   Chapter 14, Problem 118E

Chapter
Section
Textbook Problem

# Calculate the concentrations of all species present in a 0.25-M solution of ethylammonium chloride (C2H5NH3Cl).

Interpretation Introduction

Interpretation: The concentration of all the species present in the given 0.25M solution of ethylammonium chloride (C2H5NH3Cl) is to be calculated.

Concept introduction: The value of Ka is calculated by the formula,

Ka=KwKb

Explanation

Explanation

To determine: The concentration of all the species present in the given 0.25M solution of ethylammonium chloride (C2H5NH3Cl) .

The major species present in the given solution of C2H5NH3Cl in water are C2H5NH3+ , H+ , Cl and H2O .

C2H5NH3+ is a stronger acid than H2O .

The dominant equilibrium reaction is,

C2H5NH3+(aq)C2H5NH2(aq)+H+(aq)

The major species present in the given solution of C2H5NH3Cl in water are C2H5NH3+ , H+ , Cl and H2O .

The equilibrium constant expression for the given reaction is, Ka=[C2H5NH2][H+][C2H5NH3+]

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

Where,

• Ka is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

Ka=[C2H5NH2][H+][C2H5NH3+] (1)

The Ka value is 1.78×10-11M_ .

The value, Kw=KaKb

The value of Kb for ethyl amine is 5.6×104 .

The value of Ka is calculated by the formula,

Ka=KwKb

Substitute the value of Kb in the above expression.

Ka=1.0×10145.6×104=1.78×10-11_

The [C2H5NH2] is 2.10×10-6M_ .

The change in concentration of C2H5NH3+ is assumed to be x .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

C2H5NH3+(aq)C2H5NH2(aq)+H+(aq)Inititialconcentration0

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