SKOOG - SOLUCIONÁRIO CAPÍTULO 13

Prévia do material em texto

Fundamentals of Analytical Chemistry: 8
th
 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
particles
1002.6
mmol1000
mole
mole
particles
1002.6 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: 8
th
 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: 8
th
 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) 
3
3
3
3
AgNOg37.6
mole
AgNOg87.169
mole0375.0
mole0375.0mL500
mL1000
L
L
AgNOmole0750.0
AgNOM0750.0


 
Dissolve 6.37 g AgNO3 in water and bring to 500 mL total volume. 
(b) 
reagentL108.0
reagentmole00.6
L
mole650.0
mole650.0L00.2
L
HClmole325.0
HClM325.0


 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
Dilute 108 mL of the 6.00 M HCl reagent in enough water to yield 2.00 L total volume. 
(c) 
64
6464 )CN(FeKg22.6
mole
)CN(FeKg35.368
Kmoles4
)CN(FeKmole
Kmole0675.0
Kmole0675.0mL750
mL1000
L
L
Kmole0900.0
KM0900.0







 
Dissolve 6.22 g K4Fe(CN)6 in water and bring to 750 mL total volume. 
(d) 
2
2
2
2
22
2
BaClL115.0
BaClmole500.0
L
BaClmole0576.0
BaClmole0576.0mL600
g23.208
BaClmole
solutionmL100
BaClg00.2
BaCl)v/w(%00.2


 
Dilute 115 mL of 0.500M BaCl2 in enough water to yield 600 mL total volume. 
(e) 
reagentL025.0
HClOmole55.9
reagentL
HClOmole240.0reagent.vol
reagentL
HClOmole55.9
g5.100
HClOmole
reagentg100
HClOg60
reagentL
reagentg1060.1
HClOmole240.0L00.2
L
HClOmole120.0
HClOM120.0
4
4
444
3
4
4
4




 
Dilute 25 mL HClO4 reagent in enough water to yield 2.00 L total volume. 
(f) 
42
42422
2
SONag67.1
mole
SONag0.142
Namoles2
SONamole
g99.22
Namole
mg1000
g
Namg104.5
Namg1040.5solnL00.9
solnL
Namg60
Nappm0.60








 
Dissolve 1.67 g Na2SO4 in water and bring to 9.00 L total volume. 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
13-9 (a) 
4
4
4
4
4
4
KMnOg7.23
mole
KMnOg03.158
KMnOmole150.0
KMnOmole150.0L00.1
L
KMnOmole150.0
KMnOM150.0


 
Dissolve 23.7 g KMnO4 in water and bring to 1.00 L total volume. 
(b) 
reagentHClOL139.0
HClOmole00.9
L
HClOmole25.1
HClOmole25.1L50.2
L
HClOmole500.0
HClOM500.0
4
4
4
4
4
4


 
Dilute 139 mL HClO4 reagent in enough water to yield 2.50 L total volume. 
(c) 
2
22 MgIg78.2
mole
MgIg11.278
Imoles2
MgImole
Imole0200.0
Imole0200.0mL400
mL1000
L
L
Imole0500.0
IM0500.0







 
Dissolve 2.78 g MgI2 in water and bring to 400 mL total volume. 
(d) 
4
4
4
4
44
4
CuSOL0575.0
CuSOmole218.0
L
CuSOmole0125.0
CuSOmole0125.0mL200
g61.159
CuSOmole
mL100
CuSOg00.1
CuSO)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: 8
th
 ed. Chapter 13 
(e) 
reagentL0169.0
NaOHmole10906.1
reagentL
NaOHmole3225.0reagent.vol
reagentL
NaOHmole10906.1
g00.40
NaOHmole
reagentg100
NaOHg50
reagentL
reagentg10525.1
NaOHmole3225.0L50.1
L
NaOHmole215.0
NaOHM215.0
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
g
Kmg108.1
Kmg108.1solnL50.1
solnL
Kmg12
Kppm12









 
Dissolve 42.4 mg K4Fe(CN)6 in water and bring to 1.50 L total volume. 
13-10 
mole
g
59.216HgO M
 
4
44
2
42
HClOM08190.0
mL51.46
mole
HClOmmol1000
OHmole1
HClOmole1
HgOmole
OHmole2
g59.216
HgOmole1
HgOg4125.0
OH2HgBrOHBr4)(HgO







s
 
Fundamentalsof Analytical Chemistry: 8
th
 ed. Chapter 13 
13-11 
mole
g
99.105
32CONa
M
 
42
4242
32
32
32
22
2
3
SOHM1168.0
mL44.36
mole
SOHmmol1000
Hmole2
SOHmole1
CONamole
Hmole2
g99.105
CONamole1
CONag4512.0
)(COOHH2CO






 g
 
13-12 
mole
g
04.142
42SONa
M
 
2
42
24242
4
2
4
2
BaClM06581.0
mL25.41
mole
mmol1000
SONamole1
BaClmole1
g04.142
SONamole1
sampleg100
SONag4.96
sampleg4000.0
)(BaSOSOBa



 s
 
13-13 (Note: In the first printing of the text, the answer in the back of the book was in error.) 
NaOHmL
HClOmL
0972.1
NaOHmL00.25
HClOmL43.27
V
V
44
NaOH
HClO4 
 
The volume of HClO4 required to titrate 0.3125 g Na2CO3 is 
NaOHM2239.0
HClOmole
NaOHmole1
NaOHmL
HClOmL0972.1
L
HClOmole2041.0
M2041.0
V
V
cc
and
HClOM2041.0
mole
mmol1000
CONamole1
HClOmole2
g99.105
CONamole1
HClOmL896.28
CONag3125.0
,Thus
HClOmL896.28
NaOHmL
HClOmL0972.1
NaOHmL12.10HClOmL00.40
4
44
NaOH
HClO
HClONaOH
4
32
432
4
32
4
4
4
4
4










 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
13-14 
OH8)(CO10Mn2H6OCH5MnO2 22
2
4224 


 g
 
4
422
4422
422
KMnOM02858.0
mL75.36
mole
mmol1000
OCNamole5
KMnOmole2
mL1000
L
L
OCNamole05251.0
OCNamL00.50


 
13-15 
mole
g
00.214
3KIO
M
 







2
64
2
322
223
OSI2OS2I
OH3I3H6I5IO 
322
2
322
3
23
3
OSNaM09537.0
mL72.30
mole
mmol1000
Imole1
OSNamole2
KIOmole1
Imole3
g00.214
KIOmole1
KIOg1045.0


 
13-16 
)(AgSCNNHSCNNHAg
,SCNNHwithtitratedisAgunreactedThe
)(AgClHCOOHHOCHOHAgCOOHClCH
44
4
222
s
s





 
SCNNHM098368.0
mL98.22
mole
mmol1000
AgNOmole1
SCNNHmole1
mL1000
L
L
AgNOmole04521.0
mL00.50
4
3
43


 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
 
COOHClCHmg7.116
g
mg1000
mole
g50.94
AgClmole1
COOHClCHmole1
AgClmole102345.1
AgClmmol2345.1
mmol02598.1mL00.50
mL
mmol04521.0
edprecipitat)s(AgClmmol
SCNNHmmol02598.1mL43.10
mL
SCNNHmmol098368.0
SCNNHmmol
2
23
4
4
4












 
13-17 
)(AgSCNSCNAg
OH5)(Ag8BOHOH8Ag8BH 2324
s
s



 
mmol excess Ag
+
 equals mmol KSCN, 
 
%5.11%100
materialg213.3
KBHg371.0
KBHpurity%
KBHg371.0
mole
KBHg941.53
BHmole1
KBHmole1
mL500
mL1000
L
L
BHmole0138.0
BHM0138.0
Agmmol8
BHmmol1
mL100
Agmmol1010.1
Agmmol1010.1mmol133.01011.1Agmmolreacted
AgNOmmol1011.1mL00.50
mL
AgNOmmol2221.0
AgNOmmol
Agmmol133.0
KSCNmmol1
Agmmol1
mL36.3
mL
KSCNmmol0397.0
Agexcessmmol
4
4
4
4
4
44
4
4
1
11
3
13
3
















 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
13-18 
)(AsOAgH3Ag3AsOH 4343 s

 
3
3
3 AgNOmmol4888.2mL00.40
mL
AgNOmmol06222.0
addedAgNOmmol
Agmmol0760.1mL76.10
KSCNmmol1
Agmmol1
mL
KSCNmmol1000.0
Agexcessmmol
,KSCNmmolequalsAgexcessmmol

 



 
32
32
43
3243
32
OAs%612.4
100
sampleg010.1
mmol1000
OAsg84.197
AsOAgmmol2
OAsmmol1
Agmmol3
AsOAgmmol1
Agmmol4128.1
sampleinOAs%
Agmmol4128.1mmol)0760.14888.2(reactedAgmmol














 
13-19 
mole
g
32.373
7510 ClHC
M
 
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
33.37mLmL
heptachlor%
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, 
100
mmol1000
ClHCg32.373
AgNOmmol.no
ClHCmmol.no
mmol
g
33.37 7510
3
7510 
 
Thus, 
00.1
100ClHCg32.373
mmol1000
mmol
g
33.37
AgNOmmol.no
ClHCmmol.no
75103
7510 



 
confirming that only one of the chlorines in the heptachlor reacts with the AgNO3. 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
13-20 
  H2)(BiPOPOHBi 442
3 s
 
eulytite%90.39
%100
sampleg6423.0
mmol1000
SiO3OBi2g1112
Bimmol4
SiO3OBi2mmol1
Bimol921758.0
eulytitepurity%
Bimol921758.0
PONaHmmol1
Bimmol1
PONaHmol921758.0Bimol
PONaHmol921758.0mL36.27
mL
PONaHmmol03369.0
PONaHmol
232
3
2323
3
42
3
42
3
42
42
42







 











 
13-21 (a) 
2
56
256
56
2
)OH(BaM01190.0
mL42.40
mole
mmol1000
COOHHCmole2
)OH(Bamole1
g12.122
COOHHCmole1
COOHHCg1175.0
)OH(Baofmolarity



 
(b) 
M102.2
42.40
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
COOHHCmole1
COOHHCg0003.01175.0
E
55222
56
256
56
 















 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
The relative error, Er, in the molarity calculation resulting from this weighing error is 
 
ppt3or100.3
M10190.1
M100.3
E 3
2
5
r 


 

 
13-22 
HOAc%529.1
%100
mL00.50
mmol1000
HOAcg05.60
)OH(Bammol1
HOAcmmol2
mL17.43
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%528.1
4
1134.6
4
x
HOAcpercentagev/wx
i

 
(b) 
HOAc%1071.5
3
4
)1134.6(
34351132.9
3
4
)x(
x
s 3
22
i2
i







 
(c) 
HOAc)%007.0(528.1
2
)1063.5(35.2
528.1
4
ts
xCI
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: 8
th
 ed. Chapter 13 
001.0
mL00.50
mL05.0
HOAcV
HOAcV
,1sampleFor 

 
The results for the remaining samples are found in the following spreadsheet. 
00125.0
4
005.0
n
x
errorsystematicrelativemean 


 
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 xi
2
 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 1.521342732.31448370 -0.002 
11 4 50.00 43.33 1.535 1.53516024 2.35671695 -0.001 
12 
13 (xi) 6.11339959 
14 (xi
2
) 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: 8
th
 ed. Chapter 13 
13-23 
3
3
3
3
AgNOmmol5204.1mL81.2
KSCNmmol1
AgNOmmol1
mL
KSCNmmol04124.0
mL00.20
mL
AgNOmmol08181.0
samplebyconsumedAgNOmmol.no














 
tablet
saccharinmg
60.15
tablets20
g
mg1000
mmol1000
saccharing17.205
AgNOmmol1
saccharinmmol1
AgNOmmol5204.1
tablet/saccharinmg
3
3









 
13-24 (a) 
3
3
3
3
100533.2
mL3.502
mole
mmol1000
AgNOmole1
Agmole1
g87.169
AgNOmole1
AgNOg1752.0
Agmolarityweight






 
(b) 
3
3
3
109386.1
mL171.25
mole
mmol1000
mL765.23
mL1000
AgNOmole100533.2
KSCNmolarityweight





 
(c) 
mole
g
26.244OH2BaCl 22 M
 
mmol026653.0
mL543.7
KSCNmmol1
AgNOmmol1
mL
KSCNmmol109386.1
mL102.20
mL
AgNOmmol100533.2
consumedAgNOmmol
3
3
3
3
3




















 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
 
%4572.0
%100
sampleg7120.0
mmol1000
g26.244
AgNOmmol2
OH2BaClmmol1
AgNOmmol026653.0
OH2BaCl% 3
22
3
22






 
13-25 (a) 
OH6MgClKClM01821.0
L000.2
g85.277
OH6MgClKClmole1
OH6MgClKClg12.10
22
22
22



(b) 
      2622 MgM01821.0OH6MgClKClMg
 
(c) 
  

 

 ClM05463.0
OH6MgClKClmole1
Clmole3
OH6MgClKClmole01821.0Cl
22
22
 
(d) 
%506.0%100
mL1000
L
L000.2
g12.10
OH6MgClKCl)%v/w( 22 
 
(e) 


 Clmmol37.1mL0.25
mL
Clmmol05463.0
 
(f) 







Kppm0.712
g
mg1000
mole1
Kg10.39
OH6MgClKClmole1
Kmole1
L
OH6MgClKClmole01821.0
22
22
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
13-26 
mole
g
03.30OCH2 M
 
OCH%5.21%100
mL500
mL0.25
sampleg00.5
mmol1000
OCHg03.30
OCHmmol787.1
OCHmmol787.1mL1.16
mL
SCNNHmmol134.0
mL0.40
mL
AgNOmmol100.0
mL0.30
mL
KCNmmol121.0
reactedKCNmmolOCHmmol
2
2
2
2
43
2









































 
13-27 
mole
g
34.308
41619 OHC
M
 
41619
41619
41619
41619
3
41619
3
3
341619
3
3
3
OHC%4348.0%100
sampleg96.13
mmol1000
OHCg34.308
OHCmmol1968.0
OHCmmol1968.0
CHImmol1
OHCmmol1
AgNOmmol3
CHImmol1
AgNOmmol5905.0OHCmmol
AgNOmmol5905.0mL85.2
mL
KSCNmmol05411.0
mL00.25
mL
AgNOmmol02979.0
reactedAgNOmmol
























 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
13-28 




4322223
32333
NH6)(SeOAg)(SeAg2OH3)(Se3)NH(Ag6
NO)NH(AgNH2AgNO
sss
 
samplemL/Semg94.7
mL00.5
mmol
Semg96.78
Semmol503.0
Semmol503.0
)(SeAgmmol2
)(Semmol3
)NH(Agmmol2
)(SeAgmmol1
AgNOmmol1
)NH(Agmmol1
AgNOmmol6707.0)(SeAgfromSemmol
AgNOmmol6707.0mL74.16
mL
KSCNmmol01370.0
mL00.25
mL
AgNOmmol0360.0
)(SeAgformtoreactedAgNOmmol
223
2
3
23
32
3
3
23


























s
ss
s
s
 
13-29 













































4
4
4
4
4
3
43
4
3
3
ClO%65.55%100
mL0.250
mL00.50
sampleg998.1
mmol1000
ClOg45.99
ClOmmol236.2
ClO%
Cl%60.10%100
mL0.250
mL00.50
sampleg998.1
mmol1000
Clg453.35
Clmmol195.1
Cl%
ClOmmol236.2
AgNOmmol1
ClOmmol1
)mL97.13mL12.40(
mL
AgNOmmol08551.0
ClOmmol
Clmmol195.1
AgNOmmol1
Clmmol1
g97.13
mL
AgNOmmol08551.0
Clmmol
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 13 
13-30 (a) The equivalence point occurs at 50.0 mL, 
mL00.50
SCNNHmmol02500.0
mL1
Agmmol1
SCNNHmmol1
Agmmol250.1SCNmL
Agmmol250.1mL00.25
mL
AgNOmmol05000.0
Agmmol
4
4
3





 
At 30.00 mL, 
 
 
 


















SCNM102.11009.9/101.11009.9/K]SCN[
04.21009.9logpAg
AgM1009.9
mL00.30mL00.25
mL00.30
mL
SCNmmol0250.0
Agmmol250.1
]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[
M1029.3
mL00.25mL00.51
mmol250.1mL00.51
mL
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: 8
th
 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: 8
th
 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-031.38E-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: 8
th
 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: 8
th
 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
mL
SOmmol10400.1PbmL
SOmmol10400.1mL00.35
mL
SONammol4000.0
SOmmol




 
At 50.00 mL, 
 
47.6)104.3log(pPb
PbM104.310706.4/106.1]Pb[
SOM10706.4
)mL00.50mL00.35(
mL00.50
mL
)NO(Pbmmol2000.0
mmol10400.1
]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[
PbM10887.1
mL00.71mL00.35
SOmmol10400.1mL00.71
mL
)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: 8
th
 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 Pb
2+
 
4 Vol. Na2SO4 35.00 
5 Conc. Pb(NO3)2 0.2000 
6 Ksp 1.60E-08 
7 Vol. Pb
2+
 [SO4
2-
] [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: 8
th
 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 SO4
2-
 
4 Vol. BaCl2 40.00 
5 Conc. Na2SO4 0.0500 
6 Ksp 1.10E-10 
7 Vol. SO4
2-
 [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: 8
th
 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: 8
th
 ed. Chapter 13 
KBrmmol00.2mL0.50
mL
KBrmmol0400.0
KBrmmol 
 
At 5.00 mL, 
 
80.10)106.1log(pAg
AgM106.11018.3/100.5]Br/[K]Ag[
M1018.3
mL00.5mL0.50
mL00.5
mL
AgNOmmol0500.0
mmol00.2
]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
AgM1049.5
mL00.41mL0.50
Brmmol00.2mL00.41
mL
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: 8
th
 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 50.005 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: 8
th
 ed. Chapter 13 
Challenge Problem 
]SCN][Fe[
])SCN(Fe[
1005.1)SCN(FeSCNFe
3
2
3
f
23




  K
 
For part (a) we find, 
%81.0%100
Agmol101588.1
SCNmol
Agmol1
)SCN(Femol
SCNmol1
)SCN(Femol104030.9
Error%
)SCN(Femol104030.9
L106353.4
mL1000
L
mL00.50
L
)SCN(Femol10759.9
)SCN(Femol
10759.9
1005.1
101
SCNL106353.4
mol025.0
L
Agmol
SCNmol1
Agmol101588.1SCNL
Agmol101588.1
g8682.107
Agmol1
Agg125.0Agmol
Agg125.0mL00.50
mL100
g250.0
%250.0Agmass
3
2
26
26
2
25
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: 8
th
 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