Chapter 4

Fundamentals of Analytical Chemistry: 8th

ed. Chapter 4

Chapter 4

4-1 (a) The millimole is an amount of a chemical species, such as an atom, an ion, a molecule

or an electron. A millimole contains

millimole

particles1002.6

millimole

mole10

mole

particles1002.6 20323

×=∗×−

(b) The molar mass is the mass in grams of one mole of a chemical species.

(c) The millimolar mass is the mass in grams of one millimole of a chemical species.

(d) Parts per million, cppm, is a term expressing the concentration of very dilute solutions.

Thus,

cppm ppm10solution of mass

solute of mass 6×=

The units of mass in the numerator and the denominator must be the same.

4-2 The species molarity of a solution expresses the equilibrium concentration of a chemical

species in terms of moles per liter. The analytical molarity of a solution gives the total

number of moles of a solute in one liter. The species molarity takes into account

chemical reactions that occur in solution. The analytical molarity specifies how the

solution was prepared, but does not account for any subsequent reactions.

4-3 33

33

m10cm100

m

mL

cm1

L1

mL1000 1L −

=

××=

3333 m10

mole1

m10

L

L

mole1M1

−−=×=


ed. Chapter 4

4-4 (a) kHz 320Hz1000

kHzHz102.3 5

=××

(b) ng 6.45g

ng10g1056.4

98

=××−

(c) mmol 843mol10

mmolmol1043.8

3

5=

µ×µ×

(d) Ms5.6s10

Mss105.6

6

6=××

(e) m 6.89nm10

mnm1096.8

3

4µ=

µ××

(f) kg 72g1000

kgg000,72 =×

4-5 +

+

++

×=×

××× Na1098.5Namol

Na1002.6

PONamol

Namol3

g94.163

PONamol1PONag43.5 22

23

43

4343

4-6 +

+

++

×=×

×× K1022.1Kmol

K1002.6

POKmol

Kmol3POKmol76.6 25

23

43

43

4-7 (a) 32

32

3232 OBmol0712.0

OBg62.69

OBmolOBg96.4 =×

(b)

O10HOBNamol1073.8

381.37g

O10HOBNamol

mg1000

gOH10OBNamg333

2742

4

27422742

⋅×=

⋅××•

−

(c) 43

43

4343 OMnmol0382.0

OMng81.228

OMnmolOMng75.8 =×

(d) 42

3

42

4242 OCaCmol1031.1

OCaCg128.10

OCaCmol

mg1000

gOCaCmg2.167 −

×=××


ed. Chapter 4

4-8 (a) 52

52

5252 OPmmol40.0

mol

mmol1000

OPg94.141

OPmol

mg1000

gOPmg57 =×××

(b) 2

2

22 COmmol6.293

mol

mmol1000

COg01.44

COmolCOg92.12 =××

(c) 3

3

33 NaHCOmmol476

mol

mmol1000

NaHCOg01.84

NaHCOmolNaHCOg0.40 =××

(d)

44

44

4444

POMgNHmmol2.6

mol

mmol1000

POMgNHg137.32

POMgNHmol

mg1000

gPOMgNHmg850

=

×××

4-9 (a)

4

4

3

4

3

KMnOmmol50.6

L00.2mol

mmol1000

L

KMnOmol1025.3M KMnO1025.3

=

×××

≡×

−

−

(b)

KSCNmmol6.41

mL750mL1000

L

mol

mmol1000

L

KSCNmol0555.0 M KSCN0555.0

=

×××≡

(c)

4

3

4

444

CuSOmmol1047.8mL250mL1000

L

mol

mmol1000

CuSOg61.159

CuSOmol

mg1000

g

L

CuSOmg41.5CuSO ppm 41.5

−×=××

×××≡

(d) KClmmol6.1165L50.3mol

mmol1000

L

KClmol333.0 M KCl333.0 =××≡


ed. Chapter 4

4-10 (a)

4

44

HClOmmol0.56

mL175mL1000

L

mol

mmol1000

L

HClOmol320.0HClOM320.0

=

×××≡

(b)

42

42

3

42

3

CrOKmmol121

L0.15mol

mmol1000

L

CrOKmol1005.8CrOKM1005.8

=

×××

≡×

−

−

(c)

3

3

333

AgNOmmol199.0L00.5

mol

mmol1000

AgNOg87.169

AgNOmol

mg1000

g

L

AgNOmg75.6AgNOppm75.6

=

××××≡

(d)

KOHmmol0.17

mL851mL1000

L

mol

mmol1000

L

KOHmol0200.0KOHM0200.0

=

×××≡

4-11 (a) 3

4

3

33 HNOmg1090.4

g

mg1000

HNOmol

HNOg01.63HNOmol777.0 ×=××

(b) MgOmg10015.2g

mg1000

MgOmol

MgOg40.30

mmol1000

molMgOmmol500 4

×=×××

(c) 34

6

34

3434 NONHmg1080.1

g

mg1000

NONHmol

NONHg04.80NONHmol5.22 ×=××

(d)

6324

6

6324

63246324

)NO(Ce)NH(mg1037.2

g

mg1000

)NO(Ce)NH(mol

)NO(Ce)NH(g23.548)NO(Ce)NH(mol32.4

×=

××


ed. Chapter 4

4-12 (a) KBrg840KBrmol

KBrg0.119KBrmol1.7 =×

(b) PbOg49.4PbOmol

PbOg20.223

mmol1000

molPbOmmol1.20 =××

(c) 4

4

44 MgSOg452

MgSOmol

MgSOg37.120MgSOmol76.3 =×

(d)

OH6)SO()NH(Feg8.3

OH6)SO()NH(Femol

OH6)SO()NH(Feg23.392

mmol1000

molOH6)SO()NH(Femmol6.9

22424

22424

2242422424

⋅=

⋅

⋅××⋅

4-13 (a)

sucrosemg1022.2

mL0.26g

mg1000

sucrosemol

sucroseg342

mL1000

L

L

sucrosemol250.0sucroseM250.0

3×=

××××≡

(b)

22

22

2222

3

22

3

OHmg8.472

L92.2g

mg1000

OHmol

OHg02.34

L

OHmol1076.4OHM1076.4

=

××××

≡×

−

−

(c)

2323

23 )NO(Pbmg25.3mL656mL1000

L

L

)NO(Pbmg96.4)NO(Pbppm96.4 =××≡

(d)

3

333

KNOmg2.42

mL75.6mL1000

L

g

mg1000

mol

KNOg10.101

L

KNOmol0619.0KNOM0619.0

=

××××≡


ed. Chapter 4

4-14 (a)

22

22

222222

OHg51.2

mL450OHmol

OHg34.02

mL1000

L

L

OHmol0.164OHM164.0

=

×××≡

(b)

acidbenzoicg1088.2mL0.27

acidbenzoicmol

acidbenzoicg122

mL1000

L

L

acidbenzoicmol1075.8acidbenzoicM1075.8

3

44

−

−

−

×=

××××

≡×

(c) 22

2 SnClg0760.0L50.3mg1000

g

L

SnClmg7.21SnClppm7.21 =××≡

(d)

3

3

333

KBrOg0453.0

mL7.21KBrOmol

KBrOg167

mL1000

L

L

KBrOmol0125.0KBrOM0125.0

=

×××≡

4-15 (a)

077.1)2(923.0

)10log()38.8log(

)M1038.8log()M0838.0log()M0503.0M0335.0log(pNa

2

2

=−−−=

−−=

×−=−=+−=

−

−

475.1)2(525.0

)10log()35.3log(

)M1035.3log()M0335.0log(pCl

2

2

=−−−=

−−=

×−=−=

−

−

298.1)2(702.0

)10log()03.5log(

)M1003.5log()M0503.0log(pOH

2

2

=−−−=

−−=

×−=−=

−

−


ed. Chapter 4

(b)

116.2)3(884.0

)10log()65.7log(

)M1065.7log(pBa

3

3

=−−−=

−−=

×−=

−

−

188.0)M54.1log(pMn −=−=

490.0)M08.3log())M54.12(M1065.7log(pCl 3−=−=×+×−=

−

(c)

222.0)1(778.0

)10log()00.6log(

)M1000.6log()M600.0log(pH

1

1

=−−−=

−−=

×−=−=

−

−

096.0)1(904.0

)10log()02.8log(

)M1002.8log()M802.0log())M101.02(M600.0log(pCl

1

1

=−−=

−−=

×−=−=×+−=

−

−

996.0)1(00432.0

)10log()01.1log(

)M1001.1log()M101.0log(pZn

1

1

=−−−=

−−=

×−=−=

−

−

(d)

320.1)2(679.0

)10log()78.4log(

)M1078.4log(pCu

2

2

=−−−=

−−=

×−=

−

−

983.0)1(0170.0

)10log()04.1log(

)M1004.1log()M104.0log(pZn

1

1

=−−−=

−−=

×−=−=

−

−

517.0)1(483.0

)10log()04.3log(

)M1004.3log()M304.0log())M104.02()M0478.02log((pNO

1

1

3

=−−−=

−−=

×−=−=×+×−=

−

−


ed. Chapter 4

(e)

836.5)6(164.0

)10log()46.1log(

)M1046.1log()M1012.4))M1062.2(4log((pK

6

677

=−−−=

−−=

×−=×+××−=

−

−−−

385.6)7(615.0

)10log()12.4log(

)M1012.4log(pOH

7

7

=−−−=

−−=

×−=

−

−

582.6)7(418.0

)10log()62.2log(

)M1062.2log()CN(pFe

7

7

6

=−−−=

−−=

×−=

−

−

(f)

171.3)4(829.0

)10log()75.6log(

)M1075.6log(pH

4

4

=−−−=

−−=

×−=

−

−+

475.3)4(525.0

)10log()35.3log(

)M1035.3log(pBa

4

4

=−−−=

−−=

×−=

−

−

873.2)3(127.0

)10log()34.1log(

)M1034.1log()M1075.6)M1035.3(2log(pClO

3

344

4

=−−−=

−−=

×−=×+××−=

−

−−−

4-16 (a) M107.1)5(antilog)240.0(antilog]OH[ 5

3

−+×=−×=

as in part (a)

(b) M106.2]OH[ 5

3

−+×=

(c) M30.0]OH[ 3 =+


ed. Chapter 4

(d) M104.2]OH[ 14

3

−+×=

(e) M108.4]OH[ 8

3

−+×=

(f) M107.1]OH[ 6

3

−+×=

(g) M04.2]OH[ 3 =+

(h) M3.3]OH[ 3 =+

4-17 (a)

699.1)2(301.0

)10log()00.2log()M0200.0log(pBrpNa 2

=−−−=

−−=−==−−

pH = pOH = - log(1.0×10-7

M) = 7.00

(b)

000.2)2(000.0

)10log()00.1log()M0100.0log(pBa 2

=−−=

−−=−=−

699.1)2(301.0

)10log()00.2log()M0100.02log(pBr 2

=−−−=

−−=×−=−

pH = pOH = - log(1.0×10-7

M) = 7.00

(c)

46.2)3(54.0

)10log()5.3log()M105.3log(pBa 33

=−−−=

−−=×−=−−

15.2)3(84.0

)10log()0.7log()M100.7log()M105.3(2log(pOH 333

=−−−=

−−=×−=××−=−−−

pH = 14.00 – pOH = (14.00 – 2.15) = 11.85


ed. Chapter 4

(d)

40.1)2(60.0

)10log()0.4log()M100.4log()M040.0log(pH 22

=−−−=

−−=×−=−=−−

70.1)2(30.0

)10log()0.2log()M100.2log()M020.0log(pNa 22

=−−−=

−−=×−=−=−−

22.1)2(78.0

)10log()0.6log(

)M100.6log()M060.0log()M020.0M040.0log(pCl

2

2

=−−−=

−−=

×−=−=+−=

−

−

pOH = 14.00 – 1.40 = 12.60

(e)

17.2)3(83.0

)10log()7.6log()M107.6log(pCa 33

=−−−=

−−=×=−−

12.2)3(88.0

)10log()6.7log()M106.7log(pBa 33

=−−−=

−−=×−=−−

54.1)2(46.0

)10log()9.2log(

)M109.2log()))M106.7(2())M107.6(2log((pCl

2

233

=−−−=

−−=

×−=××+××−=

−

−−−

pH = pOH = - log(1.0×10-7

M) = 7.00

(f)

32.7)8(68.0

)10log()8.4log()M108.4log(pZn 88

=−−−=

−−=×−=−−

25.6)7(75.0

)10log()6.5log()M106.5log(pCd 77

=−−−=

−−=×−=−−


ed. Chapter 4

92.5)6(79.0

)10log()2.1log(

)M102.1log()))M106.5(2())M108.4(2log((pNO

6

678

3

=−−−=

−−=

×−=××+××−=

−

−−−

pH = pOH = - log(1.0×10-7

M) = 7.00

4-18 (a) M1014.2)10(antilog)33.0(antilog]OH[ 10

3

−+×=−×=

as in part (a),

(b) M733.0]OH[ =−

(c) M92.0]Br[ =−

(d) M105.4]Ca[ 132 −+×=

(e) [Li+] = 1.66 M

(f) M107.1]NO[ 8

3

−−×=

(g) M99.0]Mn[ 2=

+

(h) [Cl-] = 0.0955 M

4-19 (a)

+

+

+=××××× NaM0479.0

g99.22

Namol

L

mL1000

mL

g02.1

ppm10

1Nappm1008.1

6

3

−−

−

−×=××××

2

4

3

3

4

6

2

4 SOM1087.2g96.06

SOmol

L

mL1000

mL

g1.02

ppm10

1SOppm270

(b)

320.1)2(680.0

)10log()79.4log()M1079.4log(pNa 22

=−−−=

−−=×−=−−


ed. Chapter 4

543.2)3(458.0

)10log()87.2log()M1087.2log(pSO 33

4

=−−−=

−−=×−=−−

4-20 (a)

+−

++

×=××× KM106.4g10.39

Kmol

mg1000

g

L

mL1000

mL100

Kmg18 3

−

−−

=××× ClM103.0g45.35

Clmol

mg1000

g

L

mL1000

mL100

Clmg365

(b)

34.2)3(66.0

)10log()6.4log()M106.4log(pK 33

=−−−=

−−=×−=−−

pCl = –log(1.03×10-1

M) = –log(1.03)–log(10-1

)

= –0.0133–(–1) = 0.987

4-21 (a)

OH6MgClKClM01037.0g85.277

OH6MgClKClmol

L00.2

OH6MgClKClg76.522

2222 ⋅⋅=⋅⋅

×⋅⋅

(b)

+

+

=⋅⋅

×⋅⋅2

22

2

22 MgM01037.0OH6MgClKClmol

MgmolOH6MgClKClM01037.0

(c) −

−

=⋅⋅

×⋅⋅ ClM0311.0OH6MgClKClmol

Clmol3OH6MgClKClM01037.0

22

22

(d) (w/v)%288.0%100mL1000

L

L00.2

OH6MgClKClg76.5 22 =××⋅⋅


ed. Chapter 4

(e)

−

−

−=×××≡ Clmmol777.0mL25.0

mol

mmol1000

mL1000

L

L

Clmol0.0311ClM0311.0

(f)

+

+

++

≡=

××⋅⋅

×⋅⋅

Kppm405L

mg405

g

mg1000

Kmol

Kg10.39

OH6MgClKClmol1

Kmol1OH6MgClKClM01037.0

22

22

(g) 984.1)2(0170.0)10log()04.1log()M1004.1log(pMg 22=−−−=−−=×−=

−−

(h) 507.1)2(494.0)10log()12.3log()M1012.3log(pCl 22=−−−=−−=×−=

−−

4-22 (a)

63

3636363 )CN(FeKM1074.4g2.329

)CN(FeKmol

L775

)CN(FeKg1210

mL775

)CN(FeKmg1210 −×=×≡

(b) +

+

−=×× KM0142.0

)CN(FeKmol

Kmol3)CN(FeKM1074.4

63

63

3

(c) −−

−

−×=××

3

6

3

63

3

663

3 )CN(FeM1074.4)CN(FeKmol

)CN(Femol)CN(FeKM1074.4

(d) (w/v)%156.0%100mg1000

g

mL775

)CN(FemgK1210 63 =××

(e)

+

+

++−

+−

=

××××

≡×

Kmmol710.0

mL50.0Kmol

mmolK1000

mL1000

L

L

Kmol101.42KM1042.1

22


ed. Chapter 4

(f)

−−

−

−

−−

−

×≡=××

××

≡×

3

6

33

63

6

3

6

63

3

663

3

63

3

Fe(CN)ppm1000.1Fe(CN)L

mg1000

g

mg1000

Fe(CN)mol

Fe(CN)g211.95

Fe(CN)Kmol

Fe(CN)mol

L

Fe(CN)Kmol104.74Fe(CN)KM1074.4

(g) 848.1)2(152.0)10log()42.1log()M1042.1log(pK 22=−−−=−−=×−=

−−

(h) 324.2)3(676.0)10log()74.4log()M1074.4log()CN(pFe 333

6 =−−−=−−=×−=−−−

4-23 (a)

33

333333

)NO(FeM281.0

g86.241

)NO(Femol

L

mL1000

mL

g059.1

100

1

solutiong

)NO(Feg42.6)NO(Fe%42.6

=

××××≡

(b)

−

−

=×≡ 3

33

33333 NOM844.0

)NO(Femol

NOmol3

L

)NO(Femol281.0)NO(FeM281.0

(c)

L

)NO(Feg0.68

mol

)NO(Feg86.241

L

)NO(Femol281.0)NO(FeM281.0 333333

33 =×≡

4-24 (a)

222

2 NiClM11.1g61.129

NiClmol

L

mL1000

mL

g149.1

100

1

solutiong

NiClg5.12NiCl%5.12 =××××≡

(b) −

−

=×≡ ClM22.2NiClmol

Clmol2

L

NiClmol11.1NiClM11.1

2

22

(c) 222

3 NiClg144Lmol

NiClg61.129

L

NiClmol11.1NiCl11.1 =××≡


ed. Chapter 4

4-25 (a)

OHHCg8.23solnmL500100

1

solnmL

OHHCg75.4OHHC)v/w(%75.4 52

5252 =××≡

Weigh 23.8 g ethanol and add enough water to give a final volume of 500 mL.

(b)

waterg2.476OHHCg8.23solng500waterg

watergOHHCg23.8solng500

OHHCg8.23solng500100

1

solng

OHHCg75.4OHHC)w/w(%75.4

52

52

5252

52

=−=

+=

=××≡

x

x

Mix 23.8 g ethanol with 476.2 g water.

(c)

OHHCmL8.23solnmL500100

1

solnmL

OHHCmL75.4OHHC)v/v(%75.4 52

5252 =××≡

Dilute 23.8 mL ethanol with enough water to give a final volume of 500 mL.

4-26 (a)

383383

383 OHCg525solnL50.2L

mL1000

100

1

solnmL

OHCg0.21OHC)v/w(%0.21 =×××≡

Weigh 525 g glycerol and add enough water to give a final volume of 2.50 L.

(b)

waterkg98.1OHCkg525.0solnkg50.2waterkg

waterkgOHCgk0.525solnkg50.2

OHCg525solnkg50.2kg

g1000

100

1

solng

OHCg0.21OHC)w/w(%0.21

383

383

383383

383

=−=

+=

=×××≡

x

x

Mix 525 g glycerol with 1.98 kg water.


ed. Chapter 4

(c)

383383

383 OHCmL525solnL50.2L

mL1000

100

1

solnmL

OHCmL0.21OHC)v/v(%0.21 =×××≡

Dilute 525 mL glycerol with enough water to give a final volume of 2.50 L.

4-27

L300.0POHmol0.15

LPOHmol50.4requiredPOH)w/w(%86volume

L

POHmol0.15

g0.98

POHmol

L

mL1000

mL

waterg

waterg

reagentg71.1

100

1

reagentg

POHg86POH)w/w(%86

POHmol50.4mL750mL1000

L

L

POHmol00.6POHM00.6

43

4343

43

434343

4343

43

=×=

=

×××××≡

=××≡

Dilute 300 mL of the concentrated reagent to 750 mL using water.

4-28

L170.0HNOmol9.15

LHNOmol70.2requiredHNO%5.70volume

L

HNOmol9.15

g0.63

HNOmol

L

mL1000

mL

waterg

waterg

reagentg42.1

100

1

reagentg

HNOg5.70HNO)w/w(%5.70

HNOmol70.2mL900mL1000

L

L

HNOmol00.3HNOM00.3

3

33

3

333

33

3

=×=

=

×××××≡

=××≡

Dilute 170 mL of the concentrated reagent to 900 mL using water.

4-29 (a)

3

333

AgNOg37.6

mL500mL1000

L

mol

AgNOg87.169

L

AgNOmol0750.0AgNOM0750.0

=

×××≡

Dissolve 6.37 g AgNO3 in enough water to give a final volume of 500 mL.


ed. Chapter 4

(b)

HClL0475.0HClmol00.6

LHClmol285.0

HClmol285.0L1L

HClmol285.0HClM285.0

=×

=×≡

Take 47.5 mL of the 6.00 M HCl and dilute to 1.00 L using water.

(c)

6464642

2

)CN(FeKg98.2mol

)CN(FeKg43.368

Kmol4

)CN(FeKmolKmol1024.3

Kmol1024.3mL400mL1000

L

L

Kmol0810.0KM0810.0

=×××

×=××≡

+

+−

+−

+

+

Dissolve 2.98 g K4Fe(CN)6 in enough water to give a final volume of 400 mL.

(d)

L216.0BaClmol400.0

L

g23.208

BaClmolBaClg0.18

BaClg0.18mL600100

1

solnmL

BaClg00.3BaCl)v/w(%00.3

2

22

22

2

=××

=××≡

Take 216 mL of the 0.400 M BaCl2 solution and dilute to 600 mL using water.

(e)

L0203.0HClOmol8.11

LHClOmol240.0requiredHClO)w/w(%71volume

L

HClOmol8.11

g46.100

HClOmol

L

mL1000

mL

waterg

waterg

reagentg67.1

100

1

reagentg

HClOg71HClO)w/w(%71

HClOmol240.0L00.2L


4

44

4

444

44

4

=×=

=

×××××≡

=×≡

Take 20.3 mL of the concentrated reagent and dilute to a final volume of 2.00 L using

water.


ed. Chapter 4

(f)

424242 SONag67.1

mol

SONag04.142

Namol2

SONamolNamol02348.0

Namol02348.0g99.22

Namol

mg1000

gNamg540

Namg540L00.9solnL

Namg60Nappm60

=××

=××

=×≡

+

+

+

+

+

+

+

+

Dissolve 1.67 g Na2SO4 in enough water to give a final volume of 9.00 L.

4-30 (a)

444

4 KMnOg5.39mol

KMnOg03.158L00.5

L

KMnOmol0500.0KMnOM0500.0 =××≡

Dissolve 39.5 g KMnO4 in enough water to give a final volume of 5.00 L.

(b)

reagentL125.0reagentmol00.8

LHClOmol00.1

HClOmol00.1L00.4L


4

44

4

=×

=×≡

Take 125 mL of the 8.00 M reagent and dilute a final of volume of 4.00 L using water.

(c)

2222 MgIg39.1

mol

MgIg11.278

Imol2

MgImolImol1000.1

Imol0100.0mL400mL1000

L

L

Imol0250.0IM0250.0

=×××

=××≡

−

−−

−

−

−

Dissolve 1.39 g MgI2 in enough water to give a final volume of 400 mL.

(d)


ed. Chapter 4

L0343.0CuSOmol365.0

L

g60.159

CuSOmolCuSOg00.2

CuSOg00.2mL200100

1

solnmL

CuSOg00.1CuSO)v/w(%00.1

4

44

44

4

=××

=××≡

Take 34.3 mL of the 0.365 M CuSO4 and dilute to a final volume of 200 mL using water.

(e)

L0169.0NaOHmol062.19

LNaOHmol3225.0requiredNaOH)w/w(%50volume

L

NaOHmol062.19

g00.40

NaOHmol

L

mL1000

mL

waterg

waterg

reagentg525.1

100

1

reagentg

NaOHg50NaOH)w/w(%50

NaOHmol3225.0L50.1L

NaOHmol215.0NaOHM215.0

=×=

=

×××××≡

=×≡

Take 16.9 mL of the concentrated reagent and dilute to a final volume of 1.50 L using

water.

(f)

6464644

41

1

)CN(FeKg0424.0mol

)CN(FeKg35.368

Kmol4

)CN(FeKmolKmol1060.4

Kmol1060.4g10.39

Kmol

mg1000

gKmg108.1

Kmg108.1L50.1solnL

Kmg12Kppm0.12

=×××

×=×××

×=×≡

+

+−

+−

+

+

+

+

+

Dissolve 42.4 mg K4Fe(CN)6 in enough water to give a final volume of 1.50 L.

4-31

−−

−

−

+−

+

+

×=××≡

×=××≡

3

233

323

3

IOmol1027.2mL0.75mL1000

L

L

IOmol302.0IOM302.0

Lamol1025.1mL0.50mL1000

L

L

Lamol250.0LaM250.0

Because each mole of La(IO3)3 requires three moles IO3-, IO3

- becomes the limiting


ed. Chapter 4

reagent. Thus,

formed)IO(Lag01.5mol

)IO(Lag6.663

IOmol3

)IO(LamolIOmol1027.2 33

33

3

333

2=×××

−

−−

4-32

−−

−

−

+−

+

+

×=××≡

×=××≡

Clmol1000.7mL400mL1000

L

L

Clmol175.0ClM175.0

Pbmol1050.2mL200mL1000

L

L

Pbmol125.0PbM125.0

2

222

2

Because each mole of PbCl2 requires two moles Cl-, Pb

2+ becomes the limiting reagent.

Thus,

formedPbClg95.6mol

PbClg10.278

Pbmol

PbClmolPbmol1050.2 2

2

2

222=×××

+

+−

4-33 A balanced chemical equation can be written as shown below.

)(COOHNaCl2HCl2CONa 2232 g++→+

(a)

HClmol1031.7mL0.100mL1000

L

L

HClmol0731.0HClM0731.0

CONamol10094.2g99.105

CONamolCONag2220.0

3

32

33232

−

−

×=××≡

×=×

Because one mole of CO2 is evolved for every mole Na2CO3 reacted, Na2CO3 is the

limiting reagent. Thus,

evolvedCOg09218.0mol

COg01.44

CONamol

COmolCONamol10094.2 2

2

32

232

3=×××

−


ed. Chapter 4

(b)

HClM0312.0L

mL1000

mL0.100

HClmol1012.3

mol1012.3))mol10094.2(2(mol1031.7leftHClmol

3

333

=××

×=××−×=

−

−−−


433343 POHgNaNO3HgNO3PONa +→+

(a)

33

3

42

34343

HgNOmol05151.0mL0.100mL1000

L

L

HgNOmol5151.0HgNOM5151.0

PONamol1039.9mL0.25mL1000

L

L

PONamol3757.0PONaM3757.0

=××≡

×=××≡−

The limiting reagent is Na2PO4. Thus,

formedPOHgg54.6mol

POHgg74.696

PONamol

POHgmolPONamol1039.9 43

43

43

4342

3=×××

−

(b)

33

3

32

3

HgNOM187.0L

mL1000

mL0.125

HgNOmol0233.0

HgNOmol0233.0))mol1039.9(3(mol10151.5unreactedHgNOmol

=×

=××−×=−−


)(SOOHNaClO2HClO2SONa 224432 g++→+

(a)

3232

32 SONa mol 23490.0mL 00.75mL 1000

L

L

SONa mol 3132.0SONa M 3132.0 =××≡

44

4 HClOmol 06038.0mL 0.150mL 1000

L

L

HClOmol 4025.0 M HClO4025.0 =××≡

Because one mole SO2 is evolved per mole Na2SO3, Na2SO3 is the limiting reagent.


ed. Chapter 4

Thus,

evolved SO g 505.1moL

SO g 64.06

SONa mol

SO molSONa mol 02349.0 2

2

32

232 =××

(b)

44

44

M HClO0595.0L

mL 1000

mL 225.0

HClOmol 0.01340

HClOmol 13400.0mol)) (0.023492-mol (0.06038unreacted HClOmol

=×

=×=


−++++→++ Cl2Na2)(POMgNHNHPONaMgCl 444422 s

2

222

MgClmol02101.0

g21.95

MgClmolmL0.200

100

1

mL

MgClg000.1MgCl)v/w(%000.1

=

×××≡

42

34242 PONamol1001.7mL0.40

mL1000

L

L

PONamol1753.0PONaM1753.0 −

×=××≡

The Na2PO4 is the limiting reagent. Thus,

22

2

3

2

44443

44

MgClM0583.0L

mL1000

mL240.0

MgClmol0.0140

MgClmol0140.0)1001.702101.0(unreactedMgClmol

PO MgNHg 9628.0mol

PO MgNHg 137.351001.7ppt PO MgNHmass

=×

=×−=

=××=

−

−


−+++→+ 33 NOK)(AgIKIAgNO s

3

3

3

3

AgNOmL2930L

mL1000

AgNOmol0100.0

L

KImol

AgNOmolKImol02930.0

KImol02930.0g0.166

KImolmL0.200

mL

g

ppt10

1KIppt31.24

=×××

=××××

2930 mL of 0.0100 M AgNO3 would be required to precipitate all I- as AgI.


ed. Chapter 4


−+++→+ 3

3

434223 NO6Al2BaSO3)SO(Al)NO(Ba3

(a)

342

3

342342

23

3

23

623

)SO(Almol1018.6

mL0.200mL1000

L

L

)SO(Almol03090.0)SO(AlM03090.0

)NO(Bamol1038.1

g34.261

)NO(BamolmL0.750

mL

g

ppm10

1)NO(Bappm4.480

−

−

×=

××≡

×=

××××

The Ba(NO3)2 is the limiting reagent. Thus,

formedBaSOg322.0mol

BaSOg39.233

)NO(Bamol3

BaSOmol3)NO(Bamol1038.1 4

4

23

423

3=×××

−

(b)

342

3342

3

342

333

342

)SO(AlM1002.6L

mL1000

mL0.950

)SO(Almol1072.5

)SO(Almol1072.5))1038.1(3

1(1018.6(unreacted)SO(Almol

−

−

−−−

×=××

×=××−×=

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Chapter 4

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