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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Chapter 14, Problem 60E

(a)

Interpretation Introduction

Interpretation: The titration of H2NNH2 with different volumes of HNO3 is given. The pH of each solution is to be calculated.

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 value of pH of solution when 0.0mL HNO3 has been added to it.

(a)

Expert Solution
Check Mark

Answer to Problem 60E

The value of pH of solution when 0.0mL HNO3 has been added is. 10.74_ .

Explanation of Solution

Explanation

The value of pH of solution when 0.0mL HNO3 has been added is. 10.74_ .

Given:

The concentration of H2NNH2 is 0.100M

The concentration of HNO3 is 0.200M .

The volume of H2NNH2 is 100.0mL .

The volume of HNO3 is 0.0mL .

The value of Kb of H2NNH2 is 3.0×106 .

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,

H2NNH2+H2OH2NNH3++OH

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

H2NNH2+H2OH2NNH3++OHInitial moles:0.10000Change:x+xxFinalmoles:0.100xxx

The equilibrium ratio for the given reaction is,

Kb=[H2NNH3+][OH][H2NNH2]

Substitute the calculated concentration values in the above expression.

Kb=[H2NNH3+][OH][H2NNH2]3.0×106=(x)(x)(0.100x)M

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

Simplify the above equation,

3.0×106=(x)(x)(0.100)Mx=5.4×104M

It is the concentration of OH .

The pOH of the solution is calculated by the formula,

pOH=log[OH]

Where,

  • [OH] is the concentration of Hydroxide ions.

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

pOH=log[OH]=log(5.4×104)=3.26

The relationship between pOH is given as,

pH+pOH=14

Substitute the value of pOH in the above equation as,

pH+pOH=14pH+3.26=14pH=10.74_

The value of pH of solution when 0.0mL HNO3 has been added is. 10.74_ .

(b)

Interpretation Introduction

Interpretation: The titration of H2NNH2 with different volumes of HNO3 is given. The pH of each solution is to be calculated.

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 value of pH of solution when 20.0mL HNO3 has been added to it.

(b)

Expert Solution
Check Mark

Answer to Problem 60E

The value of pH of solution when 20.0mL HNO3 has been added is 8.7_ .

Explanation of Solution

Explanation

The value of pH of solution when 20.0mL HNO3 has been added is 8.7_ .

Given

The volume of HNO3 is 20.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 20mL into L is done as,

20mL=20×0.001L=0.02L

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 value of concentration and volume of HNO3 in equation (4) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.200M×0.02L=0.004moles

Make the ICE table for the reaction between H2NNH2 and HNO3 .

H2NNH2+HNO3H2NNH3++NO3Initial moles:0.010.0040Change:0.0040.004+0.004Finalmoles:0.00600.004

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

Total volume of solution =VolumeofH2NNH2+VolumeofHNO3=0.1L+0.052L=0.12L

Substitute the value of number of moles of H2NNH2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.006moles0.12L=0.05M

Substitute the value of number of moles of H2NNH3+ and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.004moles0.12L=0.033M

Make the ICE table for the dissociation reaction of H2NNH2 .

H2NNH2+H2OH2NNH3++OHInitial moles:0.050.033Change:x+xxFinalmoles:0.05x0.033+xx

The equilibrium ratio for the given reaction is,

Kb=[H2NNH3+][OH][H2NNH2]

Substitute the calculated concentration values in the above expression.

Kb=[H2NNH3+][OH][H2NNH2]3.0×106=(0.033+x)(x)(0.05x)M

Since, value of Kb is very small, hence, (0.05x) is taken as (0.05) and (0.033+x) is taken as 0.033 .

Simplify the above equation,

3.0×106=(0.033)(x)(0.05)Mx=4.5×106M

It is the concentration of OH .

The pOH of the solution is calculated by the formula,

pOH=log[OH]

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

pOH=log[OH]=log(4.5×106)=5.35

The relationship between pOH is given as,

pH+pOH=14

Substitute the value of pOH in the above equation as,

pH+pOH=14pH+5.35=14pH=8.7_

The value of pH of solution when 20.0mL HNO3 has been added is. 8.7_ .

(c)

Interpretation Introduction

Interpretation: The titration of H2NNH2 with different volumes of HNO3 is given. The pH of each solution is to be calculated.

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 value of pH of solution when 25.0mL HNO3 has been added to it.

(c)

Expert Solution
Check Mark

Answer to Problem 60E

The value of pH of solution when 25.0mL HNO3 has been added is. 8.48_ .

Explanation of Solution

Explanation

The value of pH of solution when 25.0mL HNO3 has been added is. 8.48_ .

Given

The volume of HNO3 is 25.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 25mL into L is done as,

25mL=25×0.001L=0.025L

Substitute the value of concentration and volume of HNO3 in equation (4) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.200M×0.025L=0.005moles

Make the ICE table for the reaction between H2NNH2 and HNO3 .

H2NNH2+HNO3H2NNH3++NO3Initial moles:0.010.0050Change:0.0050.005+0.005Finalmoles:0.00500.005

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

Total volume of solution =VolumeofH2NNH2+VolumeofHNO3=0.1L+0.025L=0.125L

Substitute the value of number of moles of H2NNH2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.005moles0.125L=0.04M

Substitute the value of number of moles of H2NNH3+ and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.005moles0.125L=0.04M

Make the ICE table for the dissociation reaction of H2NNH2 .

H2NNH2+H2OH2NNH3++OHInitial moles:0.040.040Change:x+xxFinalmoles:0.04x0.04+xx

The equilibrium ratio for the given reaction is,

Kb=[H2NNH3+][OH][H2NNH2]

Substitute the calculated concentration values in the above expression.

Kb=[H2NNH3+][OH][H2NNH2]3.0×106=(0.04+x)(x)(0.04x)M

Since, value of Kb is very small, hence, (0.04x) is taken as (0.04) and (0.04+x) is taken as 0.04 .

Simplify the above equation,

3.0×106=(0.04)(x)(0.04)Mx=3.0×106M

It is the concentration of OH .

The pOH of the solution is calculated by the formula,

pOH=log[OH]

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

pOH=log[OH]=log(3.0×106)=5.52

Substitute the value of pOH in the above equation as,

pH+pOH=14pH+5.52=14pH=8.48_

The value of pH of solution when 25.0mL HNO3 has been added is. 8.48_ .

(d)

Interpretation Introduction

Interpretation: The titration of H2NNH2 with different volumes of HNO3 is given. The pH of each solution is to be calculated.

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 value of pH of solution when 40.0mL HNO3 has been added to it.

(d)

Expert Solution
Check Mark

Answer to Problem 60E

The value of pH of solution when 40.0mL HNO3 has been added is. 7.87_ .

Explanation of Solution

Explanation

The value of pH of solution when 40.0mL HNO3 has been added is. 7.87_ .

Given

The volume of HNO3 is 40.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 40mL into L is done as,

40mL=40×0.001L=0.04L

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

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.200M×0.04L=0.008moles

Make the ICE table for the reaction between H2NNH2 and HNO3 .

H2NNH2+HNO3H2NNH3++NO3Initial moles:0.010.0080Change:0.0080.008+0.008Finalmoles:0.0020.0020.008

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

Total volume of solution =VolumeofH2NNH2+VolumeofHNO3=0.1L+0.04L=0.14L

Substitute the value of number of moles of H2NNH2 and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.002moles0.14L=0.014M

Substitute the value of number of moles of H2NNH3+ and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.008moles0.14L=0.057M

Make the ICE table for the dissociation reaction of H2NNH2 .

H2NNH2+H2OH2NNH3++OHInitial moles:0.0140.0571Change:x+xxFinalmoles:0.014x0.0571+xx

The equilibrium ratio for the given reaction is,

Kb=[H2NNH3+][OH][H2NNH2]

Substitute the calculated concentration values in the above expression.

Kb=[H2NNH3+][OH][H2NNH2]3.0×106=(0.0571+x)(x)(0.014x)M

Since, value of Kb is very small, hence, (0.014x) is taken as (0.014) and (0.0571+x) is taken as 0.0571 .

Simplify the above equation,

3.0×106=(0.0571)(x)(0.014)Mx=0.74×106M

It is the concentration of OH .

The pOH of the solution is calculated by the formula,

pOH=log[OH]

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

pOH=log[OH]=log(0.74×106)=6.13

The relationship between pOH is given as,

pH+pOH=14

Substitute the value of pOH in the above equation as,

pH+pOH=14pH+6.13=14pH=7.87_

The value of pH of solution when 40.0mL HNO3 has been added is. 7.87_ .

(e)

Interpretation Introduction

Interpretation: The titration of H2NNH2 with different volumes of HNO3 is given. The pH of each solution is to be calculated.

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 value of pH of solution when 50.0mL HNO3 has been added to it.

(e)

Expert Solution
Check Mark

Answer to Problem 60E

The value of pH of solution when 50.0mL HNO3 has been added is. 4.85_ .

Explanation of Solution

Explanation

The value of pH of solution when 50.0mL HNO3 has been added is. 4.85_ .

Given

The volume of HNO3 is 50.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 50mL into L is done as,

50mL=50×0.001L=0.05L

Substitute the value of concentration and volume of HNO3 in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.200M×0.05L=0.01moles

Make the ICE table for the reaction between H2NNH2 and HNO3 .

H2NNH2+HNO3H2NNH3++NO3Initial moles:0.010.010Change:0.010.01+0.01Finalmoles:000.01

Total volume of solution =VolumeofH2NNH2+VolumeofHNO3=0.1L+0.05L=0.15L

Substitute the value of number of moles of H2NNH3+ and final volume of solution in equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.01moles0.15L=0.067M

Make the ICE table for the dissociation reaction of H2NNH3+ .

H2NNH3+H2NNH2H+Initial(M):0.06700Change(M):xxxEquilibrium(M):0.067xxx

The conjugate acid-base pair are related to each other through a formula as shown below.

Kw=Ka×Kb

Where,

  • Ka is the dissociation constant of an acid.
  • Kb is the dissociation constant of a base.
  • Kw is the ionic product of water (1014) .

Rearrange the equation to obtain the value of Ka as,

Ka=KwKb

Substitute the value of Kw and Kb in the above equation as,

Ka=KwKb=10143×106=3.3×109

The equilibrium ratio for the given reaction is,

Ka=[H2NNH2][H+][H2NNH3+]

Substitute the calculated concentration values in the above expression.

Ka=[H2NNH2][H+][H2NNH3+]3.3×109=(x)(x)(0.067x)M

Since, value of Ka is very small, hence, (0.067x) is taken as 0.067 .

Simplify the above equation,

3.3×109=(x)(x)(0.067)Mx=1.41×105M

The pH of the solution is calculated by the formula,

pH=log[H+]

Where,

  • [H+] is the concentration of Hydrogen ions.

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

pH=log[H+]=log(1.41×105)=4.85_

The value of pH of solution when 50.0mL HNO3 has been added is. 4.85_ .

(f)

Interpretation Introduction

Interpretation: The titration of H2NNH2 with different volumes of HNO3 is given. The pH of each solution is to be calculated.

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 value of pH of solution when 100.0mL HNO3 has been added to it.

(f)

Expert Solution
Check Mark

Answer to Problem 60E

The value of pH of solution when 100.0mL HNO3 has been added is 1.3_ .

Explanation of Solution

Explanation

The value of pH of solution when 100.0mL HNO3 has been added is 1.3_ .

Given

The volume of HNO3 is 100.0mL .

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 100.0mL into L is done as,

100mL=100×0.001L=0.1L

Substitute the value of concentration and volume of HNO3 in equation (2) as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.200M×0.1L=0.02moles

Make the ICE table for the reaction between H2NNH2 and HNO3 .

H2NNH2+HNO3H2NNH3++NO3Initial moles:0.010.020Change:0.010.01+0.01Finalmoles:00.010.01

There is excess of Hydrogen ions in the solution

Total volume of solution =VolumeofH2NNH2+VolumeofHNO3=0.1L+0.1L=0.2L

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

Concentration=NumberofmolesVolumeofsolutioninlitres=0.01moles0.2L=0.05M

The pH of the solution is calculated by the formula,

pH=log[H+]

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

pH=log[H+]=log(0.05)=1.3_

The value of pH of solution when 100.0mL HNO3 has been added is 1.3_ .

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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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