Chapter 8, Problem 28Q

(d)

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion corresponds to given concentration of hydrogen ion has to be calculated.

Concept Introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
• $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• $\text{pH}\text{+}\text{pOH}\text{=}\text{14}$

Explanation of Solution

Given,

    ${\text{[H}}^{\text{+}}\text{] = 1 ×}{\text{10}}^{\text{-8}}\text{M}$

The concentration of hydroxide ion is calculated as,

    $\begin{array}{l}\text{pH}\text{=}\text{-log}\text{(1 ×}{\text{10}}^{\text{-8}}\text{M)}\text{=}\text{8}\\ \\ \text{pH}\text{+}\text{pOH}\text{=}\text{14}\Rightarrow \text{pOH}\text{=}\text{14}\text{-8}\text{=}\text{6}\\ \\ \text{pOH}\text{=}\text{6}\\ \\ {\text{-log([OH}}^{\text{-}}\text{])}\text{=6}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\text{1 ×}{\text{10}}^{\text{-6}}\text{M}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion corresponds to given concentration of hydrogen ion has to be calculated.

Concept Introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
• $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• $\text{pH}\text{+}\text{pOH}\text{=}\text{14}$

Explanation of Solution

Given,

    ${\text{[H}}^{\text{+}}\text{] = 5 ×}{\text{10}}^{\text{-7}}\text{M}$

The concentration of hydroxide ion is calculated as,

    $\begin{array}{l}\text{pH}\text{=}\text{-log}\text{(5 ×}{\text{10}}^{\text{-7}}\text{M)}\text{=}6\\ \\ \text{pH}\text{+}\text{pOH}\text{=}\text{14}\Rightarrow \text{pOH}\text{=}\text{14}\text{-6}\text{=}8\\ \\ \text{pOH}\text{=}8\\ \\ {\text{-log([OH}}^{\text{-}}\text{])}\text{=8}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\text{1 ×}{\text{10}}^{\text{-8}}\text{M}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The concentration of hydrogen ion corresponds to given concentration of hydroxide ion has to be calculated.

Concept Introduction:

• $\text{pOH}$ is the logarithm of the reciprocal of the concentration  of ${\text{OH}}^{\text{-}}$  in a solution.
• $\text{pOH}$ is used to determine the basicity of an aqueous solution.
• $\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$
• $\text{pH}\text{+}\text{pOH}\text{=}\text{14}$

Explanation of Solution

Given,

    ${\text{[OH}}^{\text{-}}\text{] = 1 ×}{\text{10}}^{\text{-2}}\text{M}$

The concentration of hydrogen ion is calculated as,

    $\begin{array}{l}\text{pOH}\text{=}\text{-log}\text{(1 ×}{\text{10}}^{\text{-2}}\text{M)}\text{=}2\\ \\ \text{pH}\text{+}\text{pOH}\text{=}\text{14}\Rightarrow \text{pH}\text{=}\text{14}\text{-2}\text{=}12\\ \\ \text{pH}\text{=}12\\ \\ {\text{-log([H}}^{+}\text{])}\text{=12}\\ \\ {\text{[H}}^{+}\text{]}\text{=}\text{1 ×}{\text{10}}^{\text{-12}}\text{M}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The concentration of hydrogen ion corresponds to given concentration of hydroxide ion has to be calculated.

Concept Introduction:

• $\text{pOH}$ is the logarithm of the reciprocal of the concentration  of ${\text{OH}}^{\text{-}}$  in a solution.
• $\text{pOH}$ is used to determine the basicity of an aqueous solution.
• $\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{]}$
• $\text{pH}\text{+}\text{pOH}\text{=}\text{14}$

Explanation of Solution

Given,

    ${\text{[OH}}^{\text{-}}\text{] = 1 ×}{\text{10}}^{\text{-12}}\text{M}$

The concentration of hydrogen ion is calculated as,

    $\begin{array}{l}\text{pOH}\text{=}\text{-log}\text{(1 ×}{\text{10}}^{\text{-12}}\text{M)}\text{=}12\\ \\ \text{pH}\text{+}\text{pOH}\text{=}\text{14}\Rightarrow \text{pH}\text{=}\text{14}\text{-12}\text{=}2\\ \\ \text{pH}\text{=}2\\ \\ {\text{-log([H}}^{+}\text{])}\text{=2}\\ \\ {\text{[H}}^{+}\text{]}\text{=}\text{1 ×}{\text{10}}^{\text{-2}}\text{M}\end{array}$