Chemistry
Chemistry
12th Edition
ISBN: 9780078021510
Author: Raymond Chang Dr., Kenneth Goldsby Professor
Publisher: McGraw-Hill Education
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Chapter 15, Problem 15.59QP

Write the equation relating Ka for a weak acid and Kb for its conjugate base. Use NH3 and its conjugate acid to derive the relationship between Ka and Kb.

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Interpretation Introduction

Interpretation:

  • The equation relating Ka  for a weak acid and Kb for its conjugate base has to be written.
  • Using NH3 and its conjugate acid NH-4, the relationship between Ka and Kb has to be derived.

Concept Information:

When a Bronsted acid donates a proton, what remains of the acid is known as a conjugate base; when a Bronsted base accepts a proton, the newly formed protonated species is known as a conjugate acid.  This can be given by the below equation as shown in Figure 1.

Chemistry, Chapter 15, Problem 15.59QP

Figure 1

Acid ionization constant Ka

Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent.

The ionization of a weak acid HA can be given as follows,

HA(aq)H+(aq)+A-(aq)

The equilibrium expression for the above reaction is given below.

Ka=[H+][A-][HA]

Where,

Ka is acid ionization constant,

[H+]  is concentration of hydrogen ion

[A-]  is concentration of acid anion

[HA] is concentration of the acid

Base ionization constant Kb

Strong base dissociates into its constituent ions fully. It produces more of hydroxide ions while dissolved in water. Weak bases partially dissociates into its constituent ions.

Since, the ionization of a weak base is incomplete, it is treated in the same way as the ionization of a weak acid.

The ionization of a weak base B is given by the below equation.

B(aq)+ H2O(l)HB+(aq)+OH-(aq)

The equilibrium expression for the ionization of weak base B will be,

Kb=[HB+][OH-][B]

Where,

Kb is base ionization constant,

[OH] is concentration of hydroxide ion

[HB+]  is concentration of conjugate acid

[B] is concentration of the base

To Write: The equation relating Ka  for a weak acid and Kb for its conjugate base

Answer to Problem 15.59QP

The equation relating Ka  for a weak acid and Kb for its conjugate base is Ka×Kb=Kw

Relationship between Ka and  Kb using NH3 and its conjugate acid NH-4 is,

Kw=Ka×Kb,Kw=[NH4+][OH-][NH3]×[NH3][H3O+][NH4+]=[OH-][H3O+]

Explanation of Solution

A simple relationship between the ionization constant of a weak acid (Ka) and the ionization constant of its conjugate base (Kb) can be derived as follows,

Consider, weak acid HA

HA(aq)H+(aq)+A-(aq)conjugatebase

The acid ionization constant will be

Ka=[H+][A-][HA]

The conjugate base  reacts with water as follows,

A-(aq)+H2O(l)HA(aq)+OH-(aq)

The base ionization constant will be

Kb=[HA][OH-][A-]

The product of the two ionization constants will be Kw (autoionization of water)

Ka×Kb=[H+][A-][HA]×[HA][OH-][A-]On cancelling the common terms, we getKa×Kb=[H+][OH-]The autoionization of waterKw=[H+][OH-]Therefore,Ka×Kb=Kw

This is the simple relation relating Ka and Kb of weak acid.

To derive: Using NH3 and its conjugate acid NH-4, the relationship between Ka and Kb

Using NH3 and its conjugate acid NH+4 to derive the above relationship gives:

NH3(aq)+H2ONH4+(aq)+OH-(aq)Kb=[NH4+][OH-][NH3]NH4+(aq)+H2ONH3(aq)+H3O+(aq)Ka=[NH3][H3O+][NH4+]Since,Kw=Ka×Kb,wehave:Kw=[NH4+][OH-][NH3]×[NH3][H3O+][NH4+]=[OH-][H3O+]

Conclusion

àThe equation relating Ka  for a weak acid and Kb for its conjugate base was written.

àUsing NH3 and its conjugate acid NH-4, the relationship between Ka and Kb was derived.

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Chapter 15 Solutions

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