Chapter 7, Problem 41E

(a)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ and ${\text{[OH}}^{\text{-}}\text{]}$ for the solution with pH 7.40 needs to be determined and the solution needs to be identified as neutral, acidic and basic at 25 °C.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

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

On the contrary, a weak acid ionized partially and reaches to equilibrium.

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

The pH value depends on the concentration of H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution. The mathematical relation between pH and H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions can be written as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}]$

Answer to Problem 41E

• ${\text{[OH}}^{\text{-}}\text{]= }2.5\times {10}^{-7}\text{ M}$
• ${\text{[H}}^{\text{+}}\text{] =4}{\text{.0 ×10}}^{\text{-8}}\text{M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

Explanation of Solution

Given:

pH= 7.40

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

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-7}\text{.40}}\\ {\text{[H}}^{\text{+}}\text{] =4}{\text{.0×10}}^{\text{-8}}\text{M}\end{array}$

Calculate ${\text{[OH}}^{\text{-}}\text{]}$

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[4}{\text{.0×10}}^{\text{-8}}\text{M]}}\\ {\text{[OH}}^{\text{-}}\text{]= }2.5\times {10}^{-7}\text{ M}\end{array}$

• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

(b)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ and ${\text{[OH}}^{\text{-}}\text{]}$ for the solution with pH 15.3 needs to be determined and solution needs to be identified as neutral, acidic and basic at 25 °C.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

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

On the contrary, a weak acid ionized partially and reaches to equilibrium.

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

The pH value depends on the concentration of H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution. The mathematical relation between pH and H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions can be written as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}]$

Answer to Problem 41E

• ${\text{[OH}}^{\text{-}}\text{]= }20\text{ M}$
• ${\text{[H}}^{\text{+}}\text{] =5}{\text{.0×10}}^{\text{-16}}\text{M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

Explanation of Solution

Given:

pH=15.3

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

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-15}\text{.3}}\\ {\text{[H}}^{\text{+}}\text{] =5}{\text{.0×10}}^{\text{-16}}\text{M}\end{array}$

Calculate ${\text{[OH}}^{\text{-}}\text{]}$

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[5}{\text{.0×10}}^{\text{-16}}\text{M]}}\\ {\text{[OH}}^{\text{-}}\text{]= }20\text{ M}\end{array}$

• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

(c)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ and ${\text{[OH}}^{\text{-}}\text{]}$ for the solution with pH -1.0 needs to be determined and the solution needs to be identified as neutral, acidic and basic at 25 °C.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

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

On the contrary, a weak acid ionized partially and reaches to equilibrium.

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

The pH value depends on the concentration of H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution. The mathematical relation between pH and H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions can be written as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}]$

Answer to Problem 41E

• ${\text{[OH}}^{\text{-}}\text{]= 1}{\text{.0×10}}^{\text{-15}}\text{ M}$
• ${\text{[H}}^{\text{+}}\text{] = 10 M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$, the solution must be acidic.

Explanation of Solution

Given:

pH=-1.0

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

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-(-1}\text{.0)}}\\ {\text{[H}}^{\text{+}}\text{] =10 M}\end{array}$

Calculate ${\text{[OH}}^{\text{-}}\text{]}$

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[10 M]}}\\ {\text{[OH}}^{\text{-}}\text{]= 1}{\text{.0×10}}^{\text{-15}}\text{ M}\end{array}$

• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$, the solution must be acidic.

(d)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ and ${\text{[OH}}^{\text{-}}\text{]}$ for the solution with pH 3.20 needs to be determined and the solution needs to be identified as neutral, acidic and basic at 25 °C.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

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

On the contrary, a weak acid ionized partially and reaches to equilibrium.

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

The pH value depends on the concentration of H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution. The mathematical relation between pH and H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions can be written as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}]$

Answer to Problem 41E

• ${\text{[OH}}^{\text{-}}\text{]= 1}{\text{.6×10}}^{\text{-11}}\text{ M}$
• ${\text{[H}}^{\text{+}}\text{] =6}\text{.3}\times {\text{10}}^{\text{-4 }}\text{M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$, the solution must be acidic.

Explanation of Solution

Given:

pH=3.20

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

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-3}\text{.20}}\\ {\text{[H}}^{\text{+}}\text{] =6}\text{.3}\times {\text{10}}^{\text{-4 }}\text{M}\end{array}$

Calculate ${\text{[OH}}^{\text{-}}\text{]}$

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[6}\text{.3}\times {\text{10}}^{\text{-4 }}\text{M]}}\\ {\text{[OH}}^{\text{-}}\text{]= 1}{\text{.6×10}}^{\text{-11}}\text{ M}\end{array}$

• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$, the solution must be acidic.

(e)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ and ${\text{[OH}}^{\text{-}}\text{]}$ for the solution with pOH 5.0 needs to be determined and the solution needs to be identified as neutral, acidic and basic at 25 °C.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

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

On the contrary, a weak acid ionized partially and reaches to equilibrium.

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

The pH value depends on the concentration of H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution. The mathematical relation between pH and H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions can be written as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}]$

Answer to Problem 41E

• ${\text{[OH}}^{\text{-}}\text{]= 1}{\text{.0×10}}^{\text{-5}}\text{ M}$
• ${\text{[H}}^{\text{+}}\text{] = 1}\text{.0 }\times {\text{10}}^{\text{-9 }}\text{M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

Explanation of Solution

Given:

pOH=5.0

Calculate pH:

pH = 14 − pOH = 14 − 5.0 = 9.0

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

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-9}}\\ {\text{[H}}^{\text{+}}\text{] = 1}\text{.0 }\times {\text{10}}^{\text{-9 }}\text{M}\end{array}$

Calculate ${\text{[OH}}^{\text{-}}\text{]}$

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[1}\text{.0}\times {\text{10}}^{\text{-9 }}\text{M]}}\\ {\text{[OH}}^{\text{-}}\text{]= 1}{\text{.0×10}}^{\text{-5}}\text{ M}\end{array}$

• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

(f)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ and ${\text{[OH}}^{\text{-}}\text{]}$ for the solution with pOH 9.60 needs to be determined and the solution needs to be identified as neutral, acidic and basic at 25 °C.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. A strong acid is ionized completely and gives the respective ions.

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

On the contrary, a weak acid ionized partially and reaches to equilibrium.

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

The pH value depends on the concentration of H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution. The mathematical relation between pH and H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions can be written as:

  ${\text{pH = - log [H}}_{\text{3}}{\text{O}}^{\text{+}}]$

Answer to Problem 41E

• ${\text{[OH}}^{\text{-}}\text{]= 2}{\text{.5×10}}^{\text{-10}}\text{ M}$
• ${\text{[H}}^{\text{+}}\text{] = 4}\text{.0 }\times {\text{10}}^{\text{-5 }}\text{M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$, the solution must be acidic.

Explanation of Solution

Given:

pOH=9.60

Calculate pH:

pH = 14 − pOH = 14 − 9.60 = 4.4

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

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-4}\text{.4}}\\ {\text{[H}}^{\text{+}}\text{] = 4}\text{.0 }\times {\text{10}}^{\text{-5 }}\text{M}\end{array}$

Calculate ${\text{[OH}}^{\text{-}}\text{]}$

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[4}\text{.0 }\times {\text{10}}^{\text{-5 }}\text{M]}}\\ {\text{[OH}}^{\text{-}}\text{]= 2}{\text{.5×10}}^{\text{-10}}\text{ M}\end{array}$

• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$, the solution must be acidic.