Chapter 19.2, Problem 10SP

(a)

Interpretation Introduction

Interpretation: The given solution with a concentration of hydrogen ion $6.0\times {10}^{-10}\text{ M}$ needs to be classified as acidic, basic, or neutral.

Concept Introduction: For a given solution, pH can be calculated as follows:

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

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

Explanation of Solution

The given concentration of hydroxide ion is $6.0\times {10}^{-10}\text{ M}$ . From hydrogen ion concentration, the pH of the solution can be calculated as follows:

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

Substitute the values,

  $\begin{array}{c}\text{pH}=-\log \left(6.0\times {10}^{-10}\right)\\ =9.22\end{array}$

Since the pH of the solution is more than 7, thus, the solution is basic in nature.

(b)

Interpretation Introduction

Interpretation: The given solution with a concentration of hydroxide ion $3.0\times {10}^{-2}\text{ M}$ needs to be classified as acidic, basic, or neutral.

Concept Introduction: For a given solution, pH can be calculated as follows:

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

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

Similarly, pOH can be calculated as follows:

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

Here, $\left[{\text{OH}}^{-}\right]$ is the hydroxide ion concentration.

The relation between pH and pOH of the solution is represented as follows:

  $\text{pH}+\mathrm{pOH}=14$

Explanation of Solution

The given concentration of hydroxide ion is $3.0\times {10}^{-2}\text{ M}$ . From hydroxide ion concentration, pH is calculated from the pOH value. Here, pOH is calculated as follows:

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

Substitute the values,

  $\begin{array}{c}\text{pOH}=-\log \left(3.0\times {10}^{-2}\text{ M}\right)\\ =1.52\end{array}$

Now, pH is calculated from pOH as follows:

  $\text{pH}=14-\text{pOH}$

Substitute the value,

  $\begin{array}{c}\text{pH}=14-\text{1}\text{.52}\\ \text{=12}\text{.48}\end{array}$

Since the pH of the solution is more than 7, the solution will be basic in nature.

(c)

Interpretation Introduction

Interpretation: The given solution with a concentration of hydrogen ion $2.0\times {10}^{-7}\text{ M}$ needs to be classified as acidic, basic, or neutral.

Concept Introduction: For a given solution, pH can be calculated as follows:

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

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

Explanation of Solution

The given concentration of hydrogen ion is $2.0\times {10}^{-7}\text{ M}$ . From hydrogen ion concentration, the pH of the solution can be calculated as follows:

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

Substitute the values,

  $\begin{array}{c}\text{pH}=-\log \left(2.0\times {10}^{-7}\right)\\ =6.7\end{array}$

Since the pH of the solution is less than 7, thus, the solution is acidic in nature.

(d)

Interpretation Introduction

Interpretation: The given solution with concentration of hydroxide ion $1.0\times {10}^{-7}\text{ M}$ needs to be classified as acidic, basic, or neutral.

Concept Introduction: For a given solution, pH can be calculated as follows:

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

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

Similarly, pOH can be calculated as follows:

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

Here, $\left[{\text{OH}}^{-}\right]$ is the hydroxide ion concentration.

The relation between pH and pOH of the solution is represented as follows:

  $\text{pH}+\mathrm{pOH}=14$

Explanation of Solution

The given concentration of hydroxide ion is $1.0\times {10}^{-7}\text{ M}$ . From hydroxide ion concentration, pH is calculated from the pOH value. Here, pOH is calculated as follows:

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

Substitute the values,

  $\begin{array}{c}\text{pOH}=-\log \left(1.0\times {10}^{-7}\right)\\ =7\end{array}$

Now, pH is calculated from pOH as follows:

  $\text{pH}=14-\text{pOH}$

Substitute the value,

  $\begin{array}{c}\text{pH}=14-\text{7}\\ \text{=7}\end{array}$

Since the pH of the solution is equal to 7, the solution will be neutral in nature.