Fundamentals of Analytical Chemistry: 8th ed.
Chapter 13
Chapter 13
13-1
amount A (mmol) = volume (mL )  cA (mmol A / mL )
amount A (mole) = volume ( L)  cA (mol A / L)
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
6.02  10 23
particles
mole
particles

 6.02  10 20
mole
1000 mmol
mmol
(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 AgCl s   SCN 


AgSCN ( s )  Cl  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)
1 mole H 2 NNH 2
2 moles I 2
(b)
5 moles H 2 O 2

2 moles MnO 4
Fundamentals of Analytical Chemistry: 8th ed.
13-6
(c)
1 mole Na 2 B4 O7  10H 2 O
2 moles H 
(d)
2 moles S
3 moles KIO3
Chapter 13
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)
0.0750 mole AgNO 3
L

 500 mL  0.0375 mole
L
1000 mL
169.87 g AgNO 3
0.0375 mole 
 6.37 g AgNO 3
mole
0.0750 M AgNO 3 
Dissolve 6.37 g AgNO3 in water and bring to 500 mL total volume.
(b)
0.325 mole HCl
 2.00 L  0.650 mole
L
L
0.650 mole 
 0.108 L reagent
6.00 mole reagent
0.325 M HCl 
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)
0.0900 M K  
0.0900 mole K 
L

 750 mL  0.0675 mole K 
L
1000 mL
0.0675 mole K  
mole K 4 Fe(CN ) 6 368.35 g K 4 Fe(CN) 6

 6.22 g K 4 Fe(CN) 6
4 moles K 
mole
Dissolve 6.22 g K4Fe(CN)6 in water and bring to 750 mL total volume.
(d)
2.00% ( w / v ) BaCl 2 
0.0576 mole BaCl 2 
2.00 g BaCl 2
mole BaCl 2

 600 mL  0.0576 mole BaCl 2
100 mL solution
208.23 g
L
 0.115 L BaCl 2
0.500 mole BaCl 2
Dilute 115 mL of 0.500M BaCl2 in enough water to yield 600 mL total volume.
(e)
0.120 mole HClO 4
 2.00 L  0.240 mole HClO 4
L
60 g HClO 4 mole HClO 4 9.55 mole HClO 4
1.60  10 3 g reagent



L reagent
100 g reagent
100.5 g
L reagent
0.120 M HClO 4 
vol. reagent  0.240 mole HClO 4 
L reagent
 0.025 L reagent
9.55 mole HClO 4
Dilute 25 mL HClO4 reagent in enough water to yield 2.00 L total volume.
(f)
60 mg Na 
60.0 ppm Na 
 9.00 L soln  5.40  10 2 mg Na 
L soln

g
mole Na  mole Na 2SO 4 142.0 g Na 2SO 4
5.4  10 mg Na 



 1.67 g Na 2SO 4
1000 mg
22.99 g
2 moles Na 
mole
2

Dissolve 1.67 g Na2SO4 in water and bring to 9.00 L total volume.
Fundamentals of Analytical Chemistry: 8th ed.
13-9
Chapter 13
(a)
0.150 mole KMnO 4
 1.00 L  0.150 mole KMnO 4
L
158.03 g KMnO 4
0.150 mole KMnO 4 
 23.7 g KMnO 4
mole
0.150 M KMnO 4 
Dissolve 23.7 g KMnO4 in water and bring to 1.00 L total volume.
(b)
0.500 mole HClO 4
 2.50 L  1.25 mole HClO 4
L
L
1.25 mole HClO 4 
 0.139 L HClO 4 reagent
9.00 mole HClO 4
0.500 M HClO 4 
Dilute 139 mL HClO4 reagent in enough water to yield 2.50 L total volume.
(c)
0.0500 mole I 
L
0.0500 M I 

 400 mL  0.0200 mole I 
L
1000 mL
mole MgI 2 278.11 g MgI 2
0.0200 mole I  

 2.78 g MgI 2
2 moles I 
mole

Dissolve 2.78 g MgI2 in water and bring to 400 mL total volume.
(d)
1.00% ( w / v ) CuSO 4 
0.0125 mole CuSO 4 
1.00 g CuSO 4 mole CuSO 4

 200 mL  0.0125 mole CuSO 4
100 mL
159.61 g
L
 0.0575 L CuSO 4
0.218 mole CuSO 4
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)
0.215 mole NaOH
 1.50 L  0.3225 mole NaOH
L
1.525  10 3 g reagent
50 g NaOH mole NaOH 1.906  101 mole NaOH



L reagent
100 g reagent
40.00 g
L reagent
L reagent
vol. reagent  0.3225 mole NaOH 
 0.0169 L reagent
1.906  101 mole NaOH
0.215 M NaOH 
Dilute 16.9 mL of the concentrated reagent to 1.50 L total volume.
(f)
12 ppm K  
12 mg K 
 1.50 L soln  1.8  101 mg K 
L soln
1.8  101 mg K  
g
mole K  mole K 4 Fe(CN ) 6 368.3 g K 4 Fe(CN ) 6



1000 mg 39.10 g
4 moles K 
mole
 0.0424 g K 4 Fe(CN ) 6
Dissolve 42.4 mg K4Fe(CN)6 in water and bring to 1.50 L total volume.
13-10 MHgO  216.59
g
mole
HgO ( s )  4 Br   H 2 O 
HgBr 4

2
 2OH 
1 mole HgO 2 mole OH  1 mole HClO 4 1000 mmol HClO 4
0.4125 g HgO 



216.59 g
mole HgO
1 mole OH 
mole
46.51 mL
 0.08190 M HClO 4
Fundamentals of Analytical Chemistry: 8th ed.
13-11 MNa2CO3  105.99
CO3
2
 2H 


Chapter 13
g
mole
H 2 O  CO 2 ( g )
0.4512 g Na 2 CO3 
1 mole Na 2 CO3
1 mole H 2SO 4 1000 mmol H 2SO 4
2 mole H 



105.99 g
mole Na 2 CO3
2 mole H 
mole
36.44 mL
 0.1168 M H 2SO 4
13-12 MNa2SO4  142.04
Ba 2  SO 4
2
g
mole
 BaSO 4 ( s )
0.4000 g sample 
96.4 g Na 2SO 4 1 mole Na 2SO 4
1 mole BaCl 2
1000 mmol



100 g sample
142.04 g
1 mole Na 2SO 4
mole
41.25 mL
 0.06581 M BaCl 2
13-13 (Note: In the first printing of the text, the answer in the back of the book was in error.)
VHClO4
VNaOH

27.43 mL HClO 4
mL HClO 4
 1.0972
25.00 mL NaOH
mL NaOH
The volume of HClO4 required to titrate 0.3125 g Na2CO3 is

1.0972 mL HClO 4 
  28.896 mL HClO 4
40.00 mL HClO 4  10.12 mL NaOH 
mL NaOH


Thus ,
0.3125 g Na 2 CO3 1 mole Na 2 CO3
2 mole HClO 4 1000 mmol



 0.2041 M HClO 4
28.896 mL HClO 4
105.99 g
1 mole Na 2 CO3
mole
and
c NaOH  c HClO4 
 0.2041 M 
VHClO4
VNaOH
0.2041 mole HClO 4 1.0972 mL HClO 4 1 mole NaOH


 0.2239 M NaOH
L
mL NaOH
mole HClO 4
Fundamentals of Analytical Chemistry: 8th ed.

13-14 2 MnO 4  5H 2 C 2 O 4  6H 
50.00 mL Na 2 C 2 O 4 


2 Mn 2  10CO2 ( g )  8H 2 O
0.05251 mole Na 2 C 2 O 4
2 mole KMnO 4 1000 mmol
L



L
1000 mL 5 mole Na 2 C 2 O 4
mole
36.75 mL
 0.02858 M KMnO 4
13-15 MKIO3  214.00
g
mole

IO 3  5I   6H 
I 2  2S2 O 3
2

3I 2  3H 2 O

 2I   S 4 O 6
0.1045 g KIO3 
2
1 mole KIO3
2 mole Na 2S2 O 3 1000 mmol
3 mole I 2



214.00 g
1 mole KIO3
1 mole I 2
mole
30.72 mL
 0.09537 M Na 2S2 O 3
13-16
ClCH 2 COOH  Ag   H 2 O  HOCH 2 COOH  H   AgCl ( s )
The unreacted Ag  is titrated with NH 4SCN,

Ag   NH 4SCN  NH 4  AgSCN ( s )
50.00 mL 
Chapter 13
0.04521 mole AgNO 3
1 mole NH 4SCN 1000 mmol
L



L
1000 mL 1 mole AgNO 3
mole
22.98 mL
 0.098368 M NH 4SCN
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 13
0.098368 mmol NH 4SCN
 10.43 mL  1.02598 mmol NH 4SCN
mL
 0.04521 mmol

mmol AgCl (s) precipitat ed  
 50.00 mL   1.02598 mmol
mL


 1.2345 mmol AgCl
1 mole ClCH 2 COOH 94.50 g 1000 mg
1.2345  10 3 mole AgCl 


1 mole AgCl
mole
g
 116.7 mg ClCH 2 COOH
mmol NH 4SCN 
13-17


BH 4  8Ag   8OH   H 2 BO 3  8Ag ( s )  5H 2 O
Ag   SCN   AgSCN ( s )
mmol excess Ag+ equals mmol KSCN,
0.0397 mmol KSCN
1 mmol Ag 
mmol excess Ag 
 3.36 mL 
 0.133 mmol Ag 
mL
1 mmol KSCN
0.2221 mmol AgNO 3
mmol AgNO 3 
 50.00 mL  1.11  101 mmol AgNO 3
mL

reacted mmol Ag  1.11  101  0.133mmol  1.10  101 mmol Ag 


1.10  101 mmol Ag  1 mmol BH 4


 0.0138 M BH 4

100 mL
8 mmol Ag

0.0138 mole BH 4
1 mole KBH 4 53.941 g KBH 4
L

 500 mL 

 0.371 g KBH 4

L
1000 mL
mole
1 mole BH 4
% purity KBH 4 
0.371 g KBH 4
 100%  11.5%
3.213 g material
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 13
13-18 H 3 AsO 4  3Ag   3H   Ag 3 AsO 4 ( s )
mmol excess Ag  equals mmol KSCN,
0.1000 mmol KSCN 1 mmol Ag 
mmol excess Ag 

 10.76 mL  1.0760 mmol Ag 
mL
1 mmol KSCN

0.06222 mmol AgNO 3
 40.00 mL  2.4888 mmol AgNO 3
mL
mmol AgNO 3 added 
mmol Ag  reacted  ( 2.4888  1.0760) mmol  1.4128 mmol Ag 
% As 2 O 3 in sample 

1 mmol Ag 3 AsO 4
1 mmol As 2 O 3
197.84 g As 2 O 3 
1.4128 mmol Ag  




3 mmol Ag
2 mmol Ag 3 AsO 4
1000 mmol 

 100
1.010 g sample
 4.612% As 2 O 3
13-19 MC1 0H 5Cl 7  373.32
g
mole
The stoichiometry of the titration of heptachlor must be one to one (i.e., one chlorine
reacts with one silver nitrate) for the calculation,
% heptachlor 
mL
unwritten units of
37.33
Ag
 cAg  mL SCN  cSCN   37.33
mass sample
, to be true. The factor 37.33 (with
g
) found in the numerator is derived from the equation below,
mmol
no.mmol C10 H 5Cl7 373.32 g C10 H 5Cl7
g


 100
mmol
no.mmol AgNO 3
1000 mmol
Thus,
g
37.33
 1000 mmol
no.mmol C10 H 5Cl7
mmol

 1.00
no.mmol AgNO 3
373.32 g C10 H 5Cl7  100
confirming that only one of the chlorines in the heptachlor reacts with the AgNO3.
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 13

13-20 Bi 3  H 2 PO 4  BiPO 4 ( s )  2 H 
0.03369 mmol NaH 2 PO 4
 27.36 mL  0.921758 mol NaH 2 PO 4
mL
1 mmol Bi 3
 0.921758 mol NaH 2 PO 4 
 0.921758 mol Bi 3
1 mmol NaH 2 PO 4
mol NaH 2 PO 4 
mol Bi 3
% purity eulytite 

1 mmol 2 Bi 2 O 3  3SiO 2 1112 g 2 Bi 2 O 3  3SiO 2 
 0.921758 mol Bi 3 


4 mmol Bi 3
1000 mmol

  100%
0.6423 g sample
 39.90% eulytite
13-21 (a)
molarity of Ba (OH ) 2 
0.1175 g C 6 H 5COOH 
1 mole C 6 H 5COOH
1 mole Ba (OH ) 2
1000 mmol


122.12 g
2 mole C 6 H 5COOH
mole
40.42 mL
 0.01190 M Ba (OH ) 2
(b)
  0.0002    0.03 
5
s y  (1.190  10 M )  
 
  2.2  10 M
0
.
1175
40
.
42

 

2
2
2
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,
E
1 mole C 6 H 5COOH
1 mole Ba (OH ) 2
1000 mmol



 0.1175  0.0003 g C 6 H 5COOH 
122.12 g
2 mole C 6 H 5COOH
mole


40.42 mL


1.190  10 2 M  1.187  10 2 M   1.190  10 2 M   2.826  10 5 M or  3.0  10 5 M






Fundamentals of Analytical Chemistry: 8th ed.
Chapter 13
The relative error, Er, in the molarity calculation resulting from this weighing error is
Er
 3.0  10

M
 3.0  10 3 or  3 ppt
1.190  10 M
5
2
13-22
w / v percentage HOAc 
0.1475 mmol Ba (OH) 2
2 mmol HOAc
60.05 g HOAc
 43.17 mL 

mL
1 mmol Ba (OH) 2
1000 mmol
 100%
50.00 mL
 1.529% HOAc
Similar calculations for samples 2 to 4 yield the results shown in the spreadsheet that
follows,
(a)
x w / v percentage HOAc 
x
4
i

6.1134
 1.528% HOAc
4
(b)
s
x
2
i

( x i ) 2
3
4

9.34351132 
3
(6.1134) 2
4
 5.71  10 3% HOAc
(c)
CI90%  x 
(d)
ts
2.35  (5.63  10 3 )
 1.528 
 1.528( 0.007)% HOAc
2
4
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)
 ( w / v )% HOAc V

( w / v )% HOAc
V
Fundamentals of Analytical Chemistry: 8th ed.
For sample 1,
Chapter 13
V HOAc
0.05 mL

 0.001
V HOAc
50.00 mL
The results for the remaining samples are found in the following spreadsheet.
mean relative systematic error 
 x   0.005   0.00125
n
4
For the mean ( w / v ) percent HOAc ,  ( w / v )% HOAc 
 0.00125  1.528  1.91  10 3% or  2  10 3% HOAc
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
B
C
D
E
F
G
xi
xi2
V/V
Problem 13-22
Conc. Ba(OH)2
0.1475
MW HOAc
60.05
t
2.35
Sample
Ba(OH)2 Vol, mL w/v % HOAc
Sample Vol, mL
1
50.00
43.17
1.529
1.52949152
2.33934429
2
49.50
42.68
1.527
1.52740511
2.33296637
3
25.00
21.47
1.521
1.52134273
2.31448370
4
50.00
43.33
1.535
1.53516024
2.35671695
(xi)
6.11339959
(xi2)
9.34351132
(a)
(b)
(c)
(d)
mean xi
std. dev. % HOAc
5.71E-03
CI90%(t=2.35)
6.70E-03
Q(expt 1.535-1.521)
Q(expt 1.527-1.521)
(e)
1.528
(V/V)
0.41
0.44
-0.005
mean relative systematic error
-1.25E-03
mean (w/v) % HOAc
-1.91E-03
Spreadsheet Documentation
D8 = (($B$3*C8*2*$B$4/1000)/B8)*100
C16 = SQRT((B14-(B13)^2/4)/3)
E8 = D8
C18 = (D11-D8)/(D11-D10)
F8 = E8^2
C19 = (D9-D10)/(D11-D10)
G8 = -0.05/B8
B13 = SUM(E8:E11)
C20 = SUM(G8:G11)
B14 = SUM(F8:F11)
C22 = C21*C15
C15 = B13/4
C21 = C20/4
-0.001
-0.001
-0.002
-0.001
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 13
13-23
 0.08181 mmol AgNO 3

no. mmol AgNO 3 consumed by sample  
 20.00 mL  
mL


 0.04124 mmol KSCN 1 mmol AgNO 3



 2.81 mL   1.5204 mmol AgNO 3
mL
1 mmol KSCN


mg saccharin / tablet 

1 mmol saccharin 205.17 g saccharin 1000 mg 
1.5204 mmol AgNO 3 



1 mmol AgNO 3
1000 mmol
g


20 tablets
 15.60
mg saccharin
tablet
13-24 (a)
1 mole AgNO 3
1 mole Ag 
1000 mmol
0.1752 g AgNO 3 


169.87 g
1 mole AgNO 3
mole
weight molarity Ag  
502.3 mL
 2.0533  10 3
(b)
2.0533  10 3 mole AgNO 3
1000 mmol
 23.765 mL 
1000 mL
mole
weight molarity KSCN 
25.171 mL
 1.9386  10 3
(c) MBaCl 2 2 H 2O  244.26
g
mole
 2.0533  10 3 mmol AgNO 3

mmol AgNO 3 consumed  
 20.102 mL  
mL


 1.9386  10 3 mmol KSCN 1 mmol AgNO 3



 7.543 mL 
mL
1 mmol KSCN


 0.026653 mmol
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 13
1 mmol BaCl 2  2 H 2 O
244.26 g

2 mmol AgNO 3
1000 mmol
 100%
0.7120 g sample
0.026653 mmol AgNO 3 
% BaCl 2  2 H 2 O 
 0.4572%
13-25 (a)
1 mole KCl  MgCl 2  6H 2 O
277.85 g
 0.01821 M KCl  MgCl 2  6H 2 O
2.000 L
10.12 g KCl  MgCl 2  6H 2 O 
Mg   KCl  MgCl
2
(b)
2
 6H 6 O  0.01821 M Mg 2
(c)
3 mole Cl 
Cl  0.01821 mole KCl  MgCl 2  6H 2 O 
 0.05463 M Cl 
1 mole KCl  MgCl 2  6H 2 O
 

(d)
( w / v )% KCl  MgCl 2  6H 2 O 
10.12 g
L

 100%  0.506%
2.000 L 1000 mL
(e)
0.05463 mmol Cl 
 25.0 mL  1.37 mmol Cl 
mL
(f)
0.01821 mole KCl  MgCl 2  6H 2 O
1 mole K 
39.10 g K  1000 mg



L
1 mole KCl  MgCl 2  6H 2 O
1 mole
g
 712.0 ppm K 
Fundamentals of Analytical Chemistry: 8th ed.
13-26 MCH 2O  30.03
Chapter 13
g
mole
 0.121 mmol KCN

mmol CH 2 O  mmol KCN reacted  
 30.0 mL  
mL


 0.100 mmol AgNO 3
  0.134 mmol NH 4SCN

 40.0 mL   
 16.1 mL   1.787 mmol CH 2 O

mL
mL

 


30.03 g CH 2 O 
1.787 mmol CH 2 O 

1000
mmol

  100%  21.5% CH O
2

25.0 mL 
 5.00 g sample 

500 mL 

13-27 MC1 9H1 6O4  308.34
g
mole
 0.02979 mmol AgNO 3

mmol AgNO 3 reacted  
 25.00 mL  
mL


 0.05411 mmol KSCN

 2.85 mL   0.5905 mmol AgNO 3

mL


1 mmol CHI 3
mmol C19 H 16O 4  0.5905 mmol AgNO 3 

3 mmol AgNO 3
1 mmol C19 H 16O 4
 0.1968 mmol C19 H 16O 4
1 mmol CHI 3

308.34 g C19 H 16O 4 
 0.1968 mmol C19 H 16O 4 

1000 mmol

  100%  0.4348% C H O
19 16 4
13.96 g sample
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 13
13-28

AgNO 3  2 NH 3  Ag ( NH 3 ) 2  NO3


6Ag ( NH 3 ) 2  3Se( s )  3H 2 O  2Ag 2Se( s )  Ag 2SeO 3 ( s )  6 NH 4

 0.0360 mmol AgNO 3

mmol AgNO 3 reacted to form Ag 2Se( s )  
 25.00 mL  
mL


 0.01370 mmol KSCN

 16.74 mL   0.6707 mmol AgNO 3

mL



mmol Se from Ag 2Se( s )  0.6707 mmol AgNO 3 
1 mmol Ag ( NH 3 ) 2

1 mmol AgNO 3
1 mmol Ag 2Se( s )
3 mmol Se( s )

 0.503 mmol Se

2 mmol Ag 2Se( s )
2 mmol Ag ( NH 3 ) 2
78.96 mg Se 

 0.503 mmol Se 

mmol

  7.94 mg Se / mL sample
5.00 mL
13-29
mmol Cl  
mmol ClO 4

0.08551 mmol AgNO 3
1 mmol Cl 
 13.97 g 
 1.195 mmol Cl 
mL
1 mmol AgNO 3

0.08551 mmol AgNO 3
1 mmol ClO 4

 ( 40.12 mL  13.97 mL ) 
mL
1 mmol AgNO 3
 2.236 mmol ClO 4


35.453 g Cl  
1.195 mmol Cl  

1000 mmol 

%Cl 
 100%  10.60% Cl 

50.00 mL 
1.998 g sample 

250.0 mL 


99.45 g ClO 4 
 2.236 mmol ClO 4 

1000 mmol 


%ClO 4 
 100%  55.65% ClO 4

50.00 mL 
1.998 g sample 

250
.
0
mL


Fundamentals of Analytical Chemistry: 8th ed.
13-30 (a)
Chapter 13
The equivalence point occurs at 50.0 mL,
0.05000 mmol AgNO 3
 25.00 mL  1.250 mmol Ag 
mL
1 mmol NH 4SCN
1 mL
mL SCN   1.250 mmol Ag  

 50.00 mL

1 mmol Ag
0.02500 mmol NH 4SCN
mmol Ag  
At 30.00 mL,
SCN
1.250 mmol Ag    0.0250 mmol
mL

[ Ag  ] 

25.00 mL  30.00 mL 


 30.00 mL 
  9.09  10 3 M Ag 
pAg   log 9.09  10 3   2.04
[SCN  ]  K sp / 9.09  10 3  1.1  10 12 / 9.09  10 3  1.2  10 10 M SCN 
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,
[Ag  ]  [SCN  ]  K sp  1.1  10 12  1.05  10 6 M
pAg   log( 1.05  10 6 )  5.98
At 51.00 mL,
 0.0250 mmol SCN 


 51.00 mL   1.250 mmol
mL

[SCN  ]  
 3.29  10 4 M
51.00 mL  25.00 mL 
[Ag  ]  1.1  10 12 / 3.29  10 4  3.3  10 9 M
pAg   log( 3.3  10 9 )  8.48
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.
A
1
B
D
E
F
Problem 13-30(a)
2
3
Conc. AgNO3
4
Vol. AgNO3
25.00
5
Conc. KSCN
0.02500
6
Ksp
1.10E-12
7
C
Chapter 13
Vol. SCN-
0.05000
[Ag+]
The equivalence point occurs at 0.05000 mmol/mL X
25.00 mL X (1 mL/0.02500 mmol) = 50.00 mL SCN-
[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$6/C12
D8 = -LOG(B8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th ed.
(b)
Chapter 13
Proceeding as in part (a), we obtain the results in the spreadsheet below.
A
1
B
D
E
F
Problem 13-30(b)
2
3
Conc. AgNO3
4
Vol. AgNO3
5
Conc. KI
0.03000
6
Ksp
8.30E-17
7
C
0.06000
The equivalence point occurs at 0.06000 mmol/mL X
20.00 mL X (1 mL/0.03000 mmol) = 40.00 mL I-
20.00
Vol. I-
[Ag+]
[I-]
pAg
8
20.00
1.50E-02
5.53E-15
1.824
9
30.00
6.00E-03
1.38E-14
2.222
10
39.00
5.08E-04
1.63E-13
3.294
11
40.00
9.11E-09
9.11E-09
8.04
12
41.00
1.69E-13
4.92E-04
12.77
13
50.00
1.94E-14
4.29E-03
13.71
14
60.00
1.11E-14
7.50E-03
13.96
15
16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(B8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th ed.
(c)
Chapter 13
Proceeding as in part (a), we obtain the results in the spreadsheet below.
A
1
B
D
E
F
Problem 13-30(c)
2
3
Conc. AgNO3
4
Vol. AgNO3
5
Conc. NaCl
0.07500
6
Ksp
1.82E-10
7
C
0.07500
The equivalence point occurs at 0.07500 mmol/mL X
30.00 mL X (1 mL/0.07500 mmol) = 30.00 mL CI-
30.00
Vol. CI-
[Ag+]
[CI-]
pAg
8
10.00
3.75E-02
4.85E-09
1.426
9
20.00
1.50E-02
1.21E-08
1.824
10
29.00
1.27E-03
1.43E-07
2.896
11
30.00
1.35E-05
1.35E-05
4.87
12
31.00
1.48E-07
1.23E-03
6.83
13
40.00
1.70E-08
1.07E-02
7.77
14
50.00
9.71E-09
1.88E-02
8.01
15
16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(B8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th ed.
(d)
Chapter 13
The equivalence point occurs at 70.00 mL,
0.4000 mmol Na 2SO 4
2
 35.00 mL  1.400  101 mmol SO 4
mL
mL
2
 1.400  101 mmol SO 4 
 70.00 mL Pb( NO3 ) 2
0.2000 mmol Pb( NO3 ) 2
mmol SO 4
mL Pb 2
2

At 50.00 mL,
Pb( NO )
1.400  10 mmol    0.2000 mmol
mL
1
2
[SO 4 ] 
3 2

(35.00 mL  50.00 mL )

 50.00 mL 
  4.706  10 2 M SO 2
4
[ Pb 2 ]  1.6  10 8 / 4.706  10 2  3.4  10 7 M Pb 2
pPb   log( 3.4  10 7 )  6.47
At 60.00 mL and 69.00 mL, the calculations are made in the same way and the results
are shown in the following spreadsheet.
At 70.00 mL,
2
[ Pb 2 ]  [SO 4 ]  K sp  1.6  10 8  1.3  10 4 M Pb 2
pPb   log( 1.3  10 4 )  3.90
At 71.00 mL,
 0.2000 mmol Pb( NO3 ) 2

2
 71.00 mL   1.400  101 mmol SO 4

mL

[ Pb 2 ]  
 1.887  10 3 M Pb 2
35.00 mL  71.00 mL 
2
[SO 4 ]  1.6  10 8 / 1.887  10 3  8.5  10 6 M SO 4
2
pPb   log( 1.887  10 3 )  2.7243
At 80.00 mL and 90.00 mL, the calculations are made in the same way and the results
are shown in spreadsheet below.
Fundamentals of Analytical Chemistry: 8th ed.
A
1
B
D
E
F
Problem 13-30(d)
2
3
Conc. Na2SO4
0.4000
4
Vol. Na2SO4
35.00
5
Conc. Pb(NO3)2
6
Ksp
7
C
Chapter 13
The equivalence point occurs at 0.4000 mmol/mL X
35.00 mL X (1 mL/0.2000 mmol) = 70.00 mL Pb2+
0.2000
1.60E-08
Vol. Pb2+
[SO42-]
[Pb2+]
pPb
8
50.00
4.71E-02
3.40E-07
6.469
9
60.00
2.11E-02
7.60E-07
6.119
10
69.00
1.92E-03
8.32E-06
5.080
11
70.00
1.26E-04
1.26E-04
3.898
12
71.00
8.48E-06
1.89E-03
2.724
13
80.00
9.20E-07
1.74E-02
1.760
14
90.00
5.00E-07
3.20E-02
1.495
15
16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(D8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th ed.
(e)
Proceeding as in part (a), we obtain the results in the spreadsheet below.
A
1
B
C
D
E
F
Problem 13-30(e)
2
3
Conc. BaCl2
0.0250
4
Vol. BaCl2
40.00
5
Conc. Na2SO4
6
Ksp
7
Chapter 13
The equivalence point occurs at 0.02500 mmol/mL X
40.00 mL X (1 mL/0.05000 mmol) = 20.00 mL SO42-
0.0500
1.10E-10
Vol. SO42-
[Ba2+]
8
0.00
2.50E-02
[SO42-]
pBa
1.602
9
10.00
1.00E-02
1.10E-08
2.000
10
19.00
8.47E-04
1.30E-07
3.072
11
20.00
1.05E-05
1.05E-05
4.979
12
21.00
1.34E-07
8.20E-04
6.872
13
30.00
1.54E-08
7.14E-03
7.812
14
40.00
8.80E-09
1.25E-02
8.056
15
16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(B8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Fundamentals of Analytical Chemistry: 8th ed.
(f)
Proceeding as in part (d), we obtain the results in the spreadsheet below.
A
1
B
C
D
E
F
Problem 13-30(f)
2
3
Conc. NaI
0.2000
4
Vol. NaI
50.00
5
Conc. TlNO3
6
Ksp
7
Chapter 13
The equivalence point occurs at 0.2000 mmol/mL X
50.00 mL X (1 mL/0.4000 mmol) = 25.00 mL Tl-
0.4000
6.50E-08
Vol. Tl+
[I-]
[Tl+]
pTl
8
5.00
1.45E-01
4.47E-07
6.350
9
15.00
6.15E-02
1.06E-06
5.976
10
24.00
5.41E-03
1.20E-05
4.920
11
25.00
2.55E-04
2.55E-04
3.594
12
26.00
1.24E-05
5.26E-03
2.279
13
35.00
1.38E-06
4.71E-02
1.327
14
45.00
7.72E-07
8.42E-02
1.075
15
16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(C8)
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
13-31 (Note: In the first printing of the text, the last answer for pAg in the back of the book was
in error.)
Fundamentals of Analytical Chemistry: 8th ed.
mmol KBr 
0.0400 mmol KBr
mL
Chapter 13
 50.0 mL  2.00 mmol KBr
At 5.00 mL,
 0.0500 mmol AgNO 3

2.00 mmol  
 5.00 mL 
mL

  3.18  10 2 M
[ Br  ] 
50.0 mL  5.00 mL 
[ Ag  ]  K sp /[ Br  ]  5.0  10 13 / 3.18  10 2  1.6  10 11 M Ag 
pAg   log( 1.6  10 11 )  10.80
At 15.00 mL, 25.00 mL, 30.00 mL, 35.00 mL and 39.00 mL, the calculations are
performed in the same way and the results are shown in the spreadsheet at the end of this
solution.
At 40.00 mL,
[Ag  ]  [ Br  ]  K sp  5.0  10 13  7.1  10 7 M Ag 
pAg   log( 7.1  10 7 )  6.15
At 41.00 mL,
 0.0500 mmol AgNO 3

 41.00 mL   2.00 mmol Br 

mL

[ Ag  ]  
 5.49  10 4 M Ag 
50.0 mL  41.00 mL 
pAg   log( 5.49  10 4 )  3.260
At 45.00 mL and 50.00 mL, the calculations are performed in the same way and the
results are shown in the spreadsheet that follows.
Fundamentals of Analytical Chemistry: 8th ed.
A
1
B
D
E
F
Problem 13-31
2
3
Conc. AgNO3
4
Vol. KBr
5
Conc. KBr
0.04000
6
Ksp
5.00E-13
7
C
Chapter 13
0.05000
The equivalence point occurs at 0.04000 mmol/mL X
50.00 mL X (1 mL/0.05000 mmol) = 40.00 mL Ag+
50.00
Vol. Ag+
[Br-]
[Ag+]
pAg
8
5.00
3.18E-02
1.57E-11
10.804
9
15.00
1.92E-02
2.60E-11
10.585
10
25.00
1.00E-02
5.00E-11
10.301
11
30.00
6.25E-03
8.00E-11
10.097
12
35.00
2.94E-03
1.70E-10
9.770
13
39.00
5.62E-04
8.90E-10
9.051
14
40.00
7.07E-07
7.07E-07
6.151
15
41.00
7.28E+01
5.49E-04
3.260
16
45.00
1.52E+01
2.63E-03
2.580
17
50.00
8.00E+00
5.00E-03
2.301
18
19
Spreadsheet Documentation
20
B8=(($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$6/C15
D8 = -LOG(C8)
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 13
Challenge Problem
3
Fe  SCN
 

Fe(SCN)
2
[ Fe(SCN) 2 ]
K f  1.05  10 
[ Fe3 ][SCN  ]
3
For part (a) we find,
mass Ag  0.250% 
0.250 g
 50.00 mL  0.125 g Ag
100 mL
mol Ag  0.125 g Ag 
1 mol Ag
 1.1588  10 3 mol Ag
107.8682 g
L SCN   1.1588  10 3 mol Ag 
c Fe (SCN )2  
1 mol SCN 
L

 4.6353  10 2 L SCN 
mol Ag
0.025 mol
1  10 5
 9.759  10 5
3
1.05  10


9.759  10 5 mol Fe(SCN) 2 
L
  4.6353  10 2 L
  50.00 mL 
L
1000 mL 


6
2
 9.4030  10 mol Fe(SCN)
mol Fe(SCN) 2 
1 mol SCN 
1 mol Ag

2
mol Fe(SCN)
mol SCN 
 100%  0.81%
1.1588  10 3 mol Ag
9.4030  10 6 mol Fe(SCN) 2 
% Error 
Proceeding the same way for parts (b) and (c), we find the results in the following spreadsheet.
Fundamentals of Analytical Chemistry: 8th ed.
A
1
B
Chapter 13
C
D
E
F
G
moles Ag
L SCN-
c SCN cmplx
mol SCN cmplx
%Error
Problem 13-32
2
3
mL taken
4
Kf
5
conc SCN
6
AW Ag
107.8682
7
min complx
1.00E-05
8
50
1.05E+03
0.025
%Ag
g Ag
9
(a)
0.25
0.125
0.0011588
0.046353
9.759E-05
9.40308E-06
0.811434
10
(b)
0.1
0.05
0.0004635
0.018541
9.759E-05
6.68893E-06
1.443046
11
(c)
0.05
0.025
0.0002318
0.009271
9.759E-05
5.78422E-06
2.495732
12
13
Spreadsheet Documentation
14
B9=$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