Chapter 14, Problem 19QRT

(a)

Interpretation Introduction

Interpretation:

Among the solutions D, E, F, G, the one that has highest ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has to be identified.

Concept Introduction:

Equilibrium constant for water can be given by the equation shown below,

    $K_w=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]$

This is also known as ionization constant for water.

    $\begin{array}{c}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ =\left(1\times {10}^{-7}\right)\left(1\times {10}^{-7}\right)\\ =1\times {10}^{-14}\end{array}$

The above equation applies to water and all aqueous solutions.  From this the solution is acidic, basic, or neutral can be identified.  This can be given as shown below,

  $\begin{array}{l}\text{Neutral}\text{solution:}[{\text{H}}_{\text{3}}{\text{O}}^{+}]=[{\text{OH}}^{-}]\\ \text{Acidic}\text{solution:}[{\text{H}}_{\text{3}}{\text{O}}^{+}]>[{\text{OH}}^{-}]\\ \text{ Basic}\text{solution:}[{\text{H}}_{\text{3}}{\text{O}}^{+}]<[{\text{OH}}^{-}]\end{array}$

Explanation of Solution

Solution D:

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is given as $2\times {10}^{-3}\text{M}$.  Using the equation for ionization constant of water, the concentration of ${\text{OH}}^{-}$ ion can be calculated.

    $\begin{array}{c}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ 1\times {10}^{-14}=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ [{\text{OH}}^{-}]=\frac{1\times {10}^{-14}}{[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}]}\\ =\frac{1\times {10}^{-14}}{2\times {10}^{-3}}=5\times {10}^{-12}\text{M}\end{array}$

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $2\times {10}^{-3}\text{M}$ and that of ${\text{OH}}^{-}$ is $5\times {10}^{-12}\text{M}$.

Solution E:

Concentration of ${\text{OH}}^{-}$ is given as $2\times {10}^{-7}\text{M}$.  Using the equation for ionization constant of water, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion can be calculated.

    $\begin{array}{c}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ 1\times {10}^{-14}=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ [{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}]=\frac{1\times {10}^{-14}}{[{\text{OH}}^{-}]}\\ =\frac{1\times {10}^{-14}}{2\times {10}^{-7}}=5\times {10}^{-8}\text{M}\end{array}$

Concentration of ${\text{OH}}^{-}$ is $2\times {10}^{-7}\text{M}$ and that of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $5\times {10}^{-8}\text{M}$.

Solution F:

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is given as $4\times {10}^{-5}\text{M}$.  Using the equation for ionization constant of water, the concentration of ${\text{OH}}^{-}$ ion can be calculated.

    $\begin{array}{c}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ 1\times {10}^{-14}=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ [{\text{OH}}^{-}]=\frac{1\times {10}^{-14}}{[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}]}\\ =\frac{1\times {10}^{-14}}{4\times {10}^{-5}}=3\times {10}^{-10}\text{M}\end{array}$

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $4\times {10}^{-5}\text{M}$ and that of ${\text{OH}}^{-}$ is $3\times {10}^{-10}\text{M}$.

Solution G:

Concentration of ${\text{OH}}^{-}$ is given as $5\times {10}^{-11}\text{M}$.  Using the equation for ionization constant of water, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion can be calculated.

    $\begin{array}{c}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ 1\times {10}^{-14}=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ [{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}]=\frac{1\times {10}^{-14}}{[{\text{OH}}^{-}]}\\ =\frac{1\times {10}^{-14}}{5\times {10}^{-11}}=2\times {10}^{-4}\text{M}\end{array}$

Concentration of ${\text{OH}}^{-}$ is $5\times {10}^{-11}\text{M}$ and that of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $2\times {10}^{-4}\text{M}$.

All the solutions can be arranged in the order or decreasing ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and increasing ${\text{OH}}^{-}$ can be given as,

$\begin{array}{c}\overline{\text{Solution}[{\text{H}}_{\text{3}}{\text{O}}^{+}]\text{M}[{\text{OH}}^{-}]\text{M}}\\ \overline{\begin{array}{l}\text{D}{\text{2×10}}^{\text{-3}}{\text{5×10}}^{\text{-12}}\\ \text{G}{\text{2×10}}^{\text{-4}}{\text{5×10}}^{\text{-11}}\\ \text{F}{\text{4×10}}^{\text{-5}}{\text{3×10}}^{\text{-10}}\\ \underset{\_}{\text{E}{\text{5×10}}^{\text{-8}}{\text{2×10}}^{\text{-7}}}\end{array}}\end{array}$

The solution that has highest concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ can be identified from the above table and it is found as Solution D.

(b)

Interpretation Introduction

Interpretation:

Among the solutions D, E, F, G, the one that has highest ${\text{OH}}^{-}$ has to be identified.

Concept Introduction:

Refer part (a).

Explanation of Solution

All the given solutions can be arranged in the order or decreasing ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and increasing ${\text{OH}}^{-}$ can be given as,

$\begin{array}{c}\overline{\text{Solution}[{\text{H}}_{\text{3}}{\text{O}}^{+}]\text{M}[{\text{OH}}^{-}]\text{M}}\\ \overline{\begin{array}{l}\text{D}{\text{2×10}}^{\text{-3}}{\text{5×10}}^{\text{-12}}\\ \text{G}{\text{2×10}}^{\text{-4}}{\text{5×10}}^{\text{-11}}\\ \text{F}{\text{4×10}}^{\text{-5}}{\text{3×10}}^{\text{-10}}\\ \underset{\_}{\text{E}{\text{5×10}}^{\text{-8}}{\text{2×10}}^{\text{-7}}}\end{array}}\end{array}$

The solution that has highest concentration of ${\text{OH}}^{-}$ can be identified from the above table and it is found as Solution E.

(c)

Interpretation Introduction

Interpretation:

Among the solutions D, E, F, G, the one that is closest to being a neutral solution has to be identified.

Concept Introduction:

Refer part (a).

Explanation of Solution

All the given solutions can be arranged in the order or decreasing ${\text{H}}_{\text{3}}{\text{O}}^{+}$ and increasing ${\text{OH}}^{-}$ can be given as,

$\begin{array}{c}\overline{\text{Solution}[{\text{H}}_{\text{3}}{\text{O}}^{+}]\text{M}[{\text{OH}}^{-}]\text{M}}\\ \overline{\begin{array}{l}\text{D}{\text{2×10}}^{\text{-3}}{\text{5×10}}^{\text{-12}}\\ \text{G}{\text{2×10}}^{\text{-4}}{\text{5×10}}^{\text{-11}}\\ \text{F}{\text{4×10}}^{\text{-5}}{\text{3×10}}^{\text{-10}}\\ \underset{\_}{\text{E}{\text{5×10}}^{\text{-8}}{\text{2×10}}^{\text{-7}}}\end{array}}\end{array}$

Neutral solution has the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ and ${\text{OH}}^{-}$ equal to $1\times {10}^{-7}$.  From table above, the solution that has the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ and ${\text{OH}}^{-}$ nearly equal to ${10}^{-7}$ is Solution E.  Therefore, Solution E is the closest one of being neutral.