Chapter 17, Problem 27PS

Phenol, C₆H₅OH, is a weak organic acid. Suppose 0.515 g of the compound is dissolved in enough water to make 125 mL of solution. The resulting solution is titrated with 0.123 M NaOH.

C₆H₅OH(aq) + OH⁻(aq) ⇄ C₆H₅O⁻(aq) + H₂O(ℓ)

1. (a) What is the pH of the original solution of phenol?
2. (b) What are the concentrations of all of the following ions at the equivalence point: Na⁺, H₃O⁺, OH⁻, and C₆H₅O⁻?
3. (c) What is the pH of the solution at the equivalence point?

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated at the various points during the titration between ${\text{C}}_6{\text{H}}_5\text{OH}$ and $\text{NaOH}$. The value of $\text{pH}$, of the original solution of ${\text{C}}_6{\text{H}}_5\text{OH}$ has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base. There are four types of acid-base titrations.

(1) Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, $\text{HCl}$ is titrated against $\text{NaOH}$.

(2) Strong acid-Weak base, in this type of titration a strong acid is titrated against a weak base for example, $\text{HCl}$ is titrated against ${\text{NH}}_4\text{OH}$.

(3) Weak acid-Strong base, in this type of titration a weak acid is titrated against a strong base for example, ${\text{CH}}_3\text{COOH}$ is titrated against $\text{NaOH}$.

(4) Weak acid-Weak base, in this type of titration a weak acid is titrated against a weak base for example, ${\text{CH}}_3\text{COOH}$ is titrated against ${\text{NH}}_4\text{OH}$.

For weak acid-strong base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{C}}_6{\text{H}}_5\text{OH}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5\text{ONa}\left(\text{aq}\right)$

Calculation of $\text{pH}$ at various points is done as follows,

(1) The $\text{pH}$ value before the titration can be calculated by using $K_{\text{a}}$ and its relation with ${\text{H}}_3{\text{O}}^{+}$ ion concentration.

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$ (1)

(2) The $\text{pH}$ calculation just before the equivalence point,

As the addition of $\text{NaOH}$ is done there will be formation of buffer solution ${\text{C}}_6{\text{H}}_5\text{OH}$/${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. The $\text{pH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{acid}\right]}$ (2)

At the midpoint of the titration, when concentration of acid and its conjugate base is equal. Therefore $\text{pH}$ value at midpoint will be given as

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{acid}\right]}$

Substitute, $\left[\text{conjugate}\text{base}\right]\text{for}\left[\text{acid}\right]$.

$\begin{array}{c}\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{base}\right]}\\ =\text{p}{K}_{\text{a}}+\log \left(1\right)\\ =\text{p}{K}_{\text{a}}+0\\ =\text{p}{K}_{\text{a}}\end{array}$

Therefore, $\text{pH}$ value at midpoint is equal to $\text{p}{K}_{\text{a}}$.

(3) The $\text{pH}$ calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only ${\text{OH}}^{-}$ ion and ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. The ${\text{OH}}^{-}$ will be produced due to the hydrolysis of phenoxide ion at equivalence point. The hydrolysis equilibrium is represented as,

${\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right)$

By using the value of $K_{\text{b}}$ for the phenoxide ion, concentration of  ${\text{OH}}^{-}$ can be calculated. Thus the value of $\text{pH}$ is greater than $7$ at equivalence point for the weak acid- strong base titrations.

The relation between $K_{\text{a}}$ and $K_{\text{b}}$ for weak acid and its conjugate base is given as,

$K_w=\left(K_{\text{a}}\right)\left(K_{\text{b}}\right)$ (3)

Explanation of Solution

The $\text{pH}$ value before the titration can be calculated by using $K_{\text{a}}$ and its relation with ${\text{H}}_3{\text{O}}^{+}$ ion concentration.

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$

Given:

Refer to the Apendix H in the textbook for the value of $K_{\text{a}}$.

The value of $K_{\text{a}}$ for phenol is $1.3\times {10}^{-10}$.

The value of $K_{\text{w}}$ for water is $1.0\times {10}^{-14}$.

The $\text{p}{K}_{\text{a}}$ value is calculated as follows;

$\text{p}{K}_{\text{a}}=-\log (K_{\text{a}})$

Substitute, $1.3\times {10}^{-10}$ for $K_{\text{a}}$.

$\begin{array}{c}\text{p}{K}_{\text{a}}=-\log (1.3\times {10}^{-10})\\ =9.88\end{array}$

Therefore, $\text{p}{K}_{\text{a}}$ value of phenol is $9.88$.

The amount of ${\text{C}}_6{\text{H}}_5\text{OH}$ dissolved is $0.515\text{ g}$.

Molar mass of phnol is $94\text{ g}\cdot {\text{mol}}^{-1}$.

$\text{Number of moles}$ of phenol are calculated by using following expression,

$\text{Number of moles = }\frac{\text{weight of solute }}{\text{molar mass of solute}}\left(\text{mol}\right)$

Substitute $0.515\text{ g}$ for $\text{weight of solute }$, $94\text{ g}\cdot {\text{mol}}^{-1}$ for $\text{molar mass of solute}$.

$\begin{array}{c}\text{Number of moles = }\frac{\text{0}\text{.515 g}}{\text{94 g}\cdot {\text{mol}}^{-1}}\\ =0.00547\text{ mol}\end{array}$

Therefore $\text{Number of moles}$ of phenol are $0.00547\text{ mol}$.

Volume of the solvent is $125\text{ mL}$.

Unit conversion of $125\text{ mL}$ into $\text{L}$.

$\left(125\text{ mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)=0.125\text{ L}$

Therefore volume of the solvent is $0.125\text{ L}$.

The initial concentration of ${\text{C}}_6{\text{H}}_5\text{OH}$ is calculated by using the expression,

$\text{Molarity}=\frac{\text{Number of moles}}{\text{volume of the solvent}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)$

Substitute $0.00547\text{ mol}$ for $\text{Number of moles}$, $0.125\text{ L}$ for volume of the solvent.$\begin{array}{c}\text{Molarity}=\frac{0.00547\text{ mol}}{0.125\text{ L}}\\ =0.0438\text{ mol}\cdot {\text{L}}^{-1}\end{array}$

Therefore initial concentration of ${\text{C}}_6{\text{H}}_5\text{OH}$ is $0.0438\text{ mol}\cdot {\text{L}}^{-1}$.

ICE table (1) gives the dissociation of ${\text{C}}_6{\text{H}}_5\text{OH}$.

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)& +& {\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.0438& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.0438-x& & & & +x& & +x\end{array}$

From ICE table (1),

Concentration of phenol left after reaction is $\left(0.0438-x\right)\text{mol}\cdot {\text{L}}^{-1}$.

Concentration of phenoxide ion produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

Concentration of ${\text{H}}_3{\text{O}}^{+}$ produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

There is an approximation, that the value of $x$ is very small as comparison to $0.0438$ thus it can be neglected with respect to it.

Therefore, Concentration of phenol left after reaction is $0.0438\text{ mol}\cdot {\text{L}}^{-1}$.

The hydronium ion concentration is calculated by expression,

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$

Substitute $x$ for ${\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}$, $0.0438$ for ${\left[\text{HA}\right]}_{\left(\text{eq}\right)}$ and $1.3\times {10}^{-10}$ for $K_{\text{a}}$.

$\begin{array}{l}1.3\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.0438\right)}\\ 1.3\times {10}^{-10}=\frac{\left(x^2\right)}{\left(0.0438\right)}\end{array}$

Rearrange for $x^2$,

$\begin{array}{c}{x}^2=\left(1.3\times {10}^{-10}\right)\left(0.0438\right)\\ x=\left(\sqrt{0.056}\right)\left(\sqrt{{10}^{-10}}\right)\\ =0.238\times {10}^{-5}\end{array}$

Therefore, concentration of hydronium ion is $0.238\times 10^{-5}$.

Calculate the $\text{pH}$ value by using following expression,

$\text{pH = }-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$

Substitute $0.238\times 10^{-5}$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$.

$\begin{array}{c}\text{pH = }-\log \left(0.238\times {10}^{-5}\right)\\ =5.62\end{array}$

The value of $\text{pH}$ for the original solution of phenol is $5.62$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated at the various points during the titration between ${\text{C}}_6{\text{H}}_5\text{OH}$ and $\text{NaOH}$. The concentration of ${\text{OH}}^{-}$,${\text{H}}_3{\text{O}}^{+}$,${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ and ${\text{Na}}^{+}$ has to be calculated at equivalence point.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base. There are four types of acid-base titrations.

(1) Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, $\text{HCl}$ is titrated against $\text{NaOH}$.

(2) Strong acid-Weak base, in this type of titration a strong acid is titrated against a weak base for example, $\text{HCl}$ is titrated against ${\text{NH}}_4\text{OH}$.

(3) Weak acid-Strong base, in this type of titration a weak acid is titrated against a strong base for example, ${\text{CH}}_3\text{COOH}$ is titrated against $\text{NaOH}$.

(4) Weak acid-Weak base, in this type of titration a weak acid is titrated against a weak base for example, ${\text{CH}}_3\text{COOH}$ is titrated against ${\text{NH}}_4\text{OH}$.

For weak acid-strong base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{C}}_6{\text{H}}_5\text{OH}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5\text{ONa}\left(\text{aq}\right)$

Calculation of $\text{pH}$ at various points is done as follows,

(1) The $\text{pH}$ value before the titration can be calculated by using $K_{\text{a}}$ and its relation with ${\text{H}}_3{\text{O}}^{+}$ ion concentration.

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$ (1)

(2) The $\text{pH}$ calculation just before the equivalence point,

As the addition of $\text{NaOH}$ is done there will be formation of buffer solution ${\text{C}}_6{\text{H}}_5\text{OH}$/${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. The $\text{pH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{acid}\right]}$ (2)

At the midpoint of the titration, concentration of acid and its conjugate base is equal. Therefore $\text{pH}$ value at midpoint will be given as

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{acid}\right]}$

Substitute, $\left[\text{conjugate}\text{base}\right]\text{for}\left[\text{acid}\right]$.

$\begin{array}{c}\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{base}\right]}\\ =\text{p}{K}_{\text{a}}+\log \left(1\right)\\ =\text{p}{K}_{\text{a}}+0\\ =\text{p}{K}_{\text{a}}\end{array}$

Therefore, $\text{pH}$ value at midpoint is equal to $\text{p}{K}_{\text{a}}$.

(3) The $\text{pH}$ calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only ${\text{OH}}^{-}$ ion and ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. The ${\text{OH}}^{-}$ will be produced due to the hydrolysis of phenoxide ion at equivalence point. The hydrolysis equilibrium is represented as,

${\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right)$

By using the value of $K_{\text{b}}$ for the phenoxide ion, concentration of  ${\text{OH}}^{-}$ can be calculated. Thus the value of $\text{pH}$ is greater than $7$ at equivalence point for the weak acid- strong base titrations.

The relation between $K_{\text{a}}$ and $K_{\text{b}}$ for weak acid and its conjugate base is given as,

$K_w=\left(K_{\text{a}}\right)\left(K_{\text{b}}\right)$ (3)

Explanation of Solution

The concentration value of ${\text{OH}}^{-}$, phenoxide ion, ${\text{H}}_3{\text{O}}^{+}$ and ${\text{Na}}^{+}$ is calculated below.

Given:

Refer to the Apendix H in the textbook for the value of $K_{\text{a}}$.

The value of $K_{\text{a}}$ for phenol is $1.3\times {10}^{-10}$.

The value of $K_{\text{w}}$ for water is $1.0\times {10}^{-14}$.

The $\text{p}{K}_{\text{a}}$ value is calculated as follows;

$\text{p}{K}_{\text{a}}=-\log (K_{\text{a}})$

Substitute, $1.3\times {10}^{-10}$ for $K_{\text{a}}$

$\begin{array}{c}\text{p}{K}_{\text{a}}=-\log (1.3\times {10}^{-10})\\ =9.88\end{array}$

Therefore, $\text{p}{K}_{\text{a}}$ value of phenol is $9.88$.

The initial concentration of ${\text{C}}_6{\text{H}}_5\text{OH}$ is $0.0438\text{ mol}\cdot {\text{L}}^{-1}$.

The initial concentration of $\text{NaOH}$ is $0.123\text{ mol}\cdot {\text{L}}^{-1}$.

The volume of ${\text{C}}_6{\text{H}}_5\text{OH}$ is $125\text{mL}$.

Conversion of $125\text{mL}$ into $\text{L}$.

$\left(125\text{mL}\right)\left(\frac{1\text{L}}{\text{1000}\text{mL}}\right)=0.125\text{L}$

The volume of ${\text{C}}_6{\text{H}}_5\text{OH}$ is $0.125\text{L}$.

The volume of $\text{NaOH}$ used to neutralize all the phenol is calculated.

Expression used for the neutralization is as follows,

${\text{C}}_{\text{1}}{\text{V}}_1={\text{C}}_{\text{2}}{\text{V}}_{\text{2}}$

Here,

• ${\text{C}}_{\text{1}}$ is the concentration of phenol
• ${\text{V}}_{\text{1}}$ is the volume of phenol.
• ${\text{C}}_{\text{2}}$ is the concentration of $\text{NaOH}$ used.
• ${\text{V}}_2$ is the volume of $\text{NaOH}$ used for the neutralization.

Substitute $0.0438\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{1}}$, $0.125\text{L}$ for ${\text{V}}_{\text{1}}$,$0.123\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{2}}$.

$\left(0.0438\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.125\text{L}\right)=\left(0.123\text{ mol}\cdot {\text{L}}^{-1}\right){\text{V}}_{\text{2}}$

Rearrange for ${\text{V}}_2$,

$\begin{array}{c}{\text{V}}_{\text{2}}=\frac{\left(0.0438\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.125\text{L}\right)}{\left(0.123\text{ mol}\cdot {\text{L}}^{-1}\right)}\\ =0.044\text{ L}\end{array}$

Therefore, volume of the $\text{NaOH}$ used is $0.044\text{ L}$ or $44\text{ mL}$.

The calculation of moles is done by using the expression,

$\text{Number}\text{of moles}=\text{concentration}\left(\text{mol}\cdot {\text{L}}^{-1}\right)\cdot \text{volume}\left(\text{L}\right)$

The ICE table (2) for the reaction between $\text{NaOH}$ and ${\text{C}}_6{\text{H}}_5\text{OH}$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right)& +& {\text{ OH}}^{-}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{C}}_6{\text{H}}_5\text{ONa}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0054& & 0.0054& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.0054& & -0.0054& & & & 0.0054\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0& & & & 0.0054\end{array}$

From ICE table (2),

Number of moles of phenol left after reaction is $0\text{mol}$.

Number of moles of phenoxide ion produced after the reaction is $0.0054\text{ mol}$.

The total volume after the reaction is calculated as,

$\text{total}\text{volume}=\text{volume}\text{of}{\text{C}}_6{\text{H}}_5\text{OH}\left(\text{L}\right)\text{ + volume of NaOH}\left(\text{L}\right)$

$\begin{array}{c}\text{total}\text{volume = }0.125\left(\text{L}\right)+0.\text{044 L}\\ =0.169\text{ L}\end{array}$

Therefore, total volume after reaction is $0.169\text{ L}$.

Concentration calculations is done by using the expression,

$\text{concentration = }\frac{\text{Number}\text{of moles}}{\text{total volume}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)$

Calculate the concentration of phenoxide ion after reaction.

Substitute, $0.0054\text{ mol}$ for $\text{Number}\text{of moles}$ and $0.169\text{L}$ for volume in equation (4).

$\begin{array}{c}\text{concentration = }\frac{0.0054}{\text{0}\text{.169}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)\\ =0.0319\text{mol}\cdot {\text{L}}^{-1}\\ =0.032\text{ mol}\cdot {\text{L}}^{-1}\end{array}$

The concentration of phenoxide ion after reaction is $0.032\text{ mol}\cdot {\text{L}}^{-1}$.

The ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ ion produced will undergo hydrolysis in presence of water and the reaction equilibrium is written as,

${\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right)$

The hydrolysis equilibrium is represented in ICE table (3).

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{OH}}^{-}\left(\text{aq}\right)& +& {\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.032& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.032-x& & & & +x& & +x\end{array}$

From ICE table (3),

Concentration of phenoxide ion left after reaction is $\left(0.032-x\right)\text{mol}\cdot {\text{L}}^{-1}$.

Concentration of phenol produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

Concentration of ${\text{OH}}^{-}$ produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

Calculate the concentration of ${\text{OH}}^{-}$ by using the equation (3).

$K_w=\left(K_{\text{a}}\right)\left(K_{\text{b}}\right)$

Rearrange it for $K_{\text{b}}$

$K_{\text{b}}=\frac{K_w}{K_{\text{a}}}$ (5)

Substitute $1.3\times {10}^{-10}$ for $K_{\text{a}}$ and $1.0\times {10}^{-14}$ for $K_{\text{w}}$

$\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{1.3\times {10}^{-10}}\\ =0.769\times {10}^{-4}\end{array}$

Therefore, $K_{\text{b}}$ value for phenoxide ion is $0.769\times {10}^{-4}$.

The expression of $K_{\text{b}}$ for phenoxide ion from the ICE table (3) will be written as,

$K_{\text{b}}=\frac{{\left[{\text{C}}_6{\text{H}}_5\text{OH}\right]}_{\left(\text{eq}\right)}{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}_{\left(\text{eq}\right)}}$ (6)

Substitute, $0.769\times {10}^{-4}$ for $K_{\text{b}}$, $x$ for ${\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{C}}_6{\text{H}}_5\text{OH}\right]}_{\left(\text{eq}\right)}$, $\left(0.032-x\right)$ for ${\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}_{\left(\text{eq}\right)}$.

$0.76\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{\left(0.032-x\right)}$

$\left(0.76\times {10}^{-4}\right)=\frac{x^2}{\left(0.032-x\right)}$

Rearrange for $x^2$,

$\begin{array}{c}{x}^2=\left(0.769\times {10}^{-4}\right)\left(0.032-x\right)\\ =\left(0.769\times {10}^{-4}\right)\left(0.032\right)-\left(0.769\times {10}^{-4}\right)\left(x\right)\end{array}$

Rearrange it in the form of quadratic equation.

$x^2+\left(0.769\times {10}^{-4}\right)\left(x\right)-\left(0.769\times {10}^{-4}\right)\left(0.032\right)=0$

The general form of quadratic equation is given as,

$\text{a}{x}^2+\text{b}x+\text{c}=0$

On comparison,

The value of $\text{a}$ is equals to $1$.

The value of $\text{b}$ is equals to $0.769\times {10}^{-4}$.

The value of $\text{c}$ is equals to $\left(-\left(0.769\times {10}^{-4}\right)\left(0.032\right)\right)$.

The value of $x$ is calculated by the using following expression.

$x=\frac{-\text{b}\pm \sqrt{{\text{b}}^2-4\text{ac}}}{2\text{a}}$

Substitute $0.769\times {10}^{-4}$ for $\text{b}$,$\left(-\left(0.769\times {10}^{-4}\right)\left(0.032\right)\right)$ for $\text{c}$ and $1$ for $\text{a}$.

$\begin{array}{c}x=\frac{-0.769\times {10}^{-4}\pm \sqrt{{\left(0.769\times {10}^{-4}\right)}^2-4\left(1\right)\left(-\left(0.769\times {10}^{-4}\right)\left(0.032\right)\right)}}{2\left(1\right)}\\ =\frac{-0.769\times {10}^{-4}\pm \sqrt{\left(0.000060\right)\times {10}^{-4}+0.099\times {10}^{-4}}}{2}\\ =\frac{-0.769\times {10}^{-4}\pm \sqrt{\left(0.000060+0.099\right)}\times {10}^{-2}}{2}\\ =\frac{-0.769\times {10}^{-4}+0.315\times {10}^{-2}}{2}\end{array}$

Further calculation for the value of $x$,

$\begin{array}{c}x=\frac{-0.769\times {10}^{-4}+0.315\times {10}^{-2}}{2}\\ =\frac{\left(-0.00769+0.315\right)\times {10}^{-2}}{2}\\ =\frac{0.3073\times {10}^{-2}}{2}\\ =0.15\times {10}^{-2}\end{array}$

Therefore value of ${\text{OH}}^{-}$ concentration is $0.15\times 10^{-2}mol\cdot L^{-1}$ or $1.5\times 10^{-3}mol\cdot L^{-1}$.

The value of ${\text{C}}_6{\text{H}}_5\text{OH}$ concentration is $1.5\times 10^{-3}mol\cdot L^{-1}$.

The concentration of ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ is $\left(0.032-x\right)\text{ mol}\cdot {\text{L}}^{-1}$,

Substitute $1.5\times {10}^{-3}$ for $x$.

$\begin{array}{c}\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]=\left(0.032-0.0015\right)\\ =0.305\end{array}$

The concentration of ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ at equivalence point is $0.305mol\cdot L^{-1}$.

The concentration of ${\text{Na}}^{+}$ at equivalence point is equal to the initial concentration of phenoxide ion before undergo hydrolysis i.e.$0.032mol\cdot L^{-1}$

Calculate the value of $\text{pOH}$ by using the expression,

$\text{pOH = }-\log \left[{\text{OH}}^{-}\right]$

Substitute, $1.5\times {10}^{-3}$ for $\left[{\text{OH}}^{-}\right]$.

$\begin{array}{c}\text{pOH = }-\log \left(1.5\times {10}^{-3}\right)\\ =-\left(-2.82\right)\\ =2.82\end{array}$

Therefore, the value of $\text{pOH}$ is $2.82$.

Thus, the value of $\text{pH}$ is calculated by using expression,

$\text{pH + pOH =}\text{14}$

Rearrange for $\text{pH}$,

$\text{pH =}\text{14}-\text{ pOH}$

Substitute, $2.82$ for $\text{pOH}$.

$\begin{array}{c}\text{pH =}\text{14}-2.82\\ \text{= }11.18\end{array}$

Therefore, the value of $\text{pH}$ at equivalence point is $11.18$.

The $\text{pH}$ value is used to calculate the concentration of hydronium ion at equivalence point.

Calculate the hydronium ion concentration by using expression,

$\text{pH = }-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$

Rearrange for $\left[{\text{H}}_3{\text{O}}^{+}\right]$,

$\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-\left(\text{pH}\right)}$

Substitute $11.18$ for $\text{pH}$.

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-\left(11.18\right)}\\ =6.6\times 10^{-12}\end{array}$

Therefore, hydronium ion concentration at equivalence point is $6.6\times 10^{-12}mol\cdot L^{-1}$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated at the various points during the titration between ${\text{C}}_6{\text{H}}_5\text{OH}$ and $\text{NaOH}$. The value of $\text{pH}$ at equivalence point has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base. There are four types of acid-base titrations.

(1) Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, $\text{HCl}$ is titrated against $\text{NaOH}$.

(2) Strong acid-Weak base, in this type of titration a strong acid is titrated against a weak base for example, $\text{HCl}$ is titrated against ${\text{NH}}_4\text{OH}$.

(3) Weak acid-Strong base, in this type of titration a weak acid is titrated against a strong base for example, ${\text{CH}}_3\text{COOH}$ is titrated against $\text{NaOH}$.

(4) Weak acid-Weak base, in this type of titration a weak acid is titrated against a weak base for example, ${\text{CH}}_3\text{COOH}$ is titrated against ${\text{NH}}_4\text{OH}$.

For weak acid-strong base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{C}}_6{\text{H}}_5\text{OH}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5\text{ONa}\left(\text{aq}\right)$

Calculation of $\text{pH}$ at various points is done as follows,

(1) The $\text{pH}$ value before the titration can be calculated by using $K_{\text{a}}$ and its relation with ${\text{H}}_3{\text{O}}^{+}$ ion concentration.

${\text{K}}_a=\frac{{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[\text{HA}\right]}_{\left(\text{eq}\right)}}$ (1)

(2) The $\text{pH}$ calculation just before the equivalence point,

As the addition of $\text{NaOH}$ is done there will be formation of buffer solution ${\text{C}}_6{\text{H}}_5\text{OH}$/${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. The $\text{pH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{acid}\right]}$ (2)

At the midpoint of the titration, concentration of acid and its conjugate base is equal. Therefore $\text{pH}$ value at midpoint will be given as

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{acid}\right]}$

Substitute, $\left[\text{conjugate}\text{base}\right]\text{for}\left[\text{acid}\right]$.

$\begin{array}{c}\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{base}\right]}\\ =\text{p}{K}_{\text{a}}+\log \left(1\right)\\ =\text{p}{K}_{\text{a}}+0\\ =\text{p}{K}_{\text{a}}\end{array}$

Therefore, $\text{pH}$ value at midpoint is equal to $\text{p}{K}_{\text{a}}$.

(3) The $\text{pH}$ calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only ${\text{OH}}^{-}$ ion and ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. The ${\text{OH}}^{-}$ will be produced due to the hydrolysis of phenoxide ion at equivalence point. The hydrolysis equilibrium is represented as,

${\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5\text{OH}\left(\text{aq}\right)$

By using the value of $K_{\text{b}}$ for the phenoxide ion, concentration of  ${\text{OH}}^{-}$ can be calculated. Thus the value of $\text{pH}$ is greater than $7$ at equivalence point for the weak acid- strong base titrations.

The relation between $K_{\text{a}}$ and $K_{\text{b}}$ for weak acid and its conjugate base is given as,

$K_w=\left(K_{\text{a}}\right)\left(K_{\text{b}}\right)$ (3)

Explanation of Solution

The addition of $44\text{ mL}$ $\text{NaOH}$ will required to reach the equivalence point of titration. At equivalence point all the phenol and hydroxide ions will be used only phenoxide will be present. The hydrolysis of phenoxide ion will produce a very small amount of ${\text{OH}}^{-}$ ions. The $\text{pH}$ calculation at equivalence point is given below.

Given:

The value of $\text{pH}$ is calculated by using expression,

$\text{pH + pOH =}\text{14}$

Rearrange for $\text{pH}$,

$\text{pH =}\text{14}-\text{ pOH}$

Substitute, $2.82$ for $\text{pOH}$

$\begin{array}{c}\text{pH =}\text{14}-\text{2}\text{.82}\\ \text{= }11.18\end{array}$

Therefore, the value of $\text{pH}$ at equivalence point is $11.18$.