Chapter 19, Problem 86A

a.

Interpretation Introduction

Interpretation : The hydroxide-ion concentration for the solution is to be calculated.

Concept Introduction : The negative logarithm of the concentration of hydrogen ions is the pH of a solution. The formula to calculate pH is given as:

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

Answer to Problem 86A

The concentration of hydroxide ion is $4\times {10}^{-10}\text{M}$ .

Explanation of Solution

The pH value is $4.60$ .

The formula to calculate pH is given as:

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

Rearrange the above equation with respect to hydrogen ions.

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

To calculate the hydrogen ion, substitute the values in the above formula:

  $\begin{array}{c}{\text{[H}}^{\text{+}}\text{] = antilog}\left(-\text{pH}\right)\\ =\text{antilog}\left(-4.60\right)\\ {\text{[H}}^{\text{+}}\text{]}=2.51\times {10}^{-5}\text{M}\end{array}$

The concentration of hydroxide ions is given by rearranging the formula of the ion product constant for water $\left({\text{K}}_{\text{w}}\right)$ .

  $[{\text{OH}}^{-}\text{]}=\frac{{\text{K}}_w}{[{\text{H}}^{\text{+}}]}$

To calculate the concentration of hydrogen ions, substitute the values in the above formula:

  $\begin{array}{c}[{\text{OH}}^{-}\text{]}=\frac{{\text{K}}_w}{[{\text{H}}^{\text{+}}]}\\ =\frac{1\times {10}^{-14}{\text{M}}^2}{2.51\times {10}^{-5}\text{M}}\\ =0.4\times {10}^{-9}\text{M}\\ [{\text{OH}}^{-}\text{]}=4\times {10}^{-10}\text{M}\end{array}$

b.

Interpretation Introduction

Interpretation : The hydroxide-ion concentration for the solution is to be calculated.

Concept Introduction : The negative logarithm of the concentration of hydrogen ions is the pH of a solution. The formula to calculate pH is given as

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

Answer to Problem 86A

The concentration of hydroxide ion is $2\times {10}^{-5}\text{M}$ .

Explanation of Solution

The pH value is $9.30$ .

The formula to calculate pH is given as

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

Rearrange the above equation with respect to hydrogen ions.

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

To calculate the hydrogen ion, substitute the values in the above formula:

  $\begin{array}{c}{\text{[H}}^{\text{+}}\text{] = antilog}\left(-\text{pH}\right)\\ =\text{antilog}\left(-9.30\right)\\ {\text{[H}}^{\text{+}}\text{]}=5.01\times {10}^{-10}\text{M}\end{array}$

The concentration of hydroxide ions is given by rearranging the formula of the ion product constant for water $\left({\text{K}}_{\text{w}}\right)$ .

  $[{\text{OH}}^{-}\text{]}=\frac{{\text{K}}_w}{[{\text{H}}^{\text{+}}]}$

To calculate the concentration of hydrogen ions, substitute the values in the above formula:

  $\begin{array}{c}[{\text{OH}}^{-}\text{]}=\frac{{\text{K}}_w}{[{\text{H}}^{\text{+}}]}\\ =\frac{1\times {10}^{-14}{\text{M}}^2}{5.01\times {10}^{-10}\text{M}}\\ =0.2\times {10}^{-4}\text{M}\\ [{\text{OH}}^{-}\text{]}=2\times {10}^{-5}\text{M}\end{array}$

c.

Interpretation Introduction

Interpretation : The pH of the solution is to be found.

Concept Introduction : The negative logarithm of the concentration of hydrogen ions is the pH of a solution. The formula to calculate pH is given as

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

Answer to Problem 86A

The pH of the solution is $12.25$ .

Explanation of Solution

The concentration of hydroxide ion is $1.8\times {10}^{-2}\text{M}$ .

The formula to calculate pH is given as

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

The concentration of hydrogen ions is given by rearranging the formula of the ion product constant for water $\left({\text{K}}_{\text{w}}\right)$ .

  $[{\text{H}}^{\text{+}}]=\frac{{\text{K}}_w}{[{\text{OH}}^{-}\text{]}}$

To calculate the concentration of hydrogen ions, substitute the values in the above formula:

  $\begin{array}{c}[{\text{H}}^{\text{+}}]=\frac{{\text{K}}_w}{[{\text{OH}}^{-}\text{]}}\\ =\frac{1\times {10}^{-14}{\text{M}}^2}{1.8\times {10}^{-2}\text{M}}\\ =0.55\times {10}^{-12}\text{M}\end{array}$

To calculate the pH of the solution, substitute the values in the pH formula:

  $\begin{array}{c}\text{pH=}-{\text{log[H}}^{\text{+}}\text{]}\\ \text{=}-\log \left(0.55\times {10}^{-12}\right)\\ \text{pH}=12.25\end{array}$

d.

Interpretation Introduction

Interpretation : The pH of the solution is to be found.

Concept Introduction : The negative logarithm of the concentration of hydrogen ions is the pH of a solution. The formula to calculate pH is given as

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

Answer to Problem 86A

The pH of the solution is $5.86$ .

Explanation of Solution

The concentration of hydroxide ion is $7.3\times {10}^{-9}\text{M}$ .

The formula to calculate pH is given as

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

The concentration of hydrogen ions is given by rearranging the formula of the ion product constant for water $\left({\text{K}}_{\text{w}}\right)$ .

  $[{\text{H}}^{\text{+}}]=\frac{{\text{K}}_w}{[{\text{OH}}^{-}\text{]}}$

To calculate the concentration of hydrogen ions, substitute the values in the above formula:

  $\begin{array}{c}[{\text{H}}^{\text{+}}]=\frac{{\text{K}}_w}{[{\text{OH}}^{-}\text{]}}\\ =\frac{1\times {10}^{-14}{\text{M}}^2}{7.3\times {10}^{-9}\text{M}}\\ =0.137\times {10}^{-5}\text{M}\end{array}$

To calculate the pH of the solution, substitute the values in the pH formula:

  $\begin{array}{c}\text{pH=}-{\text{log[H}}^{\text{+}}\text{]}\\ \text{=}-\log \left(0.137\times {10}^{-5}\right)\\ \text{pH}=5.86\end{array}$