   Chapter 14, Problem 100E

Chapter
Section
Textbook Problem

# Calculate the pH of a 0.050-M (C2H5)2NH solution(Kb = 1.3 × 10−3).

Interpretation Introduction

Interpretation: The pH of the given (C2H5)2NH 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

To determine: The pH of the given (C2H5)2NH solution.

The equilibrium constant expression for the given reaction is,

Kb=[(C2H5)2NH2+][OH][(C2H5)2NH]

The reaction involved is,

(C2H5)2NH(aq)+H2O(l)(C2H5)2NH2+(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=[(C2H5)2NH2+][OH][(C2H5)2NH] (1)

The [OH] is 8.06×10-3M_ .

The change in concentration of (C2H5)2NH 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,

(C2H5)2NH(aq)(C2H5)2NH2+(aq)+OH(aq)Inititialconcentration0.05000Changex+x+xEquilibriumconcentration0.050xxx

The equilibrium concentration of [(C2H5)2NH] is (0.050x)M .

The equilibrium concentration of [(C2H5)2NH2+] is xM

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