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What is the relative acidity of the following molecules: acetic acid (CH3COOH), ethanol (CH3CH2OH), and water (H2O)? Provide an explanation for your answer based on the properties of each molecule.

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The relative acidity of the molecules can be determined by comparing their acid dissociation constants  Ka  or pKa values. The lower the pKa value, the stronger the acid. Here are the pKa values for the given molecules:1. Acetic acid  CH3COOH : pKa  4.762. Ethanol  CH3CH2OH : pKa  15.93. Water  H2O : pKa  15.7Based on these values, the relative acidity of the molecules is as follows:Acetic acid > Water > EthanolExplanation:1. Acetic acid  CH3COOH  is the most acidic among the three molecules due to the presence of the carboxylic acid functional group  COOH . When acetic acid donates a proton  H+ , it forms the acetate ion  CH3COO-  which is resonance-stabilized. This stabilization of the conjugate base makes it easier for acetic acid to donate a proton, making it a stronger acid compared to water and ethanol.2. Water  H2O  is more acidic than ethanol but less acidic than acetic acid. Water can act as both an acid and a base  amphoteric . When it donates a proton, it forms the hydroxide ion  OH- . The oxygen atom in water has two lone pairs of electrons, which can stabilize the negative charge on the hydroxide ion to some extent.3. Ethanol  CH3CH2OH  is the least acidic among the three molecules. The hydroxyl group  OH  in ethanol is not as acidic as the carboxylic acid group in acetic acid or the oxygen atom in water. When ethanol donates a proton, it forms the ethoxide ion  CH3CH2O- , which is not stabilized by resonance or any other significant stabilizing factor. As a result, ethanol is less likely to donate a proton and is a weaker acid compared to acetic acid and water.

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