Chapter 18, Problem 18.166P

(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 $\text{pH, pOH, and pKw}$

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

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

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

(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 $\text{pH, pOH, and pKw}$

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

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

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