Chapter 8, Problem 69E

Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. Also, the titration curve needs to be drawn.

Concept introduction:

Titration curve is drawn to determine the change in pH of an acid or base with respect to the added volume of base or acid to it.

The titration curve can be drawn between a strong/weak acid and strong/weak base. The change in pH shows different patterns for different combinations of acids and bases.

Answer to Problem 69E

The change in pH due to addition of HCl is represented as follows:

Volume of HCl added (mL)	pH
0	11.1
4.0	9.97
8.0	9.58
12.5	9.25
20.0	8.65
24.0	7.87
24.9	6.9
25.0	5.28
26.0	2.71
28.0	2.24
30.0	2.04

The titration curve can be drawn as follows:

  

Explanation of Solution

Initial pH of the analyte solution; Ammonia is a weak base that forms an equilibrium when dissolved in water. The equilibrium is as follows.

  $N{H}_3+H_{2}O\rightleftharpoons N{H}_4{}^{+}+O{H}^{-}$

The molarity of ammonia at 0 mL HCl is 0.100 M. The ICE table for its dissociation can be represented as follows:

Reaction	Ammonia	Ammonium	OH-
Initial	0.1	0	0
Change	-x	+x	+x
Equilibrium	(0.1-x)	x	x

  $\begin{array}{l}Kb=\frac{[N{H}_4{}^{+}][O{H}^{-}]}{[N{H}_3]}\\ 1.8\times {10}^{-5}=\frac{[x][x]}{[0.1-x]}=\frac{x^2}{[0.1-x]}\\ x^2-1.8\times {10}^{-5}\left(0.1-x\right)=0\end{array}$

The value of x can be calculated by solving the equation as follows:

  $x=0.00133=\left[{\text{OH}}^{-}\right]$

Then the pH of the initial solution can be determined.

  $\begin{array}{l}pOH=-\log [O{H}^{-}]=-log[0.00133]\\ pH=14.00-pOH=14.00+log[0.00133]\\ pH=11.1\end{array}$

Thus, pH of solution when 0 mL of acid is added is 11.1.

Addition of $4.0mL$of acid:

When 4.0 mL of acid is added the number of moles of ammonia and HCl can be calculated from their molarity and volume as follows:

  $\begin{array}{l}{n}_{N{H}_3}=0.025\times 0.1=0.0025\text{ mol}\\ n_{HCl}=0.004\times 0.1=0.0004\text{ mol}\end{array}$

The ICE table can be represented as follows:

Reaction	Ammonia	HCl	Ammonium chloride
Initial	0.0025	0.0004	0
Change	-0.0004	-0.0004	+0.0004
Equilibrium	0.0021	0	0.0004

The pH can be calculated using the Henderson-Hesselbalch equation as follows:

Now,

In the Henderson-Hasselbalch equation, the pKa is used. therefore, the pKa for ammonia need to be calculated using its pKb.

  $\begin{array}{l}Kb=1.8\times {10}^{-5}\\ Kw=Ka\times Kb\\ Ka=\frac{Kw}{Kb}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}=5.55\times {10}^{-10}\\ pKa=-\log [Ka]=-log[5.55\times {10}^{-10}]=9.25\end{array}$

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[ammonium\text{ chloride}]}{[ammonia]}\\ pH=9.25+\log \frac{[0.0021]}{[0.0004]}=9.97\end{array}$

Thus, the pH of the solution is 9.97.

Addition of $8.0mL$ofacid:

Total amount of ammonia to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of acid added $=0.100mol/L\times 8.0\times {10}^{-3}L=8\times {10}^{-4}mol$

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Ammonia	HCl	Ammonium chloride
Initial	0.0025	0.0008	0
Change	-0.0008	-0.0008	+0.0008
Equilibrium	0.0017	0	0.0008

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[ammonium\text{ chloride}]}{[ammonia]}\\ pH=9.25+\log \frac{[0.0017]}{[0.0008]}=9.58\end{array}$

Addition of $12.5mL$of base:

Total amount of ammonia to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of acid added $=0.100mol/L\times 12.5\times {10}^{-3}L=12.5\times {10}^{-4}mol$

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Ammonia	H+	Ammonium chloride
Initial	0.0025	0	0
Change	-0.00125	-0.00125	+0.00125
Equilibrium	0.00125	0	0.00125

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[ammonium]}{[ammonia]}\\ pH=9.25+\log \frac{[0.00125]}{[0.00125]}=9.25\end{array}$

Addition of $20.0mL$of base:

Total amount of ammonia to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of acid added $=0.100mol/L\times 20.0\times {10}^{-3}L=20\times {10}^{-4}mol$

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Ammonia	H+	Ammonium
Initial	0.0025	0	0
Change	-0.002	-0.002	+0.002
Equilibrium	0.0005	0	0.002

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[ammonium]}{[ammonia]}\\ pH=9.25+\log \frac{[0.0005]}{[0.002]}=8.65\end{array}$

Addition of $24.0mL$of base:

Total amount of ammonia to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of acid added $=0.100mol/L\times 24.0\times {10}^{-3}L=24\times {10}^{-4}mol$

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Ammonia	HCl	Ammonium chloride
Initial	0.0025	0	0
Change	-0.0024	-0.0024	0.0024
Equilibrium	0.0001	0	0.0024

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[ammonium\text{ chloride}]}{[ammonia]}\\ pH=9.25+\log \frac{[0.0001]}{[0.0024]}=7.87\end{array}$

Addition of $24.9mL$of base:

Total amount of ammonia to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of acid added $=0.100mol/L\times 24.9\times {10}^{-3}L=24.9\times {10}^{-4}mol$

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Ammonia	HCl	Ammonium chloride
Initial	0.0025	0	0
Change	-0.00249	-0.00249	-0.00249
Equilibrium	0.00001	0	0.00249

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[ammonium]}{[ammonia]}\\ pH=9.25+\log \frac{[0.00001]}{[0.00249]}=6.9\end{array}$

Addition of $25.0mL$of base:

Total amount of ammonia to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of acid added $=0.100mol/L\times 25.0\times {10}^{-3}L=25\times {10}^{-4}mol$

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Ammonia	H+	Ammonium
Initial	0.0025	0	0
Change	-0.0025	-0.0025	-0.0025
Equilibrium	0.0000	0	0.0025

From the number of moles of ammonium ion and total volume that is 0.05 L, the molarity of ammonium chloride solution can be calculated as follows:

  $M=\frac{n}{V}=\frac{0.0025\text{ mol}}{0.05\text{ L}}=0.05\text{ M}$

The ammonium ion from salt can react with water to give back ammonia as follows:

  $N{H}_4{}^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+NH3\left(aq\right)$

The ICE table can be prepared as follows:

  $\begin{array}{l}N{H}_4{}^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+N{H}_3\left(aq\right)\\ I\text{ 0}\text{.05 - 0 0 }\\ C\text{ -x - +x +x}\\ E\text{ 0}\text{.05-x - x x}\end{array}$

The acid dissociation constant can be represented as follow:

  $K_a=\frac{K_w}{K_b}=\frac{\left[H_3{O}^{+}\right]\left[N{H}_3\right]}{\left[N{H}_4^{+}\right]}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}=5.6\times {10}^{-10}$

From the ICE table,

  $5.6\times {10}^{-10}=\frac{[x][x]}{[0.05-x]}=\frac{x^2}{[0.05-x]}$

Since, the value of acid dissociation constant is very low thus; value of x can be neglected from denominator.

  $x=\sqrt{0.05\times 5.6\times {10}^{-10}}=5.3\times {10}^{-6}$

This is the concentration of hydrogen ion in the solution.

The pH can be calculated as follows:

  $pH=-\log \left[{\text{H}}^{+}\right]=-\log \left(5.3\times {10}^{-6}\right)=5.28$

Addition of 26 mL of the acid:

The number of moles of ammonia and acid will be:

  $\begin{array}{l}{n}_{N{H}_3}=M\times V=\left(0.1\times 0.025\right)\text{ mol=0}\text{.0025 mol}\\ n_{HCl}=M\times V=\left(0.1\times 0.026\right)\text{ mol=0}\text{.0026 mol}\end{array}$

The reaction is represented as follows:

Reaction	Ammonia	HCl	Ammonium choride
Initial	0.0025	0.0026	0
Change	-0.0025	-0.0025	+0.0025
Equilibrium	0	0.0001	0.0025

The pH depends only on the concentration of HCl as it is a strong acid.

The molarity of HCl using total volume of the solution will be:

  $\left[HCl\right]=\left[H^{+}\right]=\left(\frac{0.0001}{0.025+0.026}\right)\text{ M}=0.00196\text{ M}$

The pH can be calculated as follows:

  $pH=-\log \left(0.00196\right)=2.71$

Addition of 28 mL of acid:

The number of moles of each species will be:

  $\begin{array}{l}{n}_{N{H}_3}=M\times V=\left(0.1\times 0.025\right)\text{ mol=0}\text{.0025 mol}\\ n_{HCl}=M\times V=\left(0.1\times 0.028\right)\text{ mol=0}\text{.0028 mol}\end{array}$

The reaction is represented as follows:

Reaction	Ammonia	HCl	Ammonium choride
Initial	0.0025	0.0028	0
Change	-0.0025	-0.0025	+0.0025
Equilibrium	0	0.0003	0.0025

The pH depends only on the concentration of HCl as it is a strong acid.

The molarity of HCl using total volume of the solution will be:

  $\left[HCl\right]=\left[H^{+}\right]=\left(\frac{0.0003}{0.025+0.028}\right)\text{ M}=0.00566\text{ M}$

The pH can be calculated as follows:

  $pH=-\log \left(0.00566\right)=2.24$

Addition of $30.0mL$ofacid:

The number of moles of ammonia and HCl can be calculated as follows:

  $\begin{array}{l}{n}_{N{H}_3}=M\times V=\left(0.1\times 0.025\right)\text{ mol=0}\text{.0025 mol}\\ n_{HCl}=M\times V=\left(0.1\times 0.030\right)\text{ mol=0}\text{.003 mol}\end{array}$

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Ammonia	H+	Ammonium	OH-
Initial	0.0025	0.003	0	0
Change	-0.0025	-0.0025	0	0
Equilibrium	0	0.0005	0	0

Concentration of hydrogen ion $=\frac{0.0005mol}{(25.0+30.0)\times {10}^{-3}L}=0.009mol/L$

  $pH=-\log [0.009]=2.04$

The data obtained from the above calculations will be:

Volume of HCl added (mL)	pH
0	11.1
4.0	9.97
8.0	9.58
12.5	9.25
20.0	8.65
24.0	7.87
24.9	6.9
25.0	5.28
26.0	2.71
28.0	2.24
30.0	2.04

The titration curve can be drawn as follows: