   Chapter 14, Problem 99E

Chapter
Section
Textbook Problem

# Calculate the pH of a 0.20-M C2H5NH2 solution (Kb = 5.6 × 10−4).

Interpretation Introduction

Interpretation: The pH of the given C2H5NH2 solution is to be calculated.

Concept introduction: The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The pOH of a solution is calculated by the formula, pOH=log[OH]

The sum, pH+pOH=14

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

Kb=ConcentrationofproductsConcentrationofreactants

Explanation

Explanation

To determine: The pH of the given C2H5NH2 solution.

The equilibrium constant expression for the given reaction is,

Kb=[C2H5NH3+][OH][C2H5NH2]

The reaction involved is,

C2H5NH2(aq)+H2O(l)C2H5NH3+(aq)+OH(aq)

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

Kb=ConcentrationofproductsConcentrationofreactants

Where,

• Kb is the base ionization constant.

The equilibrium constant expression for the given reaction is,

Kb=[C2H5NH3+][OH][C2H5NH2] (1)

The [OH] is 1.06×10-2M_ .

The change in concentration of C2H5NH2 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,

C2H5NH2(aq)C2H5NH3+(aq)+OH(aq)Inititialconcentration0.2000Changex+x+xEquilibriumconcentration0.20xxx

The equilibrium concentration of [C2H5NH2] is (0.20x)M .

The equilibrium concentration of [C2H5NH3+] is xM

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