Chapter 15, Problem 15.35QP

(a)

Interpretation Introduction

Interpretation: The values of $\text{pH}$ and $\text{pOH}$ of the given solutions are to be calculated.

Concept introduction: The potential of ${\text{H}}^{\text{+}}$ ions is called as $\text{pH}$ and the potential of ${\text{OH}}^{-}$ ions is called as $\text{pOH}$. The value of $\text{pH}$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $\text{pH}$ is $7$.

To determine: The values of $\text{pH}$ and $\text{pOH}$ of the given solution.

Answer to Problem 15.35QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{0.809}$.

The $\text{pOH}$ of the solution is $\underset{\_}{13.191}$.

Explanation of Solution

Explanation

Given

The concentration of $\text{HCl}$ is $0.155\text{M}$.

The dissociation of $\text{HCl}$ is shown by the reaction,

$\text{HCl}\stackrel{}{\to }{\text{H}}^{\text{+}}+{\text{Cl}}^{-}$

Hydrochloric acid $\left(\text{HCl}\right)$, is a monoprotic acid as it contains one ${\text{H}}^{\text{+}}$ ion per formula. On ionization, one $\text{HCl}$ gives one ${\text{H}}^{\text{+}}$ ion. Hence, the concentration of ${\text{H}}^{\text{+}}$ ion is same as the concentration of $\text{HCl}$. The ${\text{H}}^{\text{+}}$ ion is also represented as ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion. Thus, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion is $0.155\text{M}$.

The value of $\text{pH}$ is calculated by using the expression,

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

Substitute the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ in the above expression to calculate the value of $\text{pH}$.

$\begin{array}{c}\text{pH}=-\log \left(0.155\right)\\ =\underset{\_}{0.809}\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{0.809}$.

The value of $\text{pOH}$ is calculated by using the expression,

$\text{pOH}=14-\text{pH}$

Substitute the value of $\text{pH}$ to calculate $\text{pOH}$ of the solution in the above expression.

$\begin{array}{c}\text{pOH}=14-0.809\\ =\underset{\_}{13.191}\end{array}$

Hence, the $\text{pOH}$ of the solution is $\underset{\_}{13.191}$.

(b)

Interpretation Introduction

To determine: The values of $\text{pH}$ and $\text{pOH}$ of the given solution.

Answer to Problem 15.35QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{2.301}$.

The $\text{pOH}$ of the solution is $\underset{\_}{11.699}$.

Explanation of Solution

Explanation

Given

The concentration of ${\text{HNO}}_3$ is $0.00500\text{M}$.

The dissociation of ${\text{HNO}}_3$ is shown by the reaction,

${\text{HNO}}_3\stackrel{}{\to }{\text{H}}^{\text{+}}+{\text{NO}}_3^{-}$

Nitric acid $\left({\text{HNO}}_3\right)$, is a monoprotic acid as it contains one ${\text{H}}^{\text{+}}$ ion per formula. On ionization, one ${\text{HNO}}_3$ gives one ${\text{H}}^{\text{+}}$ ion. Hence, the concentration of ${\text{H}}^{\text{+}}$ ion is same as the concentration of ${\text{HNO}}_3$. The ${\text{H}}^{\text{+}}$ ion is also represented as ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion. Thus, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion is $0.00500\text{M}$.

The value of $\text{pH}$ is calculated by using the expression,

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

Substitute the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ in the above expression to calculate the value of $\text{pH}$.

$\begin{array}{c}\text{pH}=-\log \left(0.00500\right)\\ =\underset{\_}{2.301}\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{2.301}$.

The value of $\text{pOH}$ is calculated by using the expression,

$\text{pOH}=14-\text{pH}$

Substitute the value of $\text{pH}$ to calculate $\text{pOH}$ of the solution in the above expression.

$\begin{array}{c}\text{pOH}=14-2.301\\ =\underset{\_}{11.699}\end{array}$

Hence, the $\text{pOH}$ of the solution is $\underset{\_}{11.699}$.

(c)

Interpretation Introduction

To determine: The values of $\text{pH}$ and $\text{pOH}$ of the given solution.

Answer to Problem 15.35QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{1.903}$.

The $\text{pOH}$ of the solution is $\underset{\_}{12.097}$.

Explanation of Solution

Explanation

Given

The ratio of the mixture is $2:1$.

The concentration of $\text{HCl}$ is $0.0125\text{M}$.

The concentration of $\text{NaOH}$ is $0.0125\text{M}$.

Hydrochloric acid reacts with sodium hydroxide to form sodium chloride and water according to the reaction,

$\text{HCl}+\text{NaOH}\stackrel{}{\to }\text{NaCl}+{\text{H}}_{\text{2}}\text{O}$

Hydrochloric acid is a monoprotic acid as it contains one ${\text{H}}^{\text{+}}$ ion per formula. On ionization, one $\text{HCl}$ gives one ${\text{H}}^{\text{+}}$ ion. But here the mixture contains $2:1$ proportion of $\text{HCl}$ to $\text{NaOH}$. So, $2$ $\text{HCl}$ gives $2$ ${\text{H}}^{\text{+}}$ ions on ionization. Hence, the concentration of ${\text{H}}^{\text{+}}$ ions is double the concentration of $\text{HCl}$. Thus, the concentration of ${\text{H}}^{\text{+}}$ is $2\times 0.0125\text{M}=0.025\text{M}$.

Sodium hydroxide contains one ${\text{OH}}^{-}$ ion per formula that is one $\text{NaOH}$ gives $1$ ${\text{OH}}^{-}$ ion on ionization. Hence, the concentration of ${\text{OH}}^{-}$ ions is the concentration of $\text{NaOH}$. The concentration of ${\text{OH}}^{-}$ ion is $0.0125\text{M}$.

The $\text{pH}$ of the solution is due to ${\text{H}}^{\text{+}}$ ions from $\text{HCl}$ and $\text{pOH}$ of the solution is due to ${\text{OH}}^{-}$ ions from $\text{NaOH}$.

The $\text{pH}$ of $\text{HCl}$ is calculated by using the expression,

$\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the above expression to calculate the value of $\text{pH}$.

$\begin{array}{c}\text{pH}=-\log \left(0.0125\right)\\ =\underset{\_}{1.903}\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{1.903}$.

The $\text{pOH}$ of $\text{NaOH}$ is calculated by using the expression,

$\text{pOH}=14-\text{pH}$

Substitute the value of $\text{pH}$ to calculate $\text{pOH}$ of the solution in the above expression.

$\begin{array}{c}\text{pOH}=14-1.903\\ =\underset{\_}{12.097}\end{array}$

Hence, the $\text{pOH}$ of the solution is $\underset{\_}{12.097}$.

(d)

Interpretation Introduction

To determine: The values of $\text{pH}$ and $\text{pOH}$ of the given solution.

Answer to Problem 15.35QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{1.426}$.

The $\text{pOH}$ of the solution is $\underset{\_}{12.574}$.

Explanation of Solution

Explanation

Given

The ratio of the mixture is $3:1$.

The concentration of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $0.0125\text{M}$.

The concentration of $\text{KOH}$ is $0.0125\text{M}$.

Sulfuric acid reacts with potassium hydroxide to form potassium sulfate and water according to the reaction,

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}+\text{KOH}\stackrel{}{\to }{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}$

Sulfuric acid is a diprotic acid as it contains two ${\text{H}}^{\text{+}}$ ions per formula. On ionization, one ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ gives two ${\text{H}}^{\text{+}}$ ions. But here the mixture contains $3:1$ proportion of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ to $\text{KOH}$. So, $3$ ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ gives $3$ ${\text{H}}^{\text{+}}$ ions on ionization. Hence, the concentration of ${\text{H}}^{\text{+}}$ ions is thrice the concentration of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$. Thus, the concentration of ${\text{H}}^{\text{+}}$ is $3\times 0.0125\text{M}=0.0375\text{M}$.

Potassium hydroxide contains one ${\text{OH}}^{-}$ ion per formula that is one $\text{KOH}$ gives $1$ ${\text{OH}}^{-}$ ion on ionization. Hence, the concentration of ${\text{OH}}^{-}$ ions is the concentration of $\text{KOH}$. The concentration of ${\text{OH}}^{-}$ ion is $0.0125\text{M}$.

The $\text{pH}$ of the solution is due to ${\text{H}}^{\text{+}}$ ions from ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $\text{pOH}$ of the solution is due to ${\text{OH}}^{-}$ ions from $\text{KOH}$.

The $\text{pH}$ of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated by using the expression,

$\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the above expression to calculate the value of $\text{pH}$.

$\begin{array}{c}\text{pH}=-\log \left(0.0375\right)\\ =\underset{\_}{1.426}\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{1.426}$.

The $\text{pOH}$ of $\text{KOH}$ is calculated by using the expression,

$\text{pOH}=14-\text{pH}$

Substitute the value of $\text{pH}$ to calculate $\text{pOH}$ of the solution in the above expression.

$\begin{array}{c}\text{pOH}=14-1.426\\ =\underset{\_}{12.574}\end{array}$

Hence, the $\text{pOH}$ of the solution is $\underset{\_}{12.574}$.

Conclusion

1. a. The $\text{pH}$ of the solution is $\underset{\_}{0.809}$ and the $\text{pOH}$ of the solution is $\underset{\_}{13.191}$.
2. b. The $\text{pH}$ of the solution is $\underset{\_}{2.301}$ and the $\text{pOH}$ of the solution is $\underset{\_}{11.699}$.
3. c. The $\text{pH}$ of the solution is $\underset{\_}{1.903}$ and the $\text{pOH}$ of the solution is $\underset{\_}{12.097}$.
4. d. The $\text{pH}$ of the solution is $\underset{\_}{1.426}$ and the $\text{pOH}$ of the solution is $\underset{\_}{12.574}$.