Chapter 18, Problem 18.114P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ of $\text{ 0}\text{.75 M NaF}$. 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 fluoride ion is a base therefore it can accept the proton from the water. $\text{NaF}$ is water soluble. Sodium ion (${\text{Na}}^{\text{+}}$) is from base, it will not affect the $\text{pH}$. 

The balance equation is given below,

  ${\text{F}}^{\text{-}}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }\text{HF (aq) }\text{+}{\text{ OH}}^{\text{-}}\text{(aq)}$

$\text{ 0}\text{.75 M NaF}$ Solution.

Therefore,

ICE table:

  ${\text{F}}^{\text{-}}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }\text{HF (aq) }\text{+}{\text{ OH}}^{\text{-}}\text{(aq)}$

Initial concentration	0.750 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.75-x		x	x

The initial concentration is $\text{ 0}\text{.75 M NaF}$.

  $\begin{array}{l}{\text{F}}^{\text{-}}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }\text{HF (aq) }\text{+}{\text{ OH}}^{\text{-}}\text{(aq)}\\ The value{\text{ K}}_{\text{b}}of{F}^{-}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{ K}}_{\text{a}}}\\ {\text{K}}_{\text{b}}\text{ = }\frac{1\times {10}^{-14}}{\text{ 6}\text{.8}\times {10}^{-4}}\\ {\text{K}}_{\text{b}}\text{ = }1.470\times {10}^{-11}\\ {\text{K}}_b\text{ =}\frac{\left[\text{HF}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{F}}^{\text{-}}\right]}\\ \text{given,}\\ {\text{K}}_b\text{ =}\frac{\left[\text{HF}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{F}}^{\text{-}}\right]}=1.470\times {10}^{-11},\\ letconsider,\\ 0.750-x=0.750\\ \frac{\left[\text{HF}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{F}}^{\text{-}}\right]}=1.470\times {10}^{-11},\\ \frac{x^2}{(0.750)}=1.470\times {10}^{-11}\\ x^2\text{=}1.102\times {10}^{-11}\\ \text{x}\text{=}3.32\times {10}^{-6}M=\left[{\text{OH}}^{-}\right]\end{array}$

  $\begin{array}{l}\text{x is small when compared to 0}\text{.750}\text{M, therefore,}\\ \text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \frac{3.32\times {10}^{-6}M}{\text{0}\text{.750}\text{M}}\text{×100=}4.42\times {10}^{-4}\\ \text{The}\text{dissociation}\text{value}\text{is}\text{very}\text{less, and}\text{it is valid}\\ {\text{K}}_{\text{w}}\text{=}\left[{\text{OH}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{3.32\times {10}^{-6}M}\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=3}{\text{.01×10}}^{\text{-9}}\\ \text{Therefore,}\\ {}_{\text{P}}\text{H}\text{=}\text{-}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ {}_{\text{P}}\text{H}\text{=}\text{-}\text{log}\text{(3}{\text{.01×10}}^{\text{-9}}\text{)}\\ {}_{\text{P}}\text{H}\text{=}8.52\end{array}$

$\text{pH}$ of $\text{ 0}\text{.75 M NaF}$ is 8.52.

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ of $\text{0}{\text{.88 M pyridinium chloride (C}}_{\text{5}}{\text{H}}_{\text{5}}\text{NHCl)}$ 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,

Pyridinium chloride dissociates into Pyridinium ion (${\text{C}}_{\text{6}}{\text{H}}_5{\text{NH}}_{}^{+}$) and chloride ions (${\text{Cl}}^{\text{–}}$). Chloride ions (${\text{Cl}}^{\text{–}}$) is the conjugate base of a strong acid so it will not influence the pH of the solution. Pyridinium ion (${\text{C}}_{\text{6}}{\text{H}}_5{\text{NH}}_{}^{+}$) 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,

  ${\text{C}}_{\text{6}}{\text{H}}_5{\text{NH}}_{}^{+}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{C}}_{\text{6}}{\text{H}}_5\text{NH(aq) }\text{+}{\text{ H}}_{\text{3}}{\text{O}}^{+}\text{(aq)}$

$\text{0}{\text{.88 M pyridinium chloride (C}}_{\text{5}}{\text{H}}_{\text{5}}\text{NHCl)}$ Solution.

Therefore,

ICE table:

${\text{C}}_{\text{6}}{\text{H}}_5{\text{NH}}_{}^{+}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{C}}_{\text{6}}{\text{H}}_5\text{NH(aq) }\text{+}{\text{ H}}_{\text{3}}{\text{O}}^{+}\text{(aq)}$

Initial concentration	0.88 M	-		0	0
Change	-x			+ x	+ x
At equilibrium	0.88-x			x	x

The initial concentration is $\text{0}{\text{.88 M pyridinium chloride (C}}_{\text{5}}{\text{H}}_{\text{5}}\text{NHCl)}$.

  $\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_5{\text{NH}}_{}^{+}\text{(aq) }\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{C}}_{\text{6}}{\text{H}}_5\text{NH(aq) }\text{+}{\text{ H}}_{\text{3}}{\text{O}}^{+}\text{(aq)}\\ The value{\text{ K}}_{a}of{\text{ C}}_{\text{6}}{\text{H}}_5{\text{NH}}_{}^{+}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ {\text{K}}_a\text{ = }\frac{{\text{K}}_{\text{w}}}{{\text{ K}}_b}\\ {\text{K}}_a\text{ = }\frac{1\times {10}^{-14}}{\text{ 1}\text{.7}\times {10}^{-9}}\\ {\text{K}}_a\text{ = 5}\text{.88}\times {10}^{-6}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_5\text{NH}\right]\left[{\text{H}}_{\text{3}}{\text{O}}_{}^{+}\right]}{\left[{\text{ C}}_{\text{6}}{\text{H}}_5{\text{NH}}_{}^{+}\right]}\\ \text{given,}\\ {\text{K}}_a\text{ =}\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_5\text{NH}\right]\left[{\text{H}}_{\text{3}}{\text{O}}_{}^{+}\right]}{\left[{\text{ C}}_{\text{6}}{\text{H}}_5{\text{NH}}_{}^{+}\right]}=\text{5}\text{.88}\times {10}^{-6},\\ letconsider,\\ 0.880-x=0.880\\ \frac{\left[{\text{C}}_{\text{6}}{\text{H}}_5\text{NH}\right]\left[{\text{H}}_{\text{3}}{\text{O}}_{}^{+}\right]}{\left[{\text{ C}}_{\text{6}}{\text{H}}_5{\text{NH}}_{}^{+}\right]}=\text{5}\text{.88}\times {10}^{-6},\\ \frac{x^2}{(0.880)}=\text{5}\text{.88}\times {10}^{-6}\\ x^2\text{=}5.17\times {10}^{-6}\\ \text{x}\text{=}2.27\times {10}^{-3}M\end{array}$

  $\begin{array}{l}\text{x is small when compared to 0}\text{.880}\text{M, therefore,}\\ \text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \frac{2.27\times {10}^{-3}}{\text{0}\text{.880}\text{M}}\text{×100=}0.258\\ \text{The}\text{dissociation}\text{value}\text{is}\text{very}\text{less, and}\text{it is valid}\\ \text{Therefore,}\\ {}_{\text{P}}\text{H}\text{=}\text{-}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ {}_{\text{P}}\text{H}\text{=}\text{-}\text{log}\text{(}2.27\times {10}^{-3}\text{)}\\ {}_{\text{P}}\text{H}\text{=2}\text{.64}\end{array}$

$\text{pH}$ of $\text{0}{\text{.88 M pyridinium chloride (C}}_{\text{5}}{\text{H}}_{\text{5}}\text{NHCl)}$ is 2.64.