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 65E

 (a)

Interpretation Introduction

Interpretation: The halfway point and the equivalence point pH for each of the given titrations are to be calculated.

Concept introduction: The pH value is defined as a figure that is used to express the acidity or basicity of a solution by using the logarithmic scale on which 7 is a neutral. The values lower than 7 shows acidic or greater than 7 shows basic behavior of the solution.

To determine: The halfway point and the equivalence point pH for the given titration.

 (a)

Expert Solution
Check Mark

Answer to Problem 65E

Answer

The pH at the half-way equivalence point is 4.20_ .

The pH at equivalence point is 8.45_ .

Explanation of Solution

Explanation

Given

Concentration of HC7H5O2 is 0.10M .

Volume of the solution is 100mL .

Concentration of NaOH is 0.10M

According to the half-way titration, the volume of the base consumed is half of the equivalence point. Volume of the base is calculated by using the expression,

M1V1=M2V2

Where,

  • M1 is the concentration of HC7H5O2 .
  • M2 is the concentration of NaOH .
  • V1 is the volume of HC7H5O2 solution.
  • V2 is the volume of NaOH solution.

Substitute the values of M1 , V1 , and M2 in the above expression to calculate the volume of NaOH solution.

M1V1=M2V20.10M×100mL=0.10M×V2V2=100mL

The volume of the base at the equivalence point is V2=100mL .

The volume of the base at the half-way point is 50mL .

The reaction between C6H5COOH and NaOH at the half-way equivalence point is,

C6H5COOH+NaOHC6H5COONa+H2O0.01mole0.005mole0.005mole

The pH at half-way point is calculated by using the formula,

pH=pKa+log[Salt][Acid]

The pKa of benzoic acid is 4.20 .

Substitute the values of pKa and concentration of salt and acid in the above formula to calculate the pH at half-way point.

pH=4.20+log[0.005][0.005]=4.20_

Hence, the pH at the half-way equivalence point is 4.20_ .

At equivalence point, the reaction between C6H5COOH and NaOH is,

C6H5COOH+NaOHC6H5COONa+H2O0.01mole0.01mole0.01mole

Now, first Kb is calculated by using the formula,

Kb=KwKa

Where,

  • Kb is the dissociation constant of base.
  • Ka is the dissociation constant of an acid.
  • Kw is the self-ionization constant of water.

The value of Ka=6.4×105

The value of Kw=1.0×1014

Substitute the value of Ka and Kw in the above formula.

Kb=1.0×10146.4×105=1.56×1010

The molar concentration of [OH] is calculated by using the formula,

[OH]=Kb×C

Where, the concentration of (OH) C is 0.01mol(100+100)×103L=0.05

Substitute the values of Kb and C in the above formula to calculate [OH] .

[OH]=Kb×C=1.56×1010×0.05=2.79×106

The pH is calculated by using the formula,

pH=log[OH]

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

pH=log[2.79×106]=5.55

The pH at equivalence point is 145.55=8.45_ .

Hence, the pH at equivalence point is 8.45_ .

(b)

Interpretation Introduction

Interpretation: The halfway point and the equivalence point pH for each of the given titrations are to be calculated.

Concept introduction: The pH value is defined as a figure that is used to express the acidity or basicity of a solution by using the logarithmic scale on which 7 is a neutral. The values lower than 7 shows acidic or greater than 7 shows basic behavior of the solution.

To determine: The halfway point and the equivalence point pH for the given titration.

(b)

Expert Solution
Check Mark

Answer to Problem 65E

Answer

The pH at the half-way equivalence point is 10.748_ .

The pH at equivalence point is 5.96_ .

Explanation of Solution

Explanation

Given

Concentration of C2H5NH2 is 0.10M .

Volume of the solution is 100mL .

Concentration of HNO3 is 0.20M .

According to the half-way titration, the volume of the base consumed is half of the equivalence point. Volume of the base is calculated by using the expression,

M1V1=M2V2

Where,

  • M1 is the concentration of C2H5NH2 .
  • M2 is the concentration of HNO3 .
  • V1 is the volume of C2H5NH2 solution.
  • V2 is the volume of HNO3 solution.

Substitute the values of M1 , V1 , and M2 in the above expression to calculate the volume of HNO3 solution.

M1V1=M2V20.10M×100mL=0.20M×V2V2=50mL

The volume of the acid at the equivalence point is V2=50mL .

The volume of the acid at the half-way point is 25mL .

The reaction between C2H5NH2 and HNO3 at the half-way equivalence point is,

C2H5NH2+HNO3C2H5NH3++NO30.01mole0.005mole0.005mole

The pH at half-way point is calculated by using the formula,

pH=pKa+log[Salt][Acid]

The pKa of C2H5NH2 is 10.748 .

Substitute the values of pKa and concentration of salt and acid in the above formula to calculate the pH at half-way point.

pH=10.748+log[0.005][0.005]=10.748_

Hence, the pH at the half-way equivalence point is 10.748_ .

At equivalence point, the reaction between C2H5NH2 and HNO3 is,

C2H5NH2+HNO3C2H5NH3++NO30.01mole0.01mole0.01mole

Now, first Ka is calculated by using the formula,

Ka=KwKb

Where,

  • Kb is the dissociation constant of base.
  • Ka is the dissociation constant of an acid.
  • Kw is the self-ionization constant of water.

The value of Kb=5.6×104

The value of Kw=1.0×1014

Substitute the value of Kb and Kw in the above formula.

Ka=1.0×10145.6×104=1.78×1011

The molar concentration of [H+] is calculated by using the formula,

[H+]=Ka×C

Where, the concentration of (H+) C is 0.01mol(100+50)×103L=0.066

Substitute the values of Ka and C in the above formula to calculate [H+] .

[H+]=Ka×C=1.78×1011×0.066=1.091×106

The pH is calculated by using the formula,

pH=log[H+]

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

pH=log[1.091×106]=5.96_

Hence, the pH at equivalence point is 5.96_ .

(c)

Interpretation Introduction

Interpretation: The halfway point and the equivalence point pH for each of the given titrations are to be calculated.

Concept introduction: The pH value is defined as a figure that is used to express the acidity or basicity of a solution by using the logarithmic scale on which 7 is a neutral. The values lower than 7 shows acidic or greater than 7 shows basic behavior of the solution.

To determine: The halfway point and the equivalence point pH for the given titration.

(c)

Expert Solution
Check Mark

Answer to Problem 65E

Answer

The pH at the half-way equivalence point is 0.90_ .

The pH at equivalence point is 7_

Explanation of Solution

Explanation

Given

Concentration of NaOH is 0.25M .

Volume of the HCl solution is 0.1L .

Concentration of HCl is 0.50M .

According to the half-way titration, the volume of the base consumed is half of the equivalence point. Volume of the base is calculated by using the expression,

M1V1=M2V2

Where,

  • M1 is the concentration of HCl .
  • M2 is the concentration of NaOH .
  • V1 is the volume of HCl solution.
  • V2 is the volume of NaOH solution.

Substitute the values of M1 , V1 , and M2 in the above expression to calculate the volume of NaOH solution.

M1V1=M2V20.50M×0.1L=0.25M×V2V2=0.2L

The volume of the base at the equivalence point is V2=0.2L .

The volume of the base at the half-way point is 0.1L .

The concentration of [H+] is calculated by using the expression,

[H+]=M1V1M2V2V1+V2

Substitute the values of M1 , V1 , M2 and V2 in the above formula to calculate [H+] at half-way point.

[H+]=0.50M×0.1L0.25M×0.1L0.1L+0.1L=0.125M

The pH is calculated by using the formula,

pH=log[H+]

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

pH=log[0.125]=0.90_

Hence, the pH at the half-way equivalence point is 0.90_ .

The reaction is between strong acid and strong base. So, complete neutralization takes place.

H++OHH2O

For the complete neutralization, the concentration of acid [H+] is equal to the concentration of the base [OH] .

[H+]=[OH]

Also, the product of [H+] and [OH] is,

[H+]×[OH]=1.0×10142[H+]=1.0×1014[H+]=1.0×107

The pH is calculated by using the formula,

pH=log[H+]

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

pH=log[1.0×107]=7_

Hence, the pH at equivalence point is 7_ .

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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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