Chapter 13, Problem 51E

Interpretation Introduction

Interpretation: The missing information in the given table is to be stated.

Concept introduction: The $\text{pH}$ is the measure of its $[{\text{H}}^{+}]$ of any solution. The $\text{pOH}$ is the measure of the $\left[{\text{OH}}^{-}\right]$ of any solution. $\left[{\text{H}}^{+}\right]$ represents the total hydrogen ion concentration. $\left[{\text{OH}}^{-}\right]$ represents the total hydroxide ion concentration. The nature of any solution is determined by its $\text{pH}$ and $\text{pOH}$ value.

To determine: The missing values in the table for solution a.

Answer to Problem 51E

The given table has been completed as follows.

$\begin{array}{cccccc}\text{S}\text{.No}\text{.}& \text{pH}& \text{pOH}& [{\text{H}}^{+}]& [{\text{OH}}^{-}]& \text{Acidic,}\text{Basic,}\text{or}\text{Neutral}\\ \text{Solution}\text{a}& 6.88& 7.12& 1.31\times {10}^{-7}\text{M}& 7.6\times {10}^{-8}\text{M}& \text{Acidic}\\ \text{Solution}\text{b}& 0.92& 13.08& 0.119\text{M}& 8.4\times {10}^{-14}\text{M}& \text{Acidic}\\ \text{Solution}\text{c}& 10.89& 3.11& 1.28\times {10}^{-11}\text{M}& 7.8\times {10}^{-4}\text{M}& \text{Basic}\\ \text{Solution}\text{d}& 7& 7& 1.0\times {10}^{-7}\text{M}& 1.0\times {10}^{-7}\text{M}& \text{Neutral}\end{array}$

Explanation of Solution

Explanation

For solution a

Given

The $\text{pH}$ of the solution is $6.88$.

The sum,

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

Where,

• $\text{pOH}$ is the measure of  hyroxide ion concentration.
• $\text{pH}$ is the measure of hydrogen ion concentration.

Substitute the value of $\text{pH}$ in the above equation.

$\begin{array}{c}6.88+\text{pOH}=\text{14}\\ \text{pOH}=\text{14}-\text{6}\text{.88}\\ \text{pOH}=\underset{\_}{7.12}\end{array}$

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{1.31\times 10^{-7}M}$.

Given

The $\text{pH}$ of the solution is $6.88$.

The $\text{pH}$ value is calculated by the formula,

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

Where,

• $\left[{\text{H}}^{+}\right]$ is the total concentration of hydrogen ions.

Rearrange the above equation to calculate the value of $\left[{\text{H}}^{\text{+}}\right]$.

$\left[{\text{H}}^{+}\right]=\text{antilog}\left(-\text{pH}\right)$

Substitute the value of $\text{pH}$ in the above equation.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\text{antilog}\left(-\text{pH}\right)\\ \left[{\text{H}}^{+}\right]=\text{Antilog(}-\text{6}\text{.88)}\\ \left[{\text{H}}^{+}\right]=\underset{\_}{1.31\times 10^{-7}M}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{7.6\times 10^{-8}M}$.

Given

The $\left[{\text{H}}^{+}\right]$ is $1.31\times {10}^{-7}\text{M}$.

The ionic product of water is,

$\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$

Where,

• $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ion.
• $\left[{\text{OH}}^{-}\right]$ is the concentration of hydroxide ion.

Substitute the value of $[{\text{H}}^{+}]$ in the above equation.

$\begin{array}{c}1.31\times {10}^{-7}\times [{\text{OH}}^{-}]=1.0\times {10}^{-14}\\ [{\text{OH}}^{-}]=\frac{1.0\times {10}^{-14}}{1.31\times {10}^{-7}}\\ [{\text{OH}}^{-}]=\underset{\_}{7.6\times 10^{-8}M}\end{array}$

The solution is acidic.

In the given solution, the $[{\text{H}}^{+}]>[{\text{OH}}^{-}]$. Therefore, the given solution is acidic.

For solution b

Given

The $[{\text{OH}}^{-}]$ is $8.4\times {10}^{-14}$.

The ionic product of water is,

$[{\text{H}}^{+}][{\text{OH}}^{-}]=1.0\times {10}^{-14}$

Where,

• $\left[{\text{H}}^{+}\right]$ is the hydrogen ion concentration.
• $\left[{\text{OH}}^{-}\right]$ is the hydroxide ion concentration.

Substitute the value of $[{\text{OH}}^{-}]$ in the above equation.

$\begin{array}{c}[{\text{H}}^{+}]\times 8.4\times {10}^{-14}=1.0\times {10}^{-14}\\ [{\text{H}}^{+}]=\frac{1.0\times {10}^{-14}}{8.4\times {10}^{-14}}\\ [{\text{H}}^{+}]=\underset{\_}{0.119M}\end{array}$

The $\text{pH}$ is $\underset{\_}{0.92}$

Given

The $\left[{\text{H}}^{+}\right]$ is $0.119\text{M}$

The $\text{pH}$ value is calculated by the formula,

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

Where,

• $\left[{\text{H}}^{+}\right]$ is the total concentration of hydrogen ions.

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the above equation.

$\begin{array}{l}\text{pH}=-\text{log[0}\text{.119]}\\ \text{pH}=\underset{\_}{0.92}\end{array}$

The $\text{pOH}$ is $\underset{\_}{13.08}$.

Given

The $\text{pH}$ is $0.92$.

The sum, $\text{pH}+\text{pOH}=14$

Substitute the value of $\text{pH}$ in the above equation.

$\begin{array}{c}0.92+\text{pOH}=14\\ \text{pOH}=14-0.92\\ \text{pOH}=\underset{\_}{13.08}\end{array}$

The solution b is acidic.

In the given solution, the $[{\text{H}}^{+}]>[{\text{OH}}^{-}]$. Therefore, the given solution is acidic.

For solution c

Given

The $\text{pOH}$ is $3.11$.

The sum, $\text{pH}+\text{pOH}=14$

Substitute value of $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+3.11=14\\ \text{pH}=14-3.11\\ \text{pH}=\underset{\_}{10.89}\end{array}$

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{1.28\times 10^{-11}M}$.

Given

The $\text{pH}$ is $10.89$

The $\text{pH}$ value is calculated by the formula,

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

Where,

• $\left[{\text{H}}^{+}\right]$ is the total concentration of hydrogen ions.

Rearrange the above equation to calculate the value of $\left[{\text{H}}^{\text{+}}\right]$.

$\left[{\text{H}}^{+}\right]=\text{antilog}\left(-\text{pH}\right)$

Substitute the value of $\text{pH}$ in the above equation.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\text{antilog[}-\text{10}\text{.89]}\\ \left[{\text{H}}^{+}\right]=\underset{\_}{1.28\times 10^{-11}}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{7.8\times 10^{-4}M}$.

Given

The $\left[{\text{H}}^{+}\right]$ is $1.28\times {10}^{-11}\text{M}$.

The ionic product of water is,

$\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above equation.

$\begin{array}{c}1.28\times {10}^{-11}\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}\\ \left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{1.28\times {10}^{-11}}\\ \left[{\text{OH}}^{-}\right]=\underset{\_}{7.8\times 10^{-4}M}\end{array}$

The solution c is basic.

In the given case, the $[{\text{OH}}^{-}]>[{\text{H}}^{+}]$; therefore, the solution is basic.

For solution d

Given

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

The $\text{pH}$ value is calculated by the formula,

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

Where,

• $\left[{\text{H}}^{+}\right]$ is the total concentration of hydrogen ions.

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\text{log[1}\text{.0}\times {\text{10}}^{-7}]\\ \text{pH}=\underset{\_}{7}\end{array}$

The $\text{pOH}$ is $\underset{\_}{7}$.

Given

The $\text{pH}$ is $7$.

The sum, $\text{pH}+\text{pOH}=14$

Substitute value of $\text{pH}$ in the above equation.

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

The $[{\text{OH}}^{-}]$ is $\underset{\_}{1.0\times 10^{-7}}$.

Given

The $[{\text{H}}^{+}]$ is $1.0\times {10}^{-7}\text{M}$.

The ionic product of water is,

$[{\text{H}}^{+}][{\text{OH}}^{-}]=1.0\times {10}^{-14}$

Substitute the value of $[{\text{H}}^{+}]$ in the above equation.

$\begin{array}{c}1.0\times {10}^{-7}[{\text{OH}}^{-}]=1.0\times {10}^{-14}\\ [{\text{OH}}^{-}]=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-7}}\\ [{\text{OH}}^{-}]=\underset{\_}{1.0\times 10^{-7}M}\end{array}$

The solution d is neutal.

In the given solution, the $[{\text{H}}^{+}]=[{\text{OH}}^{-}]$; hence, the solution is neutral.

Conclusion

Conclusion

The given table has been completed as follows.

$\begin{array}{cccccc}\text{S}\text{.No}\text{.}& \text{pH}& \text{pOH}& [{\text{H}}^{+}]& [{\text{OH}}^{-}]& \text{Acidic,}\text{Basic,}\text{or}\text{Neutral}\\ \text{Solution}\text{a}& 6.88& 7.12& 1.31\times {10}^{-7}\text{M}& 7.6\times {10}^{-8}\text{M}& \text{Acidic}\\ \text{Solution}\text{b}& 0.92& 13.08& 0.119\text{M}& 8.4\times {10}^{-14}\text{M}& \text{Acidic}\\ \text{Solution}\text{c}& 10.89& 3.11& 1.28\times {10}^{-11}\text{M}& 7.8\times {10}^{-4}\text{M}& \text{Basic}\\ \text{Solution}\text{d}& 7& 7& 1.0\times {10}^{-7}\text{M}& 1.0\times {10}^{-7}\text{M}& \text{Neutral}\end{array}$