   # Calculate the pH of a 0.050- M Al(NO 3 ) 3 solution. The K a value for Al(H 2 O) 6 3+ is 1.4 × 10 −5 . ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 127E
Textbook Problem
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## Calculate the pH of a 0.050-M Al(NO3)3 solution. The Ka value for Al(H2O)63+ is 1.4 × 10−5.

Interpretation Introduction

Interpretation: The pH of the given 0.050M solution of Al(NO3)3 is to be calculated.

Concept introduction: The pH of a solution is define 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

The equilibrium constant for water is denoted by Kw and is expressed as,

Kw=[H+][OH]

The value of Kw is calculated by the formula,

Kw=KaKb

### Explanation of Solution

Explanation

To determine: The pH of the given 0.050M solution of Al(NO3)3 .

The equilibrium constant expression for the given reaction is,

Ka=[Al(OH)(H2O)53+][H+][Al(H2O)63+]

The dominant equilibrium reaction is,

Al(H2O)63+(aq)Al(OH)(H2O)53+(aq)+H+(aq)

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=[Al(OH)(H2O)53+][H+][Al(H2O)63+] (1)

The [H+] is 8.2×10-4M_ .

The change in concentration of Al(H2O)63+ 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,

Al(H2O)63+(aq)Al(OH)(H2O)53+(aq)+H+(aq)Inititialconcentration0.0500Changex+x+xEquilibriumconcentration0.05xxx

The equilibrium concentration of [Al(H2O)63+] is (0

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