Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 102AE
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.

Expert Solution & Answer
Check Mark

Answer to Problem 102AE

Answer

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

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

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

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

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

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

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

The value of pH of solution when 24.9mL NaOH has been added is. 3.69_ .

The value of pH of solution when 25.0mL NaOH has been added is. 7.0_ .

The value of pH of solution when 25.1mL NaOH has been added is. 10.35_ .

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 base added is shown in Figure 1.

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 .

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.00249Change                               0.00250.002490.00249Finalmoles0.000010

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.0249L=0.0499L

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

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

It is the concentration of H+ .

Explanation

The value of pH of solution when 24.9mL NaOH has been added is. 3.69_ .

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

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

The value of pH of solution when 24.9mL NaOH has been added is. 3.69_ .

Explanation

The value of pH of solution when 25.0mL NaOH has been added is. 7.0_ .

Given

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.0025Change                               0.00250.00250.0025Finalmoles0.00

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.025L=0.05L

As all the moles have been neutralized, therefore the value of pH is 7.0_ .

The value of pH of solution when 25.0mL NaOH has been added is. 7.0_ .

Explanation

The concentration of OH is 0.00022M .

Given

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.00251Change                               0.00250.00250.002510.0025Finalmoles00.00001

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.0251L=0.0451L

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.00001moles0.0451L=0.00022M

It is the concentration of OH .

Explanation

The value of pOH of solution when 25.1mL NaOH has been added is. 3.65 .

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(0.00022)=3.65

Explanation

The value of pH of solution when 25.1mL NaOH has been added is. 10.35_ .

The relationship between pOH is given as,

pH+pOH=14

Substitute the value of pOH in the above equation.

pH+pOH=14pH+3.65=14pH=10.35_

The value of pH of solution when 25.1mL NaOH has been added is. 10.35_ .

Explanation

The concentration of OH is 0.002M .

Given

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.0026Change                               0.00250.00250.00260.0025Finalmoles00.0001

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.026L=0.051L

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

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

It is the concentration of OH .

Explanation

The value of pOH of solution when 26.0mL NaOH has been added is. 2.69 .

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

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

The value of pOH of solution when 26.0mL NaOH has been added is. 2.69 .

Explanation

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

The relationship between pOH is given as,

pH+pOH=14 (5)

Substitute the value of pOH in the above equation.

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

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

Explanation

The concentration of OH is 0.006M .

Given

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.0028Change                               0.00250.00250.00280.0025Finalmoles00.0003

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.028L=0.053L

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

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

It is the concentration of OH .

Explanation

The value of pOH of solution when 28.0mL NaOH has been added is. 2.22 .

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

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

The value of pOH of solution when 28.0mL NaOH has been added is. 2.22 .

Explanation

The value of pH of solution when 25.1mL NaOH has been added is. 11.78_ .

Substitute the value of pOH in the above 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_ .

Explanation

The concentration of OH is 0.009M .

Given

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 (3) as,

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

Make the table for the reaction between HNO3 and NaOH .

H++OHH2OInitialmoles          0.00250.003Change                               0.00250.00250.0030.0025Finalmoles00.0005

Total volume of solution =VolumeofHNO3+VolumeofNaOH=0.025L+0.03L=0.055L

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

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

It is the concentration of OH .

Explanation

The value of pOH of solution when 30.0mL NaOH has been added is. 2.04 .

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

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

The value of pOH of solution when 30.0mL NaOH has been added is. 2.04 .

Explanation

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

Substitute the value of pOH in the above 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_ .

Explanation

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

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

pH Volume of NaOH in mL
1.0 0.0
1.14 4.0
1.28 8.0
1.48 12.5
1.95 20.0
2.69 24.0
3.0 24.5
3.69 24.9
7.0 25.0
10.35 25.1
11.31 26.0
11.78 28.0
11.96 30.0

Table 1

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

Chemistry, Chapter 15, Problem 102AE

Figure 1

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 15 Solutions

Chemistry

Ch. 15 - What are the major species in solution after...Ch. 15 - A friend asks the following: Consider a buffered...Ch. 15 - Mixing together solutions of acetic acid and...Ch. 15 - Sketch two pH curves, one for the titration of a...Ch. 15 - Sketch a pH curve for the titration of a weak acid...Ch. 15 - You have a solution of the weak acid HA and add...Ch. 15 - You have a solution of the weak acid HA and add...Ch. 15 - The common ion effect for weak acids is to...Ch. 15 - Prob. 12QCh. 15 - A best buffer has about equal quantities of weak...Ch. 15 - Consider the following pH curves for 100.0 mL of...Ch. 15 - An acid is titrated with NaOH. The following...Ch. 15 - Consider the following four titrations. i. 100.0...Ch. 15 - Figure 14-4 shows the pH curves for the titrations...Ch. 15 - Acidbase indicators mark the end point of...Ch. 15 - Consider the titration of 100.0 mL of 0.10 M...Ch. 15 - Consider the following two acids: pKa1 = 2.98;...Ch. 15 - How many of the following are buffered solutions?...Ch. 15 - Which of the following can be classified as buffer...Ch. 15 - A certain buffer is made by dissolving NaHCO3 and...Ch. 15 - A buffer is prepared by dissolving HONH2 and...Ch. 15 - Calculate the pH of each of the following...Ch. 15 - Calculate the pH of each of the following...Ch. 15 - Compare the percent dissociation of the acid in...Ch. 15 - Compare the percent ionization of the base in...Ch. 15 - Calculate the pH after 0.020 mole of HCl is added...Ch. 15 - Calculate the pH after 0.020 mole of HCl is added...Ch. 15 - Calculate the pH after 0.020 mole of NaOH is added...Ch. 15 - Calculate the pH after 0.020 mole of NaOH is added...Ch. 15 - Which of the solutions in Exercise 21 shows the...Ch. 15 - Prob. 34ECh. 15 - Calculate the pH of a solution that is 1.00 M HNO2...Ch. 15 - Calculate the pH of a solution that is 0.60 M HF...Ch. 15 - Calculate the pH after 0.10 mole of NaOH is added...Ch. 15 - Calculate the pH after 0.10 mole of NaOH is added...Ch. 15 - Calculate the pH of each of the following buffered...Ch. 15 - Calculate the pH of each of the following buffered...Ch. 15 - Calculate the pH of a buffered solution prepared...Ch. 15 - A buffered solution is made by adding 50.0 g NH4Cl...Ch. 15 - Calculate the pH after 0.010 mole of gaseous HCl...Ch. 15 - Calculate the pH after 0.15 mole of solid NaOH is...Ch. 15 - Some K2SO3 and KHSO3 are dissolved in 250.0 mL of...Ch. 15 - An aqueous solution contains dissolved C6H5NH3Cl...Ch. 15 - Calculate the mass of sodium acetate that must be...Ch. 15 - What volumes of 0.50 M HNO2 and 0.50 M NaNO2 must...Ch. 15 - Consider a solution that contains both C5H5N and...Ch. 15 - Calculate the ratio [NH3]/[NH4+] in...Ch. 15 - Carbonate buffers are important in regulating the...Ch. 15 - When a person exercises, muscle contractions...Ch. 15 - Consider the acids in Table 13-2. Which acid would...Ch. 15 - Consider the bases in Table 13-3. Which base would...Ch. 15 - Calculate the pH of a solution that is 0.40 M...Ch. 15 - Calculate the pH of a solution that is 0.20 M HOCl...Ch. 15 - Which of the following mixtures would result in...Ch. 15 - Which of the following mixtures would result in a...Ch. 15 - What quantity (moles) of NaOH must be added to 1.0...Ch. 15 - Calculate the number of moles of HCl(g) that must...Ch. 15 - Consider the titration of a generic weak acid HA...Ch. 15 - Sketch the titration curve for the titration of a...Ch. 15 - Consider the titration of 40.0 mL of 0.200 M HClO4...Ch. 15 - Consider the titration of 80.0 mL of 0.100 M...Ch. 15 - Consider the titration of 100.0 mL of 0.200 M...Ch. 15 - Consider the titration of 100.0 mL of 0.100 M...Ch. 15 - Lactic acid is a common by-product of cellular...Ch. 15 - Repeat the procedure in Exercise 61, but for the...Ch. 15 - Repeat the procedure in Exercise 61, but for the...Ch. 15 - Repeat the procedure in Exercise 61, but for the...Ch. 15 - Calculate the pH at the halfway point and at the...Ch. 15 - In the titration of 50.0 mL of 1.0 M methylamine,...Ch. 15 - You have 75.0 mL of 0.10 M HA. After adding 30.0...Ch. 15 - A student dissolves 0.0100 mole of an unknown weak...Ch. 15 - Two drops of indicator HIn (Ka = 1.0 109), where...Ch. 15 - Methyl red has the following structure: It...Ch. 15 - Potassium hydrogen phthalate, known as KHP (molar...Ch. 15 - A certain indicator HIn has a pKa of 3.00 and a...Ch. 15 - Which of the indicators in Fig. 14-8 could be used...Ch. 15 - Prob. 80ECh. 15 - Which of the indicators in Fig. 14-8 could be used...Ch. 15 - Prob. 82ECh. 15 - Estimate the pH of a solution in which bromcresol...Ch. 15 - Estimate the pH of a solution in which crystal...Ch. 15 - A solution has a pH of 7.0. What would be the...Ch. 15 - A solution has a pH of 4.5. What would be the...Ch. 15 - When a diprotic acid, H2A. is titrated with NaOH,...Ch. 15 - Consider die titration of 50.0 mL of 0.10 M H3A...Ch. 15 - Derive an equation analogous to the...Ch. 15 - a. Calculate the pH of a buffered solution that is...Ch. 15 - Tris(hydroxymethyl)aminomethane, commonly called...Ch. 15 - You make 1.00 L of a buffered solution (pH = 4.00)...Ch. 15 - You have the following reagents on hand: Solids...Ch. 15 - Prob. 94AECh. 15 - Phosphate buffers are important in regulating the...Ch. 15 - When a diprotic acid, H2A, is titrated with NaOH,...Ch. 15 - Consider the blood buffer system discussed in the...Ch. 15 - What quantity (moles) of HCl(g) must be added to...Ch. 15 - Prob. 99AECh. 15 - The following plot shows the pH curves for the...Ch. 15 - Calculate the volume of 1.50 102 M NaOH that must...Ch. 15 - Prob. 102AECh. 15 - A certain acetic acid solution has pH = 2.68....Ch. 15 - A 0.210-g sample of an acid (molar mass = 192...Ch. 15 - The active ingredient in aspirin is...Ch. 15 - One method for determining the purity of aspirin...Ch. 15 - A student intends to titrate a solution of a weak...Ch. 15 - A student titrates an unknown weak acid, HA, to a...Ch. 15 - A sample of a certain monoprotic weak acid was...Ch. 15 - The pigment cyanidin aglycone is one of the...Ch. 15 - Consider 1.0 L of a solution that is 0.85 M HOC6H5...Ch. 15 - What concentration of NH4Cl is necessary to buffer...Ch. 15 - Consider the following acids and bases: HCO2H Ka =...Ch. 15 - Consider a buffered solution containing CH3NH3Cl...Ch. 15 - Consider the titration of 150.0 mL of 0.100 M HI...Ch. 15 - Consider the titration of 100.0 mL of 0.100 M HCN...Ch. 15 - Consider the titration of 100.0 mL of 0.200 M...Ch. 15 - Consider the following four titrations (iiv): i....Ch. 15 - Another way to treat data from a pH titration is...Ch. 15 - A buffer is made using 45.0 mL of 0.750 M HC3H5O2...Ch. 15 - A 0.400-M solution of ammonia was titrated with...Ch. 15 - What volume of 0.0100 M NaOH must be added to 1.00...Ch. 15 - Consider a solution formed by mixing 50.0 mL of...Ch. 15 - Cacodylic acid, (CH3)2AsO2H, is a toxic compound...Ch. 15 - The titration of Na2CO3 with HCl bas the following...Ch. 15 - Consider the titration curve in Exercise 115 for...Ch. 15 - A few drops of each of the indicators shown in the...Ch. 15 - Malonic acid (HO2CCH2CO2H) is a diprotic acid. In...Ch. 15 - A buffer solution is prepared by mixing 75.0 mL of...Ch. 15 - A 10.00-g sample of the ionic compound NaA, where...Ch. 15 - Calculate the pH of a solution prepared by mixing...Ch. 15 - Consider a solution prepared by mixing the...
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
  • Text book image
    Chemistry
    Chemistry
    ISBN:9781305957404
    Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Text book image
    Chemistry: An Atoms First Approach
    Chemistry
    ISBN:9781305079243
    Author:Steven S. Zumdahl, Susan A. Zumdahl
    Publisher:Cengage Learning
    Text book image
    Chemistry
    Chemistry
    ISBN:9781133611097
    Author:Steven S. Zumdahl
    Publisher:Cengage Learning
  • Text book image
    General Chemistry - Standalone book (MindTap Cour...
    Chemistry
    ISBN:9781305580343
    Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
    Publisher:Cengage Learning
    Text book image
    Chemistry & Chemical Reactivity
    Chemistry
    ISBN:9781337399074
    Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
    Publisher:Cengage Learning
    Text book image
    Chemistry: Principles and Practice
    Chemistry
    ISBN:9780534420123
    Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
    Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Acid-Base Titration | Acids, Bases & Alkalis | Chemistry | FuseSchool; Author: FuseSchool - Global Education;https://www.youtube.com/watch?v=yFqx6_Y6c2M;License: Standard YouTube License, CC-BY