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Question: Which out of the following is \[C{H_3}CO{O^ - }\] ion? A.Weak conjugate base B.Strong conjugate ...

Which out of the following is CH3COO−C{H_3}CO{O^ - } ion?
A.Weak conjugate base
B.Strong conjugate base
C.Weak conjugate acid
D.Strong conjugate acid

Explanation

Solution

According to the Bronsted-Lowry protonic concept, acids are chemical compounds that are proton donors whereas bases are chemical compounds that are proton acceptors. The chemical compounds that behave as an acid or a base can be a molecule, a cation, or an anion.

Complete answer:
Step 1
According to Bronsted-Lowry, an acid-base reaction can be represented as
AcidAcid ->BaseBase ++ H+{H^ + }
Thus, we can say that acid and a base are related to each other. Every acid gets converted into a base after releasing a proton, and every base changes into acid after accepting a proton. So, all the acids and bases exist in pairs which are known as conjugate pairs. Therefore, the conjugate is a term for a chemical compound that differs from the other only due to a difference in the number of protons.
Step 2
According to this concept, neutralization is the process of transfer of a proton from the acid to its conjugate base.
Step 3
The acidic strength of an acid is determined by its tendency to lose a proton. Greater is the tendency of the acid to lose a proton, the stronger is the acid.
Step 4
The basic strength of a base is determined by its tendency to gain a proton. Greater is the tendency of the base to gain a proton, stronger is the base.
Step 5
The conjugate of a strong acid is always a weak base whereas the conjugate of a weak acid is always a strong base.
Step 6
Acetic acid is a weak acid which dissociates in water as follows:
C{H_3}COOH$$$$ + $$$${H_2}O->C{H_3}CO{O^ - }$$$$ + $$$${H_3}{O^ + }
The equilibrium constant for acetic acid in the given reaction is Ka=[CH3COO−][H3O][CH3COOH]=1.8×10−5{K_a} = \dfrac{{[C{H_3}CO{O^ - }][{H_3}O]}}{{[C{H_3}COOH]}} = 1.8 \times {10^{ - 5}}.
The equilibrium constant for the acetate ion would be Keq=1Ka=[CH3COOH][CH3COO−][H3O]=5.5×104{K_{eq}} = \dfrac{1}{{{K_a}}} = \dfrac{{[C{H_3}COOH]}}{{[C{H_3}CO{O^ - }][{H_3}O]}} = 5.5 \times {10^4}.
So, we can find that the conjugate base of a weak acid has a higher dissociation constant value so is a strong base.
Thus, we can see that acetate ion CH3COO−C{H_3}CO{O^ - }, the conjugate base of the weak acid, acetic acid CH3COOHC{H_3}COOH is a strong base.

Hence, the correct option for the given question is option b).

Note: As per the Arrhenius concept an acid is a substance that gives H+{H^ + } ion in aqueous solution while a base furnishes OH−O{H^ - } ions in aqueous solution.