Chapter 14, Problem 52E

Fill in the missing information in the following table.

Interpretation Introduction

Interpretation:

The missing information in the given table is to be identified.

Concept introduction:

The $\text{pH}$ is used to measure the $\left[{\text{H}}^{+}\right]$ of any solution.

The $\text{pOH}$ is used to measure the $\left[{\text{OH}}^{-}\right]$ of any solution.

The $\left[{\text{H}}^{+}\right]$ represents the total hydrogen ion concentration.

The $\left[{\text{OH}}^{-}\right]$ represents the total hydroxide ion concentration.

The nature of any solution is determined by its $\text{pH}$ and $\text{pOH}$ value.

Answer to Problem 52E

The missing information have been rightfully identified,

$\begin{array}{cccccc}& \text{pH}& \text{pOH}& [{\text{H}}^{+}]& [{\text{OH}}^{-}]& \text{Acidic}\text{or}\text{Basic}\text{or}\text{Neutral}\\ \text{Solution}\text{a}& 9.63& 4.37& 2.34\times {10}^{-10}\text{M}& 4.27\times {10}^{-5}\text{M}& \text{Basic}\\ \text{Solution}\text{b}& 8.59& 5.41& 2.56\times {10}^{-9}\text{M}& 3.9\times {10}^{-6}\text{M}& \text{Basic}\\ \text{Solution}\text{c}& 1.57& 12.43& 0.027\text{M}& 3.72\times {10}^{-13}\text{M}& \text{Acidic}\\ \text{Solution}\text{d}& 12.8& 1.22& 1.66\times {10}^{-13}\text{M}& 0.060\text{M}& \text{Basic}\end{array}$

Explanation of Solution

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

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{2.34\times 10^{-10}M}$.

Given

The $\text{pH}$ is $9.63$.

The formula to calculate $\text{pH}$ is,

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

Where

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

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

$\begin{array}{c}9.63=-\log \left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]=\text{Antilog(}-\text{pH)}\\ \left[{\text{H}}^{+}\right]=\text{Antilog(}-\text{9}\text{.63)}\\ \left[{\text{H}}^{+}\right]=2.34\times {10}^{-10}\text{M}\end{array}$

The $\text{pOH}$ is $\underset{\_}{4.37}$.

Given

The $\text{pH}$ is $9.63$

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.

$\begin{array}{c}9.63+\text{pOH}=\text{14}\\ \text{pOH}=\text{14}-9.63\\ \text{pOH}=\underset{\_}{4.37}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{4.27\times 10^{-4}M}$.

Given

The $[{\text{H}}^{+}]$ is $2.34\times {10}^{-10}\text{M}$.

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.

$\begin{array}{c}2.34\times {10}^{-10}\times \left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}\\ \left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{2.34\times {10}^{-10}}\\ \left[{\text{OH}}^{-}\right]=\underset{\_}{4.27\times 10^{-4}M}\end{array}$

The solution a is basic.

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

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

The $\text{pOH}$ is $\underset{\_}{5.41}$.

Given

The $[{\text{OH}}^{-}]$ is $3.9\times {10}^{-6}\text{M}$.

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

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

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

$\begin{array}{l}\text{pOH}=-\text{log}\left[\text{3}\text{.9}\times {\text{10}}^{-6}\right]\\ \text{pOH}=\underset{\_}{5.41}\end{array}$

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{2.57\times 10^{-9}M}$.

Given

The $\text{pH}$ is $8.59$

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

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

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

$\begin{array}{c}8.59=-\text{log}\left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]=\text{antilog}\left[\text{8}\text{.59}\right]\\ \left[{\text{H}}^{+}\right]=\underset{\_}{2.57\times 10^{-9}}\end{array}$

The solution b is basic.

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

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

The $\text{pH}$ is $\underset{\_}{1.56}$.

Given

The $\left[{\text{H}}^{+}\right]$ is $0.027\text{M}$.

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

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

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

$\begin{array}{c}\text{pH}=-\text{log}\left[\text{0}\text{.027}\right]\\ \text{pH}=\underset{\_}{1.56}\end{array}$

The $\text{pOH}$ is $\underset{\_}{12.4}$.

Given

The $\text{pH}$ is $1.56$.

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

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

$\begin{array}{c}1.56+\text{pOH}=14\\ \text{pOH}=14-1.56\\ \text{pOH}=\underset{\_}{12.4}\end{array}$

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{3.98\times 10^{-13}M}$.

Given

The $\left[{\text{H}}^{+}\right]$ is $0.027\text{M}$.

The ionic product of water is,

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

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

$\begin{array}{c}0.027\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}\\ \left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{0.027}\\ \left[{\text{OH}}^{-}\right]=\underset{\_}{3.98\times 10^{-13}M}\end{array}$

The solution is acidic.

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

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

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{0.060M}$.

Given

The $\text{pOH}$ is $1.22$

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

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

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

$\begin{array}{c}1.22=-\text{log}\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]=\text{Antilog(-1}\text{.22)}\\ \left[{\text{OH}}^{-}\right]=\underset{\_}{0.60M}\end{array}$

The $\text{pH}$ is $\underset{\_}{12.8}$.

Given

The $\text{pOH}$ is $1.22$.

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

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

$\begin{array}{c}\text{pH}+1.22=14\\ \text{pH}=14-1.22\\ \text{pH}=\underset{\_}{12.8}\end{array}$

The $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{1.58\times 10^{-13}M}$.

Given

The $\text{pH}$ is $12.8$

The $\text{pH}$ is calculated by theformual,

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

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

$\begin{array}{c}12.8=-\text{log}\left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]=\text{antilog}\left[\text{-12}\text{.8}\right]\\ \left[{\text{H}}^{+}\right]=\underset{\_}{1.58\times 10^{-13}}\end{array}$

The solution d is basic.

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

Conclusion

The given table has been completed as follows.

$\begin{array}{cccccc}\text{S}.\text{No}.& \text{pH}& \text{pOH}& [{\text{H}}^{+}]& [{\text{OH}}^{-}]& \text{Acidic}\text{or}\text{Basic}\text{or}\text{Neutral}\\ \text{Solution}\text{a}& 9.63& 4.37& 2.34\times {10}^{-10}\text{M}& 4.27\times {10}^{-5}\text{M}& \text{Basic}\\ \text{Solution}\text{b}& 8.59& 5.41& 2.56\times {10}^{-9}\text{M}& 3.9\times {10}^{-6}\text{M}& \text{Basic}\\ \text{Solution}\text{c}& 1.57& 12.43& 0.027\text{M}& 3.72\times {10}^{-13}\text{M}& \text{Acidic}\\ \text{Solution}\text{d}& 12.8& 1.22& 1.66\times {10}^{-13}\text{M}& 0.060\text{M}& \text{Basic}\end{array}$