Chapter 13, Problem 73QP

(a)

Interpretation Introduction

Interpretation:

Concentration of ${\text{OH}}^{-}$ , $\text{pH}$ and $\text{pOH}$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\times {10}^{-5}$ is to be found, and nature of solution is to be determined.

Explanation of Solution

Ionic product of water is defined as the product of concentrations of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions present in the water. It is represented as $K_{\text{w}}$ .

Expression for $K_{\text{w}}$ is as follows:

$K_{\text{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Here, $K_{\text{w}}$ is the ionic product of water, $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions and $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$ ions.

Substitute, $1.0\times {10}^{-5}$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ and $1.0\times {10}^{-14}$ for $K_{\text{w}}$ in the above equation.

$\begin{array}{c}1.0\times {10}^{-14}=\left(1.0\times {10}^{-5}\right)\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-5}}\\ \left[{\text{OH}}^{-}\right]=1.0\times {10}^{-9}\end{array}$

$\text{pH}$ of a solution gives the concentration of hydrogen ions in that solution. Range of $\text{pH}$ values is from $0$ to $14$ . Solutions having $\text{pH}$ values less than $7$ are acidic, those having greater than $7$ are basic. Solutions with $\text{pH}=7$ are neutral. $\text{pOH}$ of a solution gives the concentration of ${\text{OH}}^{-}$ ions in that solution.

$\text{pOH}$ can be found using the concentration of ${\text{OH}}^{-}$ as follows:

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

Substitute, $1.0\times {10}^{-9}$ for $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}\text{pOH}=-\text{log}\left(1.0\times {10}^{-9}\right)\\ \text{pOH}=9\end{array}$

The relation between $\text{pH}$ and $\text{pOH}$ is as follows:

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

Substitute, $9$ for $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+9=14\\ \text{pH}=14-9\\ \text{pH}=5\end{array}$

As the $\text{pH}$ is less than $7$ , so the solution is acidic in nature.

Therefore, $\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-9}$ , $\text{pOH}=9$ and $\text{pH}=5$ for the given value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ , and the nature of the solution is acidic.

(b)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}_3{\text{O}}^{+}$ , $\text{pH}$ and $\text{pOH}$ for $\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-4}$ is to be found, and nature of solution is to be determined.

Explanation of Solution

Ionic product of water is defined as the product of concentrations of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions present in the water. It is represented as $K_{\text{w}}$ .

Expression for $K_{\text{w}}$ is as follows:

$K_{\text{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Here, $K_{\text{w}}$ is the ionic product of water, $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions and $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$ ions.

Substitute, $1.0\times {10}^{-4}$ for $\left[{\text{OH}}^{-}\right]$ and $1.0\times {10}^{-14}$ for $K_{\text{w}}$ in the above equation.

$\begin{array}{c}1.0\times {10}^{-14}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left(1.0\times {10}^{-4}\right)\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-4}}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\times {10}^{-10}\end{array}$

$\text{pH}$ of a solution gives the concentration of hydrogen ions in that solution. Range of $\text{pH}$ values is from $0$ to $14$ . Solutions having $\text{pH}$ values less than $7$ are acidic, those having greater than $7$ are basic. Solutions with $\text{pH}=7$ are neutral. $\text{pOH}$ of a solution gives the concentration of ${\text{OH}}^{-}$ ions in that solution.

$\text{pOH}$ can be found using concentration of ${\text{OH}}^{-}$ as follows:

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

Substitute, $1.0\times {10}^{-4}$ for $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}\text{pOH}=-\text{log}\left(1.0\times {10}^{-4}\right)\\ \text{pOH}=4\end{array}$

The relation between $\text{pH}$ and $\text{pOH}$ is as follows:

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

Substitute, $4$ for $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+4=14\\ \text{pH}=14-4\\ \text{pH}=10\end{array}$

As the $\text{pH}$ is more than $7$ , so the solution is basic in nature.

Therefore, $\left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\times {10}^{-10}$ , $\text{pOH}=4$ and $\text{pH}=10$ for the given value of $\left[{\text{OH}}^{-}\right]$ , and the nature of the solution is basic.

(c)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}_3{\text{O}}^{+}$ and ${\text{OH}}^{-}$ and $\text{pH}$ for $\text{pOH}=8$ is to be found, and nature of solution is to be determined.

Explanation of Solution

$\text{pH}$ of a solution gives the concentration of hydrogen ions in that solution. Range of $\text{pH}$ values is from $0$ to $14$ . Solutions having $\text{pH}$ values less than $7$ are acidic, those having greater than $7$ are basic. Solutions with $\text{pH}=7$ are neutral. $\text{pOH}$ of a solution gives the concentration of ${\text{OH}}^{-}$ ions in that solution.

Concentration of ${\text{OH}}^{-}$ can be found using $\text{pOH}$ as follows:

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

Substitute, $8$ for $\text{pOH}$ in the above equation.

$8.00=-\text{log}\left[{\text{OH}}^{-}\right]$

Taking antilog on both sides and solving above equation, $\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-8}$ .

The relation between $\text{pH}$ and $\text{pOH}$ is as follows:

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

Substitute, $8$ for $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+8=14\\ \text{pH}=14-8\\ \text{pH}=6\end{array}$

As the $\text{pH}$ is less than $7$ , so the solution is acidic in nature.

Concentration of ${\text{H}}_3{\text{O}}^{+}$ can be found using $\text{pH}$ as follows:

$\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$

Substitute, $6$ for $\text{pH}$ in the above equation

$6=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$

Taking antilog on both sides and solving above equation, $\left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\times {10}^{-6}$ .

Therefore, $\left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\times {10}^{-6}$ , $\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-8}$ and $\text{pH}=6$ for the given value of $\text{pOH}$ , and the nature of the solution is acidic.

(d)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}_3{\text{O}}^{+}$ and ${\text{OH}}^{-}$ and $\text{pOH}$ for $\text{pH}=8.54$ is to be found and, nature of solution is to be determined.

Explanation of Solution

$\text{pH}$ of a solution gives the concentration of hydrogen ions in that solution. Range of $\text{pH}$ values is from $0$ to $14$ . Solutions having $\text{pH}$ values less than $7$ are acidic, those having greater than $7$ are basic. Solutions with $\text{pH}=7$ are neutral. $\text{pOH}$ of a solution gives the concentration of ${\text{OH}}^{-}$ ions in that solution.

Concentration of ${\text{H}}_3{\text{O}}^{+}$ can be found using $\text{pH}$ as follows:

$\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$

Substitute, $8.54$ for $\text{pH}$ in the above equation.

$8.54=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$

Taking antilog on both sides and solving above equation, $\left[{\text{H}}_3{\text{O}}^{+}\right]=3.5\times {10}^{-8}$ .

The relation between $\text{pH}$ and $\text{pOH}$ is as follows:

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

Substitute, $8.54$ for $\text{pH}$ in the above equation.

$\begin{array}{c}8.54+\text{pOH}=14\\ \text{pOH}=14-8.54\\ \text{pOH}=5.46\end{array}$

Concentration of ${\text{OH}}^{-}$ can be found using $\text{pOH}$ as follows:

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

Substitute, $5.46$ for $\text{pOH}$ in the above equation.

$5.46=-\text{log}\left[{\text{OH}}^{-}\right]$

Taking antilog on both sides and solving above equation, $\left[{\text{OH}}^{-}\right]=2.9\times {10}^{-5}$ .

As the $\text{pH}$ is more than $7$ , so the solution is basic in nature.

Therefore, $\left[{\text{H}}_3{\text{O}}^{+}\right]=3.5\times {10}^{-8}$ , $\left[{\text{OH}}^{-}\right]=2.9\times {10}^{-5}$ and $\text{pOH}=5.46$ for the given value of $\text{pH}$ , and the nature of the solution is basic.

(e)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}_3{\text{O}}^{+}$ , $\text{pH}$ and $\text{pOH}$ for $\left[{\text{OH}}^{-}\right]=9.0\times {10}^{-10}$ is to be found and, nature of solution is to be determined.

Explanation of Solution

Ionic product of water is defined as the product of concentration of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions present in the water. It is represented as $K_{\text{w}}$ .

Expression for $K_{\text{w}}$ is as follows:

$K_{\text{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Here, $K_{\text{w}}$ is the ionic product of water, $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions and $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$ ions.

Substitute, $9.0\times {10}^{-10}$ for $\left[{\text{OH}}^{-}\right]$ and $1.0\times {10}^{-14}$ for $K_{\text{w}}$ in the above equation.

$\begin{array}{c}1.0\times {10}^{-14}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left(9.0\times {10}^{-10}\right)\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{1.0\times {10}^{-14}}{9.0\times {10}^{-10}}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=1.1\times {10}^{-5}\end{array}$

$\text{pH}$ of a solution gives the concentration of hydrogen ions in that solution. Range of $\text{pH}$ values is from $0$ to $14$ . Solutions having $\text{pH}$ values less than $7$ are acidic, those having greater than $7$ are basic. Solutions with $\text{pH}=7$ are neutral. $\text{pOH}$ of a solution gives the concentration of ${\text{OH}}^{-}$ ions in that solution.

$\text{pOH}$ can be found using concentration of ${\text{OH}}^{-}$ as follows:

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

Substitute, $9.0\times {10}^{-10}$ for $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}\text{pOH}=-\text{log}\left(9.0\times {10}^{-10}\right)\\ \text{pOH}=9.05\end{array}$

The relation between $\text{pH}$ and $\text{pOH}$ is as follows:

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

Substitute, $9.05$ for $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+9.05=14\\ \text{pH}=14-9.05\\ \text{pH}=4.95\end{array}$

As the $\text{pH}$ is less than $7$ , so the solution is acidic in nature.

Therefore, $\left[{\text{H}}_3{\text{O}}^{+}\right]=1.1\times {10}^{-5}$ , $\text{pOH}=9.05$ and $\text{pH}=4.95$ for the given value of $\left[{\text{OH}}^{-}\right]$ , and the nature of the solution is acidic.