   Chapter 16, Problem 16.74QP General Chemistry - Standalone boo...

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

Solutions

Chapter
Section General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

What is the pH of a solution that is 0.15 M C2H5NH2 (ethylamine) and 0.10 M C2H5NH3Br (ethylammonium bromide)?

Interpretation Introduction

Interpretation:

The pH of a solution that is 0.15 M C2H5NH2 (ethylamine) and 0.10 M C2H5NH3Br (ethylammonium bromide) has to be calculated.

Concept Introduction:

The equilibrium expression for the weak base is given below.

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

Where,

Kb is base ionization constant,

[OH] is concentration of hydroxide ion

[BH+] is concentration of conjugate acid cation

[B] is concentration of the base

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH and pOH:

pH + pOH = 14

To calculate: The pH of a solution that is 0.15 M C2H5NH2 (ethylamine) and 0.10 M C2H5NH3Br (ethylammonium bromide)

Explanation

Given data:

The concentration of C2H5NH2 is 0.15 M

The concentration of C2H5NH3Br is 0.10 M

Calculation of concentration:

Construct an equilibrium table, using a initial C2H5NH2 as 0.15 M from C2H5NH2 and a initial concentration of C2H5NH3+ as 0.10 M from C2H5NH3Br

Let x be the unknown concentration.

 C2H5NH2(aq)+H2O(l)     ⇌        OH−(aq)  +   C2H5NH3+(aq) Initial (M) 0.15 -x 0.15-x 0.00 0.10 Change (M) +x +x Equilibrium (M) x 0.10+x

The Kb of C2H5NH2 is 4.7×104

Substitute the equilibrium concentrations into equilibrium-constant expression

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