Determining the Equilibrium Constant from Ph Values of Changing Concentrations of Ethanoic Acid and Water.

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Determining the equilibrium constant from pH values of changing concentrations of ethanoic acid


Research question: How will altering the concentrations of ethanoic acid affect the pH value, and, in-turn, the equilibrium constant?

Background information: When weak acids react, the reaction typically does not go to completion. Rather, the system goes to an intermediate state in which the rates of the forward and reverse reactions are equal. Such a system is said to be in chemical equilibrium. When equilibrium is reached, the reactants and the products have concentrations which do not change with time. When in equilibrium at a particular temperature, a reaction mixture obeys the Law of Chemical Equilibrium, which imposes a condition on the concentrations of reactants and products. This condition is expressed in the equilibrium constant Kc for the reaction. In this experiment, I will study the equilibrium properties of the reaction between ethanoic acid, otherwise known as acetic acid (CH3COOH) and water (H2O):

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

When solutions containing ethanoic acid and water are mixed, they react to some extent, forming CH3COO- and H3O+. As a result of the reaction, the equilibrium amounts of CH3COOH and H2O will be less; for every mole of CH3COO- formed, one mole of CH3COOH and one mole of H2O will react. The equilibrium constant expression Kc for Reaction 1 is:

Kc = [CH3COOH] / [H3O+][CH3COO-] The value of Kc is relatively constant at a given temperature. This means that mixtures containing CH3COOH and H2O will come to equilibrium with the same value of Kc, no matter what initial amounts of CH3COOH and H2O were used. The purpose of this experiment will be to find Kc for this reaction for several concentrations that have been made up in different ways, and to show that Kc indeed has the same…...

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