Chapter 14, Problem 68E

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# A solution is prepared by dissolving 0.56 g benzoic acid (C6H5CO2H, Ka = 6.4 × 10−5) in enough water to make 1.0 L of solution. Calculate [C6H5CO2H], [C6H5CO2−], [H+], [OH−], and the pH of this solution.

Interpretation Introduction

Interpretation: The concentration of [C6H5COOH] , [C6H5CO2] , [H+] , [OH] and the pH of the solution, prepared by dissolving 0.56g of benzoic acid in enough water o make 1.0L of the solution, 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

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

Kw=[H+][OH]

The percent dissociation of an acid is calculated by the formula,

Percentdissociation=Equilibriumconcentrationof[H+]Initialconcentrationoftheacid×100

Explanation

Explanation

To determine: The concentration of [C6H5COOH] , [C6H5CO2] , [H+] , [OH] and the pH of the solution.

The dissociation of C6H5COOH is, C6H5COOH(aq)  H+(aq) + C6H5COO-(aq)

C6H5COOH is a weak acid. Hence, it does not completely dissociate in water.

C6H5COOH is a comparatively stronger acid than H2O .

The dominant equilibrium reaction for the given case is,

C6H5COOH(aq)H+(aq)+C6H5COO(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+][C6H5COO][C6H5COOH] (1)

The number of moles of C6H5COOH is 4.59×10-3mol_ .

Given

The given mass of benzoic acid (C6H5COOH) is 0.56g .

The molar mass of C6H5COOH =7C+6H+2O=((7×12)+(6×1)+(2×16))g/mol=122g/mol

The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassMolarmass

Substitute the value of the given mass and the molar mass of C6H5COOH in the above expression.

Numberofmoles=0.56g122g/mol=4.59×10-3mol_

The initial [C6H5COOH] is 0.0046M_ .

The calculated number of moles of C6H5COOH is 4.59×103mol .

The total volume is 1.0L .

The concentration is calculated by the formula,

Concentration=NumberofmolesVolume(L)

Substitute the value of the number of moles of C6H5COOH and the volume in the above expression.

Concentration=4.59×103mol1L=0.0046M_

The [H+] and [C6H5COO] is 5.4×10-4M_

The change in concentration of C6H5COOH is assumed to be x .

The ICE table for the stated reaction is,

C6H5COOH(aq)H+(aq)+C6H5COO(aq)Inititialconcentration0.004600Changex+x+xEquilibriumconcentration0.0046xxx

The equilibrium concentration of [C6H5COOH] is (0

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