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
Question
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Chapter 13, Problem 13.CE

(a)

Interpretation Introduction

Interpretation:

The pH and concentrations of species in the given solution has to be determined by neglecting the activity coefficients.

(a)

Expert Solution
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Explanation of Solution

Given information,

The given 1.00L of solution containing 0.040 mol 2-aminobenzoic acid (HA), 0.020 mol dimethylamine (B), 0.015 molHCl.

The value of pH and concentrations of the species can be determined as follows,

2-aminobenzoicacid = HA fromdiproticH2A+,pK1 = 2.08,pK2=4.96.

Dimethylamine = B from monoprotic BH+ with pKa = 10.774

The chemical reaction that happened in the given solution is given by

Chemical reactions: H2A+HA+H+K1=10-2.08HAA+ H+K2= 10-4.96BH+B + H+Ka= 10-10.774H2OH+ + OH-Kw = 10-14.00

Charge balance:[H+] +[H2A+]+ [BH+] = [OH-]+ [A-] + [Cl]

Mass balances:[Cl-] = 0.015 M[BH+] + [B] = 0.020 M  FB[H2A+] + [HA]+[A-] = 0.040 M  FA

From, the above information we have eight equations and eight chemical species.

Write the fractional composition expressions

[BH+] = αBH+FB=[H+]FB[H+] + Ka[B] = αBFB=KaFB[H+] + Ka

[H2A+] = αH2A+FA=[H+]2FA[H+]2+[H+]K1 +K1K2

[HA] = αHAFA=K1[H+]FA[H+]2+[H+]K1 + K1K2[A-] = αA-FA=K1K2FA[H+]2+[H+]K1 + K1K2

Substitute the above values into the charge balance equation

[H+]+αH2A+FA+αBH+FB=KW/[H+]+αA-FA+[0.015M](A)

The value of pH and concentrations of species in the given solution are mentioned in the following spreadsheet in figure 1.

Quantitative Chemical Analysis 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card), Chapter 13, Problem 13.CE , additional homework tip  1

Figure 1

The value of pH is 4.153.

The concentration of species in the charge balance equation are given below:

[H+]=7.03×10-5M;[BH+]=2.00×10-2M;[A-]=5.36×10-3M[Cl-]=0.015M;[OH-]=1.42×10-10M;[H2A+]=2.90×10-4M

The other concentrations are,

[HA]=3.43×10-2M;[B]=4.79×10-9M.

(b)

Interpretation Introduction

Interpretation:

The fraction of HA is in each of its three forms and fraction of B is in each of its two forms has to be given.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information,

Quantitative Chemical Analysis 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card), Chapter 13, Problem 13.CE , additional homework tip  2

Figure 1

The value of pH is 4.153.

The concentration of species are given below:

[H+]=7.03×10-5M;[BH+]=2.00×10-2M;[A-]=5.36×10-3M[Cl-]=0.015M;[OH-]=1.42×10-10M;[H2A+]=2.90×10-4M

The other concentrations are,

[HA]=3.43×10-2M;[B]=4.79×10-9M.

The calculated fraction of 2-aminobenzoic acid (HA) is in each of its three forms are given below:

H2A+=0.7%;HA=85.9%;A-=13.4%.

The calculated fraction of methylamine (BH+) is in each of its two forms are given below:

BH+=100%;B=0.0%.

The simple prediction is that HCl would consume B , giving 100%BH+ .  The remaining 5mmolB consumes 5mmolHA , making 5mmolA- and leaving 35mmolHA .

predictedfractionsare:A-=540=12.5%;HA=3540=87.5%

The estimated pH is calculated as follows

pH=pK2+log([A-]/[HA])=4.96+log(5/35)=4.11.

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