   # Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has K a = 4.5 × 10 −3 and the amino group has K b = 7.4 × 10 −5 . Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with [H + ] = 1.0 M ? In a solution with [OH − ] = 1.0 M ? ### 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 14, Problem 86AE
Textbook Problem
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## Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has Ka = 4.5 × 10−3 and the amino group has Kb = 7.4 × 10−5. Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with [H+] = 1.0 M? In a solution with [OH−] = 1.0 M?

Interpretation Introduction

Interpretation:

The Ka value for the carboxylic acid group and the amino group in alanine is given. The ions that would predominate in a solution with [H+]=1.0M and in a solution with [OH]=1.0M are to be stated.

Concept introduction:

The value of pKa is calculated by the formula, pKa=log(Ka)

The pH of a solution is calculated by the formula, pH=log[H+]

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

### Explanation of Solution

Explanation

To find the Kb

given value of Ka is 4.5×103 .

The given value of Kb is 7.4×105 .

The value of pKa is calculated by the formula,

pKa=log(Ka)

Substitute the value of Ka in the above expression.

pKa=log(4.5×103)=2.35_

The value of pKb is calculated by the formula,

pKb=log(Kb)

Substitute the value of Kb in the above expression.

pKb=log(7.4×105)=4.13_

To find the pH

The pH of a solution is calculated by the formula,

pH=log[H+]

The value of [H+]=1

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