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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 63E
Textbook Problem
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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.

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.

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

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Chapter 14 Solutions

Chemistry: An Atoms First Approach
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