   # A weak base has K b = 1.5 × 10 −9 . What is the value of K a for the conjugate acid? ### Chemistry & Chemical Reactivity

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

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16, Problem 32PS
Textbook Problem
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## A weak base has Kb = 1.5 × 10−9. What is the value of Ka for the conjugate acid?

Interpretation Introduction

Interpretation:

The Ka value for conjugate acid of a weak base has to be calculated by using the Kb of weak base.

Concept introduction:

A base (B) undergoes dissociation in an aqueous medium. The base dissocition gives ions BH+(aq.) and OH(aq.). The conjugate acid BH+(aq.) may interact with H2O and forms BOH(aq.) and it can be represented as following equilibrium.

B(aq.)+ H2O(l)BH+(aq.)+ OH1(aq.)                                                 (1)

The dissociation constant for the acid is Kb,

Kb=[BH+][OH][B]

Interaction of conjugate acid BH+(aq.) with water is represented as,

BH+(aq.)+ H2O(l)B(aq.)+ H3O+(aq.)                                                 (2)

The dissociation constant for the conjugate acid is Ka,

Ka=[B][H3O+][BH+]

From equation (1) and (2) the net equilibrium will be

2H2O(l)H3O+(aq.)+ OH1(aq.)                                                             (3)

The equation (3) is known as the auto-ionization of water molecule. The dissociation constant for water is written as follows,

Kw=[H3O+][OH]

Thus from equation (1) and (2), there is an important relation established between ionization constant Kb for base and ionization constant Ka of conjugate acid i.e.

Kw=Ka×Kb                                                                                               (4)

The ionisation constant of water has a constant value i.e. Kw=1×10-14. On taking negative logarithm of equation (4), a new relation established in terms of pKa,pKb and pKw as follows,

-log(Kw)=-log(Ka×Kb)

pKw= pKa+pKb                                                                                           (5)

### Explanation of Solution

The Ka value for conjugate acid (BH+) of the given weak base (B) is calculated below.

Given:

Value of Kb for weak base is 1.5×109.

Value of Kw for water is 1.0×1014.

Using equation (4),Ka for conjugate acid is calculated as follows,

Kw=Ka×Kb

Substitute the value of Kb and Kw

1

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