Chapter 7, Problem 80E

(a)

Interpretation Introduction

Interpretation: The pH, ${\text{OH}}^{\text{-}}$ and ${\text{H}}^{\text{+}}$ concentration present in 0.40 M solution of aniline 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 80E

• ${\text{[OH}}^{\text{-}}\text{] =}1.2\times {\text{10}}^{\text{-5}}\text{ M}$
• ${\text{[H}}^{\text{+}}\text{]= 8}{\text{.3×10}}^{\text{-12}}\text{M}$
• $\text{pH = 11}\text{.08}$

Explanation of Solution

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}$

The ${\text{K}}_{\text{b}}$ for aniline = $\text{3}\text{.8 }\times {\text{ 10}}^{\text{-10}}$ .

Concentration of aniline = 0.40 M

The equilibrium and ${\text{K}}_{\text{b}}$ expression for aniline are:

  $\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}\text{]}}\end{array}$

Make ICE table:

Concentration	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$	${\text{OH}}^{\text{-}}{}_{\text{(aq)}}$	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40 −x	x	x

Substitute the values to calculate ‘x’:

  ${\text{K}}_{\text{b}}$ for aniline = $\text{3}\text{.8 }\times {\text{ 10}}^{\text{-10}}$ .

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}\text{]}}\\ \text{3}\text{.8 }\times {\text{ 10}}^{\text{-10}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.40-x]}}(0.40>>x)\\ \text{3}\text{.8 }\times {\text{ 10}}^{\text{-10}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.40 ]}}\\ {\text{[x]}}^2=\text{3}\text{.8 }\times {\text{ 10}}^{\text{-10}}\times \text{0}\text{.40}\\ \text{[x]}=1.2\times {\text{10}}^{\text{-5}}{\text{ =[OH}}^{\text{-}}{\text{] =[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}\text{] }\end{array}$

  $\begin{array}{l}{\text{[OH}}^{\text{-}}\text{] =}1.2\times {\text{10}}^{\text{-5}}\text{ M}\\ {\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}\text{] = }1.2\times {\text{10}}^{\text{-5}}\text{ M}\end{array}$

Calculate [ ${\text{H}}^{\text{+}}$ ]:

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{] ×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{[OH}}^{\text{-}}\text{]}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[}1.2\times {\text{10}}^{\text{-5}}\text{ M ] }}\\ {\text{[H}}^{\text{+}}\text{]= 8}{\text{.3×10}}^{\text{-12}}\text{M}\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ \text{pH = - log [ 8}{\text{.3×10}}^{\text{-12}}\text{M]}\\ \text{pH = 11}\text{.08}\end{array}$

(b)

Interpretation Introduction

Interpretation: The pH, ${\text{OH}}^{\text{-}}$ and ${\text{H}}^{\text{+}}$ concentration present in 0.40 M solution of methylamine 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, acids 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 80E

• ${\text{[OH}}^{\text{-}}\text{] =}1.32\times {\text{10}}^{\text{-2}}$
• ${\text{[H}}^{\text{+}}\text{]= 7}{\text{.57×10}}^{\text{-13}}\text{M}$
• $\text{pH = 12}\text{.1}$

Explanation of Solution

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

The ${\text{K}}_{\text{b}}$ for methylamine= $\text{4}\text{.38 }\times {\text{ 10}}^{\text{-4}}$ .

Concentration of methylamine = 0.40 M

The equilibrium and ${\text{K}}_{\text{b}}$ expression for methylamine are:

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[CH}}_{\text{3}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}\text{]}}\end{array}$

Make ICE table:

Concentration	${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$	${\text{OH}}^{\text{-}}{}_{\text{(aq)}}$	${\text{CH}}_{\text{3}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40 −x	x	x

Substitute the values to calculate ‘x’:

  ${\text{K}}_{\text{b}}$ for methylamine = $\text{4}\text{.38 }\times {\text{ 10}}^{\text{-4}}$ .

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[CH}}_{\text{3}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}\text{]}}\\ \text{4}\text{.38 }\times {\text{ 10}}^{\text{-4}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.40-x]}}(0.40>>x)\\ \text{4}\text{.38 }\times {\text{ 10}}^{\text{-4}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.40 ]}}\\ {\text{[x]}}^2=\text{4}\text{.38 }\times {\text{ 10}}^{\text{-4}}\times \text{0}\text{.40}\\ \text{[x]}=1.32\times {\text{10}}^{\text{-2}}{\text{ =[OH}}^{\text{-}}{\text{] =[CH}}_{\text{3}}{\text{NH}}_{{}_{\text{3}}}^{\text{+}}\text{] }\end{array}$

Calculate [ ${\text{H}}^{\text{+}}$ ]:

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{] ×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{[OH}}^{\text{-}}\text{]}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[}1.32\times {\text{10}}^{\text{-2}}\text{ ] }}\\ {\text{[H}}^{\text{+}}\text{]= 7}{\text{.57×10}}^{\text{-13}}\text{M}\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ \text{pH = - log [ 7}{\text{.57×10}}^{\text{-13}}\text{M]}\\ \text{pH = 12}\text{.1}\end{array}$