Chapter 19.2, Problem 16SP

(a)

Interpretation Introduction

Interpretation: The pH of the given solution with a concentration of hydroxide ions $4.3\times {10}^{-5}\text{ M}$ needs to be calculated.

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 $4.3\times {10}^{-5}\text{ M}$ . 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,

  $\begin{array}{c}\text{pOH}=-\log \left(4.3\times {10}^{-5}\text{ M}\right)\\ =4.36\end{array}$

Now, pH is calculated from pOH as follows:

  $\text{pH}=14-\text{pOH}$

Substitute the value,

  $\begin{array}{c}\text{pH}=14-4.36\\ =\text{9}\text{.64}\end{array}$

Thus, the pH of solution is 9.64.

(b)

Interpretation Introduction

Interpretation: The pH of the given solution with a concentration of hydroxide ions $4.5\times {10}^{-11}\text{ M}$ needs to be calculated.

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 $4.5\times {10}^{-11}\text{ M}$ . 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,

  $\begin{array}{c}\text{pOH}=-\log \left(4.5\times {10}^{-11}\text{ M}\right)\\ =10.35\end{array}$

Now, pH is calculated from pOH as follows:

  $\text{pH}=14-\text{pOH}$

Substitute the value,

  $\begin{array}{c}\text{pH}=14-10.35\\ =\text{3}\text{.65}\end{array}$

Thus, the pH of the solution is 3.65.