# Aniline hydrochloride, (C 6 H 5 NH 3 )Cl, is a weak acid. (Its conjugate base is the weak base aniline, C 6 H 5 NH 2 .) The acid can be titrated with a strong base such as NaOH. C 6 H 5 N H 3 + ( a q ) + O H − ( a q ) ⇄ C 6 H 5 N H 2 ( a q ) + H 2 O ( l ) Assume 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.185 M NaOH. ( K a for aniline hydrochloride is 2.4 × 10 −5 .) (a) What is the pH of the (C 6 H 5 NH 3 ) solution before the titration begins? (b) What is the pH at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) Which indicator in Figure 17.11 could be used to detect the equivalence point? (e) Calculate the pH of the solution after adding 10.0, 20.0, and 30.0 mL of base. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

Chapter
Section

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 105IL
Textbook Problem
1733 views

## Aniline hydrochloride, (C6H5NH3)Cl, is a weak acid. (Its conjugate base is the weak base aniline, C6H5NH2.) The acid can be titrated with a strong base such as NaOH. C 6 H 5 N H 3 + ( a q ) + O H − ( a q ) ⇄ C 6 H 5 N H 2 ( a q ) + H 2 O ( l ) Assume 50.0 mL of 0.100 M aniline hydrochloride is titrated with 0.185 M NaOH. (Ka for aniline hydrochloride is 2.4 × 10−5.) (a) What is the pH of the (C6H5NH3) solution before the titration begins? (b) What is the pH at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) Which indicator in Figure 17.11 could be used to detect the equivalence point? (e) Calculate the pH of the solution after adding 10.0, 20.0, and 30.0 mL of base. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.

(a)

Interpretation Introduction

Interpretation:

The value of pH for the original solution of (C6H5NH3)Cl has to be calculated before titration begins.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of (C6H5NH3)Cl with NaOH is represented as,

C6H5NH3+(aq)+ OH(aq)H2O(l)+C6H5NH2(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution C6H5NH3+/C6H5NH2. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugatebase][acid] (2)

At the midpoint of the titration, when concentration of acid and its conjugate base is equal. Therefore pH value at midpoint will be given as

pH=pKa+log[conjugatebase][acid]

Substitute, [conjugatebase]for[acid].

pH=pKa+log[conjugatebase][conjugatebase]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C6H5NH2. The OH will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

C6H5NH2(aq)+H2O(l)OH(aq)+C6H5NH3+(aq)

By using the value of Kb for the aniline, concentration of  OH can be calculated. Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

### Explanation of Solution

The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq)

Given:

Refer to the Apendix H in the textbook for the value of Ka.

The value of Ka for aniline hydrochloride is 2.4×105.

The value of Kw for water is 1.0×1014.

The pKa value is calculated as follows;

pKa=log(Ka)

Substitute, 2.4×105 for Ka.

pKa=log(2.4×105)=4.62

Therefore, pKa value of (C6H5NH3)Cl  is 4.62.

The initial concentration of (C6H5NH3)Cl is 0.100 M.

The initial concentration of NaOH is 0.185 M.

Volume of the solvent is 50 mL.

Therefore volume of the solvent is 0.050 L.

ICE table (1) gives the dissociation of (C6H5NH3)Cl.

EquationC6H5NH3+(aq)+H2O(l)H3O+(aq)+C6H5NH2(aq)Initial(molL)0.10000Change(molL1)x+x+xAfterreaction(molL1)0.100x+x+x

From ICE table (1),

Concentration of C6H5NH3+ left after reaction is (0

(b)

Interpretation Introduction

Interpretation:

The value of pH at equivalence point has to be calculated

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of (C6H5NH3)Cl with NaOH is represented as,

C6H5NH3+(aq)+ OH(aq)H2O(l)+C6H5NH2(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution C6H5NH3+/C6H5NH2. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugatebase][acid] (2)

At the midpoint of the titration, concentration of acid and its conjugate base is equal. Therefore pH value at midpoint will be given as

pH=pKa+log[conjugatebase][acid]

Substitute, [conjugatebase]for[acid].

pH=pKa+log[conjugatebase][conjugatebase]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C6H5NH2. The OH will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

C6H5NH2(aq)+H2O(l)OH(aq)+C6H5NH3+(aq)

By using the value of Kb for the aniline, concentration of  OH can be calculated. Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

(c)

Interpretation Introduction

Interpretation:

The value of pH at halfway point of the titration has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of (C6H5NH3)Cl with NaOH is represented as,

C6H5NH3+(aq)+ OH(aq)H2O(l)+C6H5NH2(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution C6H5NH3+/C6H5NH2. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugatebase][acid] (2)

At the midpoint of the titration, when concentration of acid and its conjugate base is equal. Therefore pH value at midpoint will be given as

pH=pKa+log[conjugatebase][acid]

Substitute, [conjugatebase]for[acid].

pH=pKa+log[conjugatebase][conjugatebase]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C6H5NH2. The OH will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

C6H5NH2(aq)+H2O(l)OH(aq)+C6H5NH3+(aq)

By using the value of Kb for the aniline, concentration of  OH can be calculated. Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

(d)

Interpretation Introduction

Interpretation:

A best indicator that can be used to detect the equivalence point has to be chosen for the titration.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of (C6H5NH3)Cl with NaOH is represented as,

C6H5NH3+(aq)+ OH(aq)H2O(l)+C6H5NH2(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution C6H5NH3+/C6H5NH2. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugatebase][acid] (2)

At the midpoint of the titration, concentration of acid and its conjugate base is equal. Therefore pH value at midpoint will be given as

pH=pKa+log[conjugatebase][acid]

Substitute, [conjugatebase]for[acid].

pH=pKa+log[conjugatebase][conjugatebase]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C6H5NH2. The OH will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

C6H5NH2(aq)+H2O(l)OH(aq)+C6H5NH3+(aq)

By using the value of Kb for the aniline, concentration of  OH can be calculated. Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

(e)

Interpretation Introduction

Interpretation:

The value of pH when 10 mL, 20 mL and 30 mL NaOH is added to the  (C6H5NH3)Cl solution has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of (C6H5NH3)Cl with NaOH is represented as,

C6H5NH3+(aq)+ OH(aq)H2O(l)+C6H5NH2(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution C6H5NH3+/C6H5NH2. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugatebase][acid] (2)

At midpoint of the titration, concentration of acid and its conjugate base is equal. Therefore pH value at midpoint will be given as

pH=pKa+log[conjugatebase][acid]

Substitute, [conjugatebase]for[acid].

pH=pKa+log[conjugatebase][conjugatebase]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C6H5NH2. The OH will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

C6H5NH2(aq)+H2O(l)OH(aq)+C6H5NH3+(aq)

By using the value of Kb for the aniline, concentration of  OH can be calculated. Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

(f)

Interpretation Introduction

Interpretation:

Titration curve has to be plotted by using pH values calculated in part (a),(b),(c) and

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of (C6H5NH3)Cl with NaOH is represented as,

C6H5NH3+(aq)+ OH(aq)H2O(l)+C6H5NH2(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution C6H5NH3+/C6H5NH2. The pH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pH=pKa+log[conjugatebase][acid] (2)

At the midpoint of the titration, concentration of acid and its conjugate base is equal. Therefore pH value at midpoint will be given as

pH=pKa+log[conjugatebase][acid]

Substitute, [conjugatebase]for[acid].

pH=pKa+log[conjugatebase][conjugatebase]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only C6H5NH2. The OH will be produced due to the hydrolysis of aniline at equivalence point. The hydrolysis equilibrium is represented as,

C6H5NH2(aq)+H2O(l)OH(aq)+C6H5NH3+(aq)

By using the value of Kb for the aniline, concentration of  OH can be calculated. Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
The theory that vitamin C prevents or cures colds is well supported by research. T F

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

What is the difference between a supernova explosion and a nova explosion?

Horizons: Exploring the Universe (MindTap Course List)

Does the cell cycle refer to mitosis as well as meiosis?

Human Heredity: Principles and Issues (MindTap Course List)

What forces cause (a) an automobile, (b) a propeller-driven airplane, and (c) a rowboat to move?

Physics for Scientists and Engineers, Technology Update (No access codes included)

Why is water a polar molecule? What properties of water derive from its polar nature?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin