Jan 05, 2016
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
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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.
Fundamentals of Analytical Chemistry: 8th
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.
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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.
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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(
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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
Fundamentals of Analytical Chemistry: 8th
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
AgNOmmol5905.0mL85.2mL
KSCNmmol05411.0
mL00.25mL
AgNOmmol02979.0reactedAgNOmmol
Fundamentals of Analytical Chemistry: 8th
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
AgNOmmol6707.0mL74.16mL
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
Fundamentals of Analytical Chemistry: 8th
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.
Fundamentals of Analytical Chemistry: 8th
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
16 Spreadsheet Documentation
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$6/C12 D8 = -LOG(B8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
(b) Proceeding as in part (a), we obtain the results in the spreadsheet below.
A B C D E F
1 Problem 13-30(b)
2 The equivalence point occurs at 0.06000 mmol/mL X
3 Conc. AgNO3 0.06000 20.00 mL X (1 mL/0.03000 mmol) = 40.00 mL I-
4 Vol. AgNO3 20.00
5 Conc. KI 0.03000
6 Ksp 8.30E-17
7 Vol. I- [Ag
+] [I
-] pAg
8 20.00 1.50E-02 5.53E-15 1.824
9 30.00 6.00E-03 1.38E-14 2.222
10 39.00 5.08E-04 1.63E-13 3.294
11 40.00 9.11E-09 9.11E-09 8.04
12 41.00 1.69E-13 4.92E-04 12.77
13 50.00 1.94E-14 4.29E-03 13.71
14 60.00 1.11E-14 7.50E-03 13.96
15
16 Spreadsheet Documentation
17 B8=(($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$6/C12 D8 = -LOG(B8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
(c) Proceeding as in part (a), we obtain the results in the spreadsheet below.
A B C D E F
1 Problem 13-30(c)
2 The equivalence point occurs at 0.07500 mmol/mL X
3 Conc. AgNO3 0.07500 30.00 mL X (1 mL/0.07500 mmol) = 30.00 mL CI-
4 Vol. AgNO3 30.00
5 Conc. NaCl 0.07500
6 Ksp 1.82E-10
7 Vol. CI- [Ag
+] [CI
-] pAg
8 10.00 3.75E-02 4.85E-09 1.426
9 20.00 1.50E-02 1.21E-08 1.824
10 29.00 1.27E-03 1.43E-07 2.896
11 30.00 1.35E-05 1.35E-05 4.87
12 31.00 1.48E-07 1.23E-03 6.83
13 40.00 1.70E-08 1.07E-02 7.77
14 50.00 9.71E-09 1.88E-02 8.01
15
16 Spreadsheet Documentation
17 B8=(($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$6/C12 D8 = -LOG(B8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
(d) The equivalence point occurs at 70.00 mL,
23
23
2
4
12
2
4
1422
4
)NO(PbmL00.70)NO(Pbmmol2000.0
mLSOmmol10400.1PbmL
SOmmol10400.1mL00.35mL
SONammol4000.0SOmmol
At 50.00 mL,
47.6)104.3log(pPb
PbM104.310706.4/106.1]Pb[
SOM10706.4)mL00.50mL00.35(
mL00.50mL
)NO(Pbmmol2000.0mmol10400.1
]SO[
7
27282
2
4
2
231
2
4
At 60.00 mL and 69.00 mL, the calculations are made in the same way and the results
are shown in the following spreadsheet.
At 70.00 mL,
90.3)103.1log(pPb
PbM103.1106.1K]SO[]Pb[
4
248
sp
2
4
2
At 71.00 mL,
7243.2)10887.1log(pPb
SOM105.810887.1/106.1]SO[
PbM10887.1mL00.71mL00.35
SOmmol10400.1mL00.71mL
)NO(Pbmmol2000.0
]Pb[
3
2
4
6382
4
23
2
4
123
2
At 80.00 mL and 90.00 mL, the calculations are made in the same way and the results
are shown in spreadsheet below.
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
A B C D E F
1 Problem 13-30(d)
2 The equivalence point occurs at 0.4000 mmol/mL X
3 Conc. Na2SO4 0.4000 35.00 mL X (1 mL/0.2000 mmol) = 70.00 mL Pb2+
4 Vol. Na2SO4 35.00
5 Conc. Pb(NO3)2 0.2000
6 Ksp 1.60E-08
7 Vol. Pb2+
[SO42-] [Pb
2+] pPb
8 50.00 4.71E-02 3.40E-07 6.469
9 60.00 2.11E-02 7.60E-07 6.119
10 69.00 1.92E-03 8.32E-06 5.080
11 70.00 1.26E-04 1.26E-04 3.898
12 71.00 8.48E-06 1.89E-03 2.724
13 80.00 9.20E-07 1.74E-02 1.760
14 90.00 5.00E-07 3.20E-02 1.495
15
16 Spreadsheet Documentation
17 B8=(($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$6/C12 D8 = -LOG(D8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
(e) Proceeding as in part (a), we obtain the results in the spreadsheet below.
A B C D E F
1 Problem 13-30(e)
2 The equivalence point occurs at 0.02500 mmol/mL X
3 Conc. BaCl2 0.0250 40.00 mL X (1 mL/0.05000 mmol) = 20.00 mL SO42-
4 Vol. BaCl2 40.00
5 Conc. Na2SO4 0.0500
6 Ksp 1.10E-10
7 Vol. SO42- [Ba
2+] [SO4
2-] pBa
8 0.00 2.50E-02 1.602
9 10.00 1.00E-02 1.10E-08 2.000
10 19.00 8.47E-04 1.30E-07 3.072
11 20.00 1.05E-05 1.05E-05 4.979
12 21.00 1.34E-07 8.20E-04 6.872
13 30.00 1.54E-08 7.14E-03 7.812
14 40.00 8.80E-09 1.25E-02 8.056
15
16 Spreadsheet Documentation
17 B8=(($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$6/C12 D8 = -LOG(B8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
(f) Proceeding as in part (d), we obtain the results in the spreadsheet below.
A B C D E F
1 Problem 13-30(f)
2 The equivalence point occurs at 0.2000 mmol/mL X
3 Conc. NaI 0.2000 50.00 mL X (1 mL/0.4000 mmol) = 25.00 mL Tl-
4 Vol. NaI 50.00
5 Conc. TlNO3 0.4000
6 Ksp 6.50E-08
7 Vol. Tl+ [I
-] [Tl
+] pTl
8 5.00 1.45E-01 4.47E-07 6.350
9 15.00 6.15E-02 1.06E-06 5.976
10 24.00 5.41E-03 1.20E-05 4.920
11 25.00 2.55E-04 2.55E-04 3.594
12 26.00 1.24E-05 5.26E-03 2.279
13 35.00 1.38E-06 4.71E-02 1.327
14 45.00 7.72E-07 8.42E-02 1.075
15
16 Spreadsheet Documentation
17 B8=(($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$6/C12 D8 = -LOG(C8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
13-31 (Note: In the first printing of the text, the last answer for pAg in the back of the book was
in error.)
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
KBrmmol00.2mL0.50mL
KBrmmol0400.0KBrmmol
At 5.00 mL,
80.10)106.1log(pAg
AgM106.11018.3/100.5]Br/[K]Ag[
M1018.3mL00.5mL0.50
mL00.5mL
AgNOmmol0500.0mmol00.2
]Br[
11
11213
sp
2
3
At 15.00 mL, 25.00 mL, 30.00 mL, 35.00 mL and 39.00 mL, the calculations are
performed in the same way and the results are shown in the spreadsheet at the end of this
solution.
At 40.00 mL,
15.6)101.7log(pAg
AgM101.7100.5K]Br[]Ag[
7
713
sp
At 41.00 mL,
260.3)1049.5log(pAg
AgM1049.5mL00.41mL0.50
Brmmol00.2mL00.41mL
AgNOmmol0500.0
]Ag[
4
4
3
At 45.00 mL and 50.00 mL, the calculations are performed in the same way and the
results are shown in the spreadsheet that follows.
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
A B C D E F
1 Problem 13-31
2 The equivalence point occurs at 0.04000 mmol/mL X
3 Conc. AgNO3 0.05000 50.00 mL X (1 mL/0.05000 mmol) = 40.00 mL Ag+
4 Vol. KBr 50.00
5 Conc. KBr 0.04000
6 Ksp 5.00E-13
7 Vol. Ag+ [Br
-] [Ag
+] pAg
8 5.00 3.18E-02 1.57E-11 10.804
9 15.00 1.92E-02 2.60E-11 10.585
10 25.00 1.00E-02 5.00E-11 10.301
11 30.00 6.25E-03 8.00E-11 10.097
12 35.00 2.94E-03 1.70E-10 9.770
13 39.00 5.62E-04 8.90E-10 9.051
14 40.00 7.07E-07 7.07E-07 6.151
15 41.00 7.28E+01 5.49E-04 3.260
16 45.00 1.52E+01 2.63E-03 2.580
17 50.00 8.00E+00 5.00E-03 2.301
18
19 Spreadsheet Documentation
20 B8=(($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$6/C15 D8 = -LOG(C8)
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
Challenge Problem
]SCN][Fe[
])SCN(Fe[1005.1)SCN(FeSCNFe
3
23
f
23
K
For part (a) we find,
%81.0%100Agmol101588.1
SCNmol
Agmol1
)SCN(Femol
SCNmol1)SCN(Femol104030.9
Error%
)SCN(Femol104030.9
L106353.4mL1000
LmL00.50
L
)SCN(Femol10759.9)SCN(Femol
10759.91005.1
101
SCNL106353.4mol025.0
L
Agmol
SCNmol1Agmol101588.1SCNL
Agmol101588.1g8682.107
Agmol1Agg125.0Agmol
Agg125.0mL00.50mL100
g250.0%250.0Agmass
3
2
26
26
225
2
5
3
5
)SCN(Fe
23
3
2
c
Proceeding the same way for parts (b) and (c), we find the results in the following spreadsheet.
Fundamentals of Analytical Chemistry: 8th
ed. Chapter 13
A B C D E F G
1 Problem 13-32
2
3 mL taken 50
4 Kf 1.05E+03
5 conc SCN 0.025
6 AW Ag 107.8682
7 min complx 1.00E-05
8 %Ag g Ag moles Ag L SCN- c SCN cmplx mol SCN cmplx %Error
9 (a) 0.25 0.125 0.0011588 0.046353 9.759E-05 9.40308E-06 0.811434
10 (b) 0.1 0.05 0.0004635 0.018541 9.759E-05 6.68893E-06 1.443046
11 (c) 0.05 0.025 0.0002318 0.009271 9.759E-05 5.78422E-06 2.495732
12
13 Spreadsheet Documentation
14 B9=$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