Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
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Chapter 9, Problem 9.45P

a)

Interpretation Introduction

Interpretation:

The pH of the solution has to be calculated and compared.

Concept Introduction:

The concentrations of HA and A- will not be equal to their formal concentrations for dilute solutions or at extreme pH .

Consider the mixing of FHA moles of HA and FA- moles of the salt Na+A- . Then, the mass and charge balances can be given as,

Mass balance: FHA+FA-=[HA]+[A-]

Charge balance: [Na+]+[H+]=[OH-]+[A-]

The substitution [Na+]=FA- leads to the equation,

[HA]=FHA-[H+]+[OH-][A-]=FA-+[H+]-[OH-]

Assumption has to be made that [HA]FHA and [A-]FA- and these values are used in Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation:

Henderson- Hasselbalch equation is rearranged form of acid dissociation expression.

For base:

Consider a basic reaction:

BH+Ka=Kw/KbB+H+

The Henderson- Hasselbalch equation for a basic reaction can be given as,

pH=pKa+log[B][BH+]

Where , pKa = acid dissociation constant for weak acid BH+

The value of Kw is calculated by the formula,

Kw=Ka×Kb

The pH of weak base is calculated using the equation,

pH=[H+]=Kw[OH-]pH=-log[H+]

a)

Expert Solution
Check Mark

Answer to Problem 9.45P

The pH of the solution is 11.48 .

Explanation of Solution

Mass balance and Charge balance is obtained if we dissolve B and BH+Br- .

Mass balance = FBH+FB=[BH+]+[B]

Charge balance = [Br-]+[OH-]=[BH+]+[H+]

If, [Br-]=FBH+ , then the charge balance is written as,

[BH+]=FBH++[OH-]-[H-] (1)

Substituting the above expression into the mass balance gives,

[B]=FB-[OH-]+[H+] (2)

Assume that

[B]=0.0100M[BH+]=0.0200M

Then, the pH is calculated as,

pH=pKa+log[B][BH+]

pH=12.00+log0.01000.0200pH=11.70

If the given below values are not assumed, then equations (1) and (2) can be used

[B]=0.0100M[BH+]=0.0200M

The solution is basic and [H+] are neglected relative to [OH-] and then [B] is written as,

[B]=0.0100-x[BH+]=0.0200+x

Where x= [OH-]

Kb takes the value of 10-2.00 = [BH+][OH-][B]=(0.0200+x)(x)(0.0100-x)

x=0.00303M

pH=-logKwx

pH=11.48

The pH of the solution = 11.48

b)

Interpretation Introduction

Interpretation:

The goal seek has to be done for the calculations done in part A.

b)

Expert Solution
Check Mark

Answer to Problem 9.45P

Quantitative Chemical Analysis, Chapter 9, Problem 9.45P , additional homework tip  1

Figure 1

Explanation of Solution

The goal seek is used to vary cell B5 till cell D4 that is equal to Ka .

Quantitative Chemical Analysis, Chapter 9, Problem 9.45P , additional homework tip  2

Figure 1

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