   # Calculate the pH of a 0.200- M solution of C 5 H 5 NHF. Hint: C 5 H 5 NHF is a salt composed of C 5 H 5 NH + and F − ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is C 5 H 5 NH + ( a q ) + F − ( a q ) ⇌ C 5 H 5 N ( a q ) + HF ( a q ) K = 8.2 × 10 − 3 ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 175CP
Textbook Problem
150 views

## Calculate the pH of a 0.200-M solution of C5H5NHF. Hint: C5H5NHF is a salt composed of C5H5NH+ and F− ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is C 5 H 5 NH + ( a q ) + F − ( a q ) ⇌ C 5 H 5 N ( a q ) + HF ( a q )   K = 8.2 × 10 − 3

Interpretation Introduction

Interpretation: The concentration of C5H5NHF is given. The pH value is to be calculated.

Concept introduction: The measure of concentration of hydrogen ions is known as pH .

### Explanation of Solution

Explanation

To determine: The pH value of C5H5NHF .

The value of Ka for C5H5NHF is 5.88×106_ .

Given

The concentration of C5H5NHF is 0.200M .

The C5H5NHF is a salt that dissociates into C5H5NH+ and F . The ion C5H5NH+ is a conjugate acid of base C5H5NH2 whereas F is the conjugate base of acid HF .

The base dissociation constant of C5H5NH2 is 1.7×109 .

The relation between Ka and Kb is,

Kw=Ka×Kb (1)

Where,

• Ka is acid dissociation constant.
• Kb is base dissociation constant.
• Kw is ion product constant of water (1.0×1014)

Substitute the values of Kb and Kw in the above equation.

Kw=Ka×KbKa=10141.7×109=5.88×106_

The value of Kb for HF is 1.38×1011_ .

The acid dissociation constant of HF is 7.2×104 .

Substitute the values of Ka and Kw in the equation (1).

Kw=Ka×KbKb=10147.2×104=1.38×1011_

The concentration of [H3O+] is 1.09×103M_ .

Since, Ka is greater than Kb , the solution will be acidic.

C5H5NH+(aq)+H2O(l)C5H5N(aq)+H3O+(aq)

The initial concentration of C5H5NHF is 0

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