# Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M HNO 3 with 0.100 M NaOH.

### Chemistry: An Atoms First Approach

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

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 92AE
Textbook Problem
1 views

## Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M HNO3 with 0.100 M NaOH.

Interpretation Introduction

Interpretation: The titration of Nitric 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 base 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.

### Explanation of Solution

Explanation

The value of pH of solution when 0.0mL NaOH has been added is. 1.0_ .

Given:

The concentration of HNO3 acid is 0.100M

The concentration of NaOH is 0.100M .

The volume of HNO3 is 25.0mL .

The volume of NaOH is 0.0mL .

When no base is added, then solution contains only the strong acid HNO3 . Therefore, pH is given by concentration of H+ only.

The pH of a solution is shown below.

pH=log[H+] (1)

Where,

• [H+] is the concentration of ions present in a solution.

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

pH=log[H+]=log(0.100)=1.0_

The value of pH of solution when 0.0mL NaOH has been added is. 1.0_ .

Explanation

The concentration of H+ is 0.0724M .

Given

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

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (2)

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

Numberofmoles=Concentration×Volumeofsolutioninlitres (3)

Substitute the value of concentration and volume of HNO3 in equation (3) as,

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

Substitute the value of concentration and volume of NaOH in equation (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.0004Change                               0.00250.00040.0004Finalmoles0.00210

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.004L=0.029L

Substitute the value of number of moles of HNO3 and final volume of solution in equation (2).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0021moles0.029L=0.0724M

It is the concentration of H+ .

Explanation

The value of pH of solution when 4.0mL NaOH has been added is. 1.14_ .

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

pH=log[H+]=log(0.0724)=1.14_

The value of pH of solution when 4.0mL NaOH has been added is. 1.14_ .

Explanation

The concentration of H+ is 0.0515M .

Given

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.0008Change                               0.00250.00080.0008Finalmoles0.00170

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.008L=0.033L

Substitute the value of number of moles of HNO3 and final volume of solution in equation (2).

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

It is the concentration of H+ .

Explanation

The value of pH of solution when 8.0mL NaOH has been added is. 1.28_ .

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

pH=log[H+]=log(0.0515)=1.28_

The value of pH of solution when 8.0mL NaOH has been added is. 1.28_ .

Explanation

The concentration of H+ is 0.033M .

Given

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.00125Change                               0.00250.001250.00125Finalmoles0.001250

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.0125L=0.0375L

Substitute the value of number of moles of HNO3 and final volume of solution in equation (2).

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

It is the concentration of H+ .

Explanation

The value of pH of solution when 12.5mL NaOH has been added is. 1.48_ .

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

pH=log[H+]=log(0.033)=1.48_

The value of pH of solution when 12.5mL NaOH has been added is. 1.48_ .

Explanation

The concentration of H+ is 0.011M .

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.002Change                               0.00250.0020.002Finalmoles0.00050

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.02L=0.045L

Substitute the value of number of moles of HNO3 and final volume of solution in equation (2).

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

It is the concentration of H+ .

Explanation

The value of pH of solution when 20.0mL NaOH has been added is. 1.95_ .

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

pH=log[H+]=log(0.011)=1.95_

The value of pH of solution when 20.0mL NaOH has been added is. 1.95_ .

Explanation

The concentration of H+ is 0.002M .

Given

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.0024Change                               0.00250.00240.0024Finalmoles0.00010

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.024L=0.049L

Substitute the value of number of moles of HNO3 and final volume of solution in equation (2).

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

It is the concentration of H+ .

Explanation

The value of pH of solution when 24.0mL NaOH has been added is. 2.69_ .

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

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

The value of pH of solution when 24.0mL NaOH has been added is. 2.69_ .

Explanation

The concentration of H+ is 0.001M .

Given

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.00245Change                               0.00250.002450.00245Finalmoles0.000050

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.0245L=0.0495L

Substitute the value of number of moles of HNO3 and final volume of solution in equation (2).

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

It is the concentration of H+ .

Explanation

The value of pH of solution when 24.5mL NaOH has been added is. 3.0_ .

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

pH=log[H+]=log(0.001)=3.0_

The value of pH of solution when 24.5mL NaOH has been added is. 3.0_ .

Explanation

The concentration of H+ is 0.0002M .

Given

The volume of NaOH is 24.9mL

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