Chapter 18, Problem 18.167P

(a)

Interpretation Introduction

Interpretation:

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{, }\left[{\text{OH}}^{\text{-}}\right]\text{, pH, and pOH}$ for $\text{0}\text{.240 M}$ acetic acid has to be calculated.

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 base B,

  $\text{B}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{BH}}^{\text{+}}\left(\text{aq}\right){\text{+OH}}^{\text{-}}\left(\text{aq}\right)$

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{BH}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[\text{B}\right]}$                                                                                                 $\left(1\right)$

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

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

Percent dissociation can be calculated by using following formula,

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

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

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

The Relationships Among $\text{pH, pOH, and pKw}$

  ${\text{K}}_{\text{w}}{\text{ = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{] = 1}\text{.0}\times {10}^{-14}$

  $\begin{array}{l}\text{pH = -}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \text{pOH = -}\text{log}{\text{(OH}}^{-}\text{)}\end{array}$

  ${\text{pK}}_{\text{w}}\text{ = pH + pOH = 14}\text{.00 }$

Explanation of Solution

Given,

Acetic acid is an acid therefore it can donate the proton to the water.

The balance equation is given below,

  ${\text{CH}}_3\text{COOH (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{CH}}_3{\text{COO}}^{-}\text{(aq) }\text{+}{\text{ H}}_{\text{3}}{\text{O}}^{+}\text{(aq)}$

$\text{0}\text{.240 M}$ acetic acid Solution.

Therefore,

ICE table:

${\text{CH}}_3\text{COOH (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{CH}}_3{\text{COO}}^{-}\text{(aq) }\text{+}{\text{ H}}_{\text{3}}{\text{O}}^{+}\text{(aq)}$

Initial concentration	0.240 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.240-x		x	x

The initial concentration is $\text{0}\text{.240 M}$ acetic acid.

  ${\text{CH}}_3\text{COOH (aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{CH}}_3{\text{COO}}^{-}\text{(aq) }\text{+}{\text{ H}}_{\text{3}}{\text{O}}^{+}\text{(aq)}$

The value of ${\text{K}}_{\text{a}}$ is calculated by using following formula,

    ${\text{K}}_a\text{ =}\frac{\left[{\text{CH}}_3{\text{COO}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{CH}}_3\text{COOH }\right]}$

Given,

  ${\text{K}}_a\text{ =}\frac{\left[{\text{CH}}_3{\text{COO}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{CH}}_3\text{COOH }\right]}=1.8\times {10}^{-5}$

Let consider,

  $\begin{array}{l}0.24-x=0.240\\ \frac{\left[{\text{CH}}_3{\text{COO}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{\text{CH}}_3\text{COOH }\right]}=1.8\times {10}^{-5},\\ \frac{x^2}{(0.240)}=1.8\times {10}^{-5}\\ x^2\text{=}4.32\times {10}^{-6}\\ \text{x}\text{=}2.07\times {10}^{-3}M=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\end{array}$

Concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]=2.07\times {10}^{-3}M$

X is small when compared to $\text{0}\text{.240}\text{M}$, therefore,

  $\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \frac{2.07\times {10}^{-3}}{\text{0}\text{.240}\text{M}}\text{×100=}0.86\end{array}$

The dissociation value is very less, and it is valid.

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{OH}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{2.07\times {10}^{-3}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=4}{\text{.83×10}}^{\text{-12}}\end{array}$

Therefore,

  $\begin{array}{l}\text{pH = -}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \text{pH}\text{=}\text{-}\text{log}\text{(}2.07\times {10}^{-3}\text{)}\\ \text{pH}\text{=}2.68\\ \\ \text{pOH = -}\text{log}{\text{(OH}}^{-}\text{)}\\ \text{pOH = -}\text{log}\text{(4}{\text{.83×10}}^{\text{-12}}\text{)}\\ \text{pOH =}11.31\end{array}$

(b)

Interpretation Introduction

Interpretation:

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{, }\left[{\text{OH}}^{\text{-}}\right]\text{, pH, and pOH}$ for $\text{0}\text{.240 M}$ ammonia has to be calculated.

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 base B,

  $\text{B}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{BH}}^{\text{+}}\left(\text{aq}\right){\text{+OH}}^{\text{-}}\left(\text{aq}\right)$

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{BH}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[\text{B}\right]}$                                                                                                 $\left(1\right)$

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

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

Percent dissociation can be calculated by using following formula,

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

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

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

The Relationships Among $\text{pH, pOH, and pKw}$

  ${\text{K}}_{\text{w}}{\text{ = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{] = 1}\text{.0}\times {10}^{-14}$

  $\begin{array}{l}\text{pH = -}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \text{pOH = -}\text{log}{\text{(OH}}^{-}\text{)}\end{array}$

  ${\text{pK}}_{\text{w}}\text{ = pH + pOH = 14}\text{.00 }$

Explanation of Solution

Given,

Ammonia is a base therefore it can accept the proton from the water.

The balance equation is given below,

${\text{NH}}_3\text{ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O (l)}\underset{}{\rightleftharpoons }{\text{NH}}_{\text{4}}^{\text{+}}\text{(aq) }\text{+}{\text{ HO}}^{-}\text{(aq)}$

$\text{0}\text{.240 M}$ ammonia Solution.

Therefore,

ICE table:

${\text{NH}}_3\text{ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O (l)}\underset{}{\rightleftharpoons }{\text{NH}}_{\text{4}}^{\text{+}}\text{(aq) }\text{+}{\text{ HO}}^{-}\text{(aq)}$

Initial concentration	0.240 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.240-x		x	x

The initial concentration is $\text{0}\text{.240 M}$ acetic acid.

    ${\text{NH}}_3\text{ (aq) }\text{+}{\text{H}}_{\text{2}}\text{O (l)}\underset{}{\rightleftharpoons }{\text{NH}}_{\text{4}}^{\text{+}}\text{(aq) }\text{+}{\text{ HO}}^{-}\text{(aq)}$

The value of ${\text{K}}_{\text{b}}$ is calculated by using following formula,

    ${\text{K}}_b\text{ =}\frac{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\left[{\text{HO}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Given,

  ${\text{K}}_b\text{ =}\frac{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\left[{\text{HO}}^{-}\right]}{\left[{\text{NH}}_3\right]}=1.8\times {10}^{-5}$

Let consider,

  $\begin{array}{l}0.24-x=0.240\\ \frac{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\left[{\text{HO}}^{-}\right]}{\left[{\text{NH}}_3\right]}=1.8\times {10}^{-5},\\ \frac{x^2}{(0.240)}=1.8\times {10}^{-5}\\ x^2\text{=}4.32\times {10}^{-6}\\ \text{x}\text{=}2.07\times {10}^{-3}M=\left[{\text{HO}}^{-}\right]\end{array}$

Concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]=2.07\times {10}^{-3}M$

X is small when compared to $\text{0}\text{.240}\text{M}$, therefore,

  $\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \frac{2.07\times {10}^{-3}}{\text{0}\text{.240}\text{M}}\text{×100=}0.86\end{array}$

The dissociation value is very less, and it is valid.

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{=}\left[{\text{OH}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{2.07\times {10}^{-3}}\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=4}{\text{.83×10}}^{\text{-12}}\end{array}$

Therefore,

  $\begin{array}{l}\text{pH = -}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \text{pH}\text{=}\text{-}\text{log}\text{(4}{\text{.83×10}}^{\text{-12}}\text{)}\\ \text{pH}\text{= }11.31\\ \\ \text{pOH = -}\text{log}{\text{(OH}}^{-}\text{)}\\ \text{pOH = -}\text{log}\text{(}2.07\times {10}^{-3}\text{)}\\ \text{pOH =}2.68\end{array}$