respostas capitulo 13, skoog!

Fundamentals of Analytical Chemistry: 8th

ed. Chapter 13

Chapter 13

13-1 amount A (mmol) = )mL/Ammol()mL(volume Ac

amount A (mole) = )L/Amol()L(volume Ac

13-2 (a) The millimole is the amount of an elementary species, such as an atom, an ion, a

molecule, or an electron. A millimole contains

mmol

particles1002.6

mmol1000

mole

mole

particles1002.6 2023

(b) A titration involves measuring the quantity of a reagent of known concentration

required to react with a measured quantity of sample of an unknown concentration. The

concentration of the sample is then determined from the quantities of reagent and sample,

the concentration of the reagent, and the stoichiometry of the reaction.

(c) The stoichiometric ratio is the molar ratio of two chemical species that appear in a

balanced chemical equation.

(d) Titration error is the error encountered in titrimetry that arises from the difference

between the amount of reagent required to give a detectable end point and the theoretical

amount for reaching the equivalence point.

13-3 (a) The equivalence point in a titration is that point at which sufficient titrant has been

added so that stoichiometrically equivalent amounts of analyte and titrant are present.

The end point in a titration is the point at which an observable physical change signals the

equivalence point.

(b) A primary standard is a highly purified substance that serves as the basis for a

titrimetric method. It is used either (i) to prepare a standard solution directly by mass or


ed. Chapter 13

(ii) to standardize a solution to be used in a titration.

A secondary standard is material or solution whose concentration is determined from the

stoichiometry of its reaction with a primary standard material. Secondary standards are

employed when a reagent is not available in primary standard quality. For example, solid

sodium hydroxide is hygroscopic and cannot be used to prepare a standard solution

directly. A secondary standard solution of the reagent is readily prepared, however, by

standardizing a solution of sodium hydroxide against a primary standard reagent such as

potassium hydrogen phthalate.

13-4 The Fajans method is a direct titration of the chloride ion, while the Volhard approach

requires two standard solutions and a filtration step to eliminate AgCl. The Fajans

method uses a fluorescein dye. At the end point, the fluoresceinate anions are absorbed

into the counter ion layer that surrounds the colloidal silver particles giving the solid an

intense red color. In the Volhard method, the silver chloride is more soluble that silver

thiocyanide such that the reaction

Cl)(AgSCNSCNAgCl ss occurs to a

significant extent as the end point is approached. The released Cl- ions cause the end

point color change to fade resulting in an over consumption of SCN- and a low value for

the chloride analysis.

13-5 (a) 2

22

Imoles2

NNHHmole1

(b)

4

22

MnOmoles2

OHmoles5


ed. Chapter 13

(c)

Hmoles2

OH10OBNamole1 2742

(d) 3KIOmoles3

Smoles2

13-6 In contrast to Ag2CO3 and AgCN, the solubility of AgI is unaffected by the acidity. In

addition, AgI is less soluble than AgSCN. The filtration step is thus unnecessary in the

determination of iodide, whereas it is needed in the determination of carbonate or

cyanide.

13-7 The ions that are preferentially absorbed on the surface of an ionic solid are generally

lattice ions. Thus, in a titration, one of the lattice ions is in excess and its charge

determines the sign of the charge of the particles. After the equivalence point, the ion of

the opposite charge is present in excess and determines the sign of the charge on the

particle. Thus, in the equivalence-point region, the charge shift from positive to negative,

or the reverse.

13-8 (a)

33

33

AgNOg37.6mole

AgNOg87.169mole0375.0

mole0375.0mL500mL1000

L

L

AgNOmole0750.0AgNOM0750.0

Dissolve 6.37 g AgNO3 in water and bring to 500 mL total volume.

(b)

reagentL108.0reagentmole00.6

Lmole650.0

mole650.0L00.2L

HClmole325.0HClM325.0


ed. Chapter 13

Dilute 108 mL of the 6.00 M HCl reagent in enough water to yield 2.00 L total volume.

(c)

646464 )CN(FeKg22.6

mole

)CN(FeKg35.368

Kmoles4

)CN(FeKmoleKmole0675.0

Kmole0675.0mL750mL1000

L

L

Kmole0900.0KM0900.0

Dissolve 6.22 g K4Fe(CN)6 in water and bring to 750 mL total volume.

(d)

2

2

2

222

2

BaClL115.0BaClmole500.0

LBaClmole0576.0

BaClmole0576.0mL600g23.208

BaClmole

solutionmL100

BaClg00.2BaCl)v/w(%00.2

Dilute 115 mL of 0.500M BaCl2 in enough water to yield 600 mL total volume.

(e)

reagentL025.0HClOmole55.9

reagentLHClOmole240.0reagent.vol

reagentL

HClOmole55.9

g5.100

HClOmole

reagentg100

HClOg60

reagentL

reagentg1060.1

HClOmole240.0L00.2L

HClOmole120.0HClOM120.0

4

4

444

3

44

4

Dilute 25 mL HClO4 reagent in enough water to yield 2.00 L total volume.

(f)

4242422

2

SONag67.1mole

SONag0.142

Namoles2

SONamole

g99.22

Namole

mg1000

gNamg104.5

Namg1040.5solnL00.9solnL

Namg60Nappm0.60

Dissolve 1.67 g Na2SO4 in water and bring to 9.00 L total volume.


ed. Chapter 13

13-9 (a)

44

4

44

4

KMnOg7.23mole

KMnOg03.158KMnOmole150.0

KMnOmole150.0L00.1L

KMnOmole150.0KMnOM150.0

Dissolve 23.7 g KMnO4 in water and bring to 1.00 L total volume.

(b)

reagentHClOL139.0HClOmole00.9

LHClOmole25.1

HClOmole25.1L50.2L

HClOmole500.0HClOM500.0

4

4

4

44

4

Dilute 139 mL HClO4 reagent in enough water to yield 2.50 L total volume.

(c)

222 MgIg78.2

mole

MgIg11.278

Imoles2

MgImoleImole0200.0

Imole0200.0mL400mL1000

L

L

Imole0500.0IM0500.0

Dissolve 2.78 g MgI2 in water and bring to 400 mL total volume.

(d)

4

4

4

444

4

CuSOL0575.0CuSOmole218.0

LCuSOmole0125.0

CuSOmole0125.0mL200g61.159

CuSOmole

mL100

CuSOg00.1CuSO)v/w(%00.1

Dilute 57.5 mL of the 0.218 M CuSO4 solution to yield 200 mL total volume.


ed. Chapter 13

(e)

reagentL0169.0NaOHmole10906.1

reagentLNaOHmole3225.0reagent.vol

reagentL

NaOHmole10906.1

g00.40

NaOHmole

reagentg100

NaOHg50

reagentL

reagentg10525.1

NaOHmole3225.0L50.1L

NaOHmole215.0NaOHM215.0

1

13

Dilute 16.9 mL of the concentrated reagent to 1.50 L total volume.

(f)

64

64641

1

)CN(FeKg0424.0

mole

)CN(FeKg3.368

Kmoles4

)CN(FeKmole

g10.39

Kmole

mg1000

gKmg108.1

Kmg108.1solnL50.1solnL

Kmg12Kppm12

Dissolve 42.4 mg K4Fe(CN)6 in water and bring to 1.50 L total volume.

13-10 mole

g59.216HgO M

4

44

2

42

HClOM08190.0

mL51.46

mole

HClOmmol1000

OHmole1

HClOmole1

HgOmole

OHmole2

g59.216

HgOmole1HgOg4125.0

OH2HgBrOHBr4)(HgO

s


ed. Chapter 13

13-11 mole

g99.105

32CONaM

42

4242

32

3232

22

2

3

SOHM1168.0

mL44.36

mole

SOHmmol1000

Hmole2

SOHmole1

CONamole

Hmole2

g99.105

CONamole1CONag4512.0

)(COOHH2CO

g

13-12 mole

g04.142

42SONaM

2

42

24242

4

2

4

2

BaClM06581.0

mL25.41

mole

mmol1000

SONamole1

BaClmole1

g04.142

SONamole1

sampleg100

SONag4.96sampleg4000.0

)(BaSOSOBa

s

13-13 (Note: In the first printing of the text, the answer in the back of the book was in error.)

NaOHmL

HClOmL0972.1

NaOHmL00.25

HClOmL43.27

V

V44

NaOH

HClO4

The volume of HClO4 required to titrate 0.3125 g Na2CO3 is

NaOHM2239.0HClOmole

NaOHmole1

NaOHmL

HClOmL0972.1

L

HClOmole2041.0M2041.0

V

Vcc

and

HClOM2041.0mole

mmol1000

CONamole1

HClOmole2

g99.105

CONamole1

HClOmL896.28

CONag3125.0

,Thus

HClOmL896.28NaOHmL

HClOmL0972.1NaOHmL12.10HClOmL00.40

4

44

NaOH

HClO

HClONaOH

4

32

432

4

32

44

4

4

4


ed. Chapter 13

13-14 OH8)(CO10Mn2H6OCH5MnO2 222

4224

g

4

422

4422422

KMnOM02858.0

mL75.36

mole

mmol1000

OCNamole5

KMnOmole2

mL1000

L

L

OCNamole05251.0OCNamL00.50

13-15 mole

g00.214

3KIOM

2

64

2

322

223

OSI2OS2I

OH3I3H6I5IO

322

2

322

3

233

OSNaM09537.0

mL72.30

mole

mmol1000

Imole1

OSNamole2

KIOmole1

Imole3

g00.214

KIOmole1KIOg1045.0

13-16

)(AgSCNNHSCNNHAg

,SCNNHwithtitratedisAgunreactedThe

)(AgClHCOOHHOCHOHAgCOOHClCH

44

4

222

s

s

SCNNHM098368.0

mL98.22

mole

mmol1000

AgNOmole1

SCNNHmole1

mL1000

L

L

AgNOmole04521.0mL00.50

4

3

43


ed. Chapter 13

COOHClCHmg7.116

g

mg1000

mole

g50.94

AgClmole1

COOHClCHmole1AgClmole102345.1

AgClmmol2345.1

mmol02598.1mL00.50mL

mmol04521.0edprecipitat)s(AgClmmol

SCNNHmmol02598.1mL43.10mL

SCNNHmmol098368.0SCNNHmmol

2

23

44

4

13-17

)(AgSCNSCNAg

OH5)(Ag8BOHOH8Ag8BH 2324

s

s

mmol excess Ag+ equals mmol KSCN,

%5.11%100materialg213.3

KBHg371.0KBHpurity%

KBHg371.0mole

KBHg941.53

BHmole1

KBHmole1mL500

mL1000

L

L

BHmole0138.0

BHM0138.0Agmmol8

BHmmol1

mL100

Agmmol1010.1

Agmmol1010.1mmol133.01011.1Agmmolreacted

AgNOmmol1011.1mL00.50mL

AgNOmmol2221.0AgNOmmol

Agmmol133.0KSCNmmol1

Agmmol1mL36.3

mL

KSCNmmol0397.0Agexcessmmol

44

44

4

44

44

1

11

3

133


ed. Chapter 13

13-18 )(AsOAgH3Ag3AsOH 4343 s

33

3 AgNOmmol4888.2mL00.40mL

AgNOmmol06222.0addedAgNOmmol

Agmmol0760.1mL76.10KSCNmmol1

Agmmol1

mL

KSCNmmol1000.0Agexcessmmol

,KSCNmmolequalsAgexcessmmol

32

32

43

3243

32

OAs%612.4

100sampleg010.1

mmol1000

OAsg84.197

AsOAgmmol2

OAsmmol1

Agmmol3

AsOAgmmol1Agmmol4128.1

sampleinOAs%

Agmmol4128.1mmol)0760.14888.2(reactedAgmmol

13-19 mole

g32.373

7510 ClHCM

The stoichiometry of the titration of heptachlor must be one to one (i.e., one chlorine

reacts with one silver nitrate) for the calculation,

samplemass

33.37mLmLheptachlor%

SCNSCNAgAg

cc, to be true. The factor 37.33 (with

unwritten units of mmol

g) found in the numerator is derived from the equation below,

100mmol1000

ClHCg32.373

AgNOmmol.no

ClHCmmol.no

mmol

g33.37 7510

3

7510

Thus,

00.1100ClHCg32.373

mmol1000mmol

g33.37

AgNOmmol.no

ClHCmmol.no

75103

7510

confirming that only one of the chlorines in the heptachlor reacts with the AgNO3.


ed. Chapter 13

13-20 H2)(BiPOPOHBi 442

3 s

eulytite%90.39

%100sampleg6423.0

mmol1000

SiO3OBi2g1112

Bimmol4

SiO3OBi2mmol1Bimol921758.0

eulytitepurity%

Bimol921758.0PONaHmmol1

Bimmol1PONaHmol921758.0Bimol

PONaHmol921758.0mL36.27mL

PONaHmmol03369.0PONaHmol

232

3

2323

3

42

3

42

3

4242

42

13-21 (a)

2

56

25656

2

)OH(BaM01190.0

mL42.40

mole

mmol1000

COOHHCmole2

)OH(Bamole1

g12.122

COOHHCmole1COOHHCg1175.0

)OH(Baofmolarity

(b)

M102.242.40

03.0

1175.0

0002.0)M10190.1(s 5

22

2

y

molarity of Ba(OH)2 taking into account the uncertainty of the two measurements can be

written 0.01190(0.00002) M.

(c) Weighing error of -0.0003g results in an absolute error, E, in the molarity calculation,

M100.3orM10826.2M10190.1M10187.1M10190.1

mL42.40

mole

mmol1000

COOHHCmole2

)OH(Bamole1

g12.122

COOHHCmole1COOHHCg0003.01175.0

E

55222

56

25656


ed. Chapter 13

The relative error, Er, in the molarity calculation resulting from this weighing error is

ppt3or100.3

M10190.1

M100.3E 3

2

5

r

13-22

HOAc%529.1

%100mL00.50

mmol1000

HOAcg05.60

)OH(Bammol1

HOAcmmol2mL17.43

mL

)OH(Bammol1475.0

HOAcpercentagev/w

2

2

Similar calculations for samples 2 to 4 yield the results shown in the spreadsheet that

follows,

(a)

HOAc%528.14

1134.6

4

xHOAcpercentagev/wx

i

(b)

HOAc%1071.53

4

)1134.6(34351132.9

3

4

)x(x

s 3

22

i2

i

(c)

HOAc)%007.0(528.12

)1063.5(35.2528.1

4

tsxCI

3

%90

(d) The values of 1.5352 and 1.5213 can be considered for rejection. Applying the Q

test we find, that both results are less than Qexpt = 0.765, so neither value should be

rejected.

(e) V

V

HOAc)%v/w(

HOAc)%v/w(


ed. Chapter 13

001.0mL00.50

mL05.0

HOAcV

HOAcV,1sampleFor

The results for the remaining samples are found in the following spreadsheet.

00125.04

005.0

n

xerrorsystematicrelativemean

HOAc%102or%1091.1528.100125.0

HOAc)%v/w(,HOAcpercent)v/w(meantheFor

33

A B C D E F G

1 Problem 13-22

2

3 Conc. Ba(OH)2 0.1475

4 MW HOAc 60.05

5 t 2.35

6

7 Sample Sample Vol, mL Ba(OH)2 Vol, mL w/v % HOAc xi xi2 V/V

8 1 50.00 43.17 1.529 1.52949152 2.33934429 -0.001

9 2 49.50 42.68 1.527 1.52740511 2.33296637 -0.001

10 3 25.00 21.47 1.521 1.52134273 2.31448370 -0.002

11 4 50.00 43.33 1.535 1.53516024 2.35671695 -0.001

12

13 (xi) 6.11339959

14 (xi2) 9.34351132

15 (a) mean xi 1.528

16 (b) std. dev. % HOAc 5.71E-03

17 (c) CI90%(t=2.35) 6.70E-03

18 (d) Q(expt 1.535-1.521) 0.41

19 Q(expt 1.527-1.521) 0.44

20 (e) (V/V) -0.005

21 mean relative systematic error -1.25E-03

22 mean (w/v) % HOAc -1.91E-03

23 Spreadsheet Documentation

24 D8 = (($B$3*C8*2*$B$4/1000)/B8)*100 C16 = SQRT((B14-(B13)^2/4)/3)

25 E8 = D8 C18 = (D11-D8)/(D11-D10)

26 F8 = E8^2 C19 = (D9-D10)/(D11-D10)

27 G8 = -0.05/B8 C20 = SUM(G8:G11)

28 B13 = SUM(E8:E11) C21 = C20/4

29 B14 = SUM(F8:F11) C22 = C21*C15

30 C15 = B13/4


ed. Chapter 13

13-23

33

33

AgNOmmol5204.1mL81.2KSCNmmol1

AgNOmmol1

mL

KSCNmmol04124.0

mL00.20mL

AgNOmmol08181.0samplebyconsumedAgNOmmol.no

tablet

saccharinmg60.15

tablets20

g

mg1000

mmol1000

saccharing17.205

AgNOmmol1

saccharinmmol1AgNOmmol5204.1

tablet/saccharinmg

3

3

13-24 (a)

3

3

33

100533.2

mL3.502

mole

mmol1000

AgNOmole1

Agmole1

g87.169

AgNOmole1AgNOg1752.0

Agmolarityweight

(b)

3

3

3

109386.1

mL171.25

mole

mmol1000mL765.23

mL1000

AgNOmole100533.2

KSCNmolarityweight

(c) mole

g26.244OH2BaCl 22 M

mmol026653.0

mL543.7KSCNmmol1

AgNOmmol1

mL

KSCNmmol109386.1

mL102.20mL

AgNOmmol100533.2consumedAgNOmmol

3

3

3

3

3


ed. Chapter 13

%4572.0

%100sampleg7120.0

mmol1000

g26.244

AgNOmmol2

OH2BaClmmol1AgNOmmol026653.0

OH2BaCl% 3

223

22

13-25 (a)

OH6MgClKClM01821.0L000.2

g85.277

OH6MgClKClmole1OH6MgClKClg12.10

22

2222

(b) 2622 MgM01821.0OH6MgClKClMg

(c)

ClM05463.0OH6MgClKClmole1

Clmole3OH6MgClKClmole01821.0Cl

22

22

(d)

%506.0%100mL1000

L

L000.2

g12.10OH6MgClKCl)%v/w( 22

(e)

Clmmol37.1mL0.25mL

Clmmol05463.0

(f)

Kppm0.712

g

mg1000

mole1

Kg10.39

OH6MgClKClmole1

Kmole1

L

OH6MgClKClmole01821.0

22

22


ed. Chapter 13

13-26 mole

g03.30OCH2 M

OCH%5.21%100

mL500

mL0.25sampleg00.5

mmol1000

OCHg03.30OCHmmol787.1

OCHmmol787.1mL1.16mL

SCNNHmmol134.0mL0.40

mL

AgNOmmol100.0

mL0.30mL

KCNmmol121.0reactedKCNmmolOCHmmol

2

22

243

2

13-27 mole

g34.308

41619 OHCM

41619

4161941619

41619

3

41619

3

3341619

3

33

OHC%4348.0%100sampleg96.13

mmol1000

OHCg34.308OHCmmol1968.0

OHCmmol1968.0CHImmol1

OHCmmol1

AgNOmmol3

CHImmol1AgNOmmol5905.0OHCmmol


KSCNmmol05411.0

mL00.25mL

AgNOmmol02979.0reactedAgNOmmol


ed. Chapter 13

13-28

4322223

32333

NH6)(SeOAg)(SeAg2OH3)(Se3)NH(Ag6

NO)NH(AgNH2AgNO

sss

samplemL/Semg94.7mL00.5

mmol

Semg96.78Semmol503.0

Semmol503.0)(SeAgmmol2

)(Semmol3

)NH(Agmmol2

)(SeAgmmol1

AgNOmmol1

)NH(Agmmol1AgNOmmol6707.0)(SeAgfromSemmol


KSCNmmol01370.0

mL00.25mL

AgNOmmol0360.0)(SeAgformtoreactedAgNOmmol

223

2

3

2332

3

323

s

ss

s

s

13-29

4

44

4

4

3

434

3

3

ClO%65.55%100

mL0.250

mL00.50sampleg998.1

mmol1000

ClOg45.99ClOmmol236.2

ClO%

Cl%60.10%100

mL0.250

mL00.50sampleg998.1

mmol1000

Clg453.35Clmmol195.1

Cl%

ClOmmol236.2

AgNOmmol1

ClOmmol1)mL97.13mL12.40(

mL

AgNOmmol08551.0ClOmmol

Clmmol195.1AgNOmmol1

Clmmol1g97.13

mL

AgNOmmol08551.0Clmmol


ed. Chapter 13

13-30 (a) The equivalence point occurs at 50.0 mL,

mL00.50SCNNHmmol02500.0

mL1

Agmmol1

SCNNHmmol1Agmmol250.1SCNmL

Agmmol250.1mL00.25mL

AgNOmmol05000.0Agmmol

4

4

3

At 30.00 mL,

SCNM102.11009.9/101.11009.9/K]SCN[

04.21009.9logpAg

AgM1009.9mL00.30mL00.25

mL00.30mL

SCNmmol0250.0Agmmol250.1

]Ag[

103123

sp

3

3

Proceeding in the same way, we obtain the data for 40.00 mL and 49.00 mL. The results

are displayed in the spreadsheet at the end of the solution.

At 50.00 mL,

98.5)1005.1log(pAg

M1005.1101.1K]SCN[]Ag[

6

612

sp

At 51.00 mL,

48.8)103.3log(pAg

M103.31029.3/101.1]Ag[

M1029.3mL00.25mL00.51

mmol250.1mL00.51mL

SCNmmol0250.0

]SCN[

9

9412

4

At 60.00 mL and 70.00 mL, pAg is calculated in the same way and the results are

displayed in the spreadsheet below.


ed. Chapter 13

A B C D E F

1 Problem 13-30(a)

2 The equivalence point occurs at 0.05000 mmol/mL X

3 Conc. AgNO3 0.05000 25.00 mL X (1 mL/0.02500 mmol) = 50.00 mL SCN-

4 Vol. AgNO3 25.00

5 Conc. KSCN 0.02500

6 Ksp 1.10E-12

7 Vol. SCN- [Ag

+] [SCN

-] pAg

8 30.00 9.09E-03 1.21E-10 2.041

9 40.00 3.85E-03 2.86E-10 2.415

10 49.00 3.38E-04 3.26E-09 3.471

11 50.00 1.05E-06 1.05E-06 5.979

12 51.00 3.34E-09 3.29E-04 8.48

13 60.00 3.74E-10 2.94E-03 9.43

14 70.00 2.09E-10 5.26E-03 9.68

15


17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8

18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)

19 B12=$B$6/C12 D8 = -LOG(B8)

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36


ed. Chapter 13

(b) Proceeding as in part (a), we obtain the results in the spreadsheet below.

A B C D E F

1 Problem 13-30(b)


3 Conc. AgNO3 0.06000 20.00 mL X (1 mL/0.03000 mmol) = 40.00 mL I-

4 Vol. AgNO3 20.00

5 Conc. KI 0.03000

6 Ksp 8.30E-17

7 Vol. I- [Ag

+] [I

-] pAg

8 20.00 1.50E-02 5.53E-15 1.824

9 30.00 6.00E-03 1.38E-14 2.222

10 39.00 5.08E-04 1.63E-13 3.294

11 40.00 9.11E-09 9.11E-09 8.04

12 41.00 1.69E-13 4.92E-04 12.77

13 50.00 1.94E-14 4.29E-03 13.71

14 60.00 1.11E-14 7.50E-03 13.96

15


17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8

18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)

19 B12=$B$6/C12 D8 = -LOG(B8)

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36


ed. Chapter 13

(c) Proceeding as in part (a), we obtain the results in the spreadsheet below.

A B C D E F

1 Problem 13-30(c)


3 Conc. AgNO3 0.07500 30.00 mL X (1 mL/0.07500 mmol) = 30.00 mL CI-

4 Vol. AgNO3 30.00

5 Conc. NaCl 0.07500

6 Ksp 1.82E-10

7 Vol. CI- [Ag

+] [CI

-] pAg

8 10.00 3.75E-02 4.85E-09 1.426

9 20.00 1.50E-02 1.21E-08 1.824

10 29.00 1.27E-03 1.43E-07 2.896

11 30.00 1.35E-05 1.35E-05 4.87

12 31.00 1.48E-07 1.23E-03 6.83

13 40.00 1.70E-08 1.07E-02 7.77

14 50.00 9.71E-09 1.88E-02 8.01

15


17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8

18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)

19 B12=$B$6/C12 D8 = -LOG(B8)

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36


ed. Chapter 13

(d) The equivalence point occurs at 70.00 mL,

23

23

2

4

12

2

4

1422

4

)NO(PbmL00.70)NO(Pbmmol2000.0

mLSOmmol10400.1PbmL

SOmmol10400.1mL00.35mL

SONammol4000.0SOmmol

At 50.00 mL,

47.6)104.3log(pPb

PbM104.310706.4/106.1]Pb[

SOM10706.4)mL00.50mL00.35(

mL00.50mL

)NO(Pbmmol2000.0mmol10400.1

]SO[

7

27282

2

4

2

231

2

4

At 60.00 mL and 69.00 mL, the calculations are made in the same way and the results

are shown in the following spreadsheet.

At 70.00 mL,

90.3)103.1log(pPb

PbM103.1106.1K]SO[]Pb[

4

248

sp

2

4

2

At 71.00 mL,

7243.2)10887.1log(pPb

SOM105.810887.1/106.1]SO[

PbM10887.1mL00.71mL00.35

SOmmol10400.1mL00.71mL

)NO(Pbmmol2000.0

]Pb[

3

2

4

6382

4

23

2

4

123

2

At 80.00 mL and 90.00 mL, the calculations are made in the same way and the results

are shown in spreadsheet below.


ed. Chapter 13

A B C D E F

1 Problem 13-30(d)


3 Conc. Na2SO4 0.4000 35.00 mL X (1 mL/0.2000 mmol) = 70.00 mL Pb2+

4 Vol. Na2SO4 35.00

5 Conc. Pb(NO3)2 0.2000

6 Ksp 1.60E-08

7 Vol. Pb2+

[SO42-] [Pb

2+] pPb

8 50.00 4.71E-02 3.40E-07 6.469

9 60.00 2.11E-02 7.60E-07 6.119

10 69.00 1.92E-03 8.32E-06 5.080

11 70.00 1.26E-04 1.26E-04 3.898

12 71.00 8.48E-06 1.89E-03 2.724

13 80.00 9.20E-07 1.74E-02 1.760

14 90.00 5.00E-07 3.20E-02 1.495

15


17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8

18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)

19 B12=$B$6/C12 D8 = -LOG(D8)

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36


ed. Chapter 13

(e) Proceeding as in part (a), we obtain the results in the spreadsheet below.

A B C D E F

1 Problem 13-30(e)


3 Conc. BaCl2 0.0250 40.00 mL X (1 mL/0.05000 mmol) = 20.00 mL SO42-

4 Vol. BaCl2 40.00

5 Conc. Na2SO4 0.0500

6 Ksp 1.10E-10

7 Vol. SO42- [Ba

2+] [SO4

2-] pBa

8 0.00 2.50E-02 1.602

9 10.00 1.00E-02 1.10E-08 2.000

10 19.00 8.47E-04 1.30E-07 3.072

11 20.00 1.05E-05 1.05E-05 4.979

12 21.00 1.34E-07 8.20E-04 6.872

13 30.00 1.54E-08 7.14E-03 7.812

14 40.00 8.80E-09 1.25E-02 8.056

15


17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8

18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)

19 B12=$B$6/C12 D8 = -LOG(B8)

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36


ed. Chapter 13

(f) Proceeding as in part (d), we obtain the results in the spreadsheet below.

A B C D E F

1 Problem 13-30(f)


3 Conc. NaI 0.2000 50.00 mL X (1 mL/0.4000 mmol) = 25.00 mL Tl-

4 Vol. NaI 50.00

5 Conc. TlNO3 0.4000

6 Ksp 6.50E-08

7 Vol. Tl+ [I

-] [Tl

+] pTl

8 5.00 1.45E-01 4.47E-07 6.350

9 15.00 6.15E-02 1.06E-06 5.976

10 24.00 5.41E-03 1.20E-05 4.920

11 25.00 2.55E-04 2.55E-04 3.594

12 26.00 1.24E-05 5.26E-03 2.279

13 35.00 1.38E-06 4.71E-02 1.327

14 45.00 7.72E-07 8.42E-02 1.075

15


17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8

18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)

19 B12=$B$6/C12 D8 = -LOG(C8)

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

13-31 (Note: In the first printing of the text, the last answer for pAg in the back of the book was

in error.)


ed. Chapter 13

KBrmmol00.2mL0.50mL

KBrmmol0400.0KBrmmol

At 5.00 mL,

80.10)106.1log(pAg

AgM106.11018.3/100.5]Br/[K]Ag[

M1018.3mL00.5mL0.50

mL00.5mL

AgNOmmol0500.0mmol00.2

]Br[

11

11213

sp

2

3

At 15.00 mL, 25.00 mL, 30.00 mL, 35.00 mL and 39.00 mL, the calculations are

performed in the same way and the results are shown in the spreadsheet at the end of this

solution.

At 40.00 mL,

15.6)101.7log(pAg

AgM101.7100.5K]Br[]Ag[

7

713

sp

At 41.00 mL,

260.3)1049.5log(pAg

AgM1049.5mL00.41mL0.50

Brmmol00.2mL00.41mL

AgNOmmol0500.0

]Ag[

4

4

3

At 45.00 mL and 50.00 mL, the calculations are performed in the same way and the

results are shown in the spreadsheet that follows.


ed. Chapter 13

A B C D E F

1 Problem 13-31


3 Conc. AgNO3 0.05000 50.00 mL X (1 mL/0.05000 mmol) = 40.00 mL Ag+

4 Vol. KBr 50.00

5 Conc. KBr 0.04000

6 Ksp 5.00E-13

7 Vol. Ag+ [Br

-] [Ag

+] pAg

8 5.00 3.18E-02 1.57E-11 10.804

9 15.00 1.92E-02 2.60E-11 10.585

10 25.00 1.00E-02 5.00E-11 10.301

11 30.00 6.25E-03 8.00E-11 10.097

12 35.00 2.94E-03 1.70E-10 9.770

13 39.00 5.62E-04 8.90E-10 9.051

14 40.00 7.07E-07 7.07E-07 6.151

15 41.00 7.28E+01 5.49E-04 3.260

16 45.00 1.52E+01 2.63E-03 2.580

17 50.00 8.00E+00 5.00E-03 2.301

18


20 B8=(($B$5*$B$4)-($B$3*A8))/($B$4+A8) C8=$B$6/B8

21 B14=SQRT($B$6) C15=(($B$3*A15)-($B$4*$B$5))/($B$4+A15)

22 B15=$B$6/C15 D8 = -LOG(C8)

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39


ed. Chapter 13

Challenge Problem

]SCN][Fe[

])SCN(Fe[1005.1)SCN(FeSCNFe

3

23

f

23

K

For part (a) we find,

%81.0%100Agmol101588.1

SCNmol

Agmol1

)SCN(Femol

SCNmol1)SCN(Femol104030.9

Error%

)SCN(Femol104030.9

L106353.4mL1000

LmL00.50

L

)SCN(Femol10759.9)SCN(Femol

10759.91005.1

101

SCNL106353.4mol025.0

L

Agmol

SCNmol1Agmol101588.1SCNL

Agmol101588.1g8682.107

Agmol1Agg125.0Agmol

Agg125.0mL00.50mL100

g250.0%250.0Agmass

3

2

26

26

225

2

5

3

5

)SCN(Fe

23

3

2

c

Proceeding the same way for parts (b) and (c), we find the results in the following spreadsheet.


ed. Chapter 13

A B C D E F G

1 Problem 13-32

2

3 mL taken 50

4 Kf 1.05E+03

5 conc SCN 0.025

6 AW Ag 107.8682

7 min complx 1.00E-05

8 %Ag g Ag moles Ag L SCN- c SCN cmplx mol SCN cmplx %Error

9 (a) 0.25 0.125 0.0011588 0.046353 9.759E-05 9.40308E-06 0.811434

10 (b) 0.1 0.05 0.0004635 0.018541 9.759E-05 6.68893E-06 1.443046

11 (c) 0.05 0.025 0.0002318 0.009271 9.759E-05 5.78422E-06 2.495732

12


14 B9=$B$3*(A9/100) E9=SQRT($B$7/$B$4)

15 C9=B9/$B$6 F9=E9*(($B$3/1000)+D9)

16 D9=C9/$B$5 G9=F9/C9*100