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Consider the reaction of acetic acid in water

CH3CO2H(aq)+H2O(l)CH3CO2-(aq)-+H3O+(aq),whereKa=1.8×10-5.

a. Which two bases are competing for the proton?

b. Which is the stronger base?

c. In light of your answer to part b, why do we classify the acetate ion CH3CO2-asa weak base?

Use an appropriate reaction to justify your answer.

Short Answer

Expert verified

If an acid has a higher Kavalue, it will be stronger as compared to another with a low value of Ka. Strong acids give a weak conjugate base.

  1. H2OandCH3COO-
  2. CH3COO-is the weak base as compared to H2O
  3. The stronger the acid, the weaker its conjugate base will be.

Step by step solution

01

Reaction of Acetic Acid with Water

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

02

Bases Competing for the Proton

The two bases competing for the proton in the given reaction are H2O and conjugate base of acetic acid, i.e.,CH3COO- , because two are the species acting as the base in the given reaction.

03

Stronger Base out of H2O and CH3COO-  

Acetate ion, i.e., CH3COO- is the weak base as compared to H2O . This means acetate ion has a lesser tendency to accept a proton.

04

Reason for Acetate ion to be Weaker as Compared to  H2O

Ka value for acetic acid is given1.8×10-5, while for water it is 1.0×10-14so, acetic acid is stronger acid as compared to water (this can be seen from the given equation as well).

The stronger the acid, the weaker its conjugate base will be.

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Most popular questions from this chapter

Hemoglobin (abbreviated Hb) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for O2 molecules. The oxygen binding is pH-dependent. The relevant equilibrium reaction is

HbH44+(aq)+O2(g)Hb(O2)4(aq)+4H+(aq)

Use Le Châtelier’s principle to answer the following.

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d. Order the following from strongest to weakest base:H2O, A, Cl. Explain your sequence.

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a.OH-=1.5M

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d.OH-=7.3×10-4M

Also, calculate the pH and pOH of each of these solutions.

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