Chapter 7, Problem 109E

(a)

Interpretation Introduction

Interpretation: The pH of 0.12 M solution of ${\text{KNO}}_{\text{2}}$ 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.

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

An acid is the substance that gives ${\text{H}}^{+}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

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

Answer to Problem 109E

  $\text{pH = 8}\text{.24}$

Explanation of Solution

[ ${\text{KNO}}_{\text{2}}$ ] = 0.12 M

  ${\text{KNO}}_{\text{2}}$  is a basic salt. The ionization of ${\text{KNO}}_{\text{2}}$ forms respective ions:

  ${\text{KNO}}_{\text{2}}\underset{}{\to }{\text{K}}_{{}_{}}^{+}+{\text{NO}}_{\text{2}}{}^{\text{-}}$

Here ${\text{NO}}_{\text{2}}{}^{\text{-}}$ can act as string conjugate base of weak acid ${\text{HNO}}_{\text{2}}$ therefore the salt is slightly basic in nature as ${\text{K}}_{{}_{}}^{+}$ is neither acidic nor basic.

The reaction of ${\text{NO}}_{\text{2}}{}^{\text{-}}$ ions with water can be written as:

  ${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{HNO}}_{\text{2}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}$

Initial concentration of ${\text{NO}}_{\text{2}}{}^{\text{-}}$ = 0.12 M

  ${\text{K}}_{\text{b}}$ for ${\text{NO}}_{\text{2}}{}^{\text{-}}$ = $\text{2}\text{.5}\times {\text{ 10}}^{\text{-11}}$

Make ICE table:

Concentration	${\text{NO}}_{\text{2}}{}^{\text{-}}$	${\text{OH}}^{\text{-}}$	${\text{HNO}}_{\text{2}}$
Initial	0.12	0	0
Change	-x	x	x
Equilibrium	0.12−x	x	x

Calculate ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ :

  $\begin{array}{l}{\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{HNO}}_{\text{2}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}{\text{][HNO}}_{\text{2}}\text{]}}{{\text{[NO}}_{\text{2}}{}^{\text{-}}\text{]}}\\ \text{2}\text{.5}\times {\text{ 10}}^{\text{-11}}\text{ = }\frac{\text{[x] [x]}}{\text{[0}\text{.12-x]}}\text{(0}\text{.12>>x)}\\ \text{2}\text{.5}\times {\text{ 10}}^{\text{-11}}\text{ = }\frac{\text{[x] [x]}}{\text{[0}\text{.12]}}\\ {\text{[x]}}^{\text{2}}\text{ =2}\text{.5}\times {\text{ 10}}^{\text{-11}}\text{ ×0}\text{.12}\\ \text{[x]= 1}\text{.73}\times {\text{10}}^{\text{-6}}{\text{ =[OH}}^{-}\text{] }\end{array}$

Calculate pOH and pH:

  $\begin{array}{l}{\text{pOH = -log[OH}}^{-}\text{] }\\ \text{pOH = -log[1}\text{.73}\times {\text{10}}^{\text{-6}}\text{ ]}\\ \text{pOH = 5}\text{.76}\\ \text{pH = 14 - pOH}\text{}\text{ = 14- 5}\text{.76 = 8}\text{.24}\end{array}$

(b)

Interpretation Introduction

Interpretation: The pH of 0.45 M solution of $\text{NaOCl}$ 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.

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

An acid is the substance that gives ${\text{H}}^{+}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

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

Answer to Problem 109E

  $\text{pH = 10}\text{.5}$

Explanation of Solution

[ $\text{NaOCl}$ ] = 0.45 M

  $\text{NaOCl}$  is a basic salt. The ionization of $\text{NaOCl}$ forms respective ions:

  $\text{NaOCl}\underset{}{\to }{\text{Na}}_{{}_{}}^{+}+{\text{OCl}}^{\text{-}}$

Here ${\text{OCl}}^{\text{-}}$ can act as strong conjugate base of weak acid $\text{HClO}$ therefore the salt is slightly basic in nature as ${\text{Na}}_{{}_{}}^{+}$ is neither acidic nor basic.

The reaction of ${\text{OCl}}^{\text{-}}$ ions with water can be written as:

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

Initial concentration of ${\text{OCl}}^{\text{-}}$ = 0.45 M

  ${\text{K}}_{\text{b}}$ for ${\text{OCl}}^{\text{-}}$ = $\text{2}\text{.8}\times {\text{ 10}}^{\text{-7}}$

Make ICE table:

Concentration	${\text{OCl}}^{\text{-}}$	${\text{OH}}^{\text{-}}$	$\text{HClO}$
Initial	0.45	0	0
Change	-x	x	x
Equilibrium	0.45−x	x	x

Calculate ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ :

  $\begin{array}{l}{\text{OCl}}^{\text{-}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{HClO}}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[OH}}^{\text{-}}\text{][HClO]}}{{\text{[OCl}}^{\text{-}}\text{]}}\\ \text{2}\text{.8}\times {\text{ 10}}^{\text{-7}}\text{ = }\frac{\text{[x] [x]}}{\text{[0}\text{.45-x]}}\text{(0}\text{.45>>x)}\\ \text{2}\text{.8}\times {\text{ 10}}^{\text{-7}}\text{ = }\frac{\text{[x] [x]}}{\text{[0}\text{.45]}}\\ {\text{[x]}}^{\text{2}}\text{ =2}\text{.8}\times {\text{ 10}}^{\text{-7}}\text{ ×0}\text{.45}\\ \text{[x]= 3}\text{.55}\times {\text{10}}^{\text{-4}}{\text{ =[OH}}^{-}\text{] }\end{array}$

Calculate pOH and pH:

  $\begin{array}{l}{\text{pOH = -log[OH}}^{-}\text{] }\\ \text{pOH = -log[3}\text{.55}\times {\text{10}}^{\text{-4}}\text{ ]}\\ \text{pOH = 3}\text{.45}\\ \text{pH = 14 - pOH}\text{}\text{ = 14- 3}\text{.45 = 10}\text{.5}\end{array}$

(c)

Interpretation Introduction

Interpretation: The pH of 0.40 M solution of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ 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.

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

An acid is the substance that gives ${\text{H}}^{+}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

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

Answer to Problem 109E

  $\text{pH = 4}\text{.8}$

Explanation of Solution

[ ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ ] = 0.40 M

  ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$  is aacidic salt. The ionization of ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ forms respective ions:

  ${\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\underset{}{\to }{\text{NH}}_{\text{4}}{}_{{}_{}}^{+}+{\text{ClO}}_{\text{4}}{}^{\text{-}}$

Here ${\text{ClO}}_{\text{4}}{}^{\text{-}}$ can act as weak conjugate base of strong acid ${\text{HClO}}_{\text{4}}$ therefore the salt is slightly acidic in nature as ${\text{NH}}_{\text{4}}{}_{{}_{}}^{+}$ is strong conjugate acid of weak base NH₃.

The reaction of ${\text{NH}}_{\text{4}}{}_{{}_{}}^{+}$ ions with water can be written as:

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

Initial concentration of ${\text{OCl}}^{\text{-}}$ = 0.40 M

  ${\text{K}}_{\text{a}}$ for ${\text{NH}}_{\text{4}}{}_{{}_{}}^{+}$ = $\text{5}\text{.6}\times {\text{ 10}}^{\text{-10}}$

Make ICE table:

Concentration	${\text{NH}}_{\text{4}}{}_{{}_{}}^{+}$	${\text{NH}}_{\text{3}}$	${\text{H}}_{\text{3}}{\text{O}}^{+}$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40−x	x	x

Calculate ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ :

  $\begin{array}{l}{\text{NH}}_{\text{4}}{{}_{{}_{}}^{+}}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{NH}}_{\text{3}}{}_{\text{(aq)}}{\text{+ H}}_{\text{3}}{\text{O}}^{+}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{+}\text{]}}{{\text{[NH}}_{\text{4}}{}_{{}_{}}^{+}\text{]}}\\ \text{5}\text{.6}\times {\text{ 10}}^{\text{-10}}\text{ = }\frac{\text{[x] [x]}}{\text{[0}\text{.40-x]}}\text{(0}\text{.40>>x)}\\ \text{5}\text{.6}\times {\text{ 10}}^{\text{-10}}\text{ = }\frac{\text{[x] [x]}}{\text{[0}\text{.40]}}\\ {\text{[x]}}^{\text{2}}\text{ =5}\text{.6}\times {\text{ 10}}^{\text{-10}}\text{×0}\text{.40}\\ \text{[x]= 1}\text{.49}\times {\text{10}}^{\text{-5}}{\text{ =[H}}_{\text{3}}{\text{O}}^{+}\text{] }\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pH = -log[H}}_{\text{3}}{\text{O}}^{+}\text{] }\\ \text{pH = -log[1}\text{.49}\times {\text{10}}^{\text{-5}}]\\ \text{pH = 4}\text{.8}\end{array}$