   # The pH of a 0.063- M solution of hypobromous acid (HOBr but usually written HBrO) is 4.95. Calculate K a . ### 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 78E
Textbook Problem
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## The pH of a 0.063-M solution of hypobromous acid (HOBr but usually written HBrO) is 4.95. Calculate Ka.

Interpretation Introduction

Interpretation: The pH of a 0.063M solution of hypobromous acid (HOBr) is given to be 4.95 . The Ka value for hypobromous acid 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+]

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

### Explanation of Solution

Explanation

To determine: The Ka value for hypobromous acid.

The [H+] for hypobromous acid is 1.12×10-5M_ .

Given

The pH of the given concentration of hypobromous acid is 4.95 .

The pH of a solution is calculated by the formula,

pH=log[H+]

Rearrange the above expression to obtain the value of [H+] .

[H+]=10pH

Substitute the pH value of hypobromous acid in the above expression.

[H+]=104.95=1.12×10-5M_

The equilibrium constant expression for the dissociation reaction of hypobromous acid is,

Ka=[H+][BrO][HOBr]

HOBr is a comparatively stronger acid than H2O .

The dominant equilibrium reaction for the given case is,

HOBr(aq)H+(aq)+BrO(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=[H+][BrO][HOBr] (1)

The Ka value for hypobromous acid is 1.98×10-9_ .

The change in concentration of HOBr is assumed to be x .

The ICE table for the stated reaction is,

HOBr(aq)H+(aq)+BrO(aq)Inititialconcentration0

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