th
Fundamentals of Analytical Chemistry: 8 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 millimole is the amount of an elementary species, such as an atom, an ion, a
molecule, or an electron. A millimole contains 6.02 1023
parti particles cles mole
mole 1000 mmol
6.02 1020
parti particles cles mmol
(b) A titration involves titration involves measuring the quantity of a reagent of known concentration
required to react with 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 The stoichiometric ratio is ratio is the molar ratio of two chemical species that appear in a
balanced chemical equation. (d) Titration error is 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 in a titration is that point at which sufficient titrant has been
added so that stoichiometrically equivalent amounts of anal yte and titrant are present. The end point in in a titration is the point at which an observable physical change signals the equivalence point. (b) A primary A primary standard is is a highly purified substance that serves as the basis for a
titrimetric method. It is used either (i) (i) to prepare a standard solution directly by mass or
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Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
(ii) to standardize a solution to be used in a titration. A secondary standard is is material or solution whose concentration is determined from the stoichiometry of its its reaction with a primary standard material. Secondary standards are employed when a reagent is not available available in primary standard 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, prepared, however, by standardizing a solution of sodium hydroxide against a p rimary standard reagent such as potassium hydrogen phthalate.
13-4
The Fajans method is a direct titration of the chloride ion, while the Volhard approach app roach requires two standard standard solutions and a filtration filtration step to eliminate AgCl. AgCl. The Fajans method uses a fluorescein dye. dye. At the end point, the fluoresceinate 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 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)
(b)
1 mole H 2 NNH 2 2 moles I 2
5 moles H 2 O2 2 moles MnO4
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Fundamentals of Analytical Chemistry: 8 ed.
(c)
(d)
13-6
Chapter 13
1 mole Na 2 B4O7 10H 2 O 2 moles H
2 moles S 3 moles KIO3
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 th e 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 M AgNO 3 0.0375 mole
0.0750 mole AgNO 3 L
169.87 g AgNO 3 mole
L 1000 mL
500 mL 0.0375 mole
6.37 g AgNO 3
Dissolve 6.37 g AgNO3 in water and bring to 500 mL total volume. (b) 0.325 M HCl 0.650 mole
0.325 mole HCl L L
6.00 mole reagent
2.00 L 0.650 mole 0.108 L reagent
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Fundamentals of Analytical Chemistry: 8 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
0.0675 mole K
L 1000 mL
mole K 4 Fe(CN) 6
4 moles K
750 mL 0.0675 mole K
368.35 g K 4 Fe(CN) 6 mole
6.22 g K 4 Fe(CN) 6
Dissolve 6.22 g K 4Fe(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
100 mL solution L
mole BaCl 2 208.23 g
600 mL 0.0576 mole BaCl 2
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 M HClO 4 1.60 10 g reagent
0.120 mole HClO 4 L
3
L reagent
60 g HClO 4 100 g reagent
vol. reagent 0.240 mole HClO 4
2.00 L 0.240 mole HClO 4 mole HClO 4 100.5 g
L reagent 9.55 mole HClO 4
9.55 mole HClO 4 L reagent
0.025 L reagent
Dilute 25 mL HClO4 reagent in enough water to yield 2.00 L total volume. (f)
60.0 ppm Na
60 mg Na
5.4 10 mg Na 2
L soln g 1000 mg
9.00 L soln 5.40 102 mg Na
mole Na 22.99 g
mole Na 2SO4 2 moles Na
142.0 g Na 2SO4 mole
Dissolve 1.67 g Na2SO4 in water and bring to 9.00 L total volume.
1.67 g Na 2SO4
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Fundamentals of Analytical Chemistry: 8 ed. 13-9
Chapter 13
(a) 0.150 M KMnO 4
0.150 mole KMnO 4
0.150 mole KMnO 4
L 158.03 g KMnO4 mole
1.00 L 0.150 mole KMnO4 23.7 g KMnO4
Dissolve 23.7 g KMnO4 in water and bring to 1.00 L total volume. (b) 0.500 M HClO 4
0.500 mole HClO 4
1.25 mole HClO 4
L L 9.00 mole HClO 4
2.50 L 1.25 mole HClO 4 0.139 L HClO 4 reagent
Dilute 139 mL HClO4 reagent in enough water to yield 2.50 L total volume. (c)
0.0500 M I
0.0500 mole I
0.0200 mole I
L mole MgI 2 2 moles I
L 1000 mL
400 mL 0.0200 mole I
278.11 g MgI 2 mole
2.78 g MgI 2
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 4 100 mL
mole CuSO 4
L 0.218 mole CuSO 4
159.61 g
200 mL 0.0125 mole CuSO 4
0.0575 L CuSO 4
Dilute 57.5 mL of the 0.218 M CuSO4 solution to yield 200 mL total volume.
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Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
(e)
0.215 M NaOH
0.215 mole NaOH
1.50 L 0.3225 mole NaOH
L
1.525 103 g reagent L reagent
50 g NaOH
100 g reagent
vol. reagent 0.3225 mole NaOH
mole NaOH 40.00 g
1.906 101 mole NaOH L reagent
L reagent 1.906 10 mole NaOH 1
0.0169 L reagent
Dilute 16.9 mL of the concentrated reagent to 1.50 L total volume. (f)
12 ppm K
12 mg K L soln g
1.8 10 mg K 1
1.50 L soln 1.8 101 mg K
1000 mg
mole K
39.10 g
mole K 4 Fe(CN) 6
4 moles K
368.3 g K 4 Fe(CN) 6 mole
0.0424 g K 4 Fe(CN) 6 Dissolve 42.4 mg K 4Fe(CN)6 in water and bring to 1.50 L total volume.
13-10 MHgO 216.59
g mole
HgO ( s) 4Br H 2 O HgBr 4 0.4125 g HgO
1 mole HgO 216.59 g
2
2OH
2 mole OH
mole HgO
1 mole HClO 4
46.51 mL
0.08190 M HClO 4
1 mole OH
1000 mmol HClO 4 mole
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Fundamentals of Analytical Chemistry: 8 ed.
13-11 M Na2CO3 105.99
CO3
2
2H
Chapter 13
g mole
H 2 O CO2 ( g )
0.4512 g Na 2 CO3
1 mole Na 2 CO3
105.99 g
2 mole H
mole Na 2 CO3
1 mole H 2SO4 2 mole H
1000 mmol H 2SO4
1000 mmol
mole
36.44 mL
0.1168 M H 2SO4
13-12 M Na2SO4 142.04 Ba
2
SO4
2
g mole
BaSO 4 ( s)
0.4000 g sample
96.4 g Na 2SO4 100 g sample
1 mole Na 2SO4 142.04 g
1 mole BaCl 2 1 mole Na 2SO4
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 V NaOH
27.43 mL HClO 4 25.00 mL NaOH
1.0972
mL HClO 4 mL NaOH
The volume of HClO4 required to titrate 0.3125 g Na2CO3 is
1.0972 mL HClO 4
mL NaOH
40.00 mL HClO 4 10.12 mL NaOH
28.896 mL HClO 4
Thus, 0.3125 g Na 2 CO3 28.896 mL HClO 4
1 mole Na 2 CO3 105.99 g
2 mole HClO 4 1 mole Na 2 CO3
1000 mmol mole
0.2041 M HClO 4
and c NaOH c HClO4
0.2041 M
VHClO4 V NaOH
0.2041 mole HClO 4 L
1.0972 mL HClO 4 mL NaOH
1 mole NaOH mole HClO 4
0.2239 M NaOH
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Fundamentals of Analytical Chemistry: 8 ed.
13-14 2MnO4 5H 2 C2 O4 6H
50.00 mL Na 2 C2 O 4
Chapter 13
2Mn2 10CO2 ( g ) 8H 2O
0.05251 mole Na 2 C2 O 4 L
L 1000 mL
2 mole KMnO 4 5 mole Na 2 C2 O 4
36.75 mL
0.02858 M KMnO4
13-15 MKIO3 214.00
IO3 5I 6H I 2 2S2 O3
2
g mole
3I 2 3H 2 O
2I S 4 O 6
0.1045 g KIO3
2
1 mole KIO3 214.00 g
3 mole I 2 1 mole KIO3
2 mole Na 2S2 O3 1 mole I 2
1000 mmol
1000 mmol
mole
30.72 mL
0.09537 M Na 2S2O 3
13-16
ClCH 2COOH Ag H 2 O HOCH 2COOH H AgCl ( s) The unreacted Ag is titrated with NH 4SCN,
Ag NH 4SCN NH 4 AgSCN ( s) 50.00 mL
0.04521 mole AgNO 3 L
L 1000 mL
22.98 mL
0.098368 M NH 4SCN
1 mole NH 4SCN 1 mole AgNO 3
mole
1000 mmol mole
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Fundamentals of Analytical Chemistry: 8 ed.
mmol NH 4SCN
Chapter 13
0.098368 mmol NH 4SCN mL
10.43 mL 1.02598 mmol NH 4SCN
0.04521 mmol 50.00 mL 1.02598 mmol mL
mmol AgCl (s) precipitat ed
1.2345 mmol AgCl 3
1.2345 10 mole AgCl
1 mole ClCH 2 COOH 1 mole AgCl
94.50 g mole
1000 mg g
116.7 mg ClCH 2 COOH
13-17
BH 4 8Ag 8OH H 2 BO3 8Ag ( s) 5H 2 O Ag SCN AgSCN ( s)
+
mmol excess Ag equals mmol KSCN, mmol excess Ag
mmol AgNO 3
0.0397 mmol KSCN mL
0.2221 mmol AgNO 3 mL
3.36 mL
1 mmol Ag
1 mmol KSCN
0.133 mmol Ag
50.00 mL 1.11 101 mmol AgNO 3
1 1 reacted mmol Ag 1.11 10 0.133 mmol 1.10 10 mmol Ag
1.10 10 mmol Ag 1
100 mL 0.0138 mole BH 4
L % purity KBH4
1 mmol BH 4 8 mmol Ag L
1000 mL
0.0138 M BH 4
500 mL
0.371 g KBH4 3.213 g material
1 mole KBH4 1 mole BH 4
100% 11.5%
53.941 g KBH4 mole
0.371 g KBH4
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Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
13-18 H 3AsO 4 3Ag 3H Ag 3AsO 4 ( s)
mmol excess Ag equals mmol KSCN, 0.1000 mmol KSCN
mmol excess Ag
mL
mmol AgNO 3 added
1 mmol Ag 1 mmol KSCN
10.76 mL 1.0760 mmol Ag
0.06222 mmol AgNO 3 40.00 mL 2.4888 mmol AgNO 3 mL
mmol Ag reacted (2.4888 1.0760) mmol 1.4128 mmol Ag % As 2O3 in sample
1 mmol Ag 3AsO 4 1 mmol As 2 O3 197.84 g As 2 O3 1.4128 mmol Ag 3 mmol Ag 2 mmol Ag AsO 1000 mmol 3 4 100 1.010 g sample
4.612% As 2 O3
13-19 MC10H5Cl7 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
g mmol
cAg mLSCN cSCN 37.33 mass sample
unwritten units of
37.33
Ag
g mmol
, to be true. The factor 37.33 (with
) found in the numerator is derived from the equation below,
no.mmol C10 H 5Cl7 no.mmol AgNO 3
373.32 g C10 H 5Cl7 1000 mmol
100
Thus,
no.mmol C10 H 5Cl7 no.mmol AgNO 3
g
1000 mmol mmol 1.00 373.32 g C10 H 5Cl7 100 37.33
confirming that only one of the chlorines in the heptachlor reacts with the AgNO3.
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Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
13-20 Bi3 H 2 PO4 BiPO4 ( s) 2H mol NaH 2 PO4 mol Bi
3
0.03369 mmol NaH 2 PO4 mL
0.921758 mol NaH 2 PO4
27.36 mL 0.921758 mol NaH 2 PO4
1 mmol Bi
3
1 mmol NaH 2 PO4
0.921758 mol Bi3
% purity eulytite
1 mmol 2 Bi 2 O3 3SiO 2 1112 g 2Bi 2 O3 3SiO 2 0.921758 mol Bi3 3 4 mmol Bi 1000 mmol 100% 0.6423 g sample
39.90% eulytite
13-21 (a) molarity of Ba(OH) 2 0.1175 g C6 H 5COOH
1 mole C6 H 5COOH 122.12 g
1 mole Ba(OH) 2 2 mole C6 H 5COOH
1000 mmol mole
40.42 mL
0.01190 M Ba(OH) 2 (b) 2
2
0.0002 0.03 5 s y (1.190 10 M ) 2.2 10 M 0.1175 40.42 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 C6 H 5COOH 1 mole Ba(OH) 2 1000 mmol 0.1175 0.0003 g C6 H 5COOH 122.12 g 2 mole C6 H 5COOH mole 40.42 mL 1.190 102 M 1.187 102 M 1.190 102 M 2.826 105 M or 3.0 105 M
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Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
The relative error, Er , in the molarity calculation resulting from this weighing error is
3.0 10 M 3.0 10 5
E r
3
2
1.190 10 M
or 3 ppt
13-22 w / v percentage HOAc 0.1475 mmol Ba(OH) 2
43.17 mL
mL
2 mmol HOAc 1 mmol Ba(OH) 2
60.05 g HOAc 1000 mmol
50.00 mL
100%
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 4
1.528% HOAc
(b)
s
x
2 i
(
x )
2
9.34351132
i
4
3
(6.1134)
3
4
2
5.71 103% HOAc
(c)
CI90% x (d)
ts 4
1.528
2.35 (5.63 103 ) 2
1.528( 0.007)% HOAc
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 Q expt = 0.765, so neither value should be rejected. (e)
( w / v )% HOAc ( w / v )% HOAc
V V
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Fundamentals of Analytical Chemistry: 8 ed.
For sample 1,
V HOAc V HOAc
0.05 mL 50.00 mL
Chapter 13
0.001
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 1.91 103% or 2 103% HOAc
A 1 2 3 4 5 6 7 8 9 10 11 12
B
C
D
E
F
G
xi
xi 2
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
-0.001
2
49.50
42.68
1.527
1.52740511
2.33296637
-0.001
3
25.00
21.47
1.521
1.52134273
2.31448370
-0.002
4
50.00
43.33
1.535
1.53516024
2.35671695
-0.001
13
( x i)
6.11339959
14
2
9.34351132
( x i
)
15 16
(a) (b)
mean x i std. dev. % HOAc
5.71E-03
17 18 19 20 21 22 23 24 25 26 27 28 29 30
(c) (d)
CI90%(t=2.35)
6.70E-03
(e)
1.528
Q(expt 1.535-1.521)
0.41
Q(expt 1.527-1.521)
0.44
(V/V)
-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
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Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
13-23
0.08181 mmol AgNO 3 20.00 mL mL 0.04124 mmol KSCN 1 mmol AgNO 3 2.81 mL 1.5204 mmol AgNO 3 mL 1 mmol KSCN no. mmol AgNO 3 consumed by sample
mg saccharin / tablet
1 mmol saccharin 205.17 g saccharin 1000 mg 1.5204 mmol AgNO 3 1 mmol AgNO 1000 mmol g 3 20 tablets
15.60
mg saccharin tablet
13-24 (a) 0.1752 g AgNO 3
1 mole AgNO 3
weight molarity Ag
169.87 g
1 mole Ag
1 mole AgNO 3
1000 mmol mole
502.3 mL
2.0533 103
(b)
2.0533 103 mole AgNO 3 1000 mL
weight molarity KSCN
23.765 mL
1000 mmol mole
25.171 mL
1.9386 103 (c) MBaCl2 2 H2O 244.26
g mole
2.0533 103 mmol AgNO 3 20.102 mL mmol AgNO 3 consumed mL 1.9386 103 mmol KSCN 1 mmol AgNO 3 7.543 mL mL 1 mmol KSCN 0.026653 mmol
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
0.026653 mmol AgNO 3 % BaCl 2 2H 2 O
1 mmol BaCl 2 2 H 2 O 2 mmol AgNO 3
244.26 g 1000 mmol
0.7120 g sample
100%
0.4572%
13-25 (a)
10.12 g KCl MgCl 2 6H 2 O
1 mole KCl MgCl 2 6H 2 O 277.85 g
0.01821 M KCl MgCl 2 6H 2 O
2.000 L Mg2 KCl MgCl 2 6H 6O 0.01821 M Mg2
(b) (c)
Cl 0.01821 mole KCl MgCl
2
6H 2 O
3 mole Cl 1 mole KCl MgCl 2 6H 2 O
0.05463 M Cl
(d)
( w / v )% KCl MgCl 2 6H 2 O
10.12 g 2.000 L
L 1000 mL
100% 0.506%
(e)
0.05463 mmol Cl mL
25.0 mL 1.37 mmol Cl
(f)
0.01821 mole KCl MgCl 2 6H 2 O L
712.0 ppm K
1 mole K 1 mole KCl MgCl 2 6H 2 O
39.10 g K 1 mole
1000 mg g
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Fundamentals of Analytical Chemistry: 8 ed.
13-26 MCH2O 30.03
Chapter 13
g mole
0.121 mmol KCN 30.0 mL mL
mmol CH2 O mmol KCN reacted
0.100 mmol AgNO 3 0.134 mmol NH 4SCN 40 . 0 mL 16 . 1 mL 1.787 mmol CH2 O mL mL 30.03 g CH2 O 1.787 mmol CH2 O 1000 mmol 100% 21.5% CH O 2 25.0 mL 5.00 g sample 500 mL
13-27 MC19H16O4 308.34
g mole
0.02979 mmol AgNO 3 25.00 mL mL 0.05411 mmol KSCN 2.85 mL 0.5905 mmol AgNO 3 mL mmol AgNO 3 reacted
mmol C19 H16O 4 0.5905 mmol AgNO 3 1 mmol C19 H16O 4 1 mmol CHI3
1 mmol CHI3 3 mmol AgNO 3
0.1968 mmol C19 H16O4
308.34 g C19 H16O 4 0.1968 mmol C19 H16O 4 1000 mmol 100% 0.4348% C H O 19 16 4 13.96 g sample
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Fundamentals of Analytical Chemistry: 8 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 2SeO3 ( s) 6 NH 4
0.0360 mmol AgNO 3 25.00 mL mL
mmol AgNO 3 reacted to form Ag 2Se( s )
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 2Se( s ) 2 mmol Ag ( NH 3 ) 2
3 mmol Se( s ) 2 mmol Ag 2Se( s )
1 mmol Ag ( NH 3 ) 2 1 mmol AgNO 3
0.503 mmol Se
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 mL 0.08551 mmol AgNO 3 mL
2.236 mmol ClO 4
13.97 g
1 mmol Cl
1 mmol AgNO 3
(40.12 mL 13.97 mL)
1.195 mmol Cl 1 mmol ClO 4
1 mmol AgNO 3
35.453 g Cl 1.195 mmol Cl 1000 mmol 100% 10.60% Cl %Cl 50.00 mL 1.998 g sample 250 . 0 mL 99.45 g ClO 4 2.236 mmol ClO 4 1000 mmol 100% 55.65% ClO 4 %ClO 4 50.00 mL 1.998 g sample 250 . 0 mL
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
The equivalence point occurs at 50.0 mL,
13-30 (a)
mmol Ag
0.05000 mmol AgNO 3
25.00 mL 1.250 mmol Ag
mL
mL SCN 1.250 mmol Ag
1 mmol NH 4SCN 1 mmol Ag
1 mL 0.02500 mmol NH 4SCN
50.00 mL
At 30.00 mL,
0.0250 mmol SCN 1.250 mmol Ag 30.00 mL mL 9.09 103 M Ag [Ag ] 25.00 mL 30.00 mL
pAg log 9.09 10
3
2.04
[SCN ] K sp / 9.09 10
3
1.1 1012 / 9.09 103 1.2 1010 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
1.05 106 M
6 pAg log(1.05 10 ) 5.98
At 51.00 mL,
0.0250 mmol SCN 51.00 mL 1.250 mmol mL 3.29 104 M [SCN ] 51.00 mL 25.00 mL
[Ag ] 1.1 10
12
/ 3.29 10
4
3.3 109 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.
th
Fundamentals of Analytical Chemistry: 8 ed.
A 1
B
D
E
F
Problem 13-30(a)
2
The equivalence point occurs at 0.05000 mmol/mL X
3
Conc. AgNO3
4
Vol. AgNO3
25.00
5
Conc. KSCN
0.02500
6
K sp
7
C
Chapter 13
0.05000
25.00 mL X (1 mL/0.02500 mmol) = 50.00 mL SCN
-
1.10E-12
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$6/C12
D8 = -LOG(B8)
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
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Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
Proceeding as in part (a), we obtain the results in the spreadsheet below.
(b)
A 1
B
D
E
F
Problem 13-30(b)
2
The equivalence point occurs at 0.06000 mmol/mL X
3
Conc. AgNO3
4
Vol. AgNO3
5
Conc. KI
6
K sp
7
C
0.06000
20.00 mL X (1 mL/0.03000 mmol) = 40.00 mL I
-
20.00 0.03000 8.30E-17
Vol. I
-
[Ag+]
[I-]
pAg
8
20.00
1.50E-02
5.53E-15
1.824
9
30.00
6.00E-03
1.38E-14
2.222
10
39.00
5.08E-04
1.63E-13
3.294
11
40.00
9.11E-09
9.11E-09
8.04
12
41.00
1.69E-13
4.92E-04
12.77
13
50.00
1.94E-14
4.29E-03
13.71
14
60.00
1.11E-14
7.50E-03
13.96
15 16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(B8)
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
Proceeding as in part (a), we obtain the results in the spreadsheet below.
(c)
A 1
B
D
E
F
Problem 13-30(c)
2
The equivalence point occurs at 0.07500 mmol/mL X
3
Conc. AgNO3
4
Vol. AgNO3
30.00
5
Conc. NaCl
0.07500
6
K sp
7
C
0.07500
30.00 mL X (1 mL/0.07500 mmol) = 30.00 mL CI
-
1.82E-10
Vol. CI
-
[Ag+]
[CI-]
pAg
8
10.00
3.75E-02
4.85E-09
1.426
9
20.00
1.50E-02
1.21E-08
1.824
10
29.00
1.27E-03
1.43E-07
2.896
11
30.00
1.35E-05
1.35E-05
4.87
12
31.00
1.48E-07
1.23E-03
6.83
13
40.00
1.70E-08
1.07E-02
7.77
14
50.00
9.71E-09
1.88E-02
8.01
15 16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(B8)
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
The equivalence point occurs at 70.00 mL,
(d)
mmol SO4 mL Pb
2
2
0.4000 mmol Na 2SO4 mL
35.00 mL 1.400 101 mmol SO4 mL
2
1.400 101 mmol SO4
0.2000 mmol Pb( NO3 ) 2
2
70.00 mL Pb( NO3 ) 2
At 50.00 mL,
1.400 10 mmol 0.2000 mmol Pb( NO ) 1
2
3 2
[SO4 ]
mL
(35.00 mL 50.00 mL)
2
[ Pb ] 1.6 10
8
/ 4.706 10
2
50.00 mL 4.706 102 M SO 2 4
3.4 107 M Pb2
7
pPb log( 3.4 10 ) 6.47
At 60.00 mL and 69.00 mL, the calculations are made in the same way and the results are shown in the following spreadsheet. At 70.00 mL, 2
[ Pb 2 ] [SO4 ]
K sp 1.6 108 1.3 104 M Pb2
pPb log(1.3 10 4 ) 3.90
At 71.00 mL,
0.2000 mmol Pb( NO3 ) 2 2 71.00 mL 1.400 101 mmol SO4 mL 1.887 103 M Pb2 [ Pb 2 ] 35.00 mL 71.00 mL 2
[SO4 ] 1.6 10
8
/ 1.887 10
3
8.5 106 M SO4
2
3
pPb log( 1.887 10 ) 2.7243
At 80.00 mL and 90.00 mL, the calculations are made in the same way and the results are shown in spreadsheet below.
th
Fundamentals of Analytical Chemistry: 8 ed.
A 1
B
D
E
F
Problem 13-30(d)
2
The equivalence point occurs at 0.4000 mmol/mL X
3
Conc. Na2SO4
4
Vol. Na2SO4
5
Conc. Pb(NO 3)2
6
K sp
7
C
Chapter 13
0.4000
35.00 mL X (1 mL/0.2000 mmol) = 70.00 mL Pb
2+
35.00 0.2000 1.60E-08
Vol. Pb
2+
[SO42-]
[Pb2+]
pPb
8
50.00
4.71E-02
3.40E-07
6.469
9
60.00
2.11E-02
7.60E-07
6.119
10
69.00
1.92E-03
8.32E-06
5.080
11
70.00
1.26E-04
1.26E-04
3.898
12
71.00
8.48E-06
1.89E-03
2.724
13
80.00
9.20E-07
1.74E-02
1.760
14
90.00
5.00E-07
3.20E-02
1.495
15 16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(D8)
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
th
Fundamentals of Analytical Chemistry: 8 ed.
Proceeding as in part (a), we obtain the results in the spreadsheet below.
(e)
A 1
B
C
D
E
F
Problem 13-30(e)
2
The equivalence point occurs at 0.02500 mmol/mL X
3
Conc. BaCl 2
4
Vol. BaCl2
5
Conc. Na2SO4
6
K sp
7
Chapter 13
0.0250
2-
40.00 mL X (1 mL/0.05000 mmol) = 20.00 mL SO 4
40.00 0.0500 1.10E-10
Vol.
SO42-
[Ba2+]
[SO42-]
pBa
8
0.00
2.50E-02
9
10.00
1.00E-02
1.10E-08
2.000
10
19.00
8.47E-04
1.30E-07
3.072
11
20.00
1.05E-05
1.05E-05
4.979
12
21.00
1.34E-07
8.20E-04
6.872
13
30.00
1.54E-08
7.14E-03
7.812
14
40.00
8.80E-09
1.25E-02
8.056
1.602
15 16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(B8)
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
th
Fundamentals of Analytical Chemistry: 8 ed.
Proceeding as in part (d), we obtain the results in the spreadsheet below.
(f)
A 1
B
C
D
E
F
Problem 13-30(f)
2
The equivalence point occurs at 0.2000 mmol/mL X
3
Conc. NaI
4
Vol. NaI
5
Conc. TlNO3
6
K sp
7
Chapter 13
0.2000
50.00 mL X (1 mL/0.4000 mmol) = 25.00 mL Tl
-
50.00 0.4000 6.50E-08
Vol. Tl
+
[I-]
[Tl+]
pTl
8
5.00
1.45E-01
4.47E-07
6.350
9
15.00
6.15E-02
1.06E-06
5.976
10
24.00
5.41E-03
1.20E-05
4.920
11
25.00
2.55E-04
2.55E-04
3.594
12
26.00
1.24E-05
5.26E-03
2.279
13
35.00
1.38E-06
4.71E-02
1.327
14
45.00
7.72E-07
8.42E-02
1.075
15 16
Spreadsheet Documentation
17
B8=(($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$6/C12
D8 = -LOG(C8)
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
13-31 (Note: In the first printing of the text, the last answer for pAg in the back of the book was
in error.)
th
Fundamentals of Analytical Chemistry: 8 ed.
mmol KBr
0.0400 mmol KBr mL
Chapter 13
50.0 mL 2.00 mmol KBr
At 5.00 mL,
0.0500 mmol AgNO 3 5.00 mL mL 3.18 102 M 50.0 mL 5.00 mL
2.00 mmol [ Br ]
[Ag ] K sp /[Br ] 5.0 1013 / 3.18 102 1.6 1011 M Ag pAg log(1.6 10 11 ) 10.80
At 15.00 mL, 25.00 mL, 30.00 mL, 35.00 mL and 39.00 mL, the calculations are performed in the same way and the results are shown in the spreadsheet at the end of this solution. At 40.00 mL, [Ag ] [ Br ]
K sp
5.0 10
13
7.1 107 M Ag
pAg log( 7.1 10 ) 6.15 7
At 41.00 mL,
0.0500 mmol AgNO 3 41.00 mL 2.00 mmol Br mL 5.49 104 M Ag [Ag ] 50.0 mL 41.00 mL pAg log( 5.49 10 ) 3.260 4
At 45.00 mL and 50.00 mL, the calculations are performed in the same way and the results are shown in the spreadsheet that follows.
th
Fundamentals of Analytical Chemistry: 8 ed.
A 1
B
D
E
F
Problem 13-31
2
The equivalence point occurs at 0.04000 mmol/mL X
3
Conc. AgNO3
4
Vol. KBr
5
Conc. KBr
6
K sp
7
C
Chapter 13
0.05000
50.00 mL X (1 mL/0.05000 mmol) = 40.00 mL Ag
+
50.00 0.04000 5.00E-13 +
[Br -]
Vol. Ag
[Ag+]
pAg
8
5.00
3.18E-02
1.57E-11
10.804
9
15.00
1.92E-02
2.60E-11
10.585
10
25.00
1.00E-02
5.00E-11
10.301
11
30.00
6.25E-03
8.00E-11
10.097
12
35.00
2.94E-03
1.70E-10
9.770
13
39.00
5.62E-04
8.90E-10
9.051
14
40.00
7.07E-07
7.07E-07
6.151
15
41.00
7.28E+01
5.49E-04
3.260
16
45.00
1.52E+01
2.63E-03
2.580
17
50.00
8.00E+00
5.00E-03
2.301
18 19
Spreadsheet Documentation
20
B8=(($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$6/C15
D8 = -LOG(C8)
23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 13
Challenge Problem 3
Fe
SCN
Fe(SCN)
2
K f 1.05 10 3
[ Fe(SCN) 2 ] [ Fe3 ][SCN ]
For part (a) we find, mass Ag 0.250%
0.250 g 100 mL
mol Ag 0.125 g Ag
50.00 mL 0.125 g Ag
1 mol Ag 107.8682 g
3
L SCN 1.1588 10 mol Ag cFe(SCN )2
1 10
5
1.05 10
mol Fe(SCN)
2
3
1.1588 103 mol Ag
1 mol SCN mol Ag
L 0.025 mol
4.6353 102 L SCN
9.759 105 5
9.759 10 mol Fe(SCN) L
2
L 4.6353 102 L 50.00 mL 1000 mL
9.4030 106 mol Fe(SCN) 2 6
9.4030 10 mol Fe(SCN) % Error
2
1 mol SCN
mol Fe(SCN) 3
1.1588 10 mol Ag
2
1 mol Ag mol SCN
100% 0.81%
Proceeding the same way for parts (b) and (c), we find the results in the following spreadsheet.
th
Fundamentals of Analytical Chemistry: 8 ed.
A 1
B
C
D
moles Ag
L SCN
Chapter 13
E
F
G
SCN cmplx
mol SCN cmplx
%Error
Problem 13-32
2 3
mL taken
4
Kf
5
conc SCN
6 7
50 1.05E+03 0.025
AW Ag
107.8682
min complx
8
%Ag
1.00E-05
g Ag
-
c
9
(a)
0.25
0.125
0.0011588
0.046353
9.759E-05
9.40308E-06
0.811434
10
(b)
0.1
0.05
0.0004635
0.018541
9.759E-05
6.68893E-06
1.443046
11
(c)
0.05
0.025
0.0002318
0.009271
9.759E-05
5.78422E-06
2.495732
12 13
Spreadsheet Documentation
14
B9=$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
