Chapter 7, Problem 40E

(a)

Interpretation Introduction

Interpretation: The ${\text{[OH}}^{\text{-}}\text{]}$ , pH and pOH with acidic or basic nature of solution at 25 °C needs to be determined if the concentration of $[{\text{H}}^{\text{+}}]$ is $1.0\times {10}^{-7}\text{ M}$ .

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 40E

• $\begin{array}{l}\text{pH = 7}\\ \text{pOH =}7\end{array}$
• ${\text{[OH}}^{\text{-}}\text{]= }1.0\times {10}^{-7}\text{ M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$ = ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be neutral.

Explanation of Solution

Given:

  $[{\text{H}}^{\text{+}}]$ = 1.0 x 10-7 M

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[}1.0\times {10}^{-7}\text{ M]}}\\ {\text{[OH}}^{\text{-}}\text{]= }1.0\times {10}^{-7}\text{ M}\end{array}$

Since:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}]\\ {\text{pOH = - log [OH}}^{-}]\end{array}$

Substitute the values of ${\text{[OH}}^{\text{-}}\text{]}$ and $[{\text{H}}^{\text{+}}]$ to calculate pH and pOH:

  $\begin{array}{l}\text{pH = - log [1}{\text{.0 × 10}}^{\text{-7}}\text{M]}\\ \text{pH = 7}\\ {\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pOH = - log [1}{\text{.0 × 10}}^{\text{-7}}\text{]}\\ \text{pOH =7}\end{array}$

Since ${\text{[H}}^{\text{+}}\text{]}$ = ${\text{[OH}}^{\text{-}}\text{]}$ , the solution must be neutral.

(b)

Interpretation Introduction

Interpretation: The ${\text{[OH}}^{\text{-}}\text{]}$ , pH and pOH with acidic or basic nature of solution at 25 °C needs to be determined if the concentration of $[{\text{H}}^{\text{+}}]$ is $8.3\times {10}^{-16}\text{ M}$ .

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 40E

• $\begin{array}{l}\text{pH = 15}\text{.8}\\ \text{pOH = - 1}\text{.08}\end{array}$
• ${\text{[OH}}^{\text{-}}\text{]=}12\text{ M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

Explanation of Solution

Given:

  $[{\text{H}}^{\text{+}}]$ = $8.3\times {10}^{-16}\text{ M}$

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[}8.3\times {10}^{-16}\text{ M]}}\\ {\text{[OH}}^{\text{-}}\text{]= }12\text{ M}\end{array}$

Since:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}]\\ {\text{pOH = - log [OH}}^{-}]\end{array}$

Substitute the values of ${\text{[OH}}^{\text{-}}\text{]}$ and $[{\text{H}}^{\text{+}}]$ to calculate pH and pOH:

  $\begin{array}{l}\text{pH = - log [}8.3\times {10}^{-16}\text{ M]}\\ \text{pH = 15}\text{.8}\\ {\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pOH = - log [12]}\\ \text{pOH = - 1}\text{.08}\end{array}$

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

(c)

Interpretation Introduction

Interpretation: The ${\text{[OH}}^{\text{-}}\text{]}$ , pH and pOH with acidic or basic nature of solution at 25 °C needs to be determined if the concentration of $[{\text{H}}^{\text{+}}]$ is $12\text{M}$ .

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 40E

• $\begin{array}{l}\text{pH =- 1}\text{.08 }\\ \text{pOH = 15}\text{.8}\end{array}$
• ${\text{[OH}}^{\text{-}}\text{]=}8.3\times {10}^{-16}\text{ M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$, the solution must be acidic.

Explanation of Solution

Given:

  $[{\text{H}}^{\text{+}}]$ = $12\text{ M}$

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[}12\text{ M]}}\\ {\text{[OH}}^{\text{-}}\text{]= }8.3\times {10}^{-16}\text{ M}\end{array}$

Since:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}]\\ {\text{pOH = - log [OH}}^{-}]\end{array}$

Substitute the values of ${\text{[OH}}^{\text{-}}\text{]}$ and $[{\text{H}}^{\text{+}}]$ to calculate pH and pOH:

  $\begin{array}{l}\text{pH = - log [}12\text{ M]}\\ \text{pH = - 1}\text{.08}\\ {\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pOH = - log [}8.3\times {10}^{-16}\text{ M]}\\ \text{pOH = 15}\text{.8}\end{array}$

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

(d)

Interpretation Introduction

Interpretation: The ${\text{[OH}}^{\text{-}}\text{]}$ , pH and pOH with acidic or basic nature of solution at 25 °C needs to be determined if the concentration of $[{\text{H}}^{\text{+}}]$ is $\text{5}{\text{.4×10}}^{\text{-5}}\text{M}$ .

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 40E

• $\begin{array}{l}\text{pH =4}\text{.27}\\ \text{pOH = 9}\text{.73}\end{array}$
• ${\text{[OH}}^{\text{-}}\text{]=}1.8\times {10}^{-12}\text{ M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$, the solution must be acidic.

Explanation of Solution

Given:

  $[{\text{H}}^{\text{+}}]$ = $\text{5}{\text{.4×10}}^{\text{-5}}\text{M}$

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[5}{\text{.4×10}}^{\text{-5}}\text{M]}}\\ {\text{[OH}}^{\text{-}}\text{]=}1.8\times {10}^{-12}\text{ M}\end{array}$

Since:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}]\\ {\text{pOH = - log [OH}}^{-}]\end{array}$

Substitute the values of ${\text{[OH}}^{\text{-}}\text{]}$ and $[{\text{H}}^{\text{+}}]$ to calculate pH and pOH:

  $\begin{array}{l}\text{pH = - log [5}{\text{.4×10}}^{\text{-5}}\text{M]}\\ \text{pH =4}\text{.27}\\ {\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pOH = - log [}1.8\times {10}^{-12}\text{ M]}\\ \text{pOH = 9}\text{.73}\end{array}$

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