Chapter 8, Problem 8.72QP

(a)

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion in an aqueous solution has to be calculated.

Concept Introduction:

Self-Ionization of water:  Self-ionization is the process in which transfer of a proton from one water molecule to another occurs, which results in formation of hydronium ion and hydroxide ion.

    $\begin{array}{l}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l) }}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\rightleftarrows }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\\ \\ {\text{K}}_{\text{w}}{\text{ = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\end{array}$

Explanation of Solution

Given,

    ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\text{1}\text{.0 ×}{\text{10}}^{\text{-6}}\text{M}$

The concentration of hydroxide ion in an aqueous solution is calculated as,

    $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\text{=}\frac{\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}}{\text{1}\text{.0}\text{×}{\text{10}}^{\text{-6}}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-8}}\text{M}\\ \\ {\text{[OH}}^{\text{-}}\text{]=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-8}}\text{M}\text{.}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion in an aqueous solution has to be calculated.

Concept Introduction:

Refer to part a.

Explanation of Solution

Given,

    ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}\text{1}\text{.0 ×}{\text{10}}^{\text{-8}}\text{M}$

The concentration of hydroxide ion in an aqueous solution is calculated as,

    $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\text{=}\frac{\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}}{\text{1}\text{.0}\text{×}{\text{10}}^{\text{-8}}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-6}}\text{M}\\ \\ {\text{[OH}}^{\text{-}}\text{]=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-6}}\text{M}\text{.}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion in an aqueous solution has to be calculated.

Concept Introduction:

Refer to part a.

Explanation of Solution

Given,

    ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\text{=}5.6\text{ ×}{\text{10}}^{\text{-4}}\text{M}$

The concentration of hydroxide ion in an aqueous solution is calculated as,

    $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\text{=}\frac{\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}}{5.6\text{ ×}{\text{10}}^{\text{-4}}}\text{=}\text{1}\text{.78}\text{×}{\text{10}}^{\text{-11}}\text{M}\\ \\ {\text{[OH}}^{\text{-}}\text{]=}\text{1}\text{.78}\text{×}{\text{10}}^{\text{-11}}\text{M}\text{.}\end{array}$