acid with NaOH and draw the species that predominate at equilibrium.

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FInal answer:

The equilibrium of the acetic acid and NaOH reaction will favor the formation of acetate ions and water, resulting in a basic solution with a pH greater than 7.00. A step-by-step process involving balance equations, tabulation of concentrations, and calculation of the equilibrium constant is used to determine the equilibrium details.

Step-by-step explanation:

The question addresses the equilibrium position of the reaction between acetic acid and NaOH, and which species predominate at equilibrium. To determine this, we follow these steps:

Calculate the initial amounts of acetic acid (CH₃CO₂H) and hydroxide ions (OH-) present before the reaction.

Write the balanced chemical equation for the reaction: CH₃CO₂H(aq) + OH-(aq) → CH₃CO₂-(aq) + H₂O(l).

Use a table to show the initial concentrations, changes, and final concentrations of the reactants and products.

If there is excess acetate (CH₃CO₂-) after the reaction, write the equation for its reaction with water to form acetic acid and OH-.

Calculate the equilibrium constant using the relationship between water dissociation constant (Kw), the acid dissociation constant (Ka) and the hydroxide ion concentration [OH-].

From [OH-], calculate the pH of the solution.

At equilibrium, the reaction will favor the formation of the acetate ion (CH₃CO₂-) and water because NaOH is a strong base that will completely dissociate and react with acetic acid, which is a weak acid. Thus, at the equivalence point, the solution will be basic, with a pH greater than 7.00.

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