Question

This is a problem of completing the net ionic equation.

The provided acid base reaction is as follows: Compare the pK, values of ethanol and water (Refer to Table 2.2 in the text book). For CH,CH,OH pK 15.9 and for water pK,-15.7 Ethanol has a higher pK, value than water. So conjugate base of ethanol, ethoxide ion must be a base and water with lower pK, should be an acid.

In this solution... why used the pKa of CH₃CH₂OH instead of the pKa of CH₃CH₂O^- ???

Water didn't use the pKa of H₃O⁺ .

P.S. Please answer with the typing not handwrithing as possible

Answer 1

The pK_a is the acid ionization constant of an acid. We are comparing the acidity of ethanol, CH₃CH₂OH and water, H₂O.

To answer the second question first, we are using H₂O as an acid here. H₂O can act as both acid and base. When H₂O is the weak acid, the conjugate base is OH^-. When H₂O is base, the conjugate acid is H₃O⁺. Since we are comparing acidity, we use H₂O as an acid.

Now, CH₃CH₂O^- is a base and in order to compare basicity, we need to work with pK_b. Again, the question asks for acidity; hence, we use pK_a of CH₃CH₂OH.

                        CH₃CH₂O^- (aq) + H₂O (l) <========> CH₃CH₂OH (aq) + OH^- (aq)

pK_a                                               15.7                                      15.9

The higher the pK_a value of a weak acid, the weaker is the acid. Since CH₃CH₂OH has a higher pK_a, hence, CH₃CH₂OH is a weaker acid as compared to H₂O. An acid-base reaction goes in the direction to produce the weaker acid; hence, the above reaction proceeds to the right to produce CH₃CH₂OH.