This is a problem of completing the net ionic equation. In this solution... why

Question

This is a problem of completing the net ionic equation.

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

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.

Explanation / Answer

The pKa 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 pKb. Again, the question asks for acidity; hence, we use pKa of CH3CH2OH.

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

pKa                                               15.7                                      15.9

The higher the pKa value of a weak acid, the weaker is the acid. Since CH3CH2OH has a higher pKa, 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.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.