Chapter 7, Problem 42E

(a)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ , pOH and ${\text{[OH}}^{\text{-}}\text{]}$ for the solution with pH 9.63 needs to be determined and the solution needs to be identified as neutral, acidic and basic.

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 42E

• $\text{pOH = 4}\text{.37}$
• ${\text{[OH}}^{\text{-}}\text{]= }4.3\times {10}^{-5}\text{ M}$
• ${\text{[H}}^{\text{+}}\text{] =2}{\text{.3 ×10}}^{\text{-10}}\text{M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

Explanation of Solution

Given:

pH=9.63

  $\text{pOH = 14 – pH = 14 -9}\text{.63 = 4}\text{.37}$

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

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-9}\text{.63}}\\ {\text{[H}}^{\text{+}}\text{] =2}{\text{.3 ×10}}^{\text{-10}}\text{M}\end{array}$

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[2}{\text{.3 ×10}}^{\text{-10}}\text{M]}}\\ {\text{[OH}}^{\text{-}}\text{]= }4.3\times {10}^{-5}\text{ M}\end{array}$

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

(b)

Interpretation Introduction

Interpretation: The $[{\text{H}}^{\text{+}}]$ , pOH and pH for the solution with ${\text{[OH}}^{\text{-}}\text{]=3}\text{.9}\times {\text{10}}^{\text{-6}}\text{M}$ needs to be determined and the solution needs to be identified as neutral, acidic and basic.

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 42E

• $\text{pH = }8.6$
• $\text{pOH = 5}\text{.4}$
• $[{\text{H}}^{\text{+}}]\text{= }2.6\times {10}^{-9}\text{ M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$< ${\text{[OH}}^{\text{-}}\text{]}$, the solution must be basic.

Explanation of Solution

Given:

  ${\text{[OH}}^{\text{-}}\text{]=3}\text{.9}\times {\text{10}}^{\text{-6}}\text{M}$

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ [{\text{H}}^{\text{+}}]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{{\text{[OH}}^{\text{-}}\text{]}}\\ [{\text{H}}^{\text{+}}]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[3}\text{.9}\times {\text{10}}^{\text{-6}}\text{M]}}\\ [{\text{H}}^{\text{+}}]\text{= }2.6\times {10}^{-9}\text{ M}\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ \text{ pH = - log [}2.6\times {10}^{-9}\text{ M}]\\ \text{ pH = }8.6\end{array}$

Calculate pOH

  $\text{pOH = 14 – pH = 14 -8}\text{.6 = 5}\text{.4}$

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

(c)

Interpretation Introduction

Interpretation: The ${\text{[OH}}^{\text{-}}\text{]}$ , pOH and pH for the solution with $[{\text{H}}^{\text{+}}]\text{=}0.027\text{M}$ needs to be determined and the solution needs to be identified as neutral, acidic and basic.

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 42E

• $\text{pH = }1.57$
• $\text{pOH = 12}\text{.4}$
• ${\text{[OH}}^{\text{-}}\text{]= 3}\text{.7 }\times {10}^{-13}\text{ M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$ , the solution must be acidic.

Explanation of Solution

Given:

  $[{\text{H}}^{\text{+}}]\text{=}0.027\text{M}$

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[}0.027\text{M]}}\\ {\text{[OH}}^{\text{-}}\text{]= 3}\text{.7 }\times {10}^{-13}\text{ M}\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ \text{ pH = - log [}0.027\text{ M}]\\ \text{ pH = }1.57\end{array}$

Calculate pOH

  $\text{pOH = 14 – pH = 14 -1}\text{.57 = 12}\text{.4}$

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

(d)

Interpretation Introduction

Interpretation: The ${\text{[OH}}^{\text{-}}\text{]}$ , $[{\text{H}}^{\text{+}}]$ and pH for the solution with pOH = 12.2 needs to be determined and the solution needs to be identified as neutral, acidic and basic.

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 42E

• $\text{pH = }1.8$
• $\text{pOH = 12}\text{.4}$
• ${\text{[H}}^{\text{+}}\text{]=1}{\text{.6×10}}^{\text{-2}}\text{M}$
• Since ${\text{[H}}^{\text{+}}\text{]}$ >${\text{[OH}}^{\text{-}}\text{]}$ , the solution must be acidic.

Explanation of Solution

Given:

pOH = 12.2

Calculate pH

  $\text{pH = 14 –pOH = 14 - 12}\text{.2 = 1}\text{.8}$

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

  $\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{]}\\ \text{ 1}{\text{.8 = - log [H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] =10}}^{\text{-pH}}{\text{=10}}^{\text{-1}\text{.8}}\\ {\text{[H}}^{\text{+}}\text{]=1}{\text{.6×10}}^{\text{-2}}\text{M}\end{array}$

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

  $\begin{array}{l}{\text{[H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]=1}{\text{.0×10}}^{\text{-14}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{[{\text{H}}^{\text{+}}]}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\text{[}0.027\text{M]}}\\ {\text{[OH}}^{\text{-}}\text{]= 3}\text{.7 }\times {10}^{-13}\text{ M}\end{array}$

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