Solution Manual for Quantitative Chemical Analysis
Solution Manual for Quantitative Chemical Analysis
9th Edition
ISBN: 9781464175633
Author: Daniel Harris
Publisher: Palgrave Macmillan Higher Ed
Question
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Chapter 8, Problem 8.IE

(a)

Interpretation Introduction

Interpretation:

The concentration of species in 0.10MNaClO4 saturated with Mn(OH)2 has to be calculated.

Concept introduction:

Charge balance:

The overall positive charges in solution equals the overall negative charges in solution.

n1[C1]+n2[C2]+......=m1[A1]+m2[A2]+......where,[C]-concentrationofacationandn-chargeofthecation.[A]-concentrationofaanionandm-magnitudeofthechargeoftheanion.

Mass balance:

The amount of all the species in a solution containing a particular atom (or a group of atoms) must equal the amount of that atom (or group) delivered to the solution.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information,

0.10MNaClO4saturatedwithMn(OH)2.Valueofµ=0.1.SizeofMnOH+is400pm.

Mn(OH)2(s)KSPMn2++2OH-KSP=10-12.8Mn2++OH-K1MnOH+K1=103.4H2OH++OH-KW=10-14.00

Write the charge balance equation

2[Mn2+]+[MnOH+]+[H+]=[OH-]

Write the mass balance equation

[OH_]+[MnOH+]SpeciescontainingOH-=2{[Mn2+]+[MnOH+]}SpeciescontainingMn2++[H+]

Mass balance is equivalent to charge balance.

The equilibrium constant expression is given by

KSP=[Mn2+]γMn2+[OH-]2γ2OH-K1=[MnOH+]γMnOH+[Mn2+]γMn2+[OH-]γOH-KW=[H+]γH+[OH-]γOH-

Solution Manual for Quantitative Chemical Analysis, Chapter 8, Problem 8.IE , additional homework tip  1

Figure 1

The concentration of all the species is given below:

[Mn2+]=5.349×10-5M[OH-]=1.130×10-4M[MnOH+]=6.016×10-6M[H+]=1.408×10-10M

(b)

Interpretation Introduction

Interpretation:

The concentration of species in 0.10MNaClO4 saturated with Mn(OH)2 has to be calculated if there is no NaClO4 in the solution.

Concept introduction:

Charge balance:

The overall positive charges in solution equals the overall negative charges in solution.

n1[C1]+n2[C2]+......=m1[A1]+m2[A2]+......where,[C]-concentrationofacationandn-chargeofthecation.[A]-concentrationofaanionandm-magnitudeofthechargeoftheanion.

Mass balance:

The amount of all the species in a solution containing a particular atom (or a group of atoms) must equal the amount of that atom (or group) delivered to the solution.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information,

0.10MNaClO4saturatedwithMn(OH)2.Valueofµ=0.1.SizeofMnOH+is400pm.

Mn(OH)2(s)KSPMn2++2OH-KSP=10-12.8Mn2++OH-K1MnOH+K1=103.4H2OH++OH-KW=10-14.00

Write the charge balance equation

2[Mn2+]+[MnOH+]+[H+]=[OH-]

Write the mass balance equation

[OH_]+[MnOH+]SpeciescontainingOH-=2{[Mn2+]+[MnOH+]}SpeciescontainingMn2++[H+]

Mass balance is equivalent to charge balance.

The equilibrium constant expression is given by

KSP=[Mn2+]γMn2+[OH-]2γ2OH-K1=[MnOH+]γMnOH+[Mn2+]γMn2+[OH-]γOH-KW=[H+]γH+[OH-]γOH-

The spreadsheet is the same as part (a), except that cell B5 contains the formula

"= 0.5*(E8Λ2*C8+E9Λ2*C9+E10Λ2*C10+E11Λ2*C11)".

Solution Manual for Quantitative Chemical Analysis, Chapter 8, Problem 8.IE , additional homework tip  2

Figure 1

The concentration of the all species was found as

[Mn2+]=3.30×10-5M;[OH-]=7.18×10-5M.[MnOH+]=5.68×10-6M;[H+]=1.43×10-10M.Thevalueofµis1.05×10-4M.

(c)

Interpretation Introduction

Interpretation:

The solubility of Mn(OH)2 is greater when NaClO4 is present. Reason for this observation has to be given.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given information,

The solubility of Mn(OH)2 is greater when NaClO4 is present.

Reason for the observation:

The solubility of Mn(OH)2 depends on the attractive force between Mn2+ and OH- ions.

Na+ and ClO4- ions create ionic atmosphere, this ionic atmospheres reduces the attraction between Mn2+ and OH- ions.  Hence, the solubility of Mn(OH)2 is greater when NaClO4 is present.

The quotient is calculated by

([Mn2+]in0.1MNaClO4)/([Mn2+]withoutNaClO4)5.35×10-5M3.30×10-5M=1.6212=1.6.

The value of quotient is 1.6 .

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