 # Calculate the concentration of pyridine, C 5 H 5 N, in a solution that is 0.15 M pyridinium bromide, C 5 H 5 NHBr. What is the pH of the solution? ### 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 16.67QP
Textbook Problem
453 views

## Calculate the concentration of pyridine, C5H5N, in a solution that is 0.15 M pyridinium bromide, C5H5NHBr. What is the pH of the solution?

Expert Solution
Interpretation Introduction

Interpretation:

The concentration of pyridine and pH of a 0.15 M pyridinium bromide solution 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

Relationship between pH and pOH:

pH + pOH = 14

### Explanation of Solution

To calculate: The concentration of pyridine and pH of a 0.15 M pyridinium bromide solution

Given data:

The concentration of pyridinium bromide solution is 0.15 M

Calculation of Ka for pyridinium ion:

The Kb of pyridine is 1.4×109

The Ka can be calculated from the Kb of its conjugate base as follows,

Ka =KwKb =1.0×10-141.4×109 =7.14×106

Therefore, the Ka of pyridinium ion is 7.14×106

Calculation of concentration:

Set up the equilibrium table for the given solution.

Let x be the unknown concentration.

Represent, pyridinium  ion as PyNH+ and pyridine as PyN

Pyridinium bromide dissociates as C5H5NH+ and Br- ions. The hydrolysis of pyridinium ion is as follows,

 PyNH+(aq)+H2O(l)     ⇌        H3O+(aq)  +   PyN(aq) Initial (M) 0.15 -x 0.15-x 0.00 0

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