Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Textbook Question
Chapter 14, Problem 62E

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.

Expert Solution & Answer
Check Mark
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.

Answer to Problem 62E

Answer

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

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

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

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

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

The value of pH of solution when 24.0mL NaOH has been added is. 6.27_ .

The value of pH of solution when 24.5mL NaOH has been added is. 6.56_ .

The value of pH of solution when 24.9mL NaOH has been added is. 7.3_ .

The value of pH of solution when 25.0mL NaOH has been added is. 8.8_ .

The value of pH of solution when 25.1mL NaOH has been added is 10.31_

The value of pH of solution when 26.0mL NaOH has been added is 11.31_

The value of pH of solution when 28.0mL NaOH has been added is 11.78_

The value of pH of solution when 30.0mL NaOH has been added is 11.96_

The graph between pH and volume of NaOH added is shown in Figure 2.

Explanation of Solution

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):0.0020x0.049+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=(0.049+x)(x)(0.0020x)M

Since, value of Ka is very small, hence, (0.0020x) is taken as (0.0020) and (0.049+x) is taken as 0.049 .

Simplify the above equation,

1.3×105=(0.049+x)(x)(0.0020x)M1.3×105=(0.049)(x)(0.0020)Mx=5.3×107M

It is the concentration of H+ .

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

pH=log[H+]=log(5.3×107)=6.27_

The value of pH of solution when 24.0mL NaOH has been added is. 6.27_ .

The volume of NaOH is 24.5mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 24.5mL into L is done as,

24.5mL=24.5×0.001L=0.0245L

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

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.0245L=0.00245moles

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

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.002450Change:0.002450.00245+0.00245Finalmoles:0.0000500.00245

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

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.0245L=0.0495L

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00005moles0.0495L=0.0010M

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00245moles0.0495L=0.049M

Make the ICE table for the dissociation reaction of HC3H5O2 .

HC3H5O2C3H5O2H+Initial(M):0.00100.0490Change(M):x+xxEquilibrium(M):0.0010x0.049+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=(0.049+x)(x)(0.0010x)M

Since, value of Ka is very small, hence, (0.0010x) is taken as (0.0010) and (0.049+x) is taken as 0.049 .

Simplify the above equation,

1.3×105=(0.049+x)(x)(0.0010x)M1.3×105=(0.049)(x)(0.0010)Mx=2.7×107M

It is the concentration of H+ .

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

pH=log[H+]=log(2.7×107)=6.56_

The value of pH of solution when 24.5mL NaOH has been added is. 6.56_ .

The volume of NaOH is 24.9mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 24.9mL into L is done as,

24.9mL=24.9×0.001L=0.0249L

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

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.0249L=0.00249moles

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

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.002490Change:0.002490.00249+0.00249Finalmoles:0.0000100.00249

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

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.0249L=0.0499L

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00001moles0.0499L=0.0002M

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00249moles0.0499L=0.05M

Make the ICE table for the dissociation reaction of HC3H5O2 .

HC3H5O2C3H5O2H+Initial(M):0.00020.050Change(M):x+xxEquilibrium(M):0.0002x0.05+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=(0.05+x)(x)(0.0002x)M

Since, value of Ka is very small, hence, (0.0002x) is taken as (0.0002) and (0.05+x) is taken as 0.05 .

Simplify the above equation,

1.3×105=(0.05+x)(x)(0.0002x)M1.3×105=(0.05)(x)(0.0002)Mx=5.2×108M

It is the concentration of H+ .

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

pH=log[H+]=log(5.2×108)=7.3_

The value of pH of solution when 24.9mL NaOH has been added is. 7.3_

The volume of NaOH is 25.0mL .

The conversion of mL into L is done as,

1mL=0.001L

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

25.0mL=25.0×0.001L=0.025L

Substitute the value of concentration and volume of NaOH in equation (2) 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:0.00250.00250Change:0.00250.0025+0.0025Finalmoles:0.00.00.0025

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

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.025L=0.050L

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0025moles0.050L=0.05M

Make the ICE table for the above reaction.

C3H5O2Na++H2OC3H5O2H+NaOHInitial moles:0.05000Change:x0+xxFinalmoles:0.05x0xx

The equilibrium ratio for the given reaction is,

Kb=[C3H5O2H][NaOH][C3H5O2Na+]

The relationship between Ka and Kb is given as,

Ka×Kb=1014

Substitute the value of Kb in above equation.

Ka×Kb=10141.3×105×Kb=1014Ka=0.8×109

Substitute the calculated concentration values in the above expression.

Kb=[C3H5O2H][NaOH][C3H5O2Na+]0.8×109=(x)(x)(0.05x)M

Since, value of Kb is very small, hence, (0.05x) is taken as (0.05) .

Simplify the above equation,

0.8×109=(x)(x)(0.05)Mx=0.63×105M

It is the concentration of OH .

The pOH of the solution is shown below.

pOH=log[OH] (4)

Where,

  • [OH] is the concentration of Hydroxide ions.

Substitute the value of [OH] in the above equation.

pOH=log[OH]=log(0.63×105)=5.2

The relationship between pOH is given as,

pH+pOH=14 (5)

Substitute the value of pOH in the above equation.

pH+pOH=14pH+5.2=14pH=8.8_

The value of pH of solution when 25.0mL NaOH has been added is 8.8_ .

The volume of NaOH is 25.1mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 25.1mL into L is done as,

25.1mL=25.1×0.001L=0.0251L

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

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.0251L=0.00251moles

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

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.002510Change:0.00250.0025+0.0025Finalmoles:0.00.000010.0025

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

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.0251L=0.0501L

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00001moles0.0501L=0.0002M

Substitute the value of [OH] in the equation (4).

pOH=log[OH]=log(0.0002)=3.69

Substitute the value of pOH in the equation (5).

pH+pOH=14pH+3.69=14pH=10.31_

The value of pH of solution when 25.1mL NaOH has been added is 10.31_ .

The volume of NaOH is 26.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 26.0mL into L is done as,

26.0mL=26.0×0.001L=0.026L

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

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.026L=0.0026moles

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

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.00260Change:0.00250.0025+0.0025Finalmoles:0.00.00010.0025

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

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.026L=0.051L

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0001moles0.051L=0.002M

Substitute the value of [OH] in the above equation.

pOH=log[OH]=log(0.002)=2.69

Substitute the value of pOH in the above equation as,

pH+pOH=14pH+2.69=14pH=11.31_

The value of pH of solution when 26.0mL NaOH has been added is 11.31_ .

The volume of NaOH is 28.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 28.0mL into L is done as,

28.0mL=28.0×0.001L=0.028L

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

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.028L=0.0028moles

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

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.00280Change:0.00250.0025+0.0025Finalmoles:0.00.00030.0025

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

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.028L=0.053L

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0003moles0.053L=0.006M

Substitute the value of [OH] in the equation (4).

pOH=log[OH]=log(0.006)=2.22

Substitute the value of pOH in the equation (5).

pH+pOH=14pH+2.22=14pH=11.78_

The value of pH of solution when 28.0mL NaOH has been added is 11.78_ .

The volume of NaOH is 30.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 30.0mL into L is done as,

30.0mL=30.0×0.001L=0.03L

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

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.100M×0.03L=0.003moles

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

HC3H5O2+NaOHHC3H5O2Na++H2OInitial moles:0.00250.0030Change:0.00250.0025+0.0025Finalmoles:0.00.00050.0025

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

Total volume of solution =VolumeofHC3H5O2+VolumeofNaOH=0.025L+0.03L=0.055L

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0005moles0.055L=0.009M

Substitute the value of [OH] in the (4) equation.

pOH=log[OH]=log(0.009)=2.04

Substitute the value of pOH in the equation (5).

pH+pOH=14pH+2.04=14pH=11.96_

The value of pH of solution when 30.0mL NaOH has been added is 11.96_ .

The values of pH obtained are shown in the below table.

Volume of NaOH in mL pH
0.0 11.11
4.0 9.85
8.0 9.61
12.5 9.26
20.0 8.66
24.0 7.86
24.5 7.56
24.9 6.86
25.0 5.25
25.1 3.69
26.0 2.69
28.0 2.22
30.0 2.04

Figure 1

The graph between pH and volume of NaOH added is shown below.

Chemistry: An Atoms First Approach, Chapter 14, Problem 62E

Figure 2

Conclusion

Conclusion

The amount of species present in the solution during titration depends on the volume of titrant added in the solution and this further defines the value of pH of the solution

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Chapter 14 Solutions

Chemistry: An Atoms First Approach

Ch. 14 - What are the major species in solution after...Ch. 14 - Prob. 2ALQCh. 14 - Prob. 3ALQCh. 14 - Prob. 4ALQCh. 14 - Sketch two pH curves, one for the titration of a...Ch. 14 - Prob. 6ALQCh. 14 - Prob. 7ALQCh. 14 - You have a solution of the weak acid HA and add...Ch. 14 - The common ion effect for weak acids is to...Ch. 14 - Prob. 10QCh. 14 - Prob. 11QCh. 14 - Consider the following pH curves for 100.0 mL of...Ch. 14 - An acid is titrated with NaOH. The following...Ch. 14 - Consider the following four titrations. i. 100.0...Ch. 14 - Prob. 15QCh. 14 - Prob. 16QCh. 14 - How many of the following are buffered solutions?...Ch. 14 - Which of the following can be classified as buffer...Ch. 14 - A certain buffer is made by dissolving NaHCO3 and...Ch. 14 - Prob. 20ECh. 14 - Calculate the pH of each of the following...Ch. 14 - Calculate the pH of each of the following...Ch. 14 - Prob. 23ECh. 14 - Compare the percent ionization of the base in...Ch. 14 - Prob. 25ECh. 14 - Calculate the pH after 0.020 mole of HCl is added...Ch. 14 - Calculate the pH after 0.020 mole of NaOH is added...Ch. 14 - Calculate the pH after 0.020 mole of NaOH is added...Ch. 14 - Which of the solutions in Exercise 21 shows the...Ch. 14 - Prob. 30ECh. 14 - Calculate the pH of a solution that is 1.00 M HNO2...Ch. 14 - Calculate the pH of a solution that is 0.60 M HF...Ch. 14 - Calculate the pH after 0.10 mole of NaOH is added...Ch. 14 - Calculate the pH after 0.10 mole of NaOH is added...Ch. 14 - Calculate the pH of each of the following buffered...Ch. 14 - Prob. 36ECh. 14 - Calculate the pH of a buffered solution prepared...Ch. 14 - A buffered solution is made by adding 50.0 g NH4Cl...Ch. 14 - Prob. 39ECh. 14 - An aqueous solution contains dissolved C6H5NH3Cl...Ch. 14 - Prob. 41ECh. 14 - Prob. 42ECh. 14 - Consider a solution that contains both C5H5N and...Ch. 14 - Calculate the ratio [NH3]/[NH4+] in...Ch. 14 - Prob. 45ECh. 14 - Prob. 46ECh. 14 - Prob. 47ECh. 14 - Prob. 48ECh. 14 - Calculate the pH of a solution that is 0.40 M...Ch. 14 - Calculate the pH of a solution that is 0.20 M HOCl...Ch. 14 - Which of the following mixtures would result in...Ch. 14 - Prob. 52ECh. 14 - Prob. 53ECh. 14 - Calculate the number of moles of HCl(g) that must...Ch. 14 - Consider the titration of a generic weak acid HA...Ch. 14 - Sketch the titration curve for the titration of a...Ch. 14 - Consider the titration of 40.0 mL of 0.200 M HClO4...Ch. 14 - Consider the titration of 80.0 mL of 0.100 M...Ch. 14 - Consider the titration of 100.0 mL of 0.200 M...Ch. 14 - Prob. 60ECh. 14 - Lactic acid is a common by-product of cellular...Ch. 14 - Repeat the procedure in Exercise 61, but for the...Ch. 14 - Repeat the procedure in Exercise 61, but for the...Ch. 14 - Repeat the procedure in Exercise 61, but for the...Ch. 14 - Prob. 65ECh. 14 - In the titration of 50.0 mL of 1.0 M methylamine,...Ch. 14 - You have 75.0 mL of 0.10 M HA. After adding 30.0...Ch. 14 - A student dissolves 0.0100 mole of an unknown weak...Ch. 14 - Prob. 69ECh. 14 - Prob. 70ECh. 14 - Potassium hydrogen phthalate, known as KHP (molar...Ch. 14 - A certain indicator HIn has a pKa of 3.00 and a...Ch. 14 - Prob. 73ECh. 14 - Prob. 74ECh. 14 - Prob. 75ECh. 14 - Prob. 76ECh. 14 - Prob. 77ECh. 14 - Estimate the pH of a solution in which crystal...Ch. 14 - Prob. 79ECh. 14 - Prob. 80ECh. 14 - Prob. 81AECh. 14 - Prob. 82AECh. 14 - Tris(hydroxymethyl)aminomethane, commonly called...Ch. 14 - Prob. 84AECh. 14 - You have the following reagents on hand: Solids...Ch. 14 - Prob. 86AECh. 14 - Prob. 87AECh. 14 - What quantity (moles) of HCl(g) must be added to...Ch. 14 - Calculate the value of the equilibrium constant...Ch. 14 - The following plot shows the pH curves for the...Ch. 14 - Calculate the volume of 1.50 102 M NaOH that must...Ch. 14 - Prob. 92AECh. 14 - A certain acetic acid solution has pH = 2.68....Ch. 14 - A 0.210-g sample of an acid (molar mass = 192...Ch. 14 - The active ingredient in aspirin is...Ch. 14 - One method for determining the purity of aspirin...Ch. 14 - A student intends to titrate a solution of a weak...Ch. 14 - Prob. 98AECh. 14 - Prob. 99AECh. 14 - Consider 1.0 L of a solution that is 0.85 M HOC6H5...Ch. 14 - Prob. 101CWPCh. 14 - Consider the following acids and bases: HCO2H Ka =...Ch. 14 - Prob. 103CWPCh. 14 - Prob. 104CWPCh. 14 - Consider the titration of 100.0 mL of 0.100 M HCN...Ch. 14 - Consider the titration of 100.0 mL of 0.200 M...Ch. 14 - Prob. 107CWPCh. 14 - Prob. 108CPCh. 14 - A buffer is made using 45.0 mL of 0.750 M HC3H5O2...Ch. 14 - A 0.400-M solution of ammonia was titrated with...Ch. 14 - Prob. 111CPCh. 14 - Consider a solution formed by mixing 50.0 mL of...Ch. 14 - When a diprotic acid, H2A, is titrated with NaOH,...Ch. 14 - Consider the following two acids: In two separate...Ch. 14 - The titration of Na2CO3 with HCl bas the following...Ch. 14 - Prob. 116CPCh. 14 - A few drops of each of the indicators shown in the...Ch. 14 - Malonic acid (HO2CCH2CO2H) is a diprotic acid. In...Ch. 14 - A buffer solution is prepared by mixing 75.0 mL of...Ch. 14 - A 10.00-g sample of the ionic compound NaA, where...Ch. 14 - Prob. 121IPCh. 14 - Prob. 122MP
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