Chapter 8, Problem 67E

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 calculated as follows:

Lactic acid is a weak acid that forms an equilibrium mixture when dissolved in water. The equilibrium is as follows.

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

The initial molarity of lactic acid is 0.1 M.

The amount of lactic acid at the beginning can be calculated from. By constructing an ICE table, the concentration of lactate ion in the solution after the acid dissociation can be determined.

Reaction	Lactic acid	Lactate	H+
Initial	0.1	0	0
Change	-x	+x	+x
Equilibrium	(0.1-x)	x	x

The acid dissociation constant can be represented as follows:

  $\begin{array}{l}Ka=\frac{[C_5{H}_5{O}_3{}^{-}][H^{+}]}{[H{C}_5{H}_5{O}_3]}\\ 1.38\times {10}^{-4}=\frac{[x][x]}{[0.1-x]}=\frac{x^2}{[0.1-x]}\\ 1.38\times {10}^{-5}-1.38\times {10}^{-4}x=x^2\\ 0=x^2+1.38\times {10}^{-4}x-1.38\times {10}^{-5}\end{array}$

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

  $\begin{array}{l}0=x^2+1.38\times {10}^{-4}x-1.38\times {10}^{-5}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(1.38\times {10}^{-4})\pm \sqrt{{(1.38\times {10}^{-4})}^2-4(1)(-1.38\times {10}^{-5})}}{2(1)}\\ x=\frac{-(1.38\times {10}^{-4})\pm \sqrt{(1.90\times {10}^{-8}+1.38\times {10}^{-5})}}{2}\\ =\frac{-(1.38\times {10}^{-4})\pm \sqrt{0.00001382}}{2}\\ =\frac{-(1.38\times {10}^{-4})\pm 0.00372}{2}\\ =\frac{0.0037}{2}=0.00185\end{array}$

Thus, the concentration of hydrogen ion is 0.00185 and pH of the solution can be calculated as follows:

  $pH=-\log [H^{+}]=-log[0.00185]=2.73$

Addition of $4.0mL$of base:

Total amount of lactic acid to be neutralized can be calculated from its molarity and volume as follows:

  $n=M\times V$

Or,

  $n=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Now, the amount of base added can be calculated as follows:

  $n=0.100mol/L\times 4.0\times {10}^{-3}L=4\times {10}^{-4}mol$

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

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

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[lactate]}{[lacticacid]}\\ pH=3.86+\log \frac{[0.014]}{[0.072]}=3.14\end{array}$

Addition of $8.0mL$of base:

Total amount of lactic 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	Lactic acid	OH-	Lactate	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 lactic acid after addition of base $=\frac{0.0017mol}{(25.0+8.0)\times {10}^{-.3}L}=0.051mol/L$

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

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[lactate]}{[lacticacid]}\\ pH=3.86+\log \frac{[0.024]}{[0.051]}=3.53\end{array}$

Addition of $12.5mL$of base:

Total amount of lactic 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	Lactic acid	OH-	Lactate	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 lactic acid after addition of base $=\frac{0.00125mol}{(25.0+12.5)\times {10}^{-.3}L}=0.033mol/L$

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

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[lactate]}{[lacticacid]}\\ pH=3.86+\log \frac{[0.033]}{[0.033]}=3.86\end{array}$

Addition of $20.0mL$of base:

Total amount of lactic 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	Lactic acid	OH-	Lactate	H+
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 lactic acid after addition of base $=\frac{0.0005mol}{(25.0+20.0)\times {10}^{-.3}L}=0.011mol/L$

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

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[lactate]}{[lacticacid]}\\ pH=3.86+\log \frac{[0.044]}{[0.011]}=4.46\end{array}$

Addition of $24.0mL$of base:

Total amount of lactic 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	Lactic acid	OH-	Lactate	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 lactic acid after addition of base $=\frac{0.0001mol}{(25.0+24.0)\times {10}^{-.3}L}=0.002mol/L$

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

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[lactate]}{[lacticacid]}\\ pH=3.86+\log \frac{[0.049]}{[0.002]}=5.24\end{array}$

Addition of $24.5mL$of base:

Total amount of lactic 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	Lactic acid	OH-	Lactate	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 lactic acid after addition of base $=\frac{0.00005mol}{(25.0+24.5)\times {10}^{-.3}L}=0.0010mol/L$

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

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[lactate]}{[lacticacid]}\\ pH=3.86+\log \frac{[0.049]}{[0.0010]}=5.55\approx 5.6\end{array}$

Addition of $24.9mL$of base:

Total amount of lactic 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	Lactic acid	OH-	Lactate	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 lactic acid after addition of base $=\frac{0.00001mol}{(25.0+24.9)\times {10}^{-.3}L}=0.0002mol/L$

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

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pH=pKa+\log \frac{[lactate]}{[lacticacid]}\\ pH=3.86+\log \frac{[0.049]}{[0.0002]}=6.3\end{array}$

Addition of $25.0mL$of base:

Total amount of lactic 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	Lactic acid	OH-	Lactate	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 lactic acid after addition of base $=0$

Concentration of lactate 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 base of the lactic acid (that is lactate ion).

  $C_5{H}_5{O}_3{}^{-}+H_{2}O\rightleftharpoons H{C}_5{H}_5{O}_3+O{H}^{-}$

Thereafter, by obtaining the Kb value for lactate ion, the amount of hydroxide ions in the solution can be determined to get the pH value at this point.

Reaction	Lactic acid	Lactate	OH-
Initial	0.05	0	0
Change	-X	x	x
Equilibrium	(0.05-x)	x	x

  $\begin{array}{l}Kb=\frac{Kw}{Ka}\\ Kb=\frac{1.0\times {10}^{-14}}{1.38\times {10}^{-4}}=7.24\times {10}^{-11}\end{array}$

Then the pH can be calculated as follows:

  $\begin{array}{l}Kb=\frac{[H{C}_5{H}_5{O}_3][O{H}^{-}]}{[C_5{H}_5{O}_3{}^{-}]}\\ 7.24\times {10}^{-11}=\frac{[x][x]}{[0.05-x]}=\frac{x^2}{[0.05-x]}\\ 3.62\times {10}^{-12}-7.24\times {10}^{-11}x=x^2\\ 0=x^2+7.24\times {10}^{-11}x-3.62\times {10}^{-12}\end{array}$

Thereafter, this quadratic equation can be solved to determine the hydroxide ion concentration, thereby, the pOH and the pH can be determined.

  $\begin{array}{l}0=x^2+7.24\times {10}^{-11}x-3.62\times {10}^{-12}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(7.24\times {10}^{-11})\pm \sqrt{{(7.24\times {10}^{-11})}^2-4(1)(-3.62\times {10}^{-12})}}{2(1)}\\ x=\frac{-(7.24\times {10}^{-11})\pm \sqrt{52.42\times {10}^{-22}+1.45\times {10}^{-11}}}{2}\\ x=\frac{-(7.24\times {10}^{-11})\pm \sqrt{1.45\times {10}^{-11}}}{2}\\ x=\frac{-(7.24\times {10}^{-11})\pm 3.80\times {10}^{-6}}{2}\\ x=\frac{3.80\times {10}^{-6}}{2}\\ x=1.90\times {10}^{-6}\end{array}$

The calculated value of x is concentration of hydroxide ion. The pOH of the solution will be:

  $pOH=-\log [1.90\times {10}^{-6}]=5.72$

  $\begin{array}{l}pH=14-pOH=14-5.72\\ pH=8.3\end{array}$

Addition of $28.0mL$of base:

Total amount of lactic 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	Lactic acid	OH-	Lactate	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 $=\frac{0.0003mol}{(25.0+28.0)\times {10}^{-3}L}=0.0056mol/L$

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

Addition of $30.0mL$of base:

Total amount of lactic 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	Lactic acid	OH-	Lactate	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 $=\frac{0.0005mol}{(25.0+30.0)\times {10}^{-3}L}=0.009mol/L$

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

Thus, the value of pH with respect to added volume of base is as follows:

Volume (in mL)	pH
0	2.73
4	3.14
8	3.53
12.5	3.86
20	4.46
24	5.24
24.5	5.6
24.9	6.3
25.0	8.28
25.1	10.3
26.0	11.30
28.0	11.75
30.0	11.96

The titration curve can be drawn as follows: