Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 16, Problem 92GQ

The butylammonium ion, C4H9NH3+, has a Ka of 2.3 × 10−11.

C4H9NH3+(aq) + H2O() ⇄ H3O+(aq) + C4H9NH2(aq)

  1. a) Calculate Kb for the conjugate base, C4H9NH2 (butyl amine).
  2. b) Place the butylammonium ion and its conjugate base in Table 16.2. Name an acid weaker than C4H9NH3+ and a base stronger than C4H9NH3.
  3. c) What is the pH of a 0.015M solution of butylammonium chloride?

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Kb has to be calculated for the conjugate base (C4H9NH2, butyl amine)

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ion product constant for wter  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation between pH and pOH

pH + pOH =14

Answer to Problem 92GQ

Kb of conjugate base is 4.34 × 10-4

Explanation of Solution

According to the Brown stead –Lowry theory the compound which is accepts a proton to be considered as a base.

A weak base accepts a proton from water molecules producing hydroxide ions in water.

The Ka of 2.3× 10-11

Ka×Kb = KwKbKwKaKw= 1.0×10-14Ka = 2.3×10-11

Substitute the values,

Kb1.0×10-142.3×10-11     = 4.34×10-4

Therefore, Kb of conjugate base is 4.34 × 10-4

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Compound should be Named for an acid weaker than C4H9NH3+ and a base stronger than C4H9NH2.

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of  the strength of the acid and bases in the water .

Ka an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ion product constant for wter  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation between pH and pOH

pH + pOH =14

Answer to Problem 92GQ

An acid weaker than C4H9NH3+ is HPO42-

A base stronger than C4H9NH2 is PO43-

Explanation of Solution

From the table,

Ka of C4H9NH3+ is 2.3×10-11

Kb of C4H9NH2 is 4.4×10-4

The butyl ammonium ion has Ka value is in between HPO42- (Ka=3.6×10-13) and [Ni(H2O)6]2+ (Ka=3.6×10-13)

The order of Ka is as follows.

HPO42- < C4H9NH3+ < [Ni(H2O)6]2+

Butyl amine has a Kb value in between [Ni(H2O)5OH]+ (Kb=4.0×10-4) and PO43- (Kb=2.8×10-2).

The Kb order is as follows.

[Ni(H2O)5OH]+< C4H9NH2< PO43-

Therefore, An acid weaker than C4H9NH3+ is HPO42- . A base stronger than C4H9NH2 is PO43-

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The pH has be calculated for 0.015M solution of butylammonium chloride?

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of  the strength of the acid and bases in the water .

Ka is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ion product constant for wter  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation between pH and pOH

pH + pOH =14

Answer to Problem 92GQ

The pH of the 0.015M solution of butylammonium chloride is 6.2.

Explanation of Solution

Butylammonium chloride dissociates into butyl ammonium ion and chloride ion. Chloride ion does not affect the solution pH. Butylammonium donate a proton to water to form hydronium ions.

The equilibrium chemical reaction of butylammonium chloride is as follows.

C4H9NH3+(aq) + H2O(l) H3O+(aq) + C4H9NH2(aq)

The equilibrium expression:

Ka[C4H9NH2][H3O+][C4H9NH3+]

The initial concentration of ammonium chloride is 0.015M.

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

      C4H9NH3+(aq) + H2O(l) H3O+(aq) + C4H9NH2(aq)I         0.015M              --                --                     --C           -x                    --              + x                  + xE        (0.015- x)           --                x                      x

Hence,

Ka[C4H9NH2][H3O+][C4H9NH3+]

Ka= 2.3×10-11

Substitute the values form the ICE table.

2.3×10-11 = (x)(x)0.015- x2.3×10-11 = (x)20.015- x(0.015-x) approximately equals to 0.0152.3×1011 = (x)2(0.015)          x2=  (0.015)(2.3×10-11)          

Therefore,

x  =  (0.015)(2.3×10-11)              = 5.9×10-7[H3O+]  =5.9×107M

Let’s calculate the pH of the solution:

pH = -log[H3O+]pH = -log[5.9×10-7]pH = 6.2

Therefore, pH of butylammonium chloride solution is 6.2.

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Chapter 16 Solutions

Chemistry & Chemical Reactivity

Ch. 16.3 - Prob. 3RCCh. 16.3 - Prob. 4RCCh. 16.3 - Prob. 5RCCh. 16.4 - For each of the following salts in water, predict...Ch. 16.4 - Prob. 1RCCh. 16.4 - Prob. 2RCCh. 16.5 - (a) Which is the stronger Bronsted acid, HCO3 or...Ch. 16.5 - Prob. 1RCCh. 16.5 - 2. In the following reaction, does the equilibrium...Ch. 16.6 - Equal amounts (moles) of HCl(aq) and NaCN(aq) are...Ch. 16.6 - 2. Equal amounts (moles) of acetic acid(aq) and...Ch. 16.6 - Prob. 3RCCh. 16.7 - A solution prepared from 0.055 mol of butanoic...Ch. 16.7 - What are the equilibrium concentrations of acetic...Ch. 16.7 - What are the equilibrium concentrations of HF, F...Ch. 16.7 - The weak base, CIO (hypochlorite ion), is used in...Ch. 16.7 - Calculate the pH after mixing 15 mL of 0.12 M...Ch. 16.7 - 1. What is [H3O+] in a 0.10 M solution of HCN at...Ch. 16.7 - 2. 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