Chapter 17, Problem 105IL

Aniline hydrochloride, (C₆H₅NH₃)Cl, is a weak acid. (Its conjugate base is the weak base aniline, C₆H₅NH₂.) The acid can be titrated with a strong base such as NaOH.

$C_6{H}_{5}N{H}_3{}^{+}(aq)+O{H}^{-}(aq)\rightleftarrows C_6{H}_{5}N{H}_2(aq)+H_{2}O(\mathcal{l})$

Assume 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.185 M NaOH. (K_a for aniline hydrochloride is 2.4 × 10⁻⁵.)

1. (a) What is the pH of the (C₆H₅NH₃) solution before the titration begins?
2. (b) What is the pH at the equivalence point?
3. (c) What is the pH at the halfway point of the titration?
4. (d) Which indicator in Figure 17.11 could be used to detect the equivalence point?
5. (e) Calculate the pH of the solution after adding 10.0, 20.0, and 30.0 mL of base.
6. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ for the original solution of $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ has to be calculated before titration begins.

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.

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 $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right)$

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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$/${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. 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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. The ${\text{OH}}^{-}$ will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

By using the value of $K_{\text{b}}$ for the aniline, 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 aniline hydrochloride is $2.4\times {10}^{-5}$.

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, $2.4\times {10}^{-5}$ for $K_{\text{a}}$.

$\begin{array}{c}\text{p}{K}_{\text{a}}=-\log (2.4\times {10}^{-5})\\ =4.62\end{array}$

Therefore, $\text{p}{K}_{\text{a}}$ value of $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$  is $4.62$.

The initial concentration of $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ is $0.100\text{ M}$.

The initial concentration of $\text{NaOH}$ is $0.185\text{ M}$.

Volume of the solvent is $50\text{ mL}$.

Therefore volume of the solvent is $0.050\text{ L}$.

ICE table (1) gives the dissociation of $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$.

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)& +& {\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot \text{L}\right)& 0.100& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.100-x& & & & +x& & +x\end{array}$

From ICE table (1),

Concentration of ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$ left after reaction is $\left(0.100-x\right)\text{mol}\cdot {\text{L}}^{-1}$.

Concentration of aniline 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 $0.100$ thus it can be neglected with respect to it.

Therefore, Concentration of ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$ left after reaction is $0.100\text{ mol}\cdot {\text{L}}^{-1}$.

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 ${\left[{\text{A}}^{-}\right]}_{\left(\text{eq}\right)}$, $0.100$ for ${\left[\text{HA}\right]}_{\left(\text{eq}\right)}$ and $2.4\times {10}^{-5}$ for $K_{\text{a}}$.

$\begin{array}{l}2.4\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.100\right)}\\ 2.4\times {10}^{-5}=\frac{\left(x^2\right)}{\left(0.100\right)}\end{array}$

Rearrange for $x^2$,

$\begin{array}{c}{x}^2=\left(2.4\times {10}^{-5}\right)\left(0.100\right)\\ x=0.00154\end{array}$

Therefore, concentration of hydronium ion is $0.00154$.

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

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

Substitute $0.00154$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$.

$\begin{array}{c}\text{pH = }-\log \left(0.00154\right)\\ =2.81\end{array}$

The value of $\text{pH}$ for the original solution of $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ is $2.81$.

Interpretation Introduction

Interpretation:

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.

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 $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right)$

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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$/${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. 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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. The ${\text{OH}}^{-}$ will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

By using the value of $K_{\text{b}}$ for the aniline, 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

Given:

The volume of $\text{NaOH}$ used to neutralize all the acid 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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$
• ${\text{V}}_{\text{1}}$ is the volume of ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$.
• ${\text{C}}_{\text{2}}$ is the concentration of $\text{NaOH}$ used.
• ${\text{V}}_2$ is the volume of $\text{NaOH}$ used for the neutralization.

Substitute $0.100\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{1}}$, $0.050\text{ L}$ for ${\text{V}}_{\text{1}}$,$0.185\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{2}}$.

$\left(0.100\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.050\text{L}\right)=\left(0.185\text{ mol}\cdot {\text{L}}^{-1}\right){\text{V}}_{\text{2}}$

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

$\begin{array}{c}{\text{V}}_{\text{2}}=\frac{\left(0.100\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.050\text{L}\right)}{\left(0.185\text{ mol}\cdot {\text{L}}^{-1}\right)}\\ =0.027\text{ L}\end{array}$

Therefore, volume of the $\text{NaOH}$ used is $0.027\text{ L}$ or $\text{27 mL}$.

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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)& +& {\text{ OH}}^{-}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.005& & 0.005& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.005& & -0.005& & & & 0.005\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0& & & & 0.005\end{array}$

From ICE table (2),

Number of moles of anilinium ion left after reaction are $0\text{mol}$.

Number of moles of aniline produced after the reaction are $0.005\text{ mol}$.

The total volume after the reaction is calculated as,

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

$\begin{array}{c}\text{total}\text{volume = }0.050\left(\text{L}\right)+0.\text{027 L}\\ =0.077\text{ L}\end{array}$

Therefore, total volume after reaction is $0.077\text{ L}$.

Concentration calculations is done by using the expression,

$\text{concentration = }\frac{\text{Number}\text{of moles}}{\text{total volume}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)$ (4)

Calculate the concentration of aniline after reaction.

Substitute, $0.005\text{ mol}$ for $\text{Number}\text{of moles}$ and $0.077\text{ L}$ for volume in equation (4).

$\begin{array}{c}\text{concentration = }\frac{0.005}{\text{0}\text{.077}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)\\ =0.0649\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The concentration of aniline after reaction is $0.0649\text{mol}\cdot {\text{L}}^{-1}$.

The aniline produced will undergo hydrolysis in presence of water and the reaction equilibrium is written as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

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

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{OH}}^{-}\left(\text{aq}\right)& +& {\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.0649& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.0649-x& & & & +x& & +x\end{array}$

From ICE table (3),

Concentration of aniline left after reaction is $\left(0.0649-x\right)\text{mol}\cdot {\text{L}}^{-1}$.

Approximation, $x$ is very small as compared to $0.0649$. Therefore it can be neglected.

So, Concentration of aniline left after reaction is $\left(0.0649\right)\text{mol}\cdot {\text{L}}^{-1}$

Concentration of anilinium ion 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 $2.4\times {10}^{-5}$ for $K_{\text{a}}$ and $1.0\times {10}^{-14}$ for $K_{\text{w}}$

$\begin{array}{c}{K}_{\text{b}}=\frac{1.0\times {10}^{-14}}{2.4\times {10}^{-5}}\\ =0.416\times {10}^{-9}\end{array}$

Therefore, $K_{\text{b}}$ value for aniline is $0.416\times {10}^{-9}$.

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

$K_{\text{b}}=\frac{{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\right]}_{\left(\text{eq}\right)}{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\right]}_{\left(\text{eq}\right)}}$ (6)

Substitute, $0.416\times {10}^{-9}$ for $K_{\text{b}}$, $x$ for ${\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\right]}_{\left(\text{eq}\right)}$, $\left(0.0649\right)$ for ${\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\right]}_{\left(\text{eq}\right)}$.

$0.416\times {10}^{-9}=\frac{\left(x\right)\left(x\right)}{\left(0.0649\right)}$

Rearrange for $x^2$,

$\begin{array}{c}x=\sqrt{\left(0.416\times {10}^{-9}\right)\left(0.0649\right)}\\ =5.19\times {10}^{-6}\end{array}$

Therefore value of ${\text{OH}}^{-}$ concentration is $5.19\times {10}^{-6}\text{ mol}\cdot {\text{L}}^{-1}$.

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

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

Substitute, $5.19\times {10}^{-6}$ for $\left[{\text{OH}}^{-}\right]$.

$\begin{array}{c}\text{pOH = }-\log \left(5.19\times {10}^{-6}\right)\\ =5.28\end{array}$

Therefore, the value of $\text{pOH}$ is $5.28$.

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, $5.28$ for $\text{pOH}$.

$\begin{array}{c}\text{pH =}\text{14}-5.28\\ \text{= }8.72\end{array}$

The value of $\text{pH}$ at equivalence point is $8.72$. Therefore, the value of $\text{pH}$ at equivalence point is $8.72$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ at halfway point of the titration 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.

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 $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right)$

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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$/${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. 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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. The ${\text{OH}}^{-}$ will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

By using the value of $K_{\text{b}}$ for the aniline, 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 at midpoint is equal to $\text{p}{K}_{\text{a}}$.

Given:

At midpoint of the titration the concentration of acid and its conjugate base will be equal.

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

Substitute $\left[\text{conjugate}\text{base}\right]\text{for}\left[\text{acid}\right]$ and $4.62$ for $\text{p}{K}_{\text{a}}$.

$\begin{array}{l}\text{pH}=4.62+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{base}\right]}\\ \text{pH}=4.62\end{array}$

The value of $\text{pH}$ at halfway point of the titration is $4.62$ .

Therefore, the value of $\text{pH}$ at midpoint is $4.62$.

Interpretation Introduction

Interpretation:

 A best indicator that can be used to detect the equivalence point has to be chosen for the titration.

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.

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 $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right)$

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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$/${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. 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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. The ${\text{OH}}^{-}$ will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

By using the value of $K_{\text{b}}$ for the aniline, 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

Phenolphthalein is the best indicator for the titration between aniline hydrochloride and sodium hydroxide. As it changes its colour in the $\text{pH}$ region $8.2-9.8$ and for the given titration value of $\text{pH}$ at equivalence point is $8.72$ .

Phenolphthalein indicator that can be used to detect the equivalence point is chosen for the titration.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ when $10\text{ mL, 20 mL and 30 mL NaOH}$ is added to the  $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ solution 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.

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 $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right)$

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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$/${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. 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 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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. The ${\text{OH}}^{-}$ will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

By using the value of $K_{\text{b}}$ for the aniline, 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

At various points before equivalence point the value of $\text{pH}$ is calculated by using Henderson-Hasselbalch equation.

Given:

When $10\text{ mL}$ solution of base is used the ICE table for the reaction between $\text{NaOH}$ and ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)& +& {\text{ OH}}^{-}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.005& & 0.00185& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.00185& & -0.00185& & & & 0.00185\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.00315& & 0& & & & 0.00185\end{array}$

After titration volume will be equal for both acid and base. The total volume will be $0.060\text{ L}$.

Calculation is done by using equation (1),

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

Substitute $\frac{0.00185}{0.060}$ for $\left[\text{conjugate}\text{base}\right]$,$\frac{0.00315}{0.060}$ for $\left[\text{acid}\right]$ and $4.62$ for $\text{p}{K}_{\text{a}}$.

$\begin{array}{c}\text{pH}=4.62+\log \frac{\left(\frac{0.00185}{0.060}\right)}{\left(\frac{0.00315}{0.060}\right)}\\ =4.62+\log \left(0.587\right)\\ =4.62-0.231\\ =4.39\end{array}$

Therefore value of $\text{pH}$ is $4.39$.

When $20\text{ mL}$ solution of base is used the ICE table for the reaction between $\text{NaOH}$ and ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)& +& {\text{ OH}}^{-}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.005& & 0.00370& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.00370& & -0.00370& & & & 0.00370\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.00130& & 0& & & & 0.00370\end{array}$

After titration volume will be equal for both acid and base. The total volume will be $0.070\text{ L}$.

Calculation is done by using equation (1),

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

Substitute $\frac{0.00370}{0.060}$ for $\left[\text{conjugate}\text{base}\right]$,$\frac{0.00130}{0.060}$ for $\left[\text{acid}\right]$ and $4.62$ for $\text{p}{K}_{\text{a}}$.

$\begin{array}{c}\text{pH}=4.62+\log \frac{\left(\frac{0.00370}{0.060}\right)}{\left(\frac{0.00130}{0.060}\right)}\\ =4.62+\log \left(2.846\right)\\ =4.62+0.45\\ =5.17\end{array}$

Therefore value of $\text{pH}$ is $5.17$.

When $30\text{ mL}$ solution of base is used the ICE table for the reaction between $\text{NaOH}$ and ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)& +& {\text{ OH}}^{-}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0050& & 0.00555& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.0050& & -0.0050& & & & +0.0050\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0.00055& & & & 0.0050\end{array}$

After titration volume will be equal for both acid and base. The total volume will be $0.080\text{ L}$

After the equivalence point, the excess concentration ${\text{OH}}^{-}$ will present at the equilibrium. The hydrolysis of acetate ion will produce a very small amount of ${\text{OH}}^{-}$ ions, that amount can be neglected.

Concentration of hydroxide ion after equivalence point is calculated.

$\begin{array}{c}\text{concentration}=\frac{0.00055\text{ mol}}{0.080\text{ L}}\\ =0.006875\text{ mol}\cdot {\text{L}}^{-1}\end{array}$

Therefore, concentration of hydroxide ions after reaction is $0.006875\text{ mol}\cdot {\text{L}}^{-1}$.

Calculate the $\text{pOH}$

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.006875\right)\\ =2.16\end{array}$

Now, $\text{pH}$ value is calculated by using $\text{pOH}$,

 $\begin{array}{c}\text{pH}=14-\text{pOH}\\ \text{= 14}-\text{2}\text{.16}\\ \text{= }11.84\end{array}$

The value of $\text{pH}$ when $10\text{ mL NaOH }$ added $\text{is }4.39$, when $20\text{ mL NaOH}$ used $5.07$  and after addition of $30\text{ mL NaOH}$$11.84$.

Therefore value of $\text{pH}$ is $11.84$ after addition of $30\text{ mL}$ sodium hydroxide.

Interpretation Introduction

Interpretation:

Titration curve has to be plotted by using $\text{pH}$ values calculated in part (a),(b),(c) and

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.

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 $\left({\text{C}}_6{\text{H}}_5{\text{NH}}_3\right)\text{Cl}$ with $\text{NaOH}$ is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+ OH}}^{-}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right)$

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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}$/${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. 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 ${\text{C}}_6{\text{H}}_5{\text{NH}}_2$. The ${\text{OH}}^{-}$ will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

${\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

By using the value of $K_{\text{b}}$ for the aniline, 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 titration curve is drawn by using the $\text{pH}$ value at different points.  The graph parameters are $x\text{ axis}$ the volume of $\text{NaOH}$ and $\text{pH}$ value on $y\text{ axis}$.

Table gives the value of $\text{pH}$ at different volume addition of $\text{NaOH}$ and corresponding point on the curve.

$\begin{array}{ccc}\text{Points }& \begin{array}{l}\text{volume of NaOH}\\ \left(\text{mL}\right)\end{array}& \text{pH}\\ \text{a}& \text{0}& \text{2}\text{.81}\\ \text{b}& 10& \text{4}\text{.39}\\ \text{c}& \text{13}\text{.5}& \text{4}\text{.62}\\ \text{d}& \text{20}& \text{5}\text{.07}\\ \text{e}& \text{27}& \text{8}\text{.72}\\ \text{f}& \text{30}& \text{11}\text{.84}\end{array}$

Graph between $\text{pH}$ and volume of $\text{NaOH}$ added.

Titration curve is plotted by using $\text{pH}$ values calculated in part (a),(b),(c) and (e).