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 63E

Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M NH3 (Kb = 1.8 × 10−5) with 0.100 M HCl.

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation: The titration of NH3 with different volumes of HCl is given. The pH of each solution is to be calculated and then graph of pH versus volume of titrant 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 solution is to be calculated and graph of pH versus volume of titrant added.

Answer to Problem 63E

Answer

The value of pH of solution when 0.0mL HCl has been added is. 11.11_ .

The value of pH of solution when 4.0mL HCl has been added is. 9.85_ .

The value of pH of solution when 8.0mL HCl has been added is. 9.61_ .

The value of pH of solution when 12.5mL HCl has been added is. 9.26_ .

The value of pH of solution when 20.0mL HCl has been added is. 8.66_ .

The value of pH of solution when 24.0mL HCl has been added is. 7.86_ .

The value of pH of solution when 24.5mL HCl has been added is. 7.56_ .

The value of pH of solution when 24.9mL HCl has been added is. 6.86_ .

The value of pH of solution when 25.0mL HCl has been added is. 5.25_ .

The value of pH of solution when 25.1mL HCl has been added is 3.69_

The value of pH of solution when 26.0mL HCl has been added is 2.69_

The value of pH of solution when 28.0mL HCl has been added is 2.22_

The value of pH of solution when 30.0mL HCl has been added is 2.04_

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

Explanation of Solution

Explanation

Given

The concentration of NH3 is 0.100M

The concentration of HCl is 0.100M .

The volume of NH3 is 25.0mL .

The volume of HCl 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 Kb of NH3 is 1.8×105 .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 100mL into L is done as,

100mL=100×0.001L=0.100L

The reaction is represented as,

NH3+H2ONH4++OH

Make the ICE table for the reaction between H2NNH2 and H2O .

NH3+H2ONH4++OHInitial moles:0.10000Change:x+xxFinalmoles:0.100xxx

The equilibrium ratio for the given reaction is,

Kb=[NH4+][OH][NH3]

Substitute the calculated concentration values in the above expression.

Kb=[NH4+][OH][NH3]

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

Simplify the above equation,

1.8×105=(x)(x)(0.100)Mx=0.0013M

It is the concentration of OH .

The pOH of the solution is shown below.

pOH=log[OH] (1)

Where,

  • [OH] is the concentration of Hydroxide ions.

Substitute the value of [OH] in the above equation as,

pOH=log[OH]=log(0.0013)=2.89

The value of pOH of the solution is 2.89 .

The relationship between pOH is given as,

pH+pOH=14 (2)

Substitute the value of pOH in the above equation.

pH+pOH=14pH+2.89=14pH=11.11_

The value of pH of solution when 0.0mL HCl has been added is. 11.11_ .

The volume of HCl 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

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (3)

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

Numberofmoles=Concentration×Volumeofsolutioninlitres (4)

Substitute the value of concentration and volume of HCl in equation (4) as,

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

Substitute the value of concentration and volume of NH3 in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.00040Change in moles                              0.00040.0004+0.0004Equilibriummoles0.00210.00040.0004

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.004L=0.029L

Substitute the value of number of moles of NH3 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0021moles0.029L=0.05M

Substitute the value of number of moles of NH4+ and final volume of solution in equation (1).

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

Make the ICE table for the dissociation reaction of NH3 .

NH3+H2ONH4++OHInitial moles:0.050.013Change:x+xxFinalmoles:0.05x0.013+xx

The equilibrium ratio for the given reaction is,

Kb=[NH4+][OH][NH3]

Substitute the calculated concentration values in the above expression.

Kb=[NH4+][OH][NH3]1.8×105=(0.013+x)(x)(0.05x)M

Since, value of Kb is very small, hence, (0.05x) is taken as (0.05) and (0.013+x) is taken as 0.013 .

Simplify the above equation,

1.8×105=(0.013+x)(x)(0.05x)M1.8×105=(0.013)(x)(0.05)Mx=7.0×105M

It is the concentration of OH .

Substitute the value of [OH] in the equation (1).

pOH=log[OH]=log(7.0×105)=4.15

The value of pOH of the solution is 4.15 .

Substitute the value of pOH in the equation (2)

pH+pOH=14pH+4.15=14pH=9.85_

The value of pH of solution when 4.0mL HCl has been added is. 9.85_ .

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.00080Change in moles                              0.00040.0008+0.0008Equilibriummoles0.00170.00080.0008

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.008L=0.033L

Substitute the value of number of moles of NH3 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0017moles0.033L=0.0515M

Substitute the value of number of moles of NH4+ and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0008moles0.033L=0.0242M

Make the ICE table for the dissociation reaction of NH3 .

NH3+H2ONH4++OHInitial moles:0.05150.02420Change:x+xxFinalmoles:0.0515x0.0242+xx

The equilibrium ratio for the given reaction is,

Kb=[NH4+][OH][NH3]

Substitute the calculated concentration values in the above expression.

Kb=[NH4+][OH][NH3]1.8×105=(0.0242+x)(x)(0.0515x)M

Since, value of Kb is very small, hence, (0.0515x) is taken as (0.0515) and (0.0242+x) is taken as 0.0242 .

Simplify the above equation,

1.8×105=(0.0242+x)(x)(0.0515x)M1.8×105=(0.0242)(x)(0.0515)Mx=4.0×105M

It is the concentration of OH .

Substitute the value of [OH] in the equation (1).

pOH=log[OH]=log(4.0×105)=4.39

The value of pOH of the solution is 4.39 .

Substitute the value of pOH in the equation (2)

pH+pOH=14pH+4.39=14pH=9.61_

The value of pH of solution when 8.0mL HCl has been added is. 9.61_ .

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.001250Change in moles                              0.001250.00125+0.00125Equilibriummoles0.001250.001250.00125

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.0125L=0.0375L

Substitute the value of number of moles of NH3 and final volume of solution in equation (1).

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

Substitute the value of number of moles of NH4+ and final volume of solution in equation (1).

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

Make the ICE table for the dissociation reaction of NH3 .

NH3+H2ONH4++OHInitial moles:0.0330.0330Change:x+xxFinalmoles:0.033x0.033+xx

The equilibrium ratio for the given reaction is,

Kb=[NH4+][OH][NH3]

Substitute the calculated concentration values in the above expression.

Kb=[NH4+][OH][NH3]1.8×105=(0.033+x)(x)(0.033x)M

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

Simplify the above equation,

1.8×105=(0.033+x)(x)(0.033x)M1.8×105=(0.033)(x)(0.033)Mx=1.8×105M

It is the concentration of OH .

Substitute the value of [OH] in the equation (1).

pOH=log[OH]=log(1.8×105)=4.74

The value of pOH of the solution is 4.74 .

Substitute the value of pOH in the above equation as,

pH+pOH=14pH+4.74=14pH=9.26_

The value of pH of solution when 12.5mL HCl has been added is. 9.26_ .

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.0020Change in moles                              0.0020.002+0.002Equilibriummoles0.00050.0020.002

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.02L=0.045L

Substitute the value of number of moles of NH3 and final volume of solution in equation (1).

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

Substitute the value of number of moles of NH4+ and final volume of solution in equation (1).

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

Make the ICE table for the dissociation reaction of NH3 .

NH3+H2ONH4++OHInitial moles:0.0110.0440Change:x+xxFinalmoles:0.011x0.044+xx

The equilibrium ratio for the given reaction is,

Kb=[NH4+][OH][NH3]

Substitute the calculated concentration values in the above expression.

Kb=[NH4+][OH][NH3]1.8×105=(0.044+x)(x)(0.011x)M

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

Simplify the above equation,

1.8×105=(0.044+x)(x)(0.011x)M1.8×105=(0.044)(x)(0.011)Mx=4.5×106M

It is the concentration of OH .

Substitute the value of [OH] in the equation (1).

pOH=log[OH]=log(4.5×106)=5.34

The value of pOH of the solution is 5.34 .

Substitute the value of pOH in the equation (2).

pH+pOH=14pH+5.34=14pH=8.66_

The value of pH of solution when 20.0mL HCl has been added is. 8.66_ .

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.00240Change in moles                              0.00240.0024+0.0024Equilibriummoles0.00010.00240.0024

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.024L=0.049L

Substitute the value of number of moles of NH3 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0001moles0.049L=0.002M

Substitute the value of number of moles of NH4+ and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0024moles0.049L=0.05M

Make the ICE table for the dissociation reaction of NH3 .

NH3+H2ONH4++OHInitial moles:0.0020.050Change:x+xxFinalmoles:0.002x0.05+xx

The equilibrium ratio for the given reaction is,

Kb=[NH4+][OH][NH3]

Substitute the calculated concentration values in the above expression.

Kb=[NH4+][OH][NH3]1.8×105=(0.05+x)(x)(0.002x)M

Since, value of Kb is very small, hence, (0.002x) is taken as (0.002) and (0.05+x) is taken as 0.05 .

Simplify the above equation,

1.8×105=(0.05+x)(x)(0.002x)M1.8×105=(0.05)(x)(0.002)Mx=7.2×107M

It is the concentration of OH .

Substitute the value of [OH] in the above equation as,

pOH=log[OH]=log(7.2×107)=6.14

The value of pOH of the solution is 6.14 .

Substitute the value of pOH in the above equation as,

pH+pOH=14pH+6.14=14pH=7.86_

The value of pH of solution when 24.0mL HCl has been added is. 7.86_ .

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.002450Change in moles    0.002450.00245+0.00245Equilibriummoles0.000050.002450.00245

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.0245L=0.0495L

Substitute the value of number of moles of NH3 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00005moles0.0495L=0.001M

Substitute the value of number of moles of NH4+ and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00245moles0.0495L=0.05M

Make the ICE table for the dissociation reaction of NH3 .

NH3+H2ONH4++OHInitial moles:0.0010.050Change:x+xxFinalmoles:0.001x0.05+xx

The equilibrium ratio for the given reaction is,

Kb=[NH4+][OH][NH3]

Substitute the calculated concentration values in the above expression.

Kb=[NH4+][OH][NH3]1.8×105=(0.05+x)(x)(0.001x)M

Since, value of Kb is very small, hence, (0.001x) is taken as (0.001) and (0.05+x) is taken as 0.05 .

Simplify the above equation,

1.8×105=(0.05+x)(x)(0.001x)M1.8×105=(0.05)(x)(0.001)Mx=3.6×107M

It is the concentration of OH .

Substitute the value of [OH] in the equation (1).

pOH=log[OH]=log(3.6×107)=6.44

The value of pOH of the solution is 6.44 .

Substitute the value of pOH in the equation (2).

pH+pOH=14pH+6.44=14pH=7.56_

The value of pH of solution when 24.5mL HCl has been added is. 7.56_ .

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.002490Change in moles   0.002490.00249+0.00249Equilibriummoles0.000010.002490.00249

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.0249L=0.0499L

Substitute the value of number of moles of NH3 and final volume of solution in equation (1).

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

Substitute the value of number of moles of NH4+ and final volume of solution in equation (1).

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

Make the ICE table for the dissociation reaction of NH3 .

NH3+H2ONH4++OHInitial moles:0.00020.050Change:x+xxFinalmoles:0.0002x0.05+xx

The equilibrium ratio for the given reaction is,

Kb=[NH4+][OH][NH3]

Substitute the calculated concentration values in the above expression.

Kb=[NH4+][OH][NH3]1.8×105=(0.05+x)(x)(0.0002x)M

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

Simplify the above equation,

1.8×105=(0.05+x)(x)(0.0002x)M1.8×105=(0.05)(x)(0.0002)Mx=7.2×108M

It is the concentration of OH .

Substitute the value of [OH] in the equation (1).

pOH=log[OH]=log(7.2×108)=7.14

The value of pOH of the solution is 7.14 .

Substitute the value of pOH in the equation (2).

pH+pOH=14pH+7.14=14pH=6.86_

The value of pH of solution when 24.9mL HCl has been added is. 6.86_ .

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.00250Change in moles    0.00250.0025+0.0025Equilibriummoles0.00.00.0025

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.025L=0.05L

Substitute the value of number of moles of NH4+ and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0025moles0.05L=0.05M

Make the ICE table for the reaction of NH4+ with water.

NH4++H2ONH3+H3O+Initial moles:0.0500Change:x+xxFinalmoles:0.05xxx

The relationship between Ka and Kb is given as,

Ka×Kb=1014

Substitute the value of Kb in above equation.

Ka×Kb=1014Ka×1.8×105=1014Ka=5.6×1010

The equilibrium ratio for the given reaction is,

Ka=[NH3][H3O+][NH4+]

Substitute the calculated concentration values in the above expression.

Ka=[NH3][H3O+][NH4+]5.6×1010=(x)(x)(0.05x)M

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

Simplify the above equation,

5.6×1010=(x)(x)(0.05x)M5.6×1010=(x)(x)(0.05)Mx=5.5×106M

It is the concentration of H3O+ .

The pH of the solution is shown below.

pH=log[H3O+] (5)

Where,

  • [H3O+] is the concentration of Hydronium ions in the solution.

Substitute the value of [H3O+] in the above equation.

pH=log[H3O+]=log(5.5×106)=5.25_

The value of pH of solution when 25.0mL HCl has been added is. 5.25_ .

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.002510Change in moles    0.00250.0025+0.0025Equilibriummoles0.00.000010.0025

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.0251L=0.0501L

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

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

The concentration of H+ is 0.0002M .

Substitute the value of [H+] in the equation (5).

pH=log[H+]=log(0.0002)=3.69_

The value of pH of solution when 25.1mL HCl has been added is 3.69_ .

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.00260Change in moles    0.00250.0025+0.0025Equilibriummoles0.00.00010.0025

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.026L=0.051L

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

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

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

pH=log[H+]=log(0.002)=2.69_

The value of pH of solution when 26.0mL HCl has been added is 2.69_

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.00280Change in moles    0.00250.0025+0.0025Equilibriummoles0.00.00030.0025

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.028L=0.053L

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

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

The concentration of H+ is 0.006M .

Substitute the value of [H+] in the equation (5).

pH=log[H+]=log(0.006)=2.22_

The value of pH of solution when 28.0mL HCl has been added is 2.22_

The volume of HCl 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 HCl in equation (4) as,

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

Make the ICE table for the reaction between NH3 and HCl .

NH3+HClNH4ClInitialmoles          0.00250.0030Change in moles    0.00250.0025+0.0025Equilibriummoles0.00.00050.0025

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution =VolumeofNH3+VolumeofHCl=0.025L+0.03L=0.055L

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

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

The concentration of H+ is 0.009M .

Substitute the value of [H+] in the equation (5).

pH=log[H+]=log(0.009)=2.04_

The value of pH of solution when 30.0mL HCl has been added is 2.04_

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

Volume of HCl 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 HCl added is shown below.

Chemistry: An Atoms First Approach, Chapter 14, Problem 63E

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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