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The weak base ethanolamine. HOCH 2 CH 2 NH 2 , can be titrated with HCl. H O C H 2 C H 2 N H 2 ( a q ) + H 3 O + ( a q ) ⇄ H O C H 2 C H 2 N H 3 + ( a q ) + H 2 O ( l ) Assume you have 25.0 mL of a 0.010 M solution of ethanolamine and titrate it with 0.0095 M HCl. ( K b for ethanolamine is 3.2 × 10 −7 .) (a) What is the pH of the ethanolamine 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 would be the best choice to detect the equivalence point? (e) Calculate the pH of the solution after adding 5.00, 10.0, 20.0, and 30.0 mL of the acid. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.

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Chemistry & Chemical Reactivity

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

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 106IL
Textbook Problem
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The weak base ethanolamine. HOCH2CH2NH2, can be titrated with HCl.

H O C H 2 C H 2 N H 2 ( a q ) + H 3 O + ( a q ) H O C H 2 C H 2 N H 3 + ( a q ) + H 2 O ( l )

Assume you have 25.0 mL of a 0.010 M solution of ethanolamine and titrate it with 0.0095 M HCl. (Kb for ethanolamine is 3.2 × 10−7.)

  1. (a) What is the pH of the ethanolamine solution before the titration begins?
  2. (b) What is the pH at the equivalence point?
  3. (c) What is the pH at the halfway point of the titration?
  4. (d) Which indicator in Figure 17.11 would be the best choice to detect the equivalence point?
  5. (e) Calculate the pH of the solution after adding 5.00, 10.0, 20.0, and 30.0 mL of the acid.
  6. (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 has to be calculated at the various points during the titration between OHCH2CH2NH2 and HCl. The value of pH for the original solution of OHCH2CH2NH2 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 strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of OHCH2CH2NH2 with HCl is represented as,

OHCH2CH2NH2(aq)+ H3O+(aq)H2O(l)+OHCH2CH2NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

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

As the addition of HCl is done there will be formation of buffer solution OHCH2CH2NH2/OHCH2CH2NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

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

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OHCH2CH2NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

OHCH2CH2NH3+(aq)+H2O(l)H3O+(aq)+OHCH2CH2NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

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

Kw=(Ka)(Kb) (3)

Explanation of Solution

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

Kb=[OH](eq)[BH+](eq)[B](eq)

Given:

Refer to the Apendix I in the textbook for the value of Kb.

The value of Kb for HOCH2CH2NH2 is 3.2×105.

The value of Kw for water is 1.0×1014.

The pKb value is calculated as follows;

pKb=log(Kb)

Substitute, 3.2×105 for Kb.

pKb=log(3.2×105)=4.49

Therefore, pKb value of HOCH2CH2NH2 is 4.49.

The initial concentration of HOCH2CH2NH2 is 0.010 M.

The initial concentration of HCl is 0.0095 M.

Volume of the solvent is 25 mL.

Therefore volume of the solvent is 0.025 L.

ICE table (1) gives the dissociation of HOCH2CH2NH2.

EquationHOCH2CH2NH2(aq)+H2O(l)OH(aq)+HOCH2CH2NH3+(aq)Initial(molL1)0.01000Change(molL1)x+x+xAfterreaction(molL1)0.010x+x+x

From ICE table (1),

Concentration of HOCH2CH2NH2 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 strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of OHCH2CH2NH2 with HCl is represented as,

OHCH2CH2NH2(aq)+ H3O+(aq)H2O(l)+OHCH2CH2NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

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

As the addition of HCl is done there will be formation of buffer solution OHCH2CH2NH2/OHCH2CH2NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

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

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OHCH2CH2NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

OHCH2CH2NH3+(aq)+H2O(l)H3O+(aq)+OHCH2CH2NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

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 midpoint 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 strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of OHCH2CH2NH2 with HCl is represented as,

OHCH2CH2NH2(aq)+ H3O+(aq)H2O(l)+OHCH2CH2NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

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

As the addition of HCl is done there will be formation of buffer solution OHCH2CH2NH2/OHCH2CH2NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

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

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OHCH2CH2NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

OHCH2CH2NH3+(aq)+H2O(l)H3O+(aq)+OHCH2CH2NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

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.

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 strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of OHCH2CH2NH2 with HCl is represented as,

OHCH2CH2NH2(aq)+ H3O+(aq)H2O(l)+OHCH2CH2NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

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

As the addition of HCl is done there will be formation of buffer solution OHCH2CH2NH2/OHCH2CH2NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

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

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OHCH2CH2NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

OHCH2CH2NH3+(aq)+H2O(l)H3O+(aq)+OHCH2CH2NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

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 5 mL, 10 mL, 20 mL and 30 mL HCl is added to the  OHCH2CH2NH2 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 strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of OHCH2CH2NH2 with HCl is represented as,

OHCH2CH2NH2(aq)+ H3O+(aq)H2O(l)+OHCH2CH2NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

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

As the addition of HCl is done there will be formation of buffer solution OHCH2CH2NH2/OHCH2CH2NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

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

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OHCH2CH2NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

OHCH2CH2NH3+(aq)+H2O(l)H3O+(aq)+OHCH2CH2NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

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.

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 strong acid-weak base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of OHCH2CH2NH2 with HCl is represented as,

OHCH2CH2NH2(aq)+ H3O+(aq)H2O(l)+OHCH2CH2NH3+(aq)

Calculation of pH at various points is done as follows,

(1) The pOH value before the titration can be calculated by using Kb and its relation with OH ion concentration.

Kb=[OH](eq)[BH+](eq)[B](eq) (1)

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

As the addition of HCl is done there will be formation of buffer solution OHCH2CH2NH2/OHCH2CH2NH3+. The pOH calculation for buffer solution is done by using Henderson-Hasselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

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

pOH=pKb+log[conjugate acid][base]

Substitute, [conjugateacid]for[base].

pOH=pKb+log[base][base]=pKb+log(1)=pKb+0=pKb

Therefore, pOH value at midpoint is equal to pKb.

(3) The pOH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OHCH2CH2NH3+. The H3O+ will be produced due to the hydrolysis of conjugate acid at equivalence point. The hydrolysis equilibrium is represented as,

OHCH2CH2NH3+(aq)+H2O(l)H3O+(aq)+OHCH2CH2NH2(aq)

By using the value of Ka for the ethylamine, concentration of  H3O+ can be calculated.

Value of pH at every point is calculated by using the relation between pH and pOH.

pH=14pOH

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

Kw=(Ka)(Kb) (3)

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Chapter 17 Solutions

Chemistry & Chemical Reactivity
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