th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
Chapter 12
12-1
(a) A colloidal precipitate consists of solid particles with dimensions that are less than -4
10 cm. A crystalline precipitate consists of solid particles with dimensions that at least -4
10 cm or greater. As a consequence, crystalline precipitates settle rapidly, whereas colloidal precipitates remain suspended in solution unless caused to agglomerate. (b) In gravimetric precipitation, the analyte is converted to a sparing soluble precipitate,
which is then filtered, washed free of impurities, and then converted into a product of known composition by suitable heat treatment. In gravimetric volatilization , the analyte is separated from other sample constituents by converting it to a gas of known composition. (c) Precipitation is the process by which a solid phase forms and is carried out of solution
when the solubility product of a chemical species is exceeded. Coprecipitation is a process in which normally soluble compounds are carried out of solution by precipitate formation. (d) Coagulation, or agglomeration, is the process by which colloidal particles coalesce to
form larger aggregates. Peptization refers to the process by which a coagulated colloid reverts to its original dispersed state. Heating, stirring and adding an electrolyte can coagulate colloidal suspensions. Washing the coagulated colloid with water often removes sufficient electrolyte to permit the re-establishment of repulsive forces that favor return to the colloidal state. (e) Occlusion is a type of coprecipitation in which a compound is trapped within a pocket
formed during rapid crystal formation. Mixed-crystal formation is also a type of coprecipitation in which a contaminant ion replaces an ion in the crystal lattice.
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
(f) Nucleation is a process in which a minimum number of atoms, ions or molecules
associate to give a stable solid. Particle growth is a process by which growth continues on existing nuclei. Precipitation by nucleation results in a large number of small particles. Precipitation by particle growth results in a smaller number of large particles.
12-2
(a) Digestion is a process in which a precipitate is heated in the presence of the solution
from which it was formed (the mother liquor ). Digestion improves the purity and filterability of the precipitate. (b) Adsorption is the process by which ions are retained on the surface of a solid. (c) In reprecipitation , the filtered solid precipitate is redissolved and reprecipitated.
Because the concentration of the impurity in the new solution is lower, the second precipitate contains less coprecipitated impurity. (d) Precipitation from a homogeneous solution is a technique by which a precipitating
agent is generated in a solution of the analyte by a slow chemical reaction. Local reagent excess does not occur and the resultant solid product is better suited for analysis than precipitate formed by direct addition of precipitating reagent. (e) The counter-ion layer describes a layer of solution containing sufficient excess
negative ions that surrounds a charged particle. This counter-ion layer balances the surface charge on the particle. (f) Mother liquor is the solution from which a precipitate is formed. (g) Supersaturation describes an unstable state in which a solution contains higher solute
concentration than a saturated solution. Supersaturation is relieved by precipitation of excess solute.
th
Fundamentals of Analytical Chemistry: 8 ed. 12-3
Chapter 12
A chelating agent is an organic compound that contains two or more electron-donor groups located in such a configuration that five- or six-membered rings are formed when the donor groups complex a cation.
12-4
Relative supersaturation can be regulated through control of reagent concentration,
temperature and the rate at which reagents are combined.
12-5
(a) There is positive charge on the surface of the coagulated colloidal particles. (b) The positive charge arises from adsorbed Ag+ ions. -
(c) NO3 ions make up the counter-ion layer.
12-6
CH 3CSNH 4
+
H 2O
→
CH 3CONH 2
←
+
H 2S
The slow hydrolysis of thioacetamide can be used to generate a source of hydrogen sulfide gas. Hydrogen sulfide gas is then involved in the equilibria below: H 2S + H 2 O HS
−
+
2-
H 2O
→ ←
→ ←
H 3O
+
H 3O
+
+
HS
+
S2
−
−
2+
The S generated can then be used to precipitate Ni in the form of NiS.
12-7
Peptization is the process by which a coagulated colloid returns to its original dispersed
state as a consequence of a decrease in the electrolyte concentration of the solution in contact with the precipitate. Peptization can be avoided by washing the coagulated colloid with an electrolyte solution rather than pure water.
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
12-8
+
Chloroplatinic acid, H 2PtCl6, forms the precipitate K 2PtCl6 when mixed with K but does +
+
not form analogous precipitates with Li and Na . Thus, chloroplatinic acid can be used +
+
+
to separate K from a mixture containing Li and Na .
12-9
Note: M stands for molar or atomic mass in the equations below:
(a) mass SO 2
=
mass BaSO 4
MBaSO
(b) mass Mg = mass Mg 2 P2 O 7
(c) mass In = mass In 2 O 3
(d) mass K
=
×
mass K 2 PtCl 6
MSO2
×
=
2 MMg
×
MMg
2 P2 O7
2 MIn MIn
2O 3
2 MK
×
MK
(e) mass Cu = mass Cu 2 (SCN ) 2
(f) mass MnCl 2
4
mass Mn 3O 4
×
2 PtCl 6
2 MCu MCu
×
2 ( SCN ) 2
3 MMnCl2 MMn O 3
(g) mass Pb 3O 4
=
(h) mass U 2 P2 O11
mass PbO 2
=
×
MPb O 3
4
4
3 MPbO2
mass P2 O 5 ×
MU 2 P2O11 MP O 2
5
(i) mass Na 2 B2 O 7 ⋅ 10H 2 O = mass B2 O 3 ×
MNa
2 B4O 7 ⋅10 H 2 O
2 MB2O3
(j) mass Na 2 O = mass NaZn ( UO 2 ) 3 (C 2 H 3O 2 ) 9 ⋅ 6H 2 O ×
MNa 2 O
2 MNaZn ( UO2 )3 ( C2H 3O2 )9 6 H 2O ⋅
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
12-10 M AgCl
=
143.32 g
M KCl
mole
=
74.55 g mole
1 mole AgCl 1 mole KCl 74.55 g KCl × × 0.2912 g AgCl × 143 . 32 g 1 mole AgCl mole
×
0.2500 g impure sample
12-11 M Al2O3
=
101.96 g
M NH
mole
100%
=
60.59%
237.15 g
= 4 Al ( SO 4 ) 2
mole
(a)
0.2001 g Al 2 O 3 ×
1 mole Al 2 O 3
×
2 mole NH 4 Al(SO 4 ) 2
101.96 g Al 2 O 3
3.925 × 10 3 mole NH 4 Al(SO 4 ) 2 −
×
0.2001 g Al 2 O 3
×
0.910 g impure sample
3.925 × 10 3 mole NH 4 Al(SO 4 ) 2
×
100%
mole Al 2 O 3
−
237.15 g NH 4 Al(SO 4 ) 2 mole NH 4 Al(SO 4 ) 2
0.910 g
(b)
=
=
102% NH 4 Al(SO 4 ) 2
100% = 22.0% Al 2 O 3
(c)
no. mole Al = no. mole NH 4 Al(SO 4 ) 2 3.925 × 10 2 mole Al × −
=
−
3
3.925 × 10 mole
26.981 g Al
mole 0.910 g impure sample
×
100% = 11.6% Al
12-12
0.500 g CuSO 4 ⋅ 5H 2 O × ×
413.35 g Cu( IO 3 ) 2 1 mole Cu( IO 3 ) 2
=
1 mole CuSO4 ⋅ 5H 2 O 249.67 g CuSO 4 ⋅ 5H 2 O
0.828 g Cu( IO 3 ) 2
×
1 mole Cu( IO3 ) 2 1 mole CuSO4 ⋅ 5H 2 O
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
12-13
1 mole CuSO4 ⋅ 5H 2 O
0.2000 g CuSO4 ⋅ 5H 2 O × ×
2 mole KIO3
×
1 mole Cu( IO 3 ) 2
×
249.67 g CuSO 4 ⋅ 5H 2 O
214 g KIO3
=
1 mole KIO3
1 mole Cu( IO3 ) 2 1 mole CuSO4 ⋅ 5H 2 O
0.342 g KIO3
12-14 (Note: In the first printing of the text, the answer in the back of the book was in error.)
no. mole AgI = 0.512 g sample × 0.201 AlI 3 × −
4
7.57 × 10 mole AgI ×
234.773 g AgI mole AgI
=
1 mole AlI 3 407.770 g AlI 3
×
3 mole AgI 1 mole AlI 3
=
0.178 g AgI
12-15 The precipitate V 2O5·2UO3 gives the greatest mass from a given quantity of uranium.
12-16 Al 2 (CO 3 ) 3
+
6HCl
0.0515 g CO 2 =
×
→ ←
3CO 2
1 mole CO 2
×
+
3H 2 O + 2AlCl3
1 mole Al 2 (CO 3 ) 3
44.01 g CO 2
3 mole CO 2
2 mole Al
×
×
1 mole Al 2 (CO 3 ) 3
26.98 g Al mole Al
0.02105 g Al 0.02105 g Al
×
0.8102 g impure sample
12-17 CdS + 2O 2
→ ←
100% = 2.60% Al
CdSO4
no. mole CdS = no. mole CdSO 4 =
−
4
5.61 × 10 mole
=
0.117 g CdSO 4
×
1 mole CdSO 4 208.47 g
×
−
4
7.57 × 10 mole AgI
1 mole CdS 1 mole CdSO 4
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
The number moles H2S is equal to number moles CdS. mass H 2S = 5.61 × 10 4 mole × −
0.0191 g H 2S
×
75.0 g impure sample
100%
34.08 g H 2S
=
1 mole H 2S =
0.0191 g
0.025% H 2S
12-18 0.6006 g BaCO 3 ×
1 mole BaCO 3 197.34 g
1 mole C
×
×
1 mole BaCO 3
12.011 g C 1 mole C
0.2121 g sample
×
100%
=
17.23% C
12-19
0.1606 g AgCl ×
1 mole AgCl 143.37 g
×
1 mole C14 H 9 Cl5 5 moles AgCl
×
354.72 g C14 H 9 Cl 5 1 mole C14 H 9 Cl 5
5.000 g sample =
×
100%
1.589% C14 H 9 Cl 5
12-20 (Note: In the first printing of the text, the answer in the back of the book was in error.)
mol Hg 2
+
=
0.4114 g Hg 5 ( IO 6 ) 2
1 mole Hg 5 ( IO 6 ) 2
×
1448.75 g Hg 5 ( IO 6 ) 2
1.4198 × 10 3 mol Hg 2 −
+
×
1 mole Hg 2 Cl 2 2 mole Hg 2
+
0.8142 g sample
×
×
5 mole Hg 2
+
=
1 mole Hg 5 ( IO 6 ) 2
1.4198 × 10 3 mol Hg 2 −
472.09 g Hg 2 Cl 2
1 mole Hg 2 Cl 2
×
100% = 41.16% Hg 2 Cl 2
+
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
12-21 MBa ( IO
487.13 g
= 3 )2
MKI
mole
0.0612 g Ba ( IO 3 ) 2
×
=
166.00 g mole
1 mole Ba ( IO 3 ) 2 2 mole KI 166.00 g KI × × 487 . 13 g 1 mole Ba ( IO ) mole 3 2 × 100% 1.97 g impure sample
=
2.12% KI
12-22 MNH
=
17.0306 g
MPt
mole
3
=
195.08 g mole
1 mole Pt 2 mole NH 3 17.0306 g NH 3 × × 0.4693 g Pt × mole 195.08 g 1 mole Pt 0.2115 g impure sample
×
100%
=
38.74% NH 3
12-23 MMnO
=
86.94 g
2
mol MnO 2
mole =
MAlCl
= 3
133.34 g mole
1 mole MnO 2 = 3.366 × 10 3 mol 86.94 g
(0.6447 g − 0.3521 g MnO 2 ) ×
3.366 × 10 3 mol MnO 2 −
×
2 mole Cl 1 mole AlCl 3 133.34 g AlCl 3 × × 1 mole MnO 3 mole Cl mole 2 1.1402 g impure sample
=
26.24% AlCl 3
−
×
100%
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
12-24 Let S w = mass of sample in grams MBaSO
= 4
M
233.39 g / mole
0.200 g BaSO 4 −
×
SO4
1 mole BaSO 4
×
233.39 g
4
8.57 × 10 mole SO 4
2−
×
=
2−
96.064 g / mole
1 mole SO 4
=
1 mole BaSO 4
96.064 g SO 4
=
S w g sample 4
8.57 × 10 mole SO 4
2−
S w g sample =
8.57 × 10 4 mole SO4 −
2−
2−
mole
−
2−
×
100% = 20% SO 4
96.064 g SO 4
2−
2−
×
mole
100% =
20%
0.412 g sample
The maximum precipitate weight expected given this sample weight, 0.412 g sample × =
55 g SO 4
2−
×
1 mole SO 4
100 g sample
2−
×
1 mole BaSO 4
96.064 g
1 g SO 4
2−
×
233.39 g BaSO 4 1 mole
0.550 g BaSO 4
12-25 Let S w = mass of sample in grams.
The higher percentage of Ni in the alloy sample is selected because this corresponds to maximum amount expected precipitate. MNi ( HC
4 H 6O 2 N 2 )
=
288.92 g / mole
0.175 g Ni( HC 4 H 6 O 2 N 2 ) × =
MNi
=
58.693 g / mole
1 mole Ni( HC 4 H 6 O 2 N 2 ) 288.92 g
1 mole Ni
×
1 mole Ni( HC 4 H 6 O 2 N 2 )
6.06 × 10 4 mole Ni −
6.06 × 10 4 mole Ni × −
58.693 g Ni mole
=
S w g sample −
4
6.06 × 10 mole Ni × S w g sample =
100% = 35% Ni 58.693 g Ni
35%
mole
×
100% =
0.102 g sample
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
12-26
Let S w = mass of sample in grams (a) MAgCl
=
MZrCl
143.32 g / mole
0.400 g AgCl ×
1 mole AgCl
×
143.32 g
= 4
233.03 g / mole
1 mole ZrCl 4
×
233.03 g ZrCl 4
4 mole AgCl
1 mole
×
S w g sample S w g sample =
0.16259 g ZrCl 4
×
100%
=
68%
100% = 68% ZrCl 4
0.239 g sample
(b)
0.239 g sample ×
84 g ZrCl 4
×
1 mole ZrCl 4
100 g sample
233.03 g
×
4 mole AgCl 1 mole ZrCl 4
(c)
% ZrCl 4
=
0.16259 g ZrCl 4
S w g sample =
×
100%
S w g sample 0.16259 g ZrCl 4 40%
×
=
40%
100%
=
0.406 g
×
143.32 g AgCl 1 mole
=
0.494 g AgCl
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
12-27
1 2 3 4 5 6 7 8 9 10 11
A Problem 12-27
B
C
D
Coefficient Matrix
Constant Matrix
1.823
1.578
1.505
1
1
0.872
Inverse Matrix
Solution Matrix
4.0816327
-6.4408200
0.526465306
-4.0816300
7.4408160
0.345534694
Mass of Sample
% KBr
12 13 14
Spreadsheet Documentation
15 16
A8:B8=MINVERSE(A4:B5) D8:D9=MMULT(A8:B9,D4:D5)
17 18
C12=100*D9/A12 D12=100*D8/A12
1
A Problem 12-28
2 3
Coefficient Matrix
0.872
% NaBr
39.6
60.4
12-28
4 5 6 7 8 9 10 11 12
B
C
D
Constant Matrix 1
1
0.443
1
0.6104698
0.3181
Inverse Matrix
Solution Matrix
-1.5671952
2.5671952
0.122357321
2.5671952
-2.5671952
0.320642679
Mass of Sample 0.6407
Mass AgCl 0.1223573
13 14
Spreadsheet Documentation
15 16
A8:B8=MINVERSE(A4:B5) D8:D9=MMULT(A8:B9,D4:D5)
17 18 19
B12=D4-D9 C12=100*D9/A12 D12=100*D8/A12
% Cl 4.72
%I 27.05
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
12-29 MPbMoO
= 4
MP O
367.14 g / mole
mol PbMoO 4
=
2
= 5
0.2554 g PbMoO 4 ×
141.94 g / mole
1 mole PbMoO 4
=
367.14 g
1 mole P 6.9565 × 10 4 mol PbMoO 4 × 12 moles PbMoO 4 −
×
−
4
6.9565 × 10 mol PbMoO 4
1 mole P2 O 5
×
141.94 g P2 O 5
2 mole P
1 mole
0.1969 g sample =
× 100%
2.089% P2 O 5
12-30 MCO
= 2
MMgCO
44.010 g / mole
mole CO 2
=
mole MgCO 3
+
= 3
84.31 g / mole
MK
2 CO 3
=
138.21 g / mole
mole K 2 CO 3
38 g MgCO 3 1 mole MgCO 3 × 1.500 g sample × + 100 g sample 84.31 g 42 g K 2 CO 3 1 mole K 2 CO 3 × 1.500 g sample × 100 g sample 138 . 21 g =
−
3
mole CO 2
=
6.76 × 10
mass CO 2
=
0.0113 mole ×
+
4.46 × 10
−
3
=
0.0113 mole
44.010 g CO 2 1 mole
=
0.498 g CO 2
12-31 MMg
2 P2 O7
=
MNaCl
222.55 g / mole
0.1796 g Mg 2 P2 O 7
×
=
1 mole Mg 2 P2 O 7 222.55 g
58.44 g / mole ×
MMgCl
2 •6 H 2O
2 moles MgCl 2 ⋅ 6H 2 O
=
203.32 g / mole
=
1 mole Mg 2 P2 O 7
1.61402 × 10 3 mole MgCl 2 ⋅ 6H 2 O −
0.5923 g AgCl × −
3
1 mole AgCl 143.32 g
1.61402 × 10 mole MgCl 2
•
=
4.13268 × 10 3 mole AgCl
6H 2 O ×
−
2 moles AgCl 1 mole MgCl 2 ⋅ 6H 2 O
=
−
3
3.22804 × 10 mole AgCl
th
Fundamentals of Analytical Chemistry: 8 ed.
( 4.13268 × 10
−
3
−
−
Chapter 12
3
3.228 × 10 ) mole AgCl ×
1.61402 × 10 3 mole MgCl 2 −
•
6H 2 O ×
1 mole NaCl
=
1 mole AgCl
−
203.32 g MgCl 2 ⋅ 6H 2 O
×
1 mole
500.0 mL 50.0 mL
6.881 g sample =
4
9.0464 × 10 mole NaCl
×
100%
47.69% MgCl 2 ⋅ 6H 2 O
9.0464 × 10 4 mole NaCl × −
58.44 g NaCl 1 mole
×
500.0 mL 50.0 mL
×
6.881 g sample
100% = 7.68% NaCl
12-32 MBaCl
2 ⋅2 H 2O
=
0.200 g BaCl 2 ⋅ 2 H 2 O × =
−
MNaIO
244.26 g / mole
4
8.188 × 10 mole Ba
0.300 g NaIO 3 ×
= 3
197.89 g / mole
1 mole BaCl 2 ⋅ 2 H 2 O
1 mole Ba 2
×
244.26 g
MBa ( IO
3 )2
=
487.13 g / mole
+
1 mole BaCl 2 ⋅ 2H 2 O
2+
1 mole NaIO 3
×
197.89 g
1 mole IO 3
−
=
1 mole NaIO 3
1.516 × 10 3 mole IO 3 −
−
−
Because IO 3 is the limiting reagent , (a)
1.516 × 10 3 mole −
moles Ba ( IO 3 ) 2 mass Ba ( IO3 ) 2
=
=
2 =
7.580 × 10 4 mole × −
−
4
7.580 × 10 mole
487.13 g Ba ( IO 3 ) 2 1 mole
=
0.369 g Ba ( IO3 ) 2
(b)
mole BaCl 2 ⋅ 2 H 2 O remaining
=
((8.188 × 10
4
−
) − (7.580 × 10 ) ) mole = 6.080 × 10 mole
mass BaCl 2 ⋅ 2H 2 O = 6.08 × 10 5 mole BaCl 2 ⋅ 2H 2 O × −
=
0.0149 g BaCl 2 ⋅ 2H 2 O
4
−
−
244.26 g BaCl 2 ⋅ 2H 2 O 1 mole
5
th
Fundamentals of Analytical Chemistry: 8 ed.
Chapter 12
12-33 MAgNO
= 3
0.500 g AgNO 3 × =
1 mole AgNO 3
2CrO4
×
331.730 g / mole
=
1 mole Ag 2 CrO 4
169.873 g
×
MK
2 CrO4
=
194.190 g / mole
331.730 g Ag 2 CrO 4
2 mole AgNO 3
1 mole
0.4882 g Ag 2 CrO 4
0.300 g K 2 CrO 4 =
MAg
169.873 g / mole
×
1 mole K 2 CrO 4
×
194.190 g
1 mole Ag 2 CrO 4
×
331.730 g Ag 2 CrO 4
1 mole K 2 CrO 4
1 mole
0.5125 g Ag 2 CrO 4 +
Because Ag is the limiting reagent , (a) mass Ag 2 CrO 4
=
0.488 g Ag 2 CrO 4
(b)
mole K 2 CrO 4 remaining mass K 2 CrO 4 =
=
=
((1.545 × 10
−
5
3
−
) − (1.472 × 10 ) ) mole = 7.331 × 10 mole
7.331 × 10 mole K 2 CrO 4
0.0142 g K 2 CrO 4
3
−
×
194.190 g K 2 CrO 4 1 mole
5
−
