Chapter 18, Problem 18.111P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ of $\text{0}\text{.100 M sodium phenolate}$. solution 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:

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

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,

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

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ of $\text{ 0}{\text{.15 M methylammonium bromide (CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Br)}$. solution 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:

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

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,

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