Chapter 17, Problem 33PS

You titrate 25.0 mL of 0.10 M NH₃ with 0.10 M HCl.

1. (a) What is the pH of the NH₃ 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) What indicator in Figure 17.11 could be used to detect the equivalence point?
5. (e) Calculate the pH of the solution after adding 5.00, 15.0, 20.0, 22.0, and 30.0 mL of the acid. Combine this information with that in parts (a)–(c) and plot the titration curve.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated at the various points during the titration between $\text{HCl}$ and ${\text{NH}}_3$. The value of $\text{pH}$ for the original solution of ${\text{NH}}_3$ 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 the equivalence point. The equilibrium established during the titration of $\text{HCl}$ with ${\text{NH}}_3$ is represented as,

  ${\text{NH}}_3\left(\text{aq}\right)\text{+ HCl}\left(\text{aq}\right)\rightleftharpoons {\text{NH}}_4{}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

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

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

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

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

As the addition of $\text{HCl}$ is done, there will be the formation of buffer solution ${\text{NH}}_4{}^{+}$/${\text{NH}}_3$. The $\text{pH}$ calculation for buffer solution is done by using the Henderson-Hasselbalch equation.

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

At the midpoint of the titration, the 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, $\text{pH}$ the value at the midpoint is equal to $\text{p}{K}_{\text{a}}$.

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

At equivalence point, all the base will be neutralized, and there will be only ${\text{NH}}_4{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of Ammonium ion at the equivalence point. The hydrolysis equilibrium is represented as,

  ${\text{NH}}_4{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{NH}}_3\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ammonium ion, the concentration of ${\text{H}}_3{\text{O}}^{+}$ can be calculated. Thus the value of $\text{pH}$ is less than $7$ at equivalence point for the strong acid-weak base titrations.

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

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

Explanation of Solution

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

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

Given:

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

The value of $K_{\text{b}}$ for ammonia is $1.8\times {10}^{-5}$.

The initial concentration of ${\text{NH}}_{\text{3}}$ is $0.10\text{}\text{M}$.

The initial concentration of $\text{HCl}$ is $0.10\text{ M}$.

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

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

ICE table (1) gives the ionization reaction of ${\text{NH}}_{\text{3}}$.

$\begin{array}{cccccccc}\text{Equation}& {\text{NH}}_3\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{OH}}^{-}\left(\text{aq}\right)& +& {\text{NH}}_4{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot \text{L}\right)& 0.100& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.100-x& & & & +x& & +x\end{array}$

From the ICE table (1),

The concentration of ${\text{NH}}_3$ left after the reaction is $\left(0.100-x\right)\text{mol}\cdot {\text{L}}^{-1}$.

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

The 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 in a comparison to $0.100$ thus it can be neglected with respect to it.

Therefore, Concentration of ${\text{NH}}_3$ left after the reaction is $0.100\text{ mol}\cdot {\text{L}}^{-1}$.

The hydroxide ion concentration is calculated by the expression,

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

Substitute $x$ for ${\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{BH}}^{\text{+}}\right]}_{\left(\text{eq}\right)}$, $0.100$ for ${\left[\text{B}\right]}_{\left(\text{eq}\right)}$ and $1.8\times {10}^{-5}$ for $K_{\text{b}}$.

$1.8\times {10}^{-5}=\frac{\left(x^2\right)}{\left(0.100\right)}$

Rearrange for $x^2$,

$\begin{array}{c}{x}^2=\left(1.8\times {10}^{-5}\right)\left(0.100\right)\\ x=1.34\times {10}^{-3}\end{array}$

Therefore, the concentration of hydroxide ion is $1.34\times {10}^{-3}$.

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

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

Substitute $1.34\times {10}^{-3}$ for $\left[{\text{OH}}^{-}\right]$.

$\begin{array}{c}\text{pOH = }-\log \left(1.34\times {10}^{-3}\right)\\ =2.87\end{array}$

The relation between $\text{pOH}$ and $\text{pH}$ is expressed as,

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

Substitute the value of $\text{pOH}$ in the above equation to calculate $\text{pH}$ of the solution,

$\begin{array}{c}\text{pH}+2.87=14\\ \text{pH}=11.13\end{array}$

The value of $\text{pH}$ for the original solution of ${\text{NH}}_{\text{3}}$ is $11.13$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated at the various points during the titration between $\text{HCl}$ and ${\text{NH}}_3$.  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 the equivalence point. The equilibrium established during the titration of $\text{HCl}$ with ${\text{NH}}_3$ is represented as,

  ${\text{NH}}_3\left(\text{aq}\right)\text{+ HCl}\left(\text{aq}\right)\rightleftharpoons {\text{NH}}_4{}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

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

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

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

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

As the addition of $\text{HCl}$ is done, there will be the formation of buffer solution ${\text{NH}}_4{}^{+}$/${\text{NH}}_3$. The $\text{pH}$ calculation for buffer solution is done by using the Henderson-Hasselbalch equation.

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

At the midpoint of the titration, the 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, $\text{pH}$ the value at the midpoint is equal to $\text{p}{K}_{\text{a}}$.

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

At equivalence point, all the base will be neutralized, and there will be only ${\text{NH}}_4{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of Ammonium ion at the equivalence point. The hydrolysis equilibrium is represented as,

  ${\text{NH}}_4{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{NH}}_3\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ammonium ion, the concentration of ${\text{H}}_3{\text{O}}^{+}$ can be calculated. Thus the value of $\text{pH}$ is less than $7$ at equivalence point for the strong acid-weak base titrations.

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

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

Explanation of Solution

The value of $\text{pH}$ at equivalence point is calculated below.

Given:

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

An 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 ${\text{NH}}_3$

 ${\text{V}}_{\text{1}}$ is the volume of ${\text{NH}}_3$.

 ${\text{C}}_{\text{2}}$ is the concentration of $\text{HCl}$ use.

 ${\text{V}}_2$ is the volume of $\text{HCl}$ used for the neutralization.

Substitute $0.100\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{1}}$, $0.025\text{ L}$ for ${\text{V}}_{\text{1}}$,$0.100\text{ mol}\cdot {\text{L}}^{-1}$ for ${\text{C}}_{\text{2}}$.

$\left(0.100\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.025\text{L}\right)=\left(0.100\text{ mol}\cdot {\text{L}}^{-1}\right){\text{V}}_{\text{2}}$

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

$\begin{array}{c}{\text{V}}_{\text{2}}=\frac{\left(0.100\text{ mol}\cdot {\text{L}}^{-1}\right)\left(0.025\text{L}\right)}{\left(0.100\text{ mol}\cdot {\text{L}}^{-1}\right)}\\ =0.025\text{ L}\end{array}$

Therefore, the volume of the $\text{HCl}$ used is $0.025\text{ L}$ or $\text{25 mL}$.

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 ${\text{NH}}_3$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{NH}}_3\left(\text{aq}\right)& +& \text{ HCl}\left(\text{aq}\right)& \rightleftharpoons & {\text{Cl}}^{-}\left(\text{aq}\right)& +& {\text{NH}}_4{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0025& & 0.0025& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.0025& & -0.0025& & & & 0.0025\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0& & & & 0.0025\end{array}$

From the ICE table (2),

A number of moles of ammonia left after reaction is $0\text{mol}$.

A number of moles of ammonium ion produced after the reaction is $0.0025\text{ mol}$.

The total volume after the reaction is calculated as,

$\begin{array}{c}\text{total}\text{volume = }0.025\text{L}+0.\text{025 L}\\ =0.050\text{ L}\end{array}$

Therefore, total volume after the reaction is $0.050\text{ L}$.

Concentration calculations are 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 ammonium ion after the reaction.

Substitute, $0.0025\text{ mol}$ for $\text{Number}\text{of moles}$ and $0.050\text{ L}$ for volume in equation (4).

$\begin{array}{c}\text{concentration = }\frac{0.0025\text{ mol}}{\text{0}\text{.050}\text{L}}\\ =0.05\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The concentration of Ammonium ion after the reaction is $0.05\text{mol}\cdot {\text{L}}^{-1}$.

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

  ${\text{NH}}_4{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}\left(\text{aq}\right)+{\text{NH}}_3\left(\text{aq}\right)$

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

$\begin{array}{cccccccc}\text{Equation}& {\text{NH}}_4{}^{+}\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)& +& {\text{NH}}_3\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.05& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.05-x& & & & +x& & +x\end{array}$

From the ICE table (3),

The concentration of ammonium ion left after the reaction is $\left(0.05-x\right)\text{mol}\cdot {\text{L}}^{-1}$.

The approximation $x$ is very small as compared to $0.05$. Therefore it can be neglected.

So, Concentration of ammonium ion left after the reaction is $\left(0.05\right)\text{mol}\cdot {\text{L}}^{-1}$.

The concentration of ammonia produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ produced after the reaction is $x\text{mol}\cdot {\text{L}}^{-1}$.

Calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ 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 $1.8\times {10}^{-5}$ for $K_{\text{b}}$ and $1.0\times {10}^{-14}$ for $K_{\text{w}}$

$\begin{array}{c}{K}_{\text{a}}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ =0.55\times {10}^{-9}\end{array}$

Therefore, $K_{\text{a}}$ value for ammonium ion is $0.55\times {10}^{-9}$.

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

$K_{\text{a}}=\frac{{\left[{\text{NH}}_3\right]}_{\left(\text{eq}\right)}{\left[{\text{H}}_3{\text{O}}^{\text{+}}\right]}_{\left(\text{eq}\right)}}{{\left[{\text{NH}}_4{}^{+}\right]}_{\left(\text{eq}\right)}}$ (6)

Substitute, $0.55\times {10}^{-9}$ for $K_{\text{a}}$, $x$ for ${\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{NH}}_3\right]}_{\left(\text{eq}\right)}$, $0.05$ for $\left[{\text{NH}}_4{}^{\text{+}}\right]$.

$0.55\times {10}^{-9}=\frac{\left(x\right)\left(x\right)}{\left(0.05\right)}$

Rearrange for $x^2$,

$\begin{array}{c}x=\sqrt{\left(0.55\times {10}^{-9}\right)\left(0.05\right)}\\ =0.524\times {10}^{-5}\end{array}$

Therefore the value of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ concentration is $0.524\times {10}^{-5}\text{ mol}\cdot {\text{L}}^{-1}$.

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

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

Substitute, $0.524\times {10}^{-5}$ for $\left[{\text{H}}_3{\text{O}}^{\text{+}}\right]$.

$\begin{array}{c}\text{pH = }-\log \left(0.524\times {10}^{-5}\right)\\ =5.28\end{array}$

Therefore, the value of $\text{pH}$ at the equivalence point is $5.28$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ is to be calculated at the various points during the titration between $\text{HCl}$ and ${\text{NH}}_3$. 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 strong acid-weak base titration, the $\text{pH}$ value can be calculated at various points before and after the equivalence point. The equilibrium established during the titration of $\text{HCl}$ with ${\text{NH}}_3$ is represented as,

  ${\text{NH}}_3\left(\text{aq}\right)\text{+ HCl}\left(\text{aq}\right)\rightleftharpoons {\text{NH}}_4{}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

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

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

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

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

As the addition of $\text{HCl}$ is done, there will be the formation of buffer solution ${\text{NH}}_4{}^{+}$/${\text{NH}}_3$. The $\text{pH}$ calculation for buffer solution is done by using the Henderson-Hasselbalch equation.

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

At the midpoint of the titration, the 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, $\text{pH}$ the value at the midpoint is equal to $\text{p}{K}_{\text{a}}$.

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

At equivalence point, all the base will be neutralized, and there will be only ${\text{NH}}_4{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of Ammonium ion at the equivalence point. The hydrolysis equilibrium is represented as,

  ${\text{NH}}_4{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{NH}}_3\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ammonium ion, the concentration of ${\text{H}}_3{\text{O}}^{+}$ can be calculated. Thus the value of $\text{pH}$ is less than $7$ at equivalence point for the strong acid-weak base titrations.

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

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

Explanation of Solution

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

Given:

The value of $K_{\text{b}}$ for ammonia is $1.8\times {10}^{-5}$.

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;

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

The relation between $K_{\text{a}}$ and $K_{\text{b}}$ is expressed as,

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

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

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

Substitute $1.8\times {10}^{-5}$ for $K_{\text{b}}$ and $1.0\times {10}^{-14}$ for $K_{\text{w}}$

$\begin{array}{c}{K}_{\text{a}}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ =0.55\times {10}^{-9}\end{array}$

Therefore, $K_{\text{a}}$ value for ammonium ion is $0.55\times {10}^{-9}$.

Substitute, $0.55\times {10}^{-9}$ for $K_{\text{a}}$.

$\begin{array}{c}\text{p}{K}_{\text{a}}=-\log (0.55\times {10}^{-9})\\ =9.25\end{array}$

Therefore, $\text{p}{K}_{\text{a}}$ value of ${\text{NH}}_{\text{3}}$ is $9.25$.

At the 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 $9.25$ for $\text{p}{K}_{\text{a}}$.

$\begin{array}{l}\text{pH}=9.25+\log \frac{\left[\text{conjugate}\text{base}\right]}{\left[\text{conjugate}\text{base}\right]}\\ \text{pH}=9.25\end{array}$

Therefore, the value of $\text{pH}$ at midpoint is $9.25$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated at the various points during the titration between $\text{HCl}$ and ${\text{NH}}_3$. The best indicator that can be used to detect the equivalence point is to be chosen for the titration has to be determined.

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 the equivalence point. The equilibrium established during the titration of $\text{HCl}$ with ${\text{NH}}_3$ is represented as,

  ${\text{NH}}_3\left(\text{aq}\right)\text{+ HCl}\left(\text{aq}\right)\rightleftharpoons {\text{NH}}_4{}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

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

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

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

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

As the addition of $\text{HCl}$ is done, there will be the formation of buffer solution ${\text{NH}}_4{}^{+}$/${\text{NH}}_3$. The $\text{pH}$ calculation for buffer solution is done by using the Henderson-Hasselbalch equation.

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

At the midpoint of the titration, the 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, $\text{pH}$ the value at the midpoint is equal to $\text{p}{K}_{\text{a}}$.

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

At equivalence point, all the base will be neutralized, and there will be only ${\text{NH}}_4{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of Ammonium ion at the equivalence point. The hydrolysis equilibrium is represented as,

  ${\text{NH}}_4{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{NH}}_3\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ammonium ion, the concentration of ${\text{H}}_3{\text{O}}^{+}$ can be calculated. Thus the value of $\text{pH}$ is less than $7$ at equivalence point for the strong acid-weak base titrations.

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

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

Explanation of Solution

Methyl red and bromcresol green are the best indicators for the titration between ammonia and hydrochloric acid. As it changes its color in the $\text{pH}$ region $9-11$ and for the given titration value of $\text{pH}$ at equivalence point is $9.13$.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ is to be calculated at the various points during the titration between $\text{HCl}$ and ${\text{NH}}_3$. The value of $\text{pH}$ when $5\text{mL,}15\text{ mL, 20 mL,}\text{22}\text{mL and 30 mL HCl}$ is added to the ${\text{NH}}_{\text{3}}$ solution. A titration curve is to be plotted by using $\text{pH}$ values calculated in part (a), (b), (c) and (d) has to be calculated and 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 the equivalence point. The equilibrium established during the titration of $\text{HCl}$ with ${\text{NH}}_3$ is represented as,

  ${\text{NH}}_3\left(\text{aq}\right)\text{+ HCl}\left(\text{aq}\right)\rightleftharpoons {\text{NH}}_4{}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

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

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

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

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

As the addition of $\text{HCl}$ is done, there will be the formation of buffer solution ${\text{NH}}_4{}^{+}$/${\text{NH}}_3$. The $\text{pH}$ calculation for buffer solution is done by using the Henderson-Hasselbalch equation.

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

At the midpoint of the titration, the 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, $\text{pH}$ the value at the midpoint is equal to $\text{p}{K}_{\text{a}}$.

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

At equivalence point, all the base will be neutralized, and there will be only ${\text{NH}}_4{}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of Ammonium ion at the equivalence point. The hydrolysis equilibrium is represented as,

  ${\text{NH}}_4{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{NH}}_3\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ammonium ion, the concentration of ${\text{H}}_3{\text{O}}^{+}$ can be calculated. Thus the value of $\text{pH}$ is less than $7$ at equivalence point for the strong acid-weak base titrations.

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

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

The value of $\text{pH}$ at different volume addition of $\text{HCl}$ and the corresponding point on the curve is given as,

$\begin{array}{ccc}\text{Points }& \begin{array}{l}\text{volume of HCl}\\ \left(\text{mL}\right)\end{array}& \text{pH}\\ \text{a}& \text{0}& 11.13\\ \text{b}& 5& 9.85\\ \text{c}& \text{15}& 9.07\\ \text{d}& \text{20}& 8.65\\ \text{e}& \text{22}& \text{8}\text{.38}\\ \text{f}& \text{30}& 2.04\end{array}$

Graph between $\text{pH}$ and volume of $\text{HCl}$ added is plotted as,

Explanation of Solution

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

Given:

When $\text{5 mL}$ solution of acid is used the ICE table for the reaction between $\text{HCl}$ and ${\text{NH}}_3$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{NH}}_3\left(\text{aq}\right)& +& \text{ HCl}\left(\text{aq}\right)& \rightleftharpoons & {\text{Cl}}^{-}\left(\text{aq}\right)& +& {\text{NH}}_4{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0025& & 0.0005& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.0005& & -0.0005& & & & 0.0005\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.002& & 0& & & & 0.0005\end{array}$

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

The calculation is done by using equation (1),

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

Substitute $\frac{0.0005}{0.030}$ for $\left[\text{conjugate}\text{acid}\right]$,$\frac{0.002}{0.030}$ for $\left[\text{Base}\right]$ and $9.25$ for $\text{p}{K}_{\text{a}}$.

$\begin{array}{c}\text{pH}=9.25+\log \frac{\frac{0.002}{0.030}}{\frac{0.0005}{0.030}}\\ =9.25+\log \left(4\right)\\ =9.25+0.602\\ =9.85\end{array}$

Therefore the value of $\text{pH}$ is $9.85$.

When $\text{15 mL}$ solution of acid is used the ICE table for the reaction between $\text{HCl}$ and ${\text{NH}}_3$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{NH}}_3\left(\text{aq}\right)& +& \text{ HCl}\left(\text{aq}\right)& \rightleftharpoons & {\text{Cl}}^{-}\left(\text{aq}\right)& +& {\text{NH}}_4{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0025& & 0.0015& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.0015& & -0.0015& & & & 0.0015\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.001& & 0& & & & 0.0015\end{array}$

After titration volume will be equal for both acid and base. The total volume will be $0.040\text{ L}$.

The calculation is done by using equation (1),

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

Substitute $\frac{0.0015}{0.040}$ for $\left[\text{conjugate}\text{acid}\right]$,$\frac{0.001}{0.040}$ for $\left[\text{Base}\right]$ and $9.25$ for $\text{p}{K}_{\text{a}}$.

$\begin{array}{c}\text{pH}=9.25+\log \frac{\frac{0.001}{0.040}}{\frac{0.0015}{0.040}}\\ =9.25+\log \left(0.66\right)\\ =9.25-0.18\\ =9.07\end{array}$

Therefore the value of $\text{pH}$ is $9.07$.

When $\text{20 mL}$ solution of acid is used the ICE table for the reaction between $\text{HCl}$ and ${\text{NH}}_3$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{NH}}_3\left(\text{aq}\right)& +& \text{ HCl}\left(\text{aq}\right)& \rightleftharpoons & {\text{Cl}}^{-}\left(\text{aq}\right)& +& {\text{NH}}_4{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0025& & 0.002& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.002& & -0.002& & & & 0.002\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.0005& & 0& & & & 0.002\end{array}$

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

The calculation is done by using equation (1),

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

Substitute $\frac{0.002}{0.045}$ for $\left[\text{conjugate}\text{acid}\right]$,$\frac{0.0005}{0.045}$ for $\left[\text{Base}\right]$ and $9.25$ for $\text{p}{K}_{\text{a}}$.

$\begin{array}{c}\text{pH}=9.25+\log \frac{\frac{0.0005}{0.045}}{\frac{0.002}{0.045}}\\ =9.25+\log \left(0.25\right)\\ =9.25-0.602\\ =8.65\end{array}$

Therefore the value of $\text{pH}$ is $8.65$.

When $\text{22 mL}$ solution of acid is used the ICE table for the reaction between $\text{HCl}$ and ${\text{NH}}_3$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{NH}}_3\left(\text{aq}\right)& +& \text{ HCl}\left(\text{aq}\right)& \rightleftharpoons & {\text{Cl}}^{-}\left(\text{aq}\right)& +& {\text{NH}}_4{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0025& & 0.0022& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.0022& & -0.0022& & & & 0.0022\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.0003& & 0& & & & 0.0022\end{array}$

After titration volume will be equal for both acid and base. The total volume will be $0.047\text{ L}$.

The calculation is done by using equation (1),

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

Substitute $\frac{0.0022}{0.047}$ for $\left[\text{conjugate}\text{acid}\right]$,$\frac{0.0003}{0.047}$ for $\left[\text{Base}\right]$ and $9.25$ for $\text{p}{K}_{\text{a}}$.

$\begin{array}{c}\text{pH}=9.25+\log \frac{\frac{0.0003}{0.047}}{\frac{0.0022}{0.047}}\\ =9.25+\log \left(0.136\right)\\ =9.25-0.866\\ =8.38\end{array}$

Therefore the value of $\text{pH}$ is $8.38$.

When $\text{30 mL}$ solution of acid is used the ICE table for the reaction between $\text{HCl}$ and ${\text{NH}}_3$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{NH}}_3\left(\text{aq}\right)& +& \text{ HCl}\left(\text{aq}\right)& \rightleftharpoons & {\text{Cl}}^{-}\left(\text{aq}\right)& +& {\text{NH}}_4{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0025& & 0.003& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.0025& & -0.0025& & & & 0.0025\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0.0005& & & & 0.0025\end{array}$

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 of $\text{HCl}$ will present at the equilibrium. The hydrolysis of ammonium ion will produce a very small amount of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions, that amount can be neglected.

The concentration of hydrogen ion after the equivalence point is calculated.

$\begin{array}{c}\text{concentration}=\frac{0.0005\text{ mol}}{0.055\text{ L}}\\ =0.0091\text{ mol}\cdot {\text{L}}^{-1}\end{array}$

Therefore, the concentration of hydrogen ions after the reaction is $0.0091\text{ mol}\cdot {\text{L}}^{-1}$.

Calculate the $\text{pH}$

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.0091\right)\\ =2.04\end{array}$

Therefore the value of $\text{pH}$ is $2.04$ after the addition of $30\text{ mL}$ hydrochloric acid.

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 $y\text{ axis}$.

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

$\begin{array}{ccc}\text{Points }& \begin{array}{l}\text{volume of HCl}\\ \left(\text{mL}\right)\end{array}& \text{pH}\\ \text{a}& \text{0}& 11.13\\ \text{b}& 5& 9.85\\ \text{c}& \text{15}& 9.07\\ \text{d}& \text{20}& 8.65\\ \text{e}& \text{22}& \text{8}\text{.38}\\ \text{f}& \text{30}& 2.04\end{array}$

Graph between $\text{pH}$ and volume of $\text{HCl}$ added is plotted as,