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 43E

Consider a solution that contains both C5H5N and C5H5NHNO3. Calculate the ratio [C5H5N]/[C5H5NH+] if the solution has the following pH values:

a. pH = 4.50

b. pH = 5.00

c. pH = 5.23

d. pH = 5.50

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Different pH values of solution of C5H5N and C5H5NH+ is given. The ratio of [C5H5N]/[C5H5NH+] is to be calculated.

Concept introduction:

The pOH value is the measure of OH ions. The relation between pOH and pKb is given by Henderson-Hassel Bach equation. According to this equation,

pOH=pKb+log[Conjugateacid][Weakbase]

Answer to Problem 43E

The ratio of [C5H5N]/[C5H5NH+] for pH=4.50 is 0.18_ .

Explanation of Solution

Explanation

To determine pOH

The pOH value of a solution of C5H5N and C5H5NH+ is 9.5_ .

The pH value of solution is 4.50 .

The relation between pH and pOH is,

pH+pOH=14

Where,

  • pH is a measure of concentration of H+ .
  • pOH is a measure of concentration of OH .

Substitute the value of pOH in above equation.

pH+pOH=144.50+pOH=14pOH=9.5_

To determine the ratio of [C5H5NH+]/[C5H5N]

The value of Kb for C5H5N is 1.7×109 .

The pOH value of solution is 9.5 .

The pOH is calculated using the Henderson-Hassel Bach equation,

pOH=pKb+log[Conjugateacid][Weakbase]

Where,

  • pOH is the measure of OH ions.
  • pKb is the measure of basic strength.

The formula of pKb is,

pKb=logKb

Where,

  • Kb is base equilibrium constant.

Substitute the value of pKb in the above equation.

pOH=logKb+log[Conjugateacid][Weakbase]

Substitute all the given values in the above equation.

pOH=logKb+log[Conjugateacid][Weakbase]9.5=log(1.7×109)+log[C5H5NH+][C5H5N]9.5=8.769+log[C5H5NH+][C5H5N]

Simplify the above equation.

9.5=8.769+log[C5H5NH+][C5H5N]log[C5H5NH+][C5H5N]=0.731[C5H5NH+][C5H5N]=5.3826_

To find the ratio of [C5H5N]/[C5H5NH+]

The ratio of [C5H5N]/[C5H5NH+] is,

[C5H5NH+][C5H5N]=5.3826[C5H5N][C5H5NH+]=15.3826=0.18_

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Different pH values of solution of C5H5N and C5H5NH+ is given. The ratio of [C5H5N]/[C5H5NH+] is to be calculated.

Concept introduction:

The pOH value is the measure of OH ions. The relation between pOH and pKb is given by Henderson-Hassel Bach equation. According to this equation,

pOH=pKb+log[Conjugateacid][Weakbase]

Answer to Problem 43E

The ratio of [C5H5N]/[C5H5NH+] for pH=5.00 is 0.59_ .

Explanation of Solution

Explanation

To determine the pOH value

The pH value of solution is 5.00 .

The relation between pH and pOH is,

pH+pOH=14

Where,

  • pH is a measure of concentration of H+ .
  • pOH is a measure of concentration of OH .

Substitute the value of pOH in above equation.

pH+pOH=145.00+pOH=14pOH=9.0_

To determine the ratio of [C5H5NH+]/[C5H5N]

The value of Kb for C5H5N is 1.7×109 .

The pOH value of solution is 9.0 .

The pOH is calculated using the Henderson-Hassel Bach equation,

pOH=pKb+log[Conjugateacid][Weakbase]

Where,

  • pOH is the measure of OH ions.
  • pKb is the measure of basic strength.

The formula of pKb is,

pKb=logKb

Where,

  • Kb is base equilibrium constant.

Substitute the value of pKb in the above equation.

pOH=logKb+log[Conjugateacid][Weakbase]

Substitute all the given values in the above equation.

pOH=logKb+log[Conjugateacid][Weakbase]9.0=log(1.7×109)+log[C5H5NH+][C5H5N]9.0=8.769+log[C5H5NH+][C5H5N]

Simplify the above equation.

9.0=8.769+log[C5H5NH+][C5H5N]log[C5H5NH+][C5H5N]=0.231[C5H5NH+][C5H5N]=1.7021_

To find the ratio of [C5H5N]/[C5H5NH+]

The ratio of [C5H5N]/[C5H5NH+] is,

[C5H5NH+][C5H5N]=1.7021[C5H5N][C5H5NH+]=11.7021=0.59_

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Different pH values of solution of C5H5N and C5H5NH+ is given. The ratio of [C5H5N]/[C5H5NH+] is to be calculated.

Concept introduction:

The pOH value is the measure of OH ions. The relation between pOH and pKb is given by Henderson-Hassel Bach equation. According to this equation,

pOH=pKb+log[Conjugateacid][Weakbase]

Answer to Problem 43E

The ratio of [C5H5N]/[C5H5NH+] for pH=5.23 is 1_ .

Explanation of Solution

Explanation

To determine the pOH

The pOH value of a solution of C5H5N and C5H5NH+ is 8.77_ .

The pH value of solution is 5.23 .

The relation between pH and pOH is,

pH+pOH=14

Where,

  • pH is a measure of concentration of H+ .
  • pOH is a measure of concentration of OH .

Substitute the value of pOH in above equation.

pH+pOH=145.23+pOH=14pOH=8.77_

To determine the ratio of [C5H5NH+]/[C5H5N]

The value of Kb for C5H5N is 1.7×109 .

The pOH value of solution is 8.77 .

The pOH is calculated using the Henderson-Hassel Bach equation,

pOH=pKb+log[Conjugateacid][Weakbase]

Where,

  • pOH is the measure of OH ions.
  • pKb is the measure of basic strength.

The formula of pKb is,

pKb=logKb

Where,

  • Kb is base equilibrium constant.

Substitute the value of pKb in the above equation.

pOH=logKb+log[Conjugateacid][Weakbase]

Substitute all the given values in the above equation.

pOH=logKb+log[Conjugateacid][Weakbase]8.77=log(1.7×109)+log[C5H5NH+][C5H5N]8.77=8.77+log[C5H5NH+][C5H5N]

Simplify the above equation.

8.77=8.77+log[C5H5NH+][C5H5N]log[C5H5NH+][C5H5N]=0[C5H5NH+][C5H5N]=1_

To determine the ratio of [C5H5N]/[C5H5NH+]

The ratio of [C5H5N]/[C5H5NH+] is,

[C5H5NH+][C5H5N]=1[C5H5N][C5H5NH+]=11=1_

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Different pH values of solution of C5H5N and C5H5NH+ is given. The ratio of [C5H5N]/[C5H5NH+] is to be calculated.

Concept introduction:

The pOH value is the measure of OH ions. The relation between pOH and pKb is given by Henderson-Hassel Bach equation. According to this equation,

pOH=pKb+log[Conjugateacid][Weakbase]

Answer to Problem 43E

Answer:

The ratio of [C5H5N]/[C5H5NH+] for pH=5.50 is 1.86_ .

Explanation of Solution

Explanation

To determine the pOH

The pH value of solution is 5.50 .

The relation between pH and pOH is,

pH+pOH=14

Where,

  • pH is a measure of concentration of H+ .
  • pOH is a measure of concentration of OH .

Substitute the value of pOH in above equation.

pH+pOH=145.50+pOH=14pOH=8.50_

To determine the ratio of [C5H5NH+]/[C5H5N]

The value of Kb for C5H5N is 1.7×109 .

The pOH value of solution is 8.50 .

The pOH is calculated using the Henderson-Hassel Bach equation,

pOH=pKb+log[Conjugateacid][Weakbase]

Where,

  • pOH is the measure of OH ions.
  • pKb is the measure of basic strength.

The formula of pKb is,

pKb=logKb

Where,

  • Kb is base equilibrium constant.

Substitute the value of pKb in the above equation.

pOH=logKb+log[Conjugateacid][Weakbase]

Substitute all the given values in the above equation.

pOH=logKb+log[Conjugateacid][Weakbase]8.50=log(1.7×109)+log[C5H5NH+][C5H5N]8.50=8.769+log[C5H5NH+][C5H5N]

Simplify the above equation.

8.50=8.769+log[C5H5NH+][C5H5N]log[C5H5NH+][C5H5N]=0.269[C5H5NH+][C5H5N]=0.538_

To determine the ratio of [C5H5N]/[C5H5NH+]

The ratio of [C5H5N]/[C5H5NH+] is,

[C5H5NH+][C5H5N]=0.538[C5H5N][C5H5NH+]=10.538=1.86_

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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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