Chapter 7, Problem 85E

(a)

Interpretation Introduction

Interpretation: The percent ionization of 0.10 M ${\text{NH}}_{\text{3}}$ solution 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:

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

The percent ionization can be calculate with the help of concentrations value:

  $\text{Percent ionization = }\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\text{×100 or}\frac{{\text{[OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\text{×100}$

Answer to Problem 85E

  $\text{Percent ionization=1}\text{.3\%}$

Explanation of Solution

The equilibrium equation can be written as:

  ${\text{NH}}_{\text{3}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{NH}}_{\text{4}}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}$

Concentration of ${\text{NH}}_{\text{3}}$ = 0.10 M

The equilibrium and ${\text{K}}_{\text{b}}$ expression for ${\text{NH}}_{\text{3}}$ are:

  $\begin{array}{l}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{NH}}_{\text{4}}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{4}}^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[NH}}_{\text{3}}{}_{\text{(aq)}}\text{]}}\end{array}$

Make ICE table:

Concentration	${\text{NH}}_{\text{3}}$	${\text{OH}}^{\text{-}}{}_{\text{(aq)}}$	${\text{NH}}_{\text{4}}^{\text{+}}$
Initial	0.10	0	0
Change	-x	x	x
Equilibrium	0.10 −x	x	x

Substitute the values to calculate ‘x’:

  ${\text{K}}_{\text{b}}$ for ${\text{NH}}_{\text{3}}$ = $\text{1}\text{.8 }\times {\text{ 10}}^{\text{-5}}$ .

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{4}}^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[NH}}_{\text{3}}{}_{\text{(aq)}}\text{]}}\\ \text{1}\text{.8 }\times {\text{ 10}}^{\text{-5}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.10-x]}}(0.10>>x)\\ \text{1}\text{.8 }\times {\text{ 10}}^{\text{-5}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.10 ]}}\\ {\text{[x]}}^2=\text{1}\text{.8 }\times {\text{ 10}}^{\text{-5}}\times \text{0}\text{.10}\\ \text{[x]}=1.3\times {\text{10}}^{\text{-3}}{\text{ =[OH}}^{\text{-}}\text{]}\end{array}$

Calculate the percent ionization:

  $\begin{array}{l}\text{Percent ionization =}\frac{{\text{[OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\text{×100}\\ \text{Percent ionization =}\frac{\text{[1}{\text{.3×10}}^{\text{-3}}\text{]}}{\text{[0}\text{.10]}}\text{×100=1}\text{.3\%}\end{array}$

(b)

Interpretation Introduction

Interpretation: The percent ionization of 0.010 M ${\text{NH}}_{\text{3}}$ solution 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:

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

The percent ionization can be calculate with the help of concentrations value:

  $\text{Percent ionization = }\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\text{×100 or}\frac{{\text{[OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\text{×100}$

Answer to Problem 85E

  $\text{Percent ionization=4}\text{.2\%}$

Explanation of Solution

The equilibrium equation can be written as:

  ${\text{NH}}_{\text{3}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{NH}}_{\text{4}}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}$

Concentration of ${\text{NH}}_{\text{3}}$ = 0.010 M

The equilibrium and ${\text{K}}_{\text{b}}$ expression for ${\text{NH}}_{\text{3}}$ are:

  $\begin{array}{l}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{NH}}_{\text{4}}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{4}}^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[NH}}_{\text{3}}{}_{\text{(aq)}}\text{]}}\end{array}$

Make ICE table:

Concentration	${\text{NH}}_{\text{3}}$	${\text{OH}}^{\text{-}}{}_{\text{(aq)}}$	${\text{NH}}_{\text{4}}^{\text{+}}$
Initial	0.010	0	0
Change	-x	x	x
Equilibrium	0.010 −x	x	x

Substitute the values to calculate ‘x’:

  ${\text{K}}_{\text{b}}$ for ${\text{NH}}_{\text{3}}$ = $\text{1}\text{.8 }\times {\text{ 10}}^{\text{-5}}$ .

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{4}}^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[NH}}_{\text{3}}{}_{\text{(aq)}}\text{]}}\\ \text{1}\text{.8 }\times {\text{ 10}}^{\text{-5}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.010-x]}}(0.010>>x)\\ \text{1}\text{.8 }\times {\text{ 10}}^{\text{-5}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.010 ]}}\\ {\text{[x]}}^2=\text{1}\text{.8 }\times {\text{ 10}}^{\text{-5}}\times \text{0}\text{.010}\\ \text{[x]}=4.2\times {\text{10}}^{\text{-4}}{\text{ =[OH}}^{\text{-}}\text{]}\end{array}$

Calculate the percent ionization:

  $\begin{array}{l}\text{Percent ionization =}\frac{{\text{[OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\text{×100}\\ \text{Percent ionization =}\frac{\text{[}4.2\times {\text{10}}^{\text{-4}}\text{]}}{\text{[0}\text{.010]}}\text{×100=4}\text{.2\%}\end{array}$

(c)

Interpretation Introduction

Interpretation: The percent ionization of 0.10 M ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ solution 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:

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

The percent ionization can be calculate with the help of concentrations value:

  $\text{Percent ionization = }\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\text{×100 or}\frac{{\text{[OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\text{×100}$

Answer to Problem 85E

  $\text{Percent ionization=6}\text{.6\%}$

Explanation of Solution

The equilibrium equation can be written as:

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

Concentration of ${\text{NH}}_{\text{3}}$ = 0.10 M

The equilibrium and ${\text{K}}_{\text{b}}$ expression for ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ are:

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l}}{}_{\text{)}}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{NH}}_3^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{NH}}_3^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[CH}}_{\text{3}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}\text{]}}\end{array}$

Make ICE table:

Concentration	${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$	${\text{OH}}^{\text{-}}{}_{\text{(aq)}}$	${\text{CH}}_{\text{3}}{\text{NH}}_3^{\text{+}}$
Initial	0.10	0	0
Change	-x	x	x
Equilibrium	0.10 −x	x	x

Substitute the values to calculate ‘x’:

  ${\text{K}}_{\text{b}}$ for ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ = $\text{4}\text{.38 }\times {\text{ 10}}^{\text{-4}}$ .

  $\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{NH}}_3^{\text{+}}{}_{\text{(aq)}}{\text{] [OH}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[CH}}_{\text{3}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}\text{]}}\\ \text{4}\text{.38 }\times {\text{ 10}}^{\text{-4}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.10-x]}}(0.10>>x)\\ \text{4}\text{.38 }\times {\text{ 10}}^{\text{-4}}\text{=}\frac{\text{[x] [x]}}{\text{[0}\text{.10 ]}}\\ {\text{[x]}}^2=\text{4}\text{.38 }\times {\text{ 10}}^{\text{-4}}\times \text{0}\text{.10}\\ \text{[x]}=6.6\times {\text{10}}^{\text{-3}}{\text{ =[OH}}^{\text{-}}\text{]}\end{array}$

Calculate the percent ionization:

  $\begin{array}{l}\text{Percent ionization =}\frac{{\text{[OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\text{×100}\\ \text{Percent ionization =}\frac{\text{[6}{\text{.6×10}}^{\text{-3}}\text{]}}{\text{[0}\text{.10]}}\text{×100=6}\text{.6\%}\end{array}$