Quantitative Chemical Analysis 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card)
Quantitative Chemical Analysis 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card)
9th Edition
ISBN: 9781319090241
Author: Daniel C. Harris, Sapling Learning
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 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card), 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 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card), Chapter 9, Problem 9.45P , additional homework tip  2

Figure 1

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