Chapter 14, Problem 20QRT

(a)

Interpretation Introduction

Interpretation:

Among the solutions W, X, Y, Z, 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 W:

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is given as $5\times {10}^{-6}\text{M}$.  Using the equation for ionization constant of water, the concentration of ${\text{OH}}^{-}$ ion can be calculated.

    $\begin{array}{c}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ 1\times {10}^{-14}=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ [{\text{OH}}^{-}]=\frac{1\times {10}^{-14}}{[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}]}\\ =\frac{1\times {10}^{-14}}{5\times {10}^{-6}}=2\times {10}^{-9}\text{M}\end{array}$

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $5\times {10}^{-6}\text{M}$ and that of ${\text{OH}}^{-}$ is $2\times {10}^{-9}\text{M}$.

Solution X:

Concentration of ${\text{OH}}^{-}$ is given as $2\times {10}^{-4}\text{M}$.  Using the equation for ionization constant of water, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion can be calculated.

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

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

Solution Y:

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is given as $4\times {10}^{-2}\text{M}$.  Using the equation for ionization constant of water, the concentration of ${\text{OH}}^{-}$ ion can be calculated.

    $\begin{array}{c}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ 1\times {10}^{-14}=[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}][{\text{OH}}^{-}]\\ [{\text{OH}}^{-}]=\frac{1\times {10}^{-14}}{[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}]}\\ =\frac{1\times {10}^{-14}}{4\times {10}^{-2}}=3\times {10}^{-13}\text{M}\end{array}$

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $4\times {10}^{-2}\text{M}$ and that of ${\text{OH}}^{-}$ is $3\times {10}^{-13}\text{M}$.

Solution Z:

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,

$\begin{array}{c}\overline{\text{Solution}[{\text{H}}_{\text{3}}{\text{O}}^{+}]\text{M}[{\text{OH}}^{-}]\text{M}}\\ \overline{\begin{array}{l}\text{Y}{\text{4×10}}^{\text{-2}}{\text{3×10}}^{\text{-13}}\\ \text{Z}{\text{2×10}}^{\text{-4}}{\text{5×10}}^{\text{-11}}\\ \text{W}{\text{5×10}}^{\text{-6}}{\text{2×10}}^{\text{-9}}\\ \underset{\_}{\text{X}{\text{5×10}}^{\text{-11}}{\text{2×10}}^{\text{-4}}}\end{array}}\end{array}$

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 Y.

(b)

Interpretation Introduction

Interpretation:

Among the solutions W, X, Y, Z, two solutions that has concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ equal to the concentration of ${\text{OH}}^{-}$ of different solution has to be identified and their concentrations has to be given.

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,

$\begin{array}{c}\overline{\text{Solution}[{\text{H}}_{\text{3}}{\text{O}}^{+}]\text{M}[{\text{OH}}^{-}]\text{M}}\\ \overline{\begin{array}{l}\text{Y}{\text{4×10}}^{\text{-2}}{\text{3×10}}^{\text{-13}}\\ \text{Z}{\text{2×10}}^{\text{-4}}{\text{5×10}}^{\text{-11}}\\ \text{W}{\text{5×10}}^{\text{-6}}{\text{2×10}}^{\text{-9}}\\ \underset{\_}{\text{X}{\text{5×10}}^{\text{-11}}{\text{2×10}}^{\text{-4}}}\end{array}}\end{array}$

The solutions that have concentration ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ equal to that of ${\text{OH}}^{-}$ in different solution is found to be Solution Z and Solution X.

Solution Z:

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is ${\text{2×10}}^{\text{-4}}$ and this is equal to the concentration of ${\text{OH}}^{-}$ in Solution X.

Solution X:

Concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is ${\text{5×10}}^{\text{-11}}$ and this is equal to the concentration of ${\text{OH}}^{-}$ in Solution Z.

(c)

Interpretation Introduction

Interpretation:

Among the solutions W, X, Y, Z, 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,

$\begin{array}{c}\overline{\text{Solution}[{\text{H}}_{\text{3}}{\text{O}}^{+}]\text{M}[{\text{OH}}^{-}]\text{M}}\\ \overline{\begin{array}{l}\text{Y}{\text{4×10}}^{\text{-2}}{\text{3×10}}^{\text{-13}}\\ \text{Z}{\text{2×10}}^{\text{-4}}{\text{5×10}}^{\text{-11}}\\ \text{W}{\text{5×10}}^{\text{-6}}{\text{2×10}}^{\text{-9}}\\ \underset{\_}{\text{X}{\text{5×10}}^{\text{-11}}{\text{2×10}}^{\text{-4}}}\end{array}}\end{array}$

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 W.  Therefore, Solution W is the closest one of being neutral.