Chapter 8, Problem 70E

Interpretation Introduction

Interpretation:

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

Concept introduction:

A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Pyridine is a weak base while hydrochloric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and an equilibrium state.

Explanation of Solution

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

  $C_5{H}_{5}N{H}_2+H_{2}O\rightleftharpoons C_5{H}_{5}N{H}_3{}^{+}+O{H}^{-}$

The molarity of pyridine is 0.1 M thus, the ICE table can be created as follows:

Reaction	Pyridine base	Pyridine ion	OH-
Initial	0.1	0	0
Change	-x	+x	+x
Equilibrium	(0.1-x)	x	x

  $\begin{array}{l}Kb=\frac{[C_5{H}_{5}N{H}_3{}^{+}][O{H}^{-}]}{[C_5{H}_{5}N{H}_2]}\\ 1.7\times {10}^{-9}=\frac{[x][x]}{[0.1-x]}=\frac{x^2}{[0.1-x]}\\ x^2-1.7\times {10}^{-9}\left(0.1-x\right)=0\end{array}$

Here,

  $x=\left[{\text{OH}}^{-}\right]=0.000013$

Thus,

  $\left[H^{+}\right]=\frac{{10}^{-14}}{0.000013}=7.69\times {10}^{-10}$

The pH of the solution will be:

  $pH=-\log \left(7.69\times {10}^{-10}\right)=9.11$

Addition of $4.0mL$ofacid:

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

Amount of acid added $=0.100mol/L\times 4.0\times {10}^{-3}L=4\times {10}^{-4}mol$

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	Pyridine base	H+	Pyridine ion
Initial	0.0025	0	0
Change	-0.0004	-0.0004	0.0004
Equilibrium	0.0021	0	0.0004

In the Henderson-Hasselbalch equation, the pKa is used. Therefore, the pKa for pyridine need to be calculated using its Kb.

  $\begin{array}{l}Kb=1.7\times {10}^{-9}\\ Kw=Ka\times Kb\\ Ka=\frac{Kw}{Kb}=\frac{1.0\times {10}^{-14}}{1.7\times {10}^{-9}}=5.88\times {10}^{-6}\\ pKa=-\log [Ka]=-log[5.88\times {10}^{-6}]=5.2\end{array}$

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[base]}{[\text{acid}]}\\ pH=5.2+\log \frac{[0.0021]}{[0.0004]}=5.92\end{array}$

Addition of $8.0mL$ofacid:

Total amount of pyridine 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	Pyridine base	H+	Pyridine ion
Initial	0.0025	0	0
Change	-0.0008	-0.0008	0.0008
Equilibrium	0.0017	0	0.0008

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[base]}{[conjugate]}\\ pH=5.2+\log \frac{[0.0017]}{[0.0008]}=5.53\end{array}$

Addition of $12.5mL$of base:

Total amount of pyridine 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	Pyridine base	H+	Pyridine ion
Initial	0.0025	0	0
Change	-0.00125	-0.00125	0.00125
Equilibrium	0.00125	0	0.00125

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[base]}{[conjugate\text{ ion}]}\\ pH=5.2+\log \frac{[0.00125]}{[0.00125]}=5.2\end{array}$

Addition of $20.0mL$of acid:

Total amount of pyridine 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	Pyridine base	H+	Pyridine base
Initial	0.0025	0	0
Change	-0.002	-0.002	0.002
Equilibrium	0.0005	0	0.002

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[base]}{[conjugate\text{ ion}]}\\ pH=5.23+\log \frac{[0.0005]}{[0.002]}=4.6\end{array}$

Addition of $24.5mL$ofacid:

Total amount of pyridine 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	Pyridine base	H+	Pyridine ion
Initial	0.0025	0	0
Change	-0.0024	-0.0024	+0.0024
Equilibrium	0.0001	0	0.0024

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[base]}{[conjugate\text{ ion}]}\\ pH=5.2+\log \frac{[0.0001]}{[0.0024]}=3.82\end{array}$

Addition of $25.0mL$ofacid:

Total amount of pyridine 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	Pyridine base	H+	Pyridine ion
Initial	0.0025	0	0
Change	-0.0025	-0.0025	+0.0025
Equilibrium	0.0000	0	0.0025

Concentration of base after addition of base $=0$

Concentration of conjugate ion $=\frac{0.0025mol}{(25.0+25.0)\times {10}^{-.3}L}=0.05mol/L$

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate acid of the pyridine base.

  $C_5{H}_{5}N{H}_3{}^{+}+H_{2}O\rightleftharpoons C_5{H}_{5}N{H}_2+H_3{O}^{+}$

Thereafter, using the Ka value for pyridine, the amount of hydrogen ions in the solution can be determined to get the pH value at this point.

Reaction	Pyridine base	Pyridine ion	H+
Initial	0.05	0	0
Change	X	x	x
Equilibrium	(0.05-x)	x	x

Then the pH can be calculated as follows:

  $\begin{array}{l}Ka=\frac{[C_5{H}_{5}N{H}_2][H^{+}]}{[C_5{H}_{5}N{H}_3{}^{+}]}=\frac{K_w}{K_b}\\ \frac{1.0\times {10}^{14}}{1.7\times {10}^{-9}}=5.88\times {10}^{-6}=\frac{[x][x]}{[0.05-x]}=\frac{x^2}{[0.05-x]}\end{array}$

The value of x can be neglected from denominator as the acid dissociation constant has very small value.

  $\begin{array}{l}{x}^2=0.05\times 5.88\times {10}^{-6}\\ x=5.42\times {10}^{-4}=\left[H^{+}\right]\end{array}$

Thus,

  $pH=-\log [H^{+}]=-log[5.42\times {10}^{-4}]=3.3$

Addition of 26.0 mL of the acid:

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

Amount of acid added $=0.100mol/L\times 26.0\times {10}^{-3}L=26\times {10}^{-4}mol$

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	Pyridine base	H+	Pyridine ion
Initial	0.0025	0.0026	0
Change	-0.0025	-0.0025	+0.0025
Equilibrium	0.0000	0.0001	0.0025

The pH of the solution only depends on the concentration of HCl thus,

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

The pH of the solution will be:

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

Addition of $28.0mL$ofacid:

Amount of acid added $=0.100mol/L\times 28.0\times {10}^{-3}L=28\times {10}^{-4}mol$

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	Pyridine base	H+	Pyridine ion
Initial	0.0025	0.0028	0
Change	-0.0025	-0.0025	+0.0025
Equilibrium	0	0.0003	0.0025

Here, the pH will only depend on the concentration of hydrogen ion.

Concentration of hydrogen ion $=\frac{0.0003mol}{(25.0+28.0)\times {10}^{-3}L}=0.0056mol/L$

  $\begin{array}{l}pH=-\log [0.0056]=2.25\\ \end{array}$

Addition of $30.0mL$ofacid:

Amount of ammonium added $=0.100mol/L\times 30.0\times {10}^{-3}L=30\times {10}^{-4}mol$

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	Pyridine base	H+	Pyridine ion
Initial	0.0025	0.0003	0
Change	-0.0025	-0.0025	+0.0025
Equilibrium	0	0.0005	0.0025

The pH will only depend on the concentration of hydrogen ion.

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

Thus,

  $pH=-\log [0.009]=2.04$

Thus, the data obtained from the calculations is as follows:

Volume of HCl added (mL)	pH
0	9.11
4.0	5.92
8.0	5.53
12.5	5.2
20.0	4.6
24.5	3.82
25.0	3.3
26.0	2.71
28.0	2.25
30.0	2.04

The titration curve can be represented as follows: