# Obtain a the K b value for NO 2 − ; b the K a value for C 5 H 5 NH + (pyridinium ion).

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

#### Solutions

Chapter
Section
Chapter 16, Problem 16.63QP
Textbook Problem

## Obtain a the Kb value for NO2−; b the Ka value for C5H5NH+ (pyridinium ion).

Expert Solution

(a)

Interpretation Introduction

Interpretation:

The Kb value for the given ions has to be calculated

Concept Introduction:

Relationship between Ka and Kb

Ka×Kb=Kw

Where Kw is the auto-ionization of water. (1.0×1014)

Autoionization of water is the reaction in which the water undergoes ionization to give a proton and a hydroxide ion.  The ionization of water is an equilibrium reaction and hence this has equilibrium rate constant.

Kw=[H3O+][OH]=1.0×1014

### Explanation of Solution

To Calculate: The Kb value for the given NO2 ion

The reaction of the NO2 ion is:

NO2 + H2HNO2 + OH

The constant Kb can be calculated as dividing Kw by the Ka of the conjugate acid, HNO

Expert Solution

(b)

Interpretation Introduction

Interpretation:

The Ka value for the given ions has to be calculated

Concept Introduction:

Relationship between Ka and Kb

Ka×Kb=Kw

Where Kw is the auto-ionization of water. (1.0×1014)

Autoionization of water is the reaction in which the water undergoes ionization to give a proton and a hydroxide ion.  The ionization of water is an equilibrium reaction and hence this has equilibrium rate constant.

Kw=[H3O+][OH]=1.0×1014

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