Chapter 7, Problem 39E

(a)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ , pH and pOH with acidic or basic nature of solution at 25 °C if the concentration of ${\text{[OH}}^{\text{-}}\text{]}$ is 1.5 M needs to be determined.

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 39E

• $\begin{array}{l}\text{pH = 14}\text{.18}\\ \text{pOH =}-0.18\end{array}$
• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

Explanation of Solution

Given:

  ${\text{[OH}}^{\text{-}}\text{]}$ = 1.5 M

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{[OH}}^{\text{-}}\text{]}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[1}\text{.5 M]}}\\ {\text{[H}}^{\text{+}}\text{]= 6}{\text{.67×10}}^{\text{-15}}\text{M}\end{array}$

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:

  $\begin{array}{l}\text{pH = - log [6}{\text{.67×10}}^{\text{-15}}\text{M]}\\ \text{pH = 14}\text{.18}\\ {\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pOH = - log [1}\text{.5M]}\\ \text{pOH =}-0.18\end{array}$

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

(b)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ , pH and pOH with acidic or basic nature of solution at 25 °C needs to be determined, if the concentration of ${\text{[OH}}^{\text{-}}\text{]}$ is 3.6 x 10-15 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 39E

• $\begin{array}{l}\text{pH = -0}\text{.44}\\ \text{pOH =}14.44\end{array}$
• ${\text{[H}}^{\text{+}}\text{]= 2}\text{.78 M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$, the solution must be acidic.

Explanation of Solution

Given:

  ${\text{[OH}}^{\text{-}}\text{]}$ = 3.6 x 10-15 M

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{[OH}}^{\text{-}}\text{]}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[}3.6\times {10}^{-15}\text{ M]}}\\ {\text{[H}}^{\text{+}}\text{]= 2}\text{.78 M}\end{array}$

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:

  $\begin{array}{l}\text{pH = - log [2}\text{.78 M]}\\ \text{pH = -0}\text{.44}\\ {\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pOH = - log [}3.6\times {10}^{-15}\text{M]}\\ \text{pOH =}14.44\end{array}$

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

(c)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ , pH and pOH with acidic or basic nature of solution at 25 °C needs to be determined if the concentration of ${\text{[OH}}^{\text{-}}\text{]}$ is 1.0 x 10-7 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 39E

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

Explanation of Solution

Given:

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

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

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

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:

  $\begin{array}{l}\text{pH = - log [}1.0\times {10}^{-7}M\text{]}\\ \text{pH = 7}\\ {\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pOH = - log [}1.0\times {10}^{-7}\text{]}\\ \text{pOH =}7\end{array}$

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

(d)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ , pH and pOH with acidic or basic nature of solution at 25 °C needs to be determined, if the concentration of ${\text{[OH}}^{\text{-}}\text{]}$ is 7.3 x 10-4 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 39E

• $\begin{array}{l}\text{pH = 10}\text{.8}\\ \text{pOH =}3.14\end{array}$
• ${\text{[H}}^{\text{+}}\text{]= }1.4\times {10}^{-11}\text{ M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

Explanation of Solution

Given:

  ${\text{[OH}}^{\text{-}}\text{]}$ = 7.3 x 10-⁴M

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{[OH}}^{\text{-}}\text{]}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[}7.3\times {10}^{-4}\text{ M]}}\\ {\text{[H}}^{\text{+}}\text{]= }1.4\times {10}^{-11}\text{ M}\end{array}$

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:

  $\begin{array}{l}\text{pH = - log [}1.4\times {10}^{-11}\text{ M]}\\ \text{pH = 10}\text{.8}\\ {\text{pOH = - log [OH}}^{\text{-}}\text{]}\\ \text{pOH = - log [}7.3\times {10}^{-4}\text{]}\\ \text{pOH =}3.14\end{array}$

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