   Chapter 14, Problem 103E

Chapter
Section
Textbook Problem

# The pH of a 0.016-M aqueous solution of p-toluidine (CH3C6H4NH2) is 8.60. Calculate Kb.

Interpretation Introduction

Interpretation: The pH of a 0.016M aqueous solution of p-toluidine (CH3C6H4NH2) is given to be 8.60 . The value of Kb 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 pH of a solution is calculated by the formula, pH=log[H+]

The sum, pH+pOH=14

Explanation

Explanation

To determine: The value of Kb for the given solution.

The pOH value is 5.4_ .

The given pH value is 8.60 .

The sum, pH+pOH=14

Substitute the value of pH in the above expression.

pOH+8.60=14pOH=5.4_

The [OH] is 3.98×10-6M_ .

The pOH of a solution is calculated by the formula,

pOH=log[OH]

Rearrange the above expression to calculate the value of [OH] .

[OH]=10pOH

Substitute the calculated value of pH in the above expression.

[OH]=105.4=3.98×10-6M_

The equilibrium constant expression for the stated ionization reaction is,

Kb=[CH3C6H4NH3+][OH][CH3C6H4NH2]

CH3C6H4NH2 is a weaker base.

The dominant equilibrium reaction for the given case is,

CH3C6H4NH2(aq)+H2O(l)CH3C6H4NH3+(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=[CH3C6H4NH3+][OH][CH3C6H4NH2] (1)

The value of Kb for the given solution is 1.0×10-9

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