Chapter 18, Problem 18.70P

(a)

Interpretation Introduction

Interpretation:

The $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{, pH, [OH}}^{\text{-}}\text{], and pOH}$ haave to be calculated for $\text{0}\text{.20 M}$ solution.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general acid HA,

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  $K_a=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$        $\left(1\right)$

An equilibrium constant $\left(K\right)$ with subscript $a$ indicate that it is an equilibrium constant of an acid in water.

  $\begin{array}{l}{\text{Acid - dissociation constants can be expressed as pK}}_a\text{ values,}\\ {\text{pK}}_a{\text{ = -log K}}_a\text{ and}\\ {\text{10}}^{{\text{ - pK}}_a}{\text{ = K}}_{\text{a }}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}$

The ${\text{K}}_a$ value is calculating by using following formula,

  ${\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}$

Explanation of Solution

$\text{3}\text{.0\%}$ acid dissociated in a $\text{0}\text{.2 M}$ solution.

The given compound is acid therefore it can donate the proton to the water.

The balance equation is given below,

  $\text{HA (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ A}}^{-}\text{ (aq)}$

Percent dissociation can be calculated by using following formula,

  $\begin{array}{c}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \\ \text{3}\text{.0 \%= }\frac{\text{x}}{0.2}\text{×100}\\ \\ \text{x = 0}\text{.006}\text{M}\end{array}$

The dissociation of acid is $\text{0}\text{.006}\text{M}$

Therefore,

Construct ICE table:

  $\text{HA (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ A}}^{-}\text{ (aq)}$

Initial concentration	0.2 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.2-x		x	x

The initial concentration is $\text{0}\text{.2 M HA}$.

  $\begin{array}{l}\text{HA (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ A}}^{-}\text{ (aq)}\\ \\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ \\ {\text{K}}_a\text{ =}\frac{\left[{\text{A}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HA}\right]}\end{array}$

  $\text{[Dissociated acid] = x =}\left[{\text{A}}^{\text{-}}\right]\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\text{0}\text{.006}\text{M}$

Therefore,

  $\begin{array}{l}\text{pH}\text{=}\text{-}\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}\text{-}\text{log}\left(\text{0}\text{.006}\right)\\ \\ \text{pH}\text{=}2.22\end{array}$

$\text{pH}$ of the solution is $2.22$

The ${\text{K}}_b$ value is calculating by using following formula,

  $\begin{array}{c}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]\\ \\ \left[{\text{OH}}^{-}\right]\text{ = }\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \\ \left[{\text{OH}}^{-}\right]\text{ = }\frac{1\times {10}^{-14}}{6.0\times {10}^{-3}}\\ \\ \left[{\text{OH}}^{-}\right]\text{ = 1}\text{.7}\times {10}^{-12}\end{array}$

Therefore,

  $\begin{array}{c}\text{pOH}\text{=}\text{-}\text{log}\left[{\text{OH}}^{-}\right]\\ \\ \text{pOH}\text{=}\text{-}\text{log}\left(\text{1}\text{.7}\times {10}^{-12}\right)\\ \\ \text{pOH}\text{=}11.77\end{array}$

$\text{pOH}$ of the solution is $11.77$

(b)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{a}}$ has to be calculated for $\text{0}\text{.20 M}$ solution.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general acid HA,

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  $K_a=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$ $\left(1\right)$

An equilibrium constant $\left(K\right)$ with subscript $a$ indicate that it is an equilibrium constant of an acid in water.

  $\begin{array}{l}{\text{Acid - dissociation constants can be expressed as pK}}_a\text{ values,}\\ {\text{pK}}_a{\text{ = -log K}}_a\text{ and}\\ {\text{10}}^{{\text{ - pK}}_a}{\text{ = K}}_{\text{a }}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}$

The ${\text{K}}_a$ value is calculating by using following formula,

  ${\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}$

Explanation of Solution

$\text{3}\text{.0\%}$ acid dissociated in a $\text{0}\text{.2 M}$ solution.

The given compound is acid therefore it can donate the proton to the water.

The balance equation is given below,

  $\text{HA (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ A}}^{-}\text{ (aq)}$

Percent dissociation can be calculated by using following formula,

  $\begin{array}{c}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \\ \text{3}\text{.0 \%= }\frac{\text{x}}{0.2}\text{×100}\\ \\ \text{x = 0}\text{.006}\text{M}\end{array}$

The dissociation of acid is $\text{0}\text{.006}\text{M}$

Therefore,

Construct ICE table:

  $\text{HA (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ A}}^{-}\text{ (aq)}$

Initial concentration	0.2 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.2-x		x	x

The initial concentration is $\text{0}\text{.2 M HA}$.

  $\begin{array}{l}\text{HA (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ A}}^{-}\text{ (aq)}\\ \\ The value{\text{ K}}_{a}iscalculatingbyu\sin gfollowingformula, \\ \\ {\text{K}}_a\text{ =}\frac{\left[{\text{A}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HA}\right]}\end{array}$

  $\text{[Dissociated acid] = x =}\left[{\text{A}}^{\text{-}}\right]\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\text{0}\text{.006}\text{M}$

  $\begin{array}{c}\text{[Dissociated acid] = x =}\left[{\text{A}}^{\text{-}}\right]\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\text{0}\text{.006}\text{M}\\ \\ \left[\text{HA}\right]\text{ = 0}\text{.2 M -}\text{0}\text{.006}\text{M}\\ \\ \left[\text{HA}\right]\text{ = 0}\text{.194 M}\end{array}$

Therefore,

  $\begin{array}{c}{\text{K}}_a\text{ =}\frac{\left[{\text{A}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[\text{HA}\right]}\\ \\ {\text{K}}_a\text{ =}\frac{\left[\text{0}\text{.006}\right]\left[\text{0}\text{.006}\right]}{\left[0.194\right]}\\ \\ {\text{K}}_a\text{ =}\text{1}\text{.9}\times {\text{10}}^{-4}\end{array}$

The ${\text{K}}_{\text{a}}$ of the acid is $\text{1}\text{.9}\times {\text{10}}^{-4}$.