Calculate the pH of each of the following solutions. a. 0.12 M KNO 2 b. 0.45 M NaOCl c. 0.40 M NH 4 ClO 4

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 120E
Textbook Problem
1 views

Calculate the pH of each of the following solutions.a. 0.12 M KNO2b. 0.45 M NaOClc. 0.40 M NH4ClO4

(a)

Interpretation Introduction

Interpretation: The pH value for each of the given solutions 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,

Ka=ConcentrationofproductsConcentrationofreactants

The value of Kw is calculated by the formula,

Kw=KaKb

To determine: The pH value for each of the given solution of 0.12M KNO2 .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, Kb=[HNO2][OH][NO2]

The dominant equilibrium reaction is,

NO2(aq)HNO2(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=[HNO2][OH][NO2] (1)

The Kb value is 2.5×10-11_ .

The value, Kw=KaKb

The value of Ka for HNO2 is 4.0×104 .

The value of Kb is calculated by the formula,

Kb=KwKa

Substitute the value of Ka in the above expression.

Kb=1.0×10144.0×104=2.5×10-11_

The [OH] is 1.73×10-6M_ .

The change in concentration of NO2 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,

NO2(aq)HNO2(aq)+OH(aq)Inititialconcentration0.1200Changex+x+xEquilibriumconcentration0.12xxx

The equilibrium concentration of [NO2] is (0

(b)

Interpretation Introduction

Interpretation: The pH value for each of the given solutions 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,

Ka=ConcentrationofproductsConcentrationofreactants

The value of Kw is calculated by the formula,

Kw=KaKb

To determine: The pH value for each of the given solution of 0.45M NaOCl .

(c)

Interpretation Introduction

Interpretation: The pH value for each of the given solutions 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,

Ka=ConcentrationofproductsConcentrationofreactants

The value of Kw is calculated by the formula,

Kw=KaKb

To determine: The pH value for each of the given solution of 0.40M NH4ClO4 .

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