Chapter 18, Problem 18.112P

(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,

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

Explanation of Solution

Given,

Sodium phenolate ion is a base therefore it can accept the proton from the water. Sodium phenolate dissociates into sodium ions (${\text{Na}}^{\text{+}}$) and phenolate ions (${\text{PhO}}^{\text{–}}$). Sodium ion is the conjugate acid of $\text{NaOH}$, so ${\text{Na}}^{\text{+}}$ does not react with water. phenolate ions (${\text{PhO}}^{\text{–}}$) is the conjugate base of $\text{PhOH}$, so it does react with a base-dissociation reaction. 

The balance equation is given below,

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH(aq) }\text{+}{\text{ OH}}^{\text{-}}\text{(aq)}$

$\text{0}\text{.100 M sodium phenolate}$ Solution.

Therefore,

ICE table:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH(aq) }\text{+}{\text{ OH}}^{\text{-}}\text{(aq)}$

Initial concentration	0.100 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0. 0-x		x	x

The initial concentration is $\text{0}\text{.100 M sodium phenolate}$.

  $\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH(aq) }\text{+}{\text{ OH}}^{\text{-}}\text{(aq)}\\ The value{\text{ K}}_{\text{b}}of{C}_6{H}_5{O}^{-}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{ K}}_{\text{a}}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{1\times {10}^{-14}}{\text{ 1}\text{.0}\times {10}^{-10}}\\ {\text{K}}_{\text{b}}\text{ = }1.0\times {10}^{-4}\\ {\text{K}}_b\text{ =}\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\right]}\\ \text{given,}\\ {\text{K}}_b\text{ =}\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\right]}=1.0\times {10}^{-4},\\ letconsider,\\ 0.100-x=0.100\\ \frac{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{OH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}\right]}=1.0\times {10}^{-4},\\ \frac{x^2}{(0.100)}=1.0\times {10}^{-4}\\ x^2\text{=}1.0\times {10}^{-5}\\ \text{x}\text{=}3.16\times {10}^{-3}M=\left[{\text{OH}}^{-}\right]\end{array}$

  $\begin{array}{l}\text{x is small when compared to 0}\text{.100}\text{M, therefore,}\\ \text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \frac{3.16\times {10}^{-3}}{\text{0}\text{.100}\text{M}}\text{×100=}3.16\\ \text{The}\text{dissociation}\text{value}\text{is}\text{very}\text{less, and}\text{it is valid}\\ {\text{K}}_{\text{w}}\text{=}\left[{\text{OH}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{3.16\times {10}^{-3}}\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=3}{\text{.16×10}}^{\text{-12}}\\ \text{Therefore,}\\ {}_{\text{P}}\text{H}\text{=}\text{-}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ {}_{\text{P}}\text{H}\text{=}\text{-}\text{log}\text{(3}{\text{.16×10}}^{\text{-12}}\text{)}\\ {}_{\text{P}}\text{H}\text{=}\text{11}\text{.49}\end{array}$

$\text{pH}$ of $\text{0}\text{.100 M sodium phenolate}$ is 11.50.

(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,

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

Explanation of Solution

Given,

Methylammonium bromide dissociates into quartnary ammonium ion (${\text{CH}}_{\text{3}}{\text{NH}}_3^{+}$) and bromide ions (${\text{Br}}^{\text{–}}$). Bromide ion (${\text{Br}}^{\text{–}}$) is the conjugate base of a strong acid so it will not influence the pH of the solution. Methylammonium ion is the conjugate acid of a weak base, so an acid-dissociation reaction determines the pH of the solution.

The balance equation is given below,

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

$\text{ 0}{\text{.15 M methylammonium bromide (CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Br)}$ Solution.

Therefore,

ICE table:

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

Initial concentration	0.15 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.15-x		x	x

The initial concentration is $\text{ 0}{\text{.15 M methylammonium bromide (CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Br)}$.

  $\begin{array}{l}{\text{ CH}}_{\text{3}}{\text{NH}}_3^{+}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{(aq) }\text{+}{\text{ H}}_{\text{3}}{\text{O}}^{+}\text{(aq)}\\ The value{\text{ K}}_{a}of{\text{ CH}}_{\text{3}}{\text{NH}}_3^{+}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ {\text{K}}_a\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{ K}}_b}\\ {\text{K}}_a\text{ = }\frac{1\times {10}^{-14}}{\text{ 4}\text{.4}\times {10}^{-4}}\\ {\text{K}}_a\text{ = }2.27\times {10}^{-11}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}_{}^{+}\right]}{\left[{\text{ CH}}_{\text{3}}{\text{NH}}_3^{+}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}_{}^{+}\right]}{\left[{\text{ CH}}_{\text{3}}{\text{NH}}_3^{+}\right]}=2.27\times {10}^{-11},\\ letconsider,\\ 0.150-x=0.150\\ \frac{\left[{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}_{}^{+}\right]}{\left[{\text{ CH}}_{\text{3}}{\text{NH}}_3^{+}\right]}=2.27\times {10}^{-11},\\ \frac{x^2}{(0.150)}=2.27\times {10}^{-11}\\ x^2\text{=}3.40\times {10}^{-12}\\ \text{x}\text{=}1.846\times {10}^{-6}M\end{array}$

$\text{pH}$ of $\text{ 0}{\text{.15 M methylammonium bromide (CH}}_{\text{3}}{\text{NH}}_{\text{3}}\text{Br)}$ is 5.73.