Chapter 17, Problem 106IL

The weak base ethanolamine. HOCH₂CH₂NH₂, can be titrated with HCl.

$HOC{H}_{2}C{H}_{2}N{H}_2(aq)+H_3{O}^{+}(aq)\rightleftarrows HOC{H}_{2}C{H}_{2}N{H}_3{}^{+}(aq)+H_{2}O(\mathcal{l})$

Assume you have 25.0 mL of a 0.010 M solution of ethanolamine and titrate it with 0.0095 M HCl. (K_b for ethanolamine is 3.2 × 10⁻⁷.)

1. (a) What is the pH of the ethanolamine 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 would be the best choice to detect the equivalence point?
5. (e) Calculate the pH of the solution after adding 5.00, 10.0, 20.0, and 30.0 mL of the acid.
6. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated at the various points during the titration between ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$ and $\text{HCl}$. The value of $\text{pH}$ for the original solution of ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$ 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 strong acid-weak base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$ with $\text{HCl}$ is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right){\text{+ H}}_3{\text{O}}^{+}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

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

(1) The $\text{pOH}$ value before the titration can be calculated by using $K_{\text{b}}$ and its relation with ${\text{OH}}^{-}$ ion concentration.

$K_{\text{b}}=\frac{{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}{\left[{\text{BH}}^{+}\right]}_{\left(\text{eq}\right)}}{{\left[\text{B}\right]}_{\left(\text{eq}\right)}}$ (1)

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

As the addition of $\text{HCl}$ is done there will be formation of buffer solution ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$/${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The $\text{pOH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

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

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

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

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

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

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

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

At equivalence point all the acid will be neutralized, and there will be only ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ethylamine, concentration of  ${\text{H}}_3{\text{O}}^{+}$ can be calculated.

Value of $\text{pH}$ at every point is calculated by using the relation between $\text{pH}$ and $\text{pOH}$.

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

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

$K_{\text{b}}=\frac{{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}{\left[{\text{BH}}^{+}\right]}_{\left(\text{eq}\right)}}{{\left[\text{B}\right]}_{\left(\text{eq}\right)}}$

Given:

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

The value of $K_{\text{b}}$ for ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ is $3.2\times {10}^{-5}$.

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

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

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

Substitute, $3.2\times {10}^{-5}$ for $K_{\text{b}}$.

$\begin{array}{c}\text{p}{K}_{\text{b}}=-\log (3.2\times {10}^{-5})\\ =4.49\end{array}$

Therefore, $\text{p}{K}_{\text{b}}$ value of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ is $4.49$.

The initial concentration of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ is $0.010\text{ M}$.

The initial concentration of $\text{HCl}$ is $0.0095\text{ M}$.

Volume of the solvent is $25\text{ mL}$.

Therefore volume of the solvent is $0.025\text{ L}$.

ICE table (1) gives the dissociation of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$.

$\begin{array}{cccccccc}\text{Equation}& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{OH}}^{-}\left(\text{aq}\right)& +& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.010& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.010-x& & & & +x& & +x\end{array}$

From ICE table (1),

Concentration of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ left after reaction is $\left(0.010-x\right)\text{mol}\cdot {\text{L}}^{-1}$.

Concentration of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$ produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

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

There is an approximation, that the value of $x$ is very small as comparison to $0.010$ thus it can be neglected with respect to it.

Therefore, Concentration of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ left after reaction is $0.010\text{ mol}\cdot {\text{L}}^{-1}$.

The hydroxide ion concentration is calculated by expression,

$K_{\text{b}}=\frac{{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}{\left[{\text{BH}}^{+}\right]}_{\left(\text{eq}\right)}}{{\left[\text{B}\right]}_{\left(\text{eq}\right)}}$

Substitute $x$ for ${\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{BH}}^{+}\right]}_{\left(\text{eq}\right)}$, $0.010$ for ${\left[\text{B}\right]}_{\left(\text{eq}\right)}$ and $3.2\times {10}^{-5}$ for $K_{\text{a}}$.

$\begin{array}{l}3.2\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.010\right)}\\ x^2=\left(3.2\times {10}^{-5}\right)\left(0.010\right)\end{array}$

$x=0.565\times {10}^{-2}$

Therefore, concentration of hydroxide ion is $0.00565$.

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

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

Substitute $0.00565$ for $\left[{\text{OH}}^{-}\right]$.

$\begin{array}{c}\text{pOH = }-\log \left(0.00565\right)\\ =2.25\end{array}$

The value of $\text{pOH}$ for the original solution of ${\text{C}}_2{\text{H}}_5{\text{NH}}_2$ is $2.25$.

$\begin{array}{c}\text{pH = 14}-2.25\\ =11.75\end{array}$

Therefore value of $\text{pH}$ for the original solution is $11.75$.

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 strong acid-weak base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$ with $\text{HCl}$ is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right){\text{+ H}}_3{\text{O}}^{+}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

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

(1) The $\text{pOH}$ value before the titration can be calculated by using $K_{\text{b}}$ and its relation with ${\text{OH}}^{-}$ ion concentration.

$K_{\text{b}}=\frac{{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}{\left[{\text{BH}}^{+}\right]}_{\left(\text{eq}\right)}}{{\left[\text{B}\right]}_{\left(\text{eq}\right)}}$ (1)

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

As the addition of $\text{HCl}$ is done there will be formation of buffer solution ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$/${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The $\text{pOH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

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

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

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

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

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

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

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

At equivalence point all the acid will be neutralized, and there will be only ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ethylamine, concentration of  ${\text{H}}_3{\text{O}}^{+}$ can be calculated.

Value of $\text{pH}$ at every point is calculated by using the relation between $\text{pH}$ and $\text{pOH}$.

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

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 $\text{HCl}$ used to neutralize the entire base ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ 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 ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$
• ${\text{V}}_{\text{1}}$ is the volume of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$.
• ${\text{C}}_{\text{2}}$ is the concentration of $\text{HCl}$ used.
• ${\text{V}}_2$ is the volume of $\text{HCl}$ used for the neutralization.

Substitute $0.010\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{1}}$, $0.025\text{ L}$ for ${\text{V}}_{\text{1}}$,$0.0095\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{2}}$.

$\left(0.010\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.025\text{L}\right)=\left(0.0095\text{ mol}\cdot {\text{L}}^{-1}\right){\text{V}}_{\text{2}}$

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

$\begin{array}{c}{\text{V}}_{\text{2}}=\frac{\left(0.010\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.025\text{L}\right)}{\left(0.0095\text{ mol}\cdot {\text{L}}^{-1}\right)}\\ =0.0263\text{ L}\end{array}$

Therefore, volume of the $\text{HCl}$ used is $0.0263\text{ L}$ or $\text{26}\text{.3 mL}$.

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 $\text{HCl}$ and ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)& +& {\text{ H}}_3{\text{O}}^{+}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.00025& & 0.00025& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.00025& & -0.00025& & & & +0.00025\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0& & & & 0.00025\end{array}$

From ICE table (2),

Number of moles of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ left after reaction are $0\text{mol}$.

Number of moles of conjugate acid ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$ produced after the reaction are $0.00025\text{ mol}$.

The total volume after the reaction is calculated as,

$\text{total}\text{volume}=\text{volume}{\text{of HOCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{L}\right)\text{ + volume of HCl}\left(\text{L}\right)$

$\begin{array}{c}\text{total}\text{volume = }0.025\left(\text{L}\right)+0.\text{0263 L}\\ =0.0513\text{ L}\end{array}$

Therefore, total volume after reaction is $0.0513\text{ L}$.

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 ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$ after reaction.

Substitute, $0.00025\text{ mol}$ for $\text{Number}\text{of moles}$ and $0.0513\text{ L}$ for volume in equation (4).

$\begin{array}{c}\text{concentration = }\frac{0.00025}{\text{0}\text{.0513}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)\\ =0.00487\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The concentration of ${\text{C}}_2{\text{H}}_5{\text{NH}}_3{}^{+}$ after reaction is $0.00487\text{mol}\cdot {\text{L}}^{-1}$.

The ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$ produced will undergo hydrolysis in presence of water and the reaction equilibrium is written as,

${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)$

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

$\begin{array}{cccccccc}\text{Equation}& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)& +& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_{\text{2}}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.00487& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.00487-x& & & & +x& & +x\end{array}$

From ICE table (3),

Concentration of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$ left after reaction is $\left(0.00487-x\right)\text{mol}\cdot {\text{L}}^{-1}$.

Approximation, $x$ is very small as compared to $0.00487$. Therefore it can be neglected.

So, Concentration of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$ left after reaction is $0.00487\text{mol}\cdot {\text{L}}^{-1}$

Concentration of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

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

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 $K_{\text{a}}$

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

Substitute $3.2\times {10}^{-5}$ for $K_{\text{b}}$ and $1.0\times {10}^{-14}$ for $K_{\text{w}}$

$\begin{array}{c}{K}_{\text{a}}=\frac{1.0\times {10}^{-14}}{3.2\times {10}^{-5}}\\ =0.3125\times {10}^{-9}\end{array}$

Therefore, $K_{\text{a}}$ value for ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$ is $0.3125\times {10}^{-9}$.

The expression of $K_{\text{a}}$ for ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$ from the ICE table (3) will be written as,

$K_{\text{a}}=\frac{{\left[{\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\right]}_{\left(\text{eq}\right)}{\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}}{{\left[{\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]}_{\left(\text{eq}\right)}}$ (6)

Substitute, $0.3125\times {10}^{-9}$ for $K_{\text{a}}$, $x$ for ${\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\right]}_{\left(\text{eq}\right)}$, $\left(0.00487\right)$ for ${\left[{\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]}_{\left(\text{eq}\right)}$.

$0.3125\times {10}^{-9}=\frac{\left(x\right)\left(x\right)}{\left(0.00487\right)}$

Rearrange for $x$,

$\begin{array}{c}x=\sqrt{\left(0.3125\times {10}^{-9}\right)\left(0.00487\right)}\\ =0.122\times {10}^{-5}\end{array}$

Therefore value of hydronium ion concentration is $0.122\times {10}^{-5}\text{ mol}\cdot {\text{L}}^{-1}$.

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

$\text{pH = }-\log \left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$

Substitute, $0.122\times {10}^{-5}$ for $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$.

$\begin{array}{c}\text{pH = }-\log \left(0.122\times {10}^{-5}\right)\\ =5.91\end{array}$

Therefore, the value of $\text{pH}$ at equivalence point is $5.91$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ at midpoint 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 strong acid-weak base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$ with $\text{HCl}$ is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right){\text{+ H}}_3{\text{O}}^{+}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

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

(1) The $\text{pOH}$ value before the titration can be calculated by using $K_{\text{b}}$ and its relation with ${\text{OH}}^{-}$ ion concentration.

$K_{\text{b}}=\frac{{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}{\left[{\text{BH}}^{+}\right]}_{\left(\text{eq}\right)}}{{\left[\text{B}\right]}_{\left(\text{eq}\right)}}$ (1)

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

As the addition of $\text{HCl}$ is done there will be formation of buffer solution ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$/${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The $\text{pOH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

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

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

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

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

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

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

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

At equivalence point all the acid will be neutralized, and there will be only ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ethylamine, concentration of  ${\text{H}}_3{\text{O}}^{+}$ can be calculated.

Value of $\text{pH}$ at every point is calculated by using the relation between $\text{pH}$ and $\text{pOH}$.

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

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 $\text{pOH}$ value at midpoint is equal to $\text{p}{K}_{\text{b}}$.

Given:

The $\text{p}{K}_{\text{b}}$ value of ethylamine is $4.49$ calculated in part (a).

Therefore, the value of $\text{pOH}$ at midpoint is $4.49$.

The value of $\text{pH}$ is calculated.

$\begin{array}{c}\text{pH = 14}-4.49\\ =9.51\end{array}$

The value of $\text{pH}$ at the midpoint of the titration is $9.51$.

Interpretation Introduction

Interpretation:

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

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 strong acid-weak base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$ with $\text{HCl}$ is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right){\text{+ H}}_3{\text{O}}^{+}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

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

(1) The $\text{pOH}$ value before the titration can be calculated by using $K_{\text{b}}$ and its relation with ${\text{OH}}^{-}$ ion concentration.

$K_{\text{b}}=\frac{{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}{\left[{\text{BH}}^{+}\right]}_{\left(\text{eq}\right)}}{{\left[\text{B}\right]}_{\left(\text{eq}\right)}}$ (1)

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

As the addition of $\text{HCl}$ is done there will be formation of buffer solution ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$/${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The $\text{pOH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

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

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

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

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

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

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

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

At equivalence point all the acid will be neutralized, and there will be only ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ethylamine, concentration of  ${\text{H}}_3{\text{O}}^{+}$ can be calculated.

Value of $\text{pH}$ at every point is calculated by using the relation between $\text{pH}$ and $\text{pOH}$.

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

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

Alizarin or Bromcresol purple are the best indicator for the titration between ethylamine and $\text{HCl}$. As Bromcresol purple shows colour change in the $\text{pH}$ region $5.2-6.6$ (approx) and Alizarin shows colour change in the $\text{pH}$ region $5.8-7.3$ (approx). The value of $\text{pH}$ at equivalence point is $5.91$. Thus both indicators can be used to detect the equivalence point.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ when $5\text{ mL, }10\text{ mL, 20 mL and 30 mL HCl}$ is added to the  ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$ 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 strong acid-weak base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$ with $\text{HCl}$ is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right){\text{+ H}}_3{\text{O}}^{+}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

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

(1) The $\text{pOH}$ value before the titration can be calculated by using $K_{\text{b}}$ and its relation with ${\text{OH}}^{-}$ ion concentration.

$K_{\text{b}}=\frac{{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}{\left[{\text{BH}}^{+}\right]}_{\left(\text{eq}\right)}}{{\left[\text{B}\right]}_{\left(\text{eq}\right)}}$ (1)

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

As the addition of $\text{HCl}$ is done there will be formation of buffer solution ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$/${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The $\text{pOH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

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

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

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

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

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

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

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

At equivalence point all the acid will be neutralized, and there will be only ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ethylamine, concentration of  ${\text{H}}_3{\text{O}}^{+}$ can be calculated.

Value of $\text{pH}$ at every point is calculated by using the relation between $\text{pH}$ and $\text{pOH}$.

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

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 $\text{pH}$ before and after equivalence point is calculated by using Henderson-Hasselbalch equation.

Given:

When $\text{5 mL}$ solution of $\text{HCl}$ is added.

The volume of $\text{HCl}$ added is  $0.005\text{ L}$.

The ICE table for the reaction between $\text{HCl}$ and ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)& +& {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.00025& & 0.0000475& & 0& & 0\\ \text{Change}\left(\text{mol}\right)& 0.0000475& & 0.0000475& & 0& & 0.0000475\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.0002025& & 0& & 0& & 0.0000475\end{array}$

After titration volume will be equal for both acid and base. The total volume will be $0.030\text{ L}$.

Calculation is done by using equation (2),

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

Substitute $\frac{0.0000475}{0.030}$ for $\left[\text{conjugate acid}\right]$,$\frac{0.0002025}{0.030}$ for $\left[\text{base}\right]$ and $4.49$ for $\text{p}{K}_{\text{b}}$.

$\begin{array}{c}\text{pOH}=4.49+\log \frac{\left(\frac{0.0000475}{0.030}\right)}{\left(\frac{0.0002025}{0.030}\right)}\\ =4.49+\log \left(0.235\right)\\ =4.49-0.63\\ =3.86\end{array}$

Therefore value of $\text{pOH}$ is $3.86$.

Hence value of $\text{pH}$ is calculated as follows,

$\begin{array}{c}\text{pH = 14}-3.86\\ =10.14\end{array}$

Therefore value of $\text{pH}$ is $10.14$ after addition of $\text{5 mL}$$\text{HCl}$ solution.

When $\text{10 mL}$ solution of $\text{HCl}$ is added.

The volume of $\text{HCl}$ added is  $0.010\text{ L}$.

The ICE table for the reaction between $\text{HCl}$ and ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)& +& {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.00025& & 0.000095& & 0& & 0\\ \text{Change}\left(\text{mol}\right)& -0.000095& & -0.000095& & 0& & +0.000095\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.000155& & 0& & 0& & 0.000095\end{array}$

After titration volume will be equal for both acid and base. The total volume will be $0.035\text{ L}$.

Calculation is done by using equation (2),

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

Substitute $\frac{0.000095}{0.035}$ for $\left[\text{conjugate acid}\right]$,$\frac{0.000155}{0.035}$ for $\left[\text{base}\right]$ and $4.49$ for $\text{p}{K}_{\text{b}}$.

$\begin{array}{c}\text{pOH}=4.49+\log \frac{\left(\frac{0.000095}{0.035}\right)}{\left(\frac{0.000155}{0.035}\right)}\\ =4.49+\log \left(0.612\right)\\ =4.49-0.21\\ =4.28\end{array}$

Therefore value of $\text{pOH}$ is $4.28$.

Hence value of $\text{pH}$ is calculated as follows,

$\begin{array}{c}\text{pH = 14}-4.28\\ =9.72\end{array}$

Therefore value of $\text{pH}$ is $9.72$ after addition of $\text{10 mL}$$\text{HCl}$ solution.

When $\text{20 mL}$ solution of $\text{HCl}$ is added.

The volume of $\text{HCl}$ added is $0.020\text{ L}$.

The ICE table for the reaction between $\text{HCl}$ and ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)& +& {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.00025& & 0.00019& & 0& & 0\\ \text{Change}\left(\text{mol}\right)& -0.00019& & -0.00019& & 0& & +0.00019\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.00006& & 0& & 0& & 0.00019\end{array}$

After titration volume will be equal for both acid and base. The total volume will be $0.045\text{ L}$.

Calculation is done by using equation (2),

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

Substitute $\frac{0.00019}{0.045}$ for $\left[\text{conjugate acid}\right]$,$\frac{0.00006}{0.045}$ for $\left[\text{base}\right]$ and $4.49$ for $\text{p}{K}_{\text{b}}$.

$\begin{array}{c}\text{pOH}=4.49+\log \frac{\left(\frac{0.00019}{0.045}\right)}{\left(\frac{0.00006}{0.045}\right)}\\ =4.49+\log \left(3.16\right)\\ =4.49+0.499\\ =4.989\end{array}$

Therefore value of $\text{pOH}$ is $4.989$.

Hence value of $\text{pH}$ is calculated as follows,

$\begin{array}{c}\text{pH = 14}-4.989\\ =9.01\end{array}$

Therefore value of $\text{pH}$ is $9.01$ after addition of $\text{20 mL}$$\text{HCl}$ solution.

When $\text{30 mL}$ solution of $\text{HCl}$ is added.

The volume of $\text{HCl}$ added is  $0.030\text{ L}$.

This volume is more than the volume required to reach the equivalence point.

The ICE table for the reaction between $\text{HCl}$ and ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)& +& {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.00025& & 0.000285& & 0& & 0\\ \text{Change}\left(\text{mol}\right)& 0.00025& & 0.00025& & 0& & 0.00025\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0.000035& & 0& & 0.00025\end{array}$

After titration volume will be equal for both acid and base. The total volume will be $0.055\text{ L}$.

After the equivalence point, the excess concentration ${\text{H}}_3{\text{O}}^{+}$ will present at the equilibrium. The hydrolysis of ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$ will produce a very small amount of ${\text{H}}_3{\text{O}}^{+}$ ions, that amount can be neglected.

Concentration of hydronium ion after equivalence point is calculated.

$\begin{array}{c}\text{concentration}=\frac{0.000035\text{ mol}}{0.055\text{ L}}\\ =0.000636\text{ mol}\cdot {\text{L}}^{-1}\end{array}$

Therefore, concentration of hydronium ions after reaction is $0.000636\text{ mol}\cdot {\text{L}}^{-1}$.

Calculate the $\text{pH}$

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]\\ =-\log \left(0.000636\right)\\ =3.19\\ \approx 3.20\end{array}$

Therefore value of $\text{pH}$ is $3.20$ after addition of $30\text{ mL}$$\text{HCl}$ solution after equivalence point.

The value of $\text{pH}$ when $5\text{ mL HCl}$ added is $10.14$ $10\text{ mL HCl}$ added $9.72$, when $20\text{ mL HCl}$ used $9.01$  and after addition of $30\text{ mL HCl}$$3.20$.

Interpretation Introduction

Interpretation:

Titration curve has to be plotted.

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 strong acid-weak base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$ with $\text{HCl}$ is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right){\text{+ H}}_3{\text{O}}^{+}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right)$

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

(1) The $\text{pOH}$ value before the titration can be calculated by using $K_{\text{b}}$ and its relation with ${\text{OH}}^{-}$ ion concentration.

$K_{\text{b}}=\frac{{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}{\left[{\text{BH}}^{+}\right]}_{\left(\text{eq}\right)}}{{\left[\text{B}\right]}_{\left(\text{eq}\right)}}$ (1)

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

As the addition of $\text{HCl}$ is done there will be formation of buffer solution ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2$/${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The $\text{pOH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

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

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

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

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

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

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

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

At equivalence point all the acid will be neutralized, and there will be only ${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ethylamine, concentration of  ${\text{H}}_3{\text{O}}^{+}$ can be calculated.

Value of $\text{pH}$ at every point is calculated by using the relation between $\text{pH}$ and $\text{pOH}$.

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

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 $\text{HCl}$ and $\text{pH}$ value on $y\text{ axis}$.

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

$\begin{array}{ccc}\text{Points }& \begin{array}{l}\text{volume of HCl}\\ \left(\text{mL}\right)\end{array}& \text{pH}\\ \text{a}& \text{0}& \text{11}\text{.75}\\ \text{b}& \text{5}& \text{10}\text{.14}\\ \text{c}& \text{10}& \text{9}\text{.72}\\ \text{d}& \text{20}& \text{9}\text{.01}\\ \text{e}& \text{26}\text{.3}& \text{5}\text{.91}\\ \text{f}& \text{30}& \text{3}\text{.20}\end{array}$

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

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