Chapter 15, Problem 62E

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M propanoic acid (HC3H5O2,Ka = 1.3 × 10−5) with 0.100 M NaOH.

Interpretation Introduction

Interpretation: The titration of Propanoic acid with different volumes of NaOH is given. The pH of each solution is to be calculated and the graph between pH and volume of NaOH added is to be plotted.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The pH of each of the given solution and the graph between pH and volume of NaOH added.

Explanation

Explanation

The concentration of Propanoic acid is 0.100M .

The concentration of NaOH is 0.100M .

The volume of Propanoic acid is 25.0mL .

The volume of NaOH is. 0.0mL , 4.0mL , 8.0mL , 12.5mL , 20.0mL , 24.0mL , 24.5mL , 24.9mL , 25.0mL , 25.1mL , 26.0mL , 28.0mL , 30.0mL

The value of Ka of Propanoic acid is 1.3×105

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 25.0mL into L is done as,

25mL=25×0.001L=0.025L

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (1)

Rearrange the above equation to obtain the value of number of moles.

Numberofmoles=Concentration×Volumeofsolutioninlitres (2)

Substitute the concentration and volume of HC3H5O2 in the above equation as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.025L=0.0025moles

Make the ICE table for the reaction between HC3H5O2 and NaOH .

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.002500Change:00+0.0025Finalmoles:0.002500.0025

The above equation shows the presence of equilibrium condition in the solution.

Make the ICE table for the dissociation reaction of HC3H5O2 .

HC3H5O2C3H5O2H+Initial(M):0.10000Change(M):xxxEquilibrium(M):0.100xxx

The equilibrium ratio for the given reaction is,

Ka=[C3H5O2][H+][HC3H5O2]

Substitute the calculated concentration values in the above expression.

Ka=[C3H5O2][H+][HC3H5O2]1.3×105=(x)(x)(0.100x)M

Since, value of Ka is very small, hence, (0.100x) is taken as (0.100) .

Simplify the above equation,

1.3×105=(x)(x)(0.100x)M1.3×105=(x)(x)0.100Mx=0.0011M

It is the concentration of H+ .

The pH of a solution is shown below.

pH=log[H+] (3)

Where,

• [H+] is the concentration of Hydrogen ions.

Substitute the value of [H+] in the above equation as,

pH=log[H+]=log(0.0011)=2.95_

The value of pH of solution when 0.0mL NaOH has been added is. 2.95_ .

The volume of NaOH is 4.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 4.0mL into L is done as,

4.0mL=4.0×0.001L=0.004L

Substitute the value of concentration and volume of NaOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.004L=0.0004moles

Make the ICE table for the reaction between HC3H5O2 and NaOH .

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.00040Change:0.00040.0004+0.0004Finalmoles:0.002100.0004

The above equation shows the presence of equilibrium condition in the solution.

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.004L=0.029L

Substitute the value of number of moles of HC3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0021moles0.029L=0.072M

Substitute the value of number of moles of C3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0004moles0.029L=0.013M

Make the ICE table for the dissociation reaction of HC3H5O2 .

HC3H5O2C3H5O2H+Initial(M):0.0720.0130Change(M):x+xxEquilibrium(M):0.072x0.013+xx

The equilibrium ratio for the given reaction is,

Ka=[C3H5O2][H+][HC3H5O2]

Substitute the calculated concentration values in the above expression.

Ka=[C3H5O2][H+][HC3H5O2]1.3×105=(0.013+x)(x)(0.072x)M

Since, value of Ka is very small, hence, (0.072x) is taken as (0.072) and (0.013+x) is taken as 0.013 .

Simplify the above equation,

1.3×105=(0.013+x)(x)(0.072x)M1.3×105=(0.013)(x)(0.072)Mx=7.2×105M

It is the concentration of H+ .

Substitute the value of [H+] in the equation (3)

pH=log[H+]=log(7.2×105)=4.14_

The value of pH of solution when 4.0mL NaOH has been added is. 4.14_ .

The volume of NaOH is 8.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 8.0mL into L is done as,

8.0mL=8.0×0.001L=0.008L

Substitute the value of concentration and volume of NaOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.008L=0.0008moles

Make the ICE table for the reaction between HC3H5O2 and NaOH .

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.00080Change:0.00080.0008+0.0008Finalmoles:0.001700.0008

The above equation shows the presence of equilibrium condition in the solution.

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.008L=0.033L

Substitute the value of number of moles of HC3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0017moles0.033L=0.051M

Substitute the value of number of moles of C3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0008moles0.033L=0.024M

Make the ICE table for the dissociation reaction of HC3H5O2 .

HC3H5O2C3H5O2H+Initial(M):0.0510.0240Change(M):x+xxEquilibrium(M):0.051x0.024+xx

The equilibrium ratio for the given reaction is,

Ka=[C3H5O2][H+][HC3H5O2]

Substitute the calculated concentration values in the above expression.

Ka=[C3H5O2][H+][HC3H5O2]1.3×105=(0.024+x)(x)(0.051x)M

Since, value of Ka is very small, hence, (0.051x) is taken as (0.051) and (0.024+x) is taken as 0.024 .

Simplify the above equation,

1.3×105=(0.024+x)(x)(0.051x)M1.3×105=(0.024)(x)(0.051)Mx=2.7×105M

It is the concentration of H+ .

Substitute the value of [H+] in the equation (3).

pH=log[H+]=log(2.7×105)=4.56_

The value of pH of solution when 8.0mL NaOH has been added is. 4.56_ .

The volume of NaOH is 12.5mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 12.5mL into L is done as,

12.5mL=12.5×0.001L=0.0125L

Substitute the value of concentration and volume of NaOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.0125L=0.00125moles

Make the ICE table for the reaction between HC3H5O2 and NaOH .

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.001250Change:0.001250.00125+0.00125Finalmoles:0.0012500.00125

The above equation shows the presence of equilibrium condition in the solution.

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.0125L=0.0375L

Substitute the value of number of moles of HC3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00125moles0.0375L=0.033M

Substitute the value of number of moles of C3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00125moles0.0375L=0.033M

Make the ICE table for the dissociation reaction of HC3H5O2 .

HC3H5O2C3H5O2H+Initial(M):0.0330.0330Change(M):x+xxEquilibrium(M):0.033x0.033+xx

The equilibrium ratio for the given reaction is,

Ka=[C3H5O2][H+][HC3H5O2]

Substitute the calculated concentration values in the above expression.

Ka=[C3H5O2][H+][HC3H5O2]1.3×105=(0.033+x)(x)(0.033x)M

Since, value of Ka is very small, hence, (0.033x) is taken as (0.033) and (0.033+x) is taken as 0.033 .

Simplify the above equation,

1.3×105=(0.033+x)(x)(0.033x)M1.3×105=(0.033)(x)(0.033)Mx=1.3×105M

It is the concentration of H+ .

Substitute the value of [H+] in the equation (3).

pH=log[H+]=log(1.3×105)=4.88_

The value of pH of solution when 12.5mL NaOH has been added is. 4.88_ .

The volume of NaOH is 20.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 20.0mL into L is done as,

20.0mL=20.0×0.001L=0.02L

Substitute the value of concentration and volume of NaOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.02L=0.002moles

Make the ICE table for the reaction between HC3H5O2 and NaOH .

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.0020Change:0.0020.002+0.002Finalmoles:0.000500.002

The above equation shows the presence of equilibrium condition in the solution.

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.02L=0.045L

Substitute the value of number of moles of HC3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0005moles0.045L=0.011M

Substitute the value of number of moles of C3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.002moles0.045L=0.044M

Make the ICE table for the dissociation reaction of HC3H5O2 .

HC3H5O2C3H5O2H+Initial(M):0.0110.0440Change(M):x+xxEquilibrium(M):0.011x0.044+xx

The equilibrium ratio for the given reaction is,

Ka=[C3H5O2][H+][HC3H5O2]

Substitute the calculated concentration values in the above expression.

Ka=[C3H5O2][H+][HC3H5O2]1.3×105=(0.044+x)(x)(0.011x)M

Since, value of Ka is very small, hence, (0.011x) is taken as (0.011) and (0.044+x) is taken as 0.044 .

Simplify the above equation,

1.3×105=(0.044+x)(x)(0.011x)M1.3×105=(0.044)(x)(0.011)Mx=3.2×106M

It is the concentration of H+ .

Substitute the value of [H+] in the equation (3).

pH=log[H+]=log(3.2×106)=5.49_

The value of pH of solution when 20.0mL NaOH has been added is. 5.49_ .

The volume of NaOH is 24.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 24.0mL into L is done as,

24.0mL=24.0×0.001L=0.024L

Substitute the value of concentration and volume of NaOH in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.024L=0.0024moles

Make the ICE table for the reaction between HC3H5O2 and NaOH .

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.00240Change:0.00240.0024+0.0024Finalmoles:0.000100.0024

The above equation shows the presence of equilibrium condition in the solution.

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.024L=0.049L

Substitute the value of number of moles of HC3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0001moles0.049L=0.0020M

Substitute the value of number of moles of C3H5O2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0024moles0.049L=0.049M

Make the ICE table for the dissociation reaction of HC3H5O2 .

HC3H5O2C3H5O2H+Initial(M):0.00200.0490Change(M):x+xxEquilibrium(M):0

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