Chapter 10, Problem 10.69EP

(a)

Interpretation Introduction

Interpretation:

The ${\text{[OH}}^{\text{-}}\text{]}$ in aqueous solutions has to be calculated with the hydronium ion concentration of $\text{5}{\text{.5×10}}^{\text{-2}}\text{ M}$.

Concept Introduction:

Dissociation constant of water:

The dissociation of water can be given as

${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}$

The concentrations of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ and OH}}^{\text{-}}$ at room temperature are $\text{1}{\text{.0×10}}^{\text{-7}}\text{ M}$ each.

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = [OH}}^{\text{-}}\text{] = 1}{\text{.0×10}}^{\text{-7}}\\ \\ {\text{K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ {\text{K}}_{\text{w}}\text{= (1}{\text{.0×10}}^{\text{-7}}\text{)(1}{\text{.0×10}}^{\text{-7}}\text{)}\\ \\ {\text{K}}_{\text{w}}\text{= 1}{\text{.0×10}}^{\text{-14}}\text{M}\end{array}$

Explanation of Solution

Given,

$\begin{array}{l}\text{Hydronium ion concentration}\text{= 5}{\text{.5×10}}^{\text{-2}}\text{ M}\\ \\ \text{Ion product constant for water = 1}\text{.0}\times {\text{10}}^{-14}\text{ M}\end{array}$

To calculate the molar hydronium ion concentration in aqueous solutions

$\begin{array}{l}{\text{K}}_{\text{w}}{\text{=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}\text{M}}{\text{5}{\text{.5×10}}^{\text{-2}}\text{M}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=1}{\text{.8×10}}^{\text{-13}}\text{ M}\end{array}$

The ${\text{[OH}}^{\text{-}}\text{]}$ in aqueous solutions with the hydronium ion concentration of $\text{5}{\text{.5×10}}^{\text{-2}}\text{ M}$ is $\text{1}{\text{.8×10}}^{\text{-13}}\text{ M}$.

(b)

Interpretation Introduction

Interpretation:

The molar hydronium ion concentration in aqueous solutions has to be calculated with the hydroxide ion concentration of $\text{9}{\text{.4×10}}^{\text{-5}}\text{ M}$.

Concept Introduction:

Dissociation constant of water:

The dissociation of water can be given as

${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}$

The concentrations of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ and OH}}^{\text{-}}$ at room temperature are $\text{1}{\text{.0×10}}^{\text{-7}}\text{ M}$ each.

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = [OH}}^{\text{-}}\text{] = 1}{\text{.0×10}}^{\text{-7}}\\ \\ {\text{K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ {\text{K}}_{\text{w}}\text{= (1}{\text{.0×10}}^{\text{-7}}\text{)(1}{\text{.0×10}}^{\text{-7}}\text{)}\\ \\ {\text{K}}_{\text{w}}\text{= 1}{\text{.0×10}}^{\text{-14}}\text{M}\end{array}$

Explanation of Solution

Given,

$\begin{array}{l}\text{Hydronium ion concentration}\text{= 9}{\text{.4×10}}^{\text{-5}}\text{ M}\\ \\ \text{Ion product constant for water = 1}\text{.0}\times {\text{10}}^{-14}\text{ M}\end{array}$

To calculate the molar hydronium ion concentration in aqueous solutions

$\begin{array}{l}{\text{K}}_{\text{w}}{\text{=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}\text{M}}{\text{9}{\text{.4×10}}^{\text{-5}}\text{M}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=1}{\text{.1×10}}^{\text{-10}}\text{ M}\end{array}$

The ${\text{[OH}}^{\text{-}}\text{]}$ in aqueous solutions with the hydronium ion concentration of $\text{9}{\text{.4×10}}^{\text{-5}}\text{ M}$ is $\text{1}{\text{.1×10}}^{\text{-10}}\text{ M}$.

(c)

Interpretation Introduction

Interpretation:

The molar hydronium ion concentration in aqueous solutions has to be calculated with the hydroxide ion concentration of $\text{2}{\text{.3×10}}^{\text{-7}}\text{ M}$.

Concept Introduction:

Dissociation constant of water:

The dissociation of water can be given as

${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}$

The concentrations of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ and OH}}^{\text{-}}$ at room temperature are $\text{1}{\text{.0×10}}^{\text{-7}}\text{ M}$ each.

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = [OH}}^{\text{-}}\text{] = 1}{\text{.0×10}}^{\text{-7}}\\ \\ {\text{K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ {\text{K}}_{\text{w}}\text{= (1}{\text{.0×10}}^{\text{-7}}\text{)(1}{\text{.0×10}}^{\text{-7}}\text{)}\\ \\ {\text{K}}_{\text{w}}\text{= 1}{\text{.0×10}}^{\text{-14}}\text{M}\end{array}$

Explanation of Solution

Given,

$\begin{array}{l}\text{Hydronium ion concentration}\text{= 2}{\text{.3×10}}^{\text{-7}}\text{ M}\\ \\ \text{Ion product constant for water = 1}\text{.0}\times {\text{10}}^{-14}\text{ M}\end{array}$

To calculate the molar hydronium ion concentration in aqueous solutions

$\begin{array}{l}{\text{K}}_{\text{w}}{\text{=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}\text{M}}{\text{2}{\text{.3×10}}^{\text{-7}}\text{ M}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=4}{\text{.3×10}}^{\text{-8}}\text{ M}\end{array}$

The ${\text{[OH}}^{\text{-}}\text{]}$ in aqueous solutions with the hydronium ion concentration of $\text{2}{\text{.3×10}}^{\text{-7}}\text{ M}$ is $\text{4}{\text{.3×10}}^{\text{-8}}\text{ M}$.

(d)

Interpretation Introduction

Interpretation:

The molar hydronium ion concentration in aqueous solutions has to be calculated with the hydroxide ion concentration of $\text{6}{\text{.6×10}}^{\text{-12}}\text{ M}$.

Concept Introduction:

Dissociation constant of water:

The dissociation of water can be given as

${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}$

The concentrations of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ and OH}}^{\text{-}}$ at room temperature are $\text{1}{\text{.0×10}}^{\text{-7}}\text{ M}$ each.

$\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = [OH}}^{\text{-}}\text{] = 1}{\text{.0×10}}^{\text{-7}}\\ \\ {\text{K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ {\text{K}}_{\text{w}}\text{= (1}{\text{.0×10}}^{\text{-7}}\text{)(1}{\text{.0×10}}^{\text{-7}}\text{)}\\ \\ {\text{K}}_{\text{w}}\text{= 1}{\text{.0×10}}^{\text{-14}}\text{M}\end{array}$

Explanation of Solution

Given,

$\begin{array}{l}\text{Hydronium ion concentration}\text{= 6}{\text{.6×10}}^{\text{-12}}\text{ M}\\ \\ \text{Ion product constant for water = 1}\text{.0}\times {\text{10}}^{-14}\text{ M}\end{array}$

To calculate the molar hydronium ion concentration in aqueous solutions

$\begin{array}{l}{\text{K}}_{\text{w}}{\text{=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}\text{M}}{\text{6}{\text{.6×10}}^{\text{-12}}\text{ M}}\\ \\ {\text{[OH}}^{\text{-}}\text{]}\text{=1}{\text{.5×10}}^{\text{-3}}\text{ M}\end{array}$

The ${\text{[OH}}^{\text{-}}\text{]}$ in aqueous solutions with the hydronium ion concentration of $\text{6}{\text{.6×10}}^{\text{-12}}\text{ M}$ is $\text{1}{\text{.5×10}}^{\text{-3}}\text{ M}$.