Chapter 16, Problem 16.36QP

(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 16.36QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{12.653}$.

The $\text{pOH}$ of the solution is $\underset{\_}{1.347}$.

Explanation of Solution

Explanation

Given

The concentration of $\text{NaOH}$ is $0.0450\text{M}$.

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

$\text{NaOH}\stackrel{}{\to }{\text{Na}}^{\text{+}}+{\text{OH}}^{-}$

Sodium hydroxide contains one ${\text{OH}}^{-}$ ion per formula. On ionization, one $\text{NaOH}$ gives one ${\text{OH}}^{-}$ ion. Hence, the concentration of ${\text{OH}}^{-}$ ion is same as the concentration of $\text{NaOH}$. Thus, the concentration of ${\text{OH}}^{-}$ ion is $0.0450\text{M}$.

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

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression to calculate the value of $\text{pOH}$.

$\begin{array}{c}\text{pOH}=-\log \left(0.0450\right)\\ =\underset{\_}{1.347}\end{array}$

Hence, the $\text{pOH}$ of the solution is $\underset{\_}{1.347}$.

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

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

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

$\begin{array}{c}\text{pH}=14-1.347\\ =\underset{\_}{12.653}\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{12.653}$.

(b)

Interpretation Introduction

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

Answer to Problem 16.36QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{12.954}$.

The $\text{pOH}$ of the solution is $\underset{\_}{1.046}$.

Explanation of Solution

Explanation

Given

The concentration of $\text{Ca}{\left(\text{OH}\right)}_2$ is $0.160\text{M}$.

The dissociation of $\text{Ca}{\left(\text{OH}\right)}_2$ is shown by the reaction,

$\text{Ca}{\left(\text{OH}\right)}_2\stackrel{}{\to }{\text{Ca}}^{\text{+}}+2{\text{OH}}^{-}$

Calcium hydroxide contains two ${\text{OH}}^{-}$ ion per formula. On ionization, one $\text{Ca}{\left(\text{OH}\right)}_2$ gives two ${\text{OH}}^{-}$ ions. Hence, the concentration of ${\text{OH}}^{-}$ ion is double the concentration of $\text{Ca}{\left(\text{OH}\right)}_2$. Thus, the concentration of ${\text{OH}}^{-}$ ion is $2\times 0.0450\text{M}=0.09\text{M}$.

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

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression to calculate the value of $\text{pOH}$.

$\begin{array}{c}\text{pOH}=-\log \left(0.09\right)\\ =\underset{\_}{1.046}\end{array}$

Hence, the $\text{pOH}$ of the solution is $\underset{\_}{1.046}$.

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

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

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

$\begin{array}{c}\text{pH}=14-1.046\\ =\underset{\_}{12.954}\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{12.954}$.

(c)

Interpretation Introduction

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

Answer to Problem 16.36QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{12.398}$.

The $\text{pOH}$ of the solution is $\underset{\_}{1.602}$.

Explanation of Solution

Explanation

Given

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

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

The concentration of $\text{Ca}{\left(\text{OH}\right)}_2$ is $0.0125\text{M}$.

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

$\text{2HCl}+\text{Ca}{\left(\text{OH}\right)}_2\stackrel{}{\to }{\text{CaCl}}_2+2{\text{H}}_{\text{2}}\text{O}$

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 $1:1$ proportion of $\text{HCl}$ to $\text{Ca}{\left(\text{OH}\right)}_2$. 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 $2\times 0.0125\text{M}=0.025\text{M}$.

Calcium hydroxide contains two ${\text{OH}}^{-}$ ions per formula that is one $\text{Ca}{\left(\text{OH}\right)}_2$ gives $2$ ${\text{OH}}^{-}$ ions on ionization. Hence, the concentration of ${\text{OH}}^{-}$ ions is double the concentration of $\text{Ca}{\left(\text{OH}\right)}_2$. The concentration of ${\text{OH}}^{-}$ ion is $2\times 0.0125\text{M}=0.025\text{M}$.

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 $\text{Ca}{\left(\text{OH}\right)}_2$.

The $\text{pOH}$ of $\text{Ca}{\left(\text{OH}\right)}_2$ is calculated by using the expression,

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression to calculate the value of $\text{pOH}$.

$\begin{array}{c}\text{pOH}=-\log \left(0.025\right)\\ =\underset{\_}{1.602}\end{array}$

Hence, the $\text{pOH}$ of the solution is $\underset{\_}{1.602}$.

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

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

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

$\begin{array}{c}\text{pH}=14-1.602\\ =\underset{\_}{12.398}\end{array}$

Hence, the $\text{pH}$ of the solution is $\underset{\_}{12.398}$.

(d)

Interpretation Introduction

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

Answer to Problem 16.36QP

Solution

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

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

Explanation of Solution

Explanation

Given

The ratio of the mixture is $2:3$.

The concentration of ${\text{HNO}}_3$ is $0.0125\text{M}$.

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

Nitric acid reacts with potassium hydroxide to form potassium nitrate and water according to the reaction,

${\text{HNO}}_3+\text{KOH}\stackrel{}{\to }{\text{KNO}}_3+{\text{H}}_{\text{2}}\text{O}$

Nitric acid 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. But here the mixture contains $2:3$ proportion of ${\text{HNO}}_3$ to $\text{KOH}$. So, $2$ ${\text{HNO}}_3$ gives $2$ ${\text{H}}^{\text{+}}$ ions on ionization. Hence, the concentration of ${\text{H}}^{\text{+}}$ ions is double the concentration of ${\text{HNO}}_3$. Thus, the concentration of ${\text{H}}^{\text{+}}$ is $2\times 0.0125\text{M}=0.025\text{M}$.

Potassium hydroxide contains one ${\text{OH}}^{-}$ ion per formula that is one $\text{KOH}$ gives $1$ ${\text{OH}}^{-}$ ion on ionization. The mixture contains $2:3$ proportion of ${\text{HNO}}_3$ to $\text{KOH}$. Hence, the concentration of ${\text{OH}}^{-}$ ions is thrice the concentration of $\text{KOH}$. The concentration of ${\text{OH}}^{-}$ ion is $3\times 0.0125\text{M}=0.0375\text{M}$.

The $\text{pH}$ of the solution is due to ${\text{H}}^{\text{+}}$ ions from ${\text{HNO}}_3$ and $\text{pOH}$ of the solution is due to ${\text{OH}}^{-}$ ions from $\text{KOH}$.

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

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression to calculate the value of $\text{pOH}$.

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

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

The $\text{pH}$ of ${\text{HNO}}_3$ is calculated by using the expression,

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

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

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

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

Conclusion

1. a. The $\text{pH}$ of the solution is $\underset{\_}{12.653}$ and the $\text{pOH}$ of the solution is $\underset{\_}{1.347}$.
2. b. The $\text{pH}$ of the solution is $\underset{\_}{12.954}$ and the $\text{pOH}$ of the solution is $\underset{\_}{1.046}$.
3. c. The $\text{pH}$ of the solution is $\underset{\_}{12.398}$ and the $\text{pOH}$ of the solution is $\underset{\_}{1.602}$.
4. d. The $\text{pH}$ of the solution is $\underset{\_}{12.574}$ and the $\text{pOH}$ of the solution is $\underset{\_}{1.426}$.