Chapter 8, Problem 68E

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.

Explanation of Solution

Initial pH of the analyte solution can be determined as follows:

Propanoic acid is a weak acid that forms equilibrium when dissolved in water. The equilibrium is as follows.

  $H{C}_3{H}_5{O}_2+H_{2}O\rightleftharpoons C_3{H}_5{O}_2{}^{-}+H_3{O}^{+}$

The amount of acid at the beginning $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$ . By constructing an ICE table, the amount of propanoic ion in the solution after the acid dissociation can be determined.

Reaction	Proanoic acid	Propanoate ion	OH-
Initial	0.1	0	0
Change	-x	+x	+x
Equilibrium	(0.1-x)	x	x

The acid dissociation constant can be represented as follows:

  $\begin{array}{l}Ka=\frac{[C_3{H}_5{O}_2{}^{-}][H^{+}]}{[H{C}_3{H}_5{O}_2]}\\ 1.3\times {10}^{-5}=\frac{[x][x]}{[0.1-x]}=\frac{x^2}{[0.1-x]}\\ 1.3\times {10}^{-6}-1.3\times {10}^{-5}x=x^2\\ 0=x^2+1.3\times {10}^{-5}x-1.3\times {10}^{-6}\end{array}$

Solving this quadratic equation gives the amount of hydrogen ions in the solution.

  $\begin{array}{l}0=x^2+1.3\times {10}^{-5}x-1.3\times {10}^{-6}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(1.38\times {10}^{-4})\pm \sqrt{{(1.3\times {10}^{-5})}^2-4(1)(-1.3\times {10}^{-6})}}{2(1)}\end{array}$

On solving the only possible value of x is $1.14\times {10}^{-3}$

Now, pH can be calculated as follows:

  $pH=-\log [H^{+}]=-log[1.14\times {10}^{-3}]=2.94$

Addition of $4.0mL$of base:

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

Amount of base 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	Propanoic acid	OH-	Propanoate ion	H+
Initial	0.0025	0	0	0
Add	0	0.0004
Change	-0.0004	-0.0004	0.0004	0.0004
Equilibrium	0.0021	0	0.0004	0.0004

Concentration of base after addition of acid $=\frac{0.0021mol}{(25.0+4.0)\times {10}^{-.3}L}=0.072mol/L$

Concentration of ammonium ion $=\frac{0.0004mol}{(25.0+4.0)\times {10}^{-.3}L}=0.014mol/L$

In the Henderson-Hasselbalch equation, the pKa is used.

  $\begin{array}{l}Ka=1.3\times {10}^{-5}\\ pKa=-\log [Ka]=-log[1.3\times {10}^{-5}]=4.88\end{array}$

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[Propanoate]}{[Propanoic]}\\ pH=4.88+\log \frac{[0.014]}{[0.072]}=4.16\end{array}$

Addition of $8.0mL$ofbase:

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

Amount of base 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	Propanoic acid	OH-	Propanoate ion	H+
Initial	0.0025	0	0	0
Add	0	0.0008
Change	-0.0008	-0.0008	0.0008	0.0008
Equilibrium	0.0017	0	0.0008	0.0008

Concentration of acid after addition of base $=\frac{0.0017mol}{(25.0+8.0)\times {10}^{-.3}L}=0.051mol/L$

Concentration of propanoate ion $=\frac{0.0008mol}{(25.0+8.0)\times {10}^{-.3}L}=0.024mol/L$

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[propanoate]}{[propanoic]}\\ pH=4.88+\log \frac{[0.024]}{[0.051]}=4.55\end{array}$

Addition of $12.5mL$of base:

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

Amount of base 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	Propanoic acid	OH-	Propanoate ion	H+
Initial	0.0025	0	0	0
Add	0	0.00125
Change	-0.00125	-0.00125	0.00125	0.00125
Equilibrium	0.00125	0	0.00125	0.00125

Concentration of acid after addition of base $=\frac{0.00125mol}{(25.0+12.5)\times {10}^{-.3}L}=0.033mol/L$

Concentration of propanoate ion $=\frac{0.00125mol}{(25.0+4.0)\times {10}^{-.3}L}=0.033mol/L$

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[propanoate]}{[propanoic]}\\ pH=4.88+\log \frac{[0.033]}{[0.033]}=4.88\end{array}$

Addition of $20.0mL$of base:

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

Amount of base 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	Propanoic acid	H+	Propanoate ion	OH-
Initial	0.0025	0	0	0
Add	0	0.002
Change	-0.002	-0.002	0.002	0.002
Equilibrium	0.0005	0	0.002	0.002

Concentration of acid after addition of base $=\frac{0.0005mol}{(25.0+20.0)\times {10}^{-.3}L}=0.011mol/L$

Concentration of propanoate ion $=\frac{0.002mol}{(25.0+20.0)\times {10}^{-.3}L}=0.044mol/L$

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[propanoate]}{[propanoic]}\\ pH=4.88+\log \frac{[0.044]}{[0.011]}=5.48\end{array}$

Addition of $24.0mL$of base:

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

Amount of base 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	Propanoic	OH-	Propanoate	H+
Initial	0.0025	0	0	0
Add	0	0.0024
Change	-0.0024	-0.0024	0.0024	0.0024
Equilibrium	0.0001	0	0.0024	0.0024

Concentration of acid after addition of base $=\frac{0.0001mol}{(25.0+24.0)\times {10}^{-.3}L}=0.002mol/L$

Concentration of propanoate ion $=\frac{0.0024mol}{(25.0+24.0)\times {10}^{-.3}L}=0.049mol/L$

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[propanoate]}{[propanoic]}\\ pH=4.88+\log \frac{[0.049]}{[0.002]}=6.26\end{array}$

Addition of $24.5mL$of base:

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

Amount of base added $=0.100mol/L\times 24.5\times {10}^{-3}L=24.5\times {10}^{-4}mol$

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	Propanoic	OH-	Propanoate	H+
Initial	0.0025	0	0	0
Add	0	0.00245
Change	-0.00245	-0.00245	-0.00245	-0.00245
Equilibrium	0.00005	0	-0.00245	-0.00245

Concentration of acid after addition of base $=\frac{0.00005mol}{(25.0+24.5)\times {10}^{-.3}L}=0.0010mol/L$

Concentration of propanoate ion $=\frac{0.00245mol}{(25.0+24.5)\times {10}^{-.3}L}=0.049mol/L$

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[propanoate]}{[propanoic]}\\ pH=4.88+\log \frac{[0.049]}{[0.0010]}=6.57\end{array}$

Addition of $24.9mL$of base:

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

Amount of base 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	Propanoic	OH-	Propanoate	H+
Initial	0.0025	0	0	0
Add	0	0.00249
Change	-0.00249	-0.00249	-0.00249	-0.00249
Equilibrium	0.00001	0	-0.00249	-0.00249

Concentration of acid after addition of base $=\frac{0.00001mol}{(25.0+24.9)\times {10}^{-.3}L}=0.0002mol/L$

Concentration of propanoate ion $=\frac{0.00249mol}{(25.0+24.9)\times {10}^{-.3}L}=0.049mol/L$

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[propanoate]}{[propanoic]}\\ pH=4.88+\log \frac{[0.049]}{[0.0002]}=7.26\end{array}$

Addition of $25.0mL$of base:

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

Amount of base 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	Propanoic	OH-	Propanoate	H+
Initial	0.0025	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	-0.0025	-0.0025
Equilibrium	0.0000	0	-0.0025	-0.0025

Concentration of acid after addition of base $=0$

Concentration of propanoate 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 propanoic acid (that is propanoate ion).

  $C_3{H}_5{O}_2{}^{-}+H_{2}O\rightleftharpoons H{C}_3{H}_5{O}_2+O{H}^{-}$

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

Reaction	Propanoate ion	Propanoic acid	OH-
Initial	0.05	0	0
Change	-X	+x	+x
Equilibrium	(0.05-x)	x	x

Then the pH can be calculated as follows:

  $\begin{array}{l}Kb=\frac{[H{C}_3{H}_5{O}_2][O{H}^{-}]}{[C_3{H}_5{O}_2{}^{-}]}\\ Kb=\frac{Kw}{Ka}=\frac{1.0\times {10}^{-14}}{1.3\times {10}^{-5}}=7.6\times {10}^{-10}\\ 7.6\times {10}^{-10}=\frac{[x][x]}{[0.05-x]}=\frac{x^2}{[0.05-x]}\\ 3.8\times {10}^{-11}-7.6\times {10}^{-10}x=x^2\\ 0=x^2+7.6\times {10}^{-10}x-3.8\times {10}^{-11}\end{array}$

On solving the equation, the only possible value of x will be:

  $x=4.4\times {10}^{-5}$

This is the concentration of hydroxide ion. The pOH value can be calculated as follows:

  $pOH=-\log \left[{\text{OH}}^{-}\right]=-\log \left(4.4\times {10}^{-5}\right)=4.35$

Thus, pH of the solution is

  $pH=14-pOH=14-4.45=9.64$

Addition of $28.0mL$ofbase:

Total amount of acid to be neutralized $=0$

Amount of base 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	Propanoic	OH-	Propanoate	H+
Initial	0.0025	0	0	0
Add	0	0.0028
Change	-0.0025	0.0025	0	0
Equilibrium	0	0.0003	0	0

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

  $\begin{array}{l}pOH=-\log [0.0056]=2.25\\ pH=14.00-2.25=11.75\end{array}$

Addition of $30.0mL$ofbase:

Total amount of acid to be neutralized $=0$

Amount of base 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	Propanoic	OH-	Propanoate	H+
Initial	0.0025	0	0	0
Add	0	0.0030
Change	-0.0025	0.0025	0	0
Equilibrium	0	0.0005	0	0

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

  $\begin{array}{l}pOH=-\log [0.009]=2.04\\ pH=14.00-2.04=11.96\end{array}$

Thus, the pH values for volume of base added are as follows:

Volume of base added (mL)	pH
0.0	2.94
4.0	4.16
8.0	4.55
12.5	4.88
20.0	5.48
24.0	6.26
24.5	6.57
24.9	7.26
25.0	9.64
28.0	11.75
30.0	11.96

The titration curve can be drawn as follows: