Chapter 15, Problem 15.32QP

(a)

Interpretation Introduction

Interpretation: The values of $\text{pH}$ and $\text{pOH}$ with the given $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ or $\left[{\text{OH}}^{-}\right]$ values are to be calculated. Also, whether the solutions are acidic, basic or neutral is to be indicated.

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 $\text{pOH}$ with the given $\left[{\text{OH}}^{-}\right]$ value and if the solution is acidic, basic or neutral.

Answer to Problem 15.32QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{3.914}$.

The $\text{pOH}$ of the solution is $\underset{\_}{10.086}$.

The given solution is acidic.

Explanation of Solution

Explanation

Given

The value of $\left[{\text{OH}}^{-}\right]$ is $8.2\times {10}^{-11}\text{M}$.

The auto-dissociation of pure water is shown by the reaction,

${\text{2H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}+{\text{OH}}^{-}$

Concentrations of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and $\left[{\text{OH}}^{-}\right]$ in the solution are related according to the following relation,

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\times \left[{\text{OH}}^{-}\right]=K_w={10}^{-14}$

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

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

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

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

Hence, $\text{pH}$ and $\text{pOH}$ of the solution are related with each other as,

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\times \left[{\text{OH}}^{-}\right]={10}^{-14}\\ -\log \left(\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\times \left[{\text{OH}}^{-}\right]\right)=-\log \left({10}^{-14}\right)\\ \left(-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\right)+\left(-\log \left[{\text{OH}}^{-}\right]\right)=14\\ \text{pH}+\text{pOH}=14\end{array}$

Now, the $\text{pOH}$ 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 $\text{pOH}$ of the solution.

$\begin{array}{c}\text{pOH}=-\log \left(8.2\times {10}^{-11}\right)\\ =\underset{\_}{10.086}\end{array}$

Hence, $\text{pOH}$ of the solution is $\underset{\_}{10.086}$.

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.

$\begin{array}{c}\text{pH}=14-10.086\\ =\underset{\_}{3.914}\end{array}$

Hence, $\text{pH}$ of the solution is $\underset{\_}{3.914}$.

The $\text{pH}$ of the solution is less than $7$, thus it is an acidic solution.

(b)

Interpretation Introduction

To determine: The values of $\text{pH}$ and $\text{pOH}$ with the given $\left[{\text{OH}}^{-}\right]$ value and if the solution is acidic, basic or neutral.

Answer to Problem 15.32QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{8.89}$.

The $\text{pOH}$ of the solution is $\underset{\_}{5.11}$.

The given solution is basic.

Explanation of Solution

Explanation

Given

The value of $\left[{\text{OH}}^{-}\right]$ is $7.7\times {10}^{-6}\text{M}$.

The $\text{pOH}$ 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 $\text{pOH}$ of the solution.

$\begin{array}{c}\text{pOH}=-\log \left(7.7\times {10}^{-6}\right)\\ =\underset{\_}{5.11}\end{array}$

Hence, $\text{pOH}$ of the solution is $\underset{\_}{5.11}$.

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.

$\begin{array}{c}\text{pH}=14-5.11\\ =\underset{\_}{8.89}\end{array}$

Hence, $\text{pH}$ of the solution is $\underset{\_}{8.89}$.

The $\text{pH}$ of the solution is greater than $7$, thus it is a basic solution.

(c)

Interpretation Introduction

To determine: The values of $\text{pH}$ and $\text{pOH}$ with the given $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ value and if the solution is acidic, basic or neutral.

Answer to Problem 15.32QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{3.49}$.

The $\text{pOH}$ of the solution is $\underset{\_}{10.51}$.

The given solution is acidic.

Explanation of Solution

Explanation

Given

The value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ is $3.2\times {10}^{-4}\text{M}$.

The $\text{pH}$ 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 $\text{pH}$ of the solution.

$\begin{array}{c}\text{pH}=-\log \left(3.2\times {10}^{-4}\right)\\ =\underset{\_}{3.49}\end{array}$

Hence, $\text{pH}$ of the solution is $\underset{\_}{3.49}$.

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.

$\begin{array}{c}\text{pOH}=14-3.49\\ =\underset{\_}{10.51}\end{array}$

Hence, $\text{pOH}$ of the solution is $\underset{\_}{10.51}$.

The $\text{pH}$ of the solution is less than $7$, thus it is an acidic solution.

(d)

Interpretation Introduction

To determine: The values of $\text{pH}$ and $\text{pOH}$ with the given $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ value and if the solution is acidic, basic or neutral.

Answer to Problem 15.32QP

Solution

The $\text{pH}$ of the solution is $\underset{\_}{7.0}$.

The $\text{pOH}$ of the solution is $\underset{\_}{7.0}$.

The given solution is neutral.

Explanation of Solution

Explanation

Given

The value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ is $1.0\times {10}^{-7}\text{M}$.

The $\text{pH}$ 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 $\text{pH}$ of the solution.

$\begin{array}{c}\text{pH}=-\log \left(1\times {10}^{-7}\right)\\ =\underset{\_}{7.0}\end{array}$

Hence, $\text{pH}$ of the solution is $\underset{\_}{7.0}$.

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.

$\begin{array}{c}\text{pOH}=14-7\\ =\underset{\_}{7.0}\end{array}$

Hence, $\text{pOH}$ of the solution is $\underset{\_}{7.0}$.

The $\text{pH}$ of the solution is $7$, thus it is a neutral solution.

Conclusion

1. a. The $\text{pH}$ of the solution is $\underset{\_}{3.914}$ and the $\text{pOH}$ of the solution is $\underset{\_}{10.51}$. The given solution is acidic.
2. b. The $\text{pH}$ of the solution is $\underset{\_}{8.89}$ and the $\text{pOH}$ of the solution is $\underset{\_}{5.11}$. The given solution is acidic.
3. c. The $\text{pH}$ of the solution is $\underset{\_}{3.49}$ and the $\text{pOH}$ of the solution is $\underset{\_}{10.51}$. The given solution is basic.
4. d. The $\text{pH}$ of the solution is $\underset{\_}{7.0}$ and the $\text{pOH}$ of the solution is $\underset{\_}{7.0}$. The given solution is neutral.