Chapter 7, Problem 71E

(a)

Interpretation Introduction

Interpretation: Thestronger base from ${\text{ClO}}_{\text{4}}{}^{\text{- }}{\text{or C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ needs to be determined.

Concept Introduction: A base is the substance that gives ${\text{OH}}^{\text{-}}$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

  ${\text{BOH}}_{\text{(aq)}}\rightleftharpoons {\text{OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+B}}^{\text{-}}{}_{\text{(aq)}}$

The base dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak base in solution. For the given weak base BOH, it can be written as:

  $\begin{array}{l}{\text{BOH}}_{\text{(aq)}}\rightleftharpoons {\text{OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+ B}}^{+}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}{\text{] [B}}^{+}\text{]}}{\text{[BOH]}}\end{array}$

Answer to Problem 71E

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ must be stronger base than ${\text{ClO}}_{\text{4}}{}^{\text{- }}$ .

Explanation of Solution

According to Bronsted-Lowery theory, the substance which can accept H+ ion show basic nature. Here ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ shows basic nature due to presence of lone pair on N atom.

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_3^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{b}}\text{=1}{\text{.7×10}}^{\text{-9}}$

  ${\text{ClO}}_{\text{4}}{}^{\text{- }}$ also a weak conjugate base of strong acid ${\text{HClO}}_{\text{4}}$

  ${\text{ClO}}_{\text{4}}{{}^{\text{- }}}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{HClO}}_{\text{4}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}$

Hence ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ must be stronger base than ${\text{ClO}}_{\text{4}}{}^{\text{- }}$ .

(b)

Interpretation Introduction

Interpretation: The stronger base between ${\text{H}}_{\text{2}}{\text{O or C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ needs to be determined.

Concept Introduction: A base is the substance that gives ${\text{OH}}^{\text{-}}$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

  ${\text{BOH}}_{\text{(aq)}}\rightleftharpoons {\text{OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+B}}^{\text{-}}{}_{\text{(aq)}}$

The base dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak base in solution. For the given weak base BOH, it can be written as:

  $\begin{array}{l}{\text{BOH}}_{\text{(aq)}}\rightleftharpoons {\text{OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+ B}}^{+}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}{\text{] [B}}^{+}\text{]}}{\text{[BOH]}}\end{array}$

Answer to Problem 71E

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ must be stronger base than ${\text{H}}_{\text{2}}\text{O}$ .

Explanation of Solution

According to Bronsted-Lowery theory, the substance which can accept H+ ion show basic nature. Here ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ shows basic nature due to presence of lone pair on N atom.

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_3^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{b}}\text{=1}{\text{.7×10}}^{\text{-9}}$

Water shows amphoteric nature due to tendency of it to act as an acid as well as base.

  ${\text{2H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}_{}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}$

Hence ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ must be stronger base than ${\text{H}}_{\text{2}}\text{O}$ .

(c)

Interpretation Introduction

Interpretation: The stronger base between ${\text{OH}}^{\text{-}}{\text{ or C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ needs to be determined.

Concept Introduction: A base is the substance that gives ${\text{OH}}^{\text{-}}$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

  ${\text{BOH}}_{\text{(aq)}}\rightleftharpoons {\text{OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+B}}^{\text{-}}{}_{\text{(aq)}}$

The base dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak base in solution. For the given weak base BOH, it can be written as:

  $\begin{array}{l}{\text{BOH}}_{\text{(aq)}}\rightleftharpoons {\text{OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+ B}}^{+}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}{\text{] [B}}^{+}\text{]}}{\text{[BOH]}}\end{array}$

Answer to Problem 71E

  ${\text{OH}}^{\text{-}}$ must be stronger base than ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ .

Explanation of Solution

According to Bronsted-Lowery theory, the substance which can accept H+ ion show basic nature. Here ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ shows basic nature due to presence of lone pair on N atom.

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_3^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{b}}\text{=1}{\text{.7×10}}^{\text{-9}}$

  ${\text{OH}}^{\text{-}}$ is stronger base as it is conjugate base of water molecule which is a weak acid.

  ${\text{2H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}_{}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}$

Hence ${\text{OH}}^{\text{-}}$ must be stronger base than ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ .

(d)

Interpretation Introduction

Interpretation: The stronger base between ${\text{CH}}_3{\text{NH}}_{\text{2}}{\text{ or C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ needs to be determined.

Concept Introduction: A base is the substance that gives ${\text{OH}}^{\text{-}}$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

  ${\text{BOH}}_{\text{(aq)}}\rightleftharpoons {\text{OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+B}}^{\text{-}}{}_{\text{(aq)}}$

The base dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak base in solution. For the given weak base BOH, it can be written as:

  $\begin{array}{l}{\text{BOH}}_{\text{(aq)}}\rightleftharpoons {\text{OH}}^{\text{-}}{}_{\text{(aq)}}{\text{+ B}}^{+}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}{\text{] [B}}^{+}\text{]}}{\text{[BOH]}}\end{array}$

Answer to Problem 71E

  ${\text{CH}}_3{\text{NH}}_{\text{2}}$ must be stronger base than ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ .

Explanation of Solution

According to Bronsted-Lowery theory, the substance which can accept H+ ion show basic nature. Here ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ shows basic nature due to presence of lone pair on N atom.

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_3^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{b}}\text{=1}{\text{.7×10}}^{\text{-9}}$

  ${\text{CH}}_3{\text{NH}}_{\text{2}}$ shows basic nature due to presence of lone pair on N atom. ${\text{CH}}_3{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{CH}}_3{\text{NH}}_3^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{b}}\text{=4}{\text{.38×10}}^{\text{-4}}$

Hence ${\text{CH}}_3{\text{NH}}_{\text{2}}$ must be stronger base than ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ .