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What is the effect of increasing the pH from 3 to 5 on the equilibrium position of the reaction between acetic acid (CH3COOH) and sodium acetate (CH3COONa)?

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The reaction between acetic acid  CH3COOH  and sodium acetate  CH3COONa  can be represented as follows:CH3COOH + OH-  CH3COO- + H2OThis is a typical acid-base reaction, where acetic acid donates a proton  H+  to the hydroxide ion  OH- . The equilibrium constant for this reaction is given by the expression:K = [CH3COO-][H2O] / [CH3COOH][OH-]When the pH of the solution is increased from 3 to 5, the concentration of H+ ions decreases, and the concentration of OH- ions increases. According to Le Chatelier's principle, the equilibrium will shift to counteract this change. In this case, the equilibrium will shift to the right, favoring the formation of CH3COO-  acetate ion  and H2O.As a result, the concentration of acetic acid  CH3COOH  will decrease, and the concentration of sodium acetate  CH3COONa  will increase. This shift in equilibrium will help to buffer the solution, maintaining a relatively constant pH in the presence of small amounts of added acid or base. This is the basis for the buffering action of the acetic acid-sodium acetate system.

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