Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
Chapter 16
16-1
Carbon dioxide is not strongly bonded by water molecules, and thus is readily volatilized
from aqueous solution by briefly boiling. On the other hand, HCl molecules are fully
dissociated into H3O+ and Cl- when dissolved in water. Neither the H3O+ nor the Clspecies is volatile.
16-2
Nitric acid is seldom used as a standard because it is an oxidizing agent and thus will
react with reducible species in titration mixtures.
16-3
Primary standard Na2CO3 can be obtained by heating primary standard grade NaHCO3
for about an hour at 270C to 300C. The reaction is
2 NaHCO 3 ( s )  Na 2 CO3 ( s)  H 2 O( g )  CO2 ( g )
16-4
Near the equivalence point in the titration of Na2CO3, the solution contains a buffer made
up of a high concentration of H2CO3 and a small amount of Na2CO3. Boiling removes
the H2CO3 as CO2, which causes the pH of the solution to rise sharply (see Figure 16-1).
Then the change in pH, when titration is resumed, is much greater than it would
otherwise be. Thus, a sharper end point results.
16-5
Let us consider the standardization of 40 mL of 0.010 M NaOH using KH(IO3)2,
1 mmol KH( IO 3 ) 2 390 g KH( IO3 ) 2
0.010 mmol NaOH
 40 mL NaOH 

 0.16 g KH( IO 3 ) 2
mL
1 mmol NaOH
1000 mmol
Now using benzoic acid,
1 mmol C 6 H 5COOH
0.010 mmol NaOH
 40 mL NaOH 

mL
1 mmol NaOH
122 g C 6 H 5COOH
 0.049 g C 6 H 5COOH
1000 mmol
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
The primary standard KH(IO3)2 is preferable because the relative weighing error would
be less with a 0.16 g sample of KH(IO3)2 as opposed to 0.049 g sample of benzoic acid.
A second reason for preferring KH(IO3)2 is because it is a strong acid and benzoic acid is
not. A smaller titration error occurs when using a strong acid as a primary standard.
16-6
If the sodium hydroxide solution is to be used for titrations with an acid-range indicator,
the carbonate in the base solution will consume two analyte hydronium ions just as would
the two hydroxides lost in the formation of Na2CO3.
16-7
Unless a reducing agent is introduced into the H2SO4 prior to digestion, nitro-, azo- and
azoxy- groups will be partially converted to N2 or nitrogen oxides that are then lost by
volatilization. Heterocyclic compounds containing nitrogen also yield low results in
many instances because these compounds tend to be incompletely decomposed under the
usual digestion procedure.
16-8
(a)
0.15 mole KOH
56.1 g KOH
 2.00 L 
 17 g KOH
L
mole
Dissolve 17 g KOH in water and dilute to 2.00 L total volume.
(b)
0.015 mole Ba(OH) 2  8H 2 O
315 g Ba(OH) 2  8H 2 O
 2.00 L 
 9.5 g Ba(OH) 2  8H 2 O
L
mole
Dissolve 9.5 g Ba(OH)2·8H2O in water and dilute to 2.00 L total volume.
(c)
0.200 mole HCl
36.46 g HCl
mL reagent
100 g reagent
 2.00 L 


 120 mL reagent
L
mole
1.0579 g reagent 11.50 g HCl
Dilute 120 mL reagent to 2.00 L total volume.
Fundamentals of Analytical Chemistry: 8th ed.
16-9
Chapter 16
(a)
0.250 mmol H 2SO 4
98.08 g H 2SO 4
mL reagent
100 g reagent
 500.0 mL 


mL
1000 mmol
1.1539 g reagent 21.8 g H 2SO 4
 48.7 mL reagent
Dilute 48.7 mL reagent to 500.0 mL total volume.
(b)
0.30 mmol NaOH
40.0 g NaOH
 500.0 mL 
 6.0 g NaOH
mL
1000 mmol
Dissolve 6.0 g NaOH in water and dilute to 500.0 mL total volume.
(c)
0.08000 mmol Na 2 CO 3
105.99 g Na 2 CO 3
 500.0 mL 
 4.24 g Na 2 CO 3
mL
1000 mmol
Dissolve 4.24 g Na2CO3 in water and dilute to 500.0 mL total volume.
16-10 For the first data set,
0.7987 g KHP 
csample1 
1000 mmol KHP 1 mmol NaOH

204.22 g
1 mmol KHP
 0.10214 M NaOH
38.29 mL NaOH
The results below were calculated in the same way.
Sample
1
csample i, M
0.10214
csample i2
1.043310-2
2
0.10250
1.050610-2
3
0.10305
1.061910-2
4
0.10281
1.057010-2
c
samplei
 0.4105
c
samplei
2
 4.2128  102
Fundamentals of Analytical Chemistry: 8th ed.
(a)
csamplei 
Chapter 16
0.4105
 0.1026 M NaOH
4
(b)
( 4.2128  10 2 )  (0.4105) 2 / 4

3
3.9  10 4
CV 
 100%  0.38%
0.1026
s
(c)
4.657  10 7
 3.9  10 4
3
Spread, w = 0.10305 – 0.10214 = 0.00091
16-11 For the first data set,
0.2068 g Na 2 CO3 
csample1 
1000 mmol Na 2 CO3 2 mmol HClO 4

105.99 g
1 mmol Na 2 CO3
 0.10747 M HClO 4
36.31 mL HClO 4
The results below were calculated in the same way.
Sample
csample i, M
csample i2
1
2
0.10747
0.10733
1.1549910-2
1.1519610-2
3
0.10862
1.1798710-2
4
0.10742
1.1538510-2
c
samplei
(a)
csamplei 
 0.43084
0.43084
 0.1077 M HClO 4
4
c
samplei
2
 4.64069  102
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
(b)
( 4.64069  10 2 )  (0.43084) 2 / 4
1.11420  10 6

 6.1  10 4
3
3
4
6.1  10
CV 
 100%  0.57%
0.1077
s
(c)
0.10862  0.10747
 0.89
0.10862  0.10733
Q crit  0.829 at the 95% confidence level
Q
Q crit  0.926 at the 99% confidence level
Thus, 0.10862 could be rejected at 95% level but must be retained at 99% level.
With phenolphthalein, the CO32- consumes 1 mmol H3O+ per mmol CO2. Thus,
16-12 (a)
the effective amount of NaOH is lowered by 11.2 mmol, and
1 mmol NaOH 
 0.1500 mmol NaOH
 

 1000 mL   11.2 mmol CO 2 

mL
1 mmol CO 2 

 
cbase 
1000 mL
 0.1388 M NaOH
(b)
When bromocresol green is the indicator,
2
CO3  2H 3O   H 2 CO3  2H 2 O
and the effective concentration of the base is unchanged. Thus,
cbase  0.1500 M
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-13 As in part (a) of Problem 16-12,
1000 mmol CO 2 1 mmol NaOH 
 0.1019 mmol NaOH
 

 500 mL    0.652 g CO 2 


mL
44.01 g
1 mmol CO 2 

 
cbase 
500 mL
 0.07227 M NaOH
0.07227  0.1019
 100%   29%
0.1019
The relative error in the determination of acetic acid will be the same as the relative error
relative carbonate error in molarity 
in the molarity or -29%.
16-14 (a)
0.6010 g AgCl 
1000 mmol AgCl 1 mmol HCl

143.32 g
1 mmol AgCl
 0.08387 M HCl
50.00 mL HCl
(b)
0.04010 mmol Ba (OH ) 2
2 mmol HCl
 25.00 mL Ba (OH ) 2 
mL
1 mmol Ba (OH ) 2
 0.1007 M HCl
19.92 mL HCl
(c)
0.2694 g Na 2 CO3 
1000 mmol Na 2 CO3
2 mmol HCl

105.99 g
1 mmol Na 2 CO3
 0.1311 M HCl
38.77 mL HCl
16-15 (a)
0.1684 g BaSO 4 
1000 mmol BaSO 4 1 mmol Ba (OH ) 2

233.39 g
1 mmol BaSO 4
 0.01443 M Ba (OH ) 2
50.00 mL Ba (OH ) 2
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
(b)
0.4815 g KHP 
1000 mmol KHP 1 mmol Ba (OH ) 2

204.22 g
2 mmol KHP
 0.04008 M Ba (OH ) 2
29.41 mL Ba (OH ) 2
(c)
mmol C 6 H 5COOH  0.3614 g C 6 H 5COOH 
1000 mmol C 6 H 5COOH
 2.9594 mmol
122.12 g
0.05317 mmol HCl
 4.13 mL HCl  0.21959 mmol
mL
total mmol acid  2.9594  0.21959  3.17899
mmol HCl 
1 mmol Ba (OH ) 2
2 mmol acid
 0.03179 M Ba (OH ) 2
50.00 mL Ba (OH ) 2
3.17899 mmol acid 
16-16 (a)
For 35 mL,
1 mmol Na 2 CO 3 105.99 g Na 2 CO 3
0.150 mmol HClO 4
 35 mL HClO 4 

 0.28 g Na 2 CO 3
mL
2 mmol HClO 4
1000 mmol
Substituting 45 mL in the equation above gives 0.36 g Na2CO3.
Thus, the range of sample masses is 0.28 to 0.36 g Na 2 CO3 .
Proceeding in the same way as in part (a), we obtain
(b)
0.18 to 0.23 g Na 2C2O4
(c)
0.85 to 1.1 g benzoic acid
(d)
0.82 to 1.1 g KH(IO3 ) 2
(e)
0.17 to 0.22 g TRIS
(f)
1.1 to 1.4 g Na 2 B4 O 7 ·10H 2 O
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-17 In Example 16-1, we found that 20.00 mL of 0.0200 M HCl requires 0.048 g TRIS, 0.021
g Na2CO3 and 0.076 g Na2B4O7·10H2O. In each case, the absolute standard deviation in
computed molarity of 0.0200 M HCl is
TRIS: s M 
0.0001
 0.0200 M  4  105 M
0.048
Na2CO3: s M 
0.0001
 0.0200 M  1  10 4 M
0.021 g
(Note: In the first printing of the text, the answer in the back of the book was in error.)
Na2B4O7·10H2O: s M 
0.0001
 0.0200 M  3  10 5 M
0.076 g
Proceeding as above, we calculate the relative standard deviation in the computed
molarity of 30.00 mL, 40.00 mL and 50.00 mL of 0.0200 M HCl and the results are
shown in the table below.
V0.0200 M HCl (mL)
Calculated masses (g)
sM (0.0200 M HCl)
30.00
TRIS
Na2CO3
Na2B4O7·10H2O
0.073
0.032
0.11
310-5
610-5
210-5
40.00
TRIS
Na2CO3
Na2B4O7·10H2O
0.097
0.042
0.15
210-5
510-5
110-5
50.00
TRIS
Na2CO3
Na2B4O7·10H2O
0.12
0.053
0.19
210-5
410-5
110-5
Fundamentals of Analytical Chemistry: 8th ed.
16-18 (a)
Chapter 16
In each case,
mmol NaOH 
0.0400 mmol NaOH
 30.00 mL  1.20 mmol NaOH
mL
For KHP :
1.20 mmol NaOH 
1 mmol KHP 204.22 g KHP

 0.245 g KHP
1 mmol NaOH
1000 mmol
For KH( IO 3 ) 2 :
1.20 mmol NaOH 
1 mmol KH( IO 3 ) 2 389.91 g KH( IO 3 ) 2

 0.468 g KH( IO 3 ) 2
1 mmol NaOH
1000 mmol
For benzoic acid :
1.20 mmol C 6 H 5COOH 
1 mmol C 6 H 5COOH 122.12 g C 6 H 5COOH

 0.147 g C 6 H 5COOH
1 mmol NaOH
1000 mmol
(b)
For KHP :
0.002 g
sM 
 100%  0.82%
0.245 g
For KH( IO 3 ) 2 :
0.002 g
 100%  0.43%
0.468 g
For benzoic acid :
0.002 g
sM 
 100%  1.4%
0.147 g
sM 
16-19
0.03776 mmol NaOH
 21.48 mL NaOH  0.81109 mol NaOH
mL

1 mmol H 2 C 4 H 4 O 6 150.09 g H 2 C 4 H 4 O 6 
 0.81109 mol NaOH 


2 mmol NaOH
1000 mmol

  100 mL
50.00 mL
 0.1217 g H 2 C 4 H 4 O 6 per 100 mL
mol NaOH 
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-20
 0.09600 mmol NaOH
1 mmol HOAc 60.05 g HOAc 


 34.88 mL NaOH 

mL
1 mmol NaOH
1000 mmol 

 100%
 50.0 mL


 25.00 mL 
 250 mL

 4.02%( w / v ) HOAc
16-21 For each part, we may write
0.1081 mmol HCl
 31.64 mL HCl
mmol HCl
mL
 4.5978
0.7439 g sample
g sample
(a)
4.5978
mmol HCl 1 mmol Na 2 B4 O 7 201.22 g Na 2 B4 O 7


 100%  46.26% Na 2 B4 O 7
g sample
2 mmol HCl
1000 mmol
Proceeding in the same way
(b)
mmol HCl 1 mmol Na 2 B4 O 7  10H 2 O 381.37 g Na 2 B4 O 7  10H 2 O


 100%
g sample
2 mmol HCl
1000 mmol
 87.67% Na 2 B4 O 7  10H 2 O
4.5978
(c)
4.5978
mmol HCl 1 mmol B2 O 3 69.62 g B2 O 3


 100%  32.01% B2 O 3
g sample
1 mmol HCl
1000 mmol
(d)
4.5978
mmol HCl
2 mmol B 10.811 g B


 100%  9.94% B
g sample 1 mmol HCl 1000 mmol
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-22
 0.1178 mmol HCl
1 mmol OH  1 mmol HgO 216.59 g HgO 


 42.59 mL HCl 


mL
1 mmol HCl 2 mmol OH 
1000 mmol 

 100%
0.6334 g sample
 85.78% HgO
16-23
 0.0996 mmol NaOH

mmol NaOH consumed  
 50.0 mL NaOH  
mL


 0.05250 mmol H 2SO 4
2 mmol NaOH 

  2.534 mmol NaOH
 23.3 mL H 2SO 4 
mL
1
mmol
H
SO
2
4 

1 mmol HCHO 30.026 g HCHO
2.534 mmol NaOH 

1 mmol NaOH
1000 mmol
 100%  24.4% HCHO
0.3124 g sample
16-24
 0.0514 mmol NaOH
1 mmol NaO 2 CC 6 H 5 144.1 g NaO 2 CC 6 H 5 


 14.76 mL NaOH 

mL
1 mmol NaOH
1000 mmol

  100%
106.3 g sample
 0.103% NaO 2 CC 6 H 5
16-25 Tetraethylthiuram disulfide, TS4
1 mmol TS4  4 mmol SO2  4 mmol H2SO4  8 mmol NaOH
 0.03736 mmol NaOH
1 mmol TS4
296.54 g TS4 


 22.13 mL NaOH 

mL
8 mmol NaOH 1000 mmol 

 100%
0.4329 g sample
 7.079% TS4
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-26
 0.2506 mmol HCl
1 mmol NH 3 17.031 g NH 3 


 40.38 mL HCl 

mL
1 mmol HCl
1000 mmol 

 100%
 50.00 mL


 25.00 mL 
250
.
0
mL


 3.447%( w / v ) NH 3
16-27
mmol HCl  mmol NaOH  2  mmol CO 3
2
 0.1140 mmol HCl
  0.09802 mmol NaOH

 50.00 mL HCl   
 24.21 mL NaOH 

mL
mL
2
 

mmol CO 3  
2
2
 1.6635 mmol CO 3
molar mass carbonate salt 
0.1401 g salt
1000 mmol
g salt

 84.22
2
2
mole
1.6635 mmol CO 3
mole CO 3
2
2

1 mole CO 3 
g CO 3
g salt
  60.01
molar mass of carbonate salt cation   84.22

2
1 mole salt 
mole
mole CO 3

g cation
 24.21
mole
MgCO 3 with a molar mass of 84.31 g / mole appears to be a likely candidate.
16-28
0.1084 mmol NaOH
1 mmol NaA
 28.62 mL NaOH 
 3.1024 mmol NaOH
mL
1 mmol NaOH
0.2110 g NaA
1000 mmol
g NaA
molar mass NaA 

 68.01
3.1024 mmol NaA
mole
mole
molar mass HA  equivalent weight  molar mass NaA  atomic mass Na  atomic mass H
g HA
 68.01  22.99  1.008  46.03
mole
mmol NaA 
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-29
mmol Ba (OH ) 2  mmol CO 2 
mmol HCl
2
 0.0108 mmol HCl

 23.6 mL HCl 

 0.0116 mmol Ba (OH ) 2
 
mL

mmol CO 2  
 50.0 mL Ba (OH ) 2  
mL
2

 





1
 4.526  10 mmol CO 2
0.4526 mmol CO 2 
44.01 g CO 2
1000 mmol
3.00 L

1 L CO 2
 10 6 ppm  3.35  10 3 ppm CO 2
1.98 g CO 2
16-30
1 mmol SO 2
64.06 g SO 2
0.00204 mmol NaOH
 11.1 mL NaOH 

 7.253  10 4 g SO 4
mL
2 mmol NaOH 1000 mmol
1.00 L SO 2
1 min
1
7.253  10 4 g SO 4 


 10 6 ppm  0.848 ppm SO 2
30.0 L sample 10.0 min 2.85 g SO 2
16-31 ( NH 4 ) 3 PO4  12MoO3 ( s)  26OH   HPO4
2
 12MoO4
2
 14H 2 O  3NH 3 ( g )
 0.2000 mmol NaOH

mmol NaOH consumed  
 50.00 mL NaOH  
mL


 0.1741 mmol HCl

 14.17 mL HCl   7.533 mmol NaOH

mL


1 mmol ( NH 4 ) 3 PO4  12 MoO 3
mmol P  7.533 mmol NaOH 

26 mmol NaOH
1 mmol P
 2.897  10 1 mmol P
1 mmol ( NH 4 ) 3 PO4  12 MoO 3
30.974 g P
1000 mmol
 100%  6.333% P
0.1417 g sample
2.897  10 1 mmol P 
Fundamentals of Analytical Chemistry: 8th ed.
16-32 C6 H 4 (COOCH 3 ) 2  2OH   C6 H 4 (COO) 2
Chapter 16
2
 2CH 3OH
C 6 H 4 (COOCH 3 ) 2  DMP
 0.1031 mmol NaOH

mmol NaOH consumed  
 50.00 mL NaOH  
mL


 0.1251 mmol HCl

 32.25 mL HCl   1.1205 mmol NaOH

mL


1 mmol DMP
mmol DMP  1.1205 mmol NaOH 
 0.5603 mmol DMP
2 mmol NaOH
194.19 g DMP
0.5603 mmol DMP 
1000 mmol
 100%  13.33% DMP
0.8160 g sample
16-33
Neohetramine, C16H 21ON 4  RN 4
1 mmol RN 4  3 mmol NH 3  4 mmol HCl
1 mmol RN 4 285.37 g RN 4
0.01522 mmol HCl
 36.65 mL HCl 

mL
4 mmol HCl
1000 mmol
 100%  25.98% RN 4
0.1532 g sample
16-34 (a)
 0.1750 mmol HCl

mmol HCl consumed  
 100.0 mL HCl  
mL


 0.1080 mmol NaOH

 11.37 mL NaOH   16.272 mmol HCl

mL


16.272 mmol HCl 
1 mmol CH 5 N 3 59.07 mg CH 5 N 3

3 mmol HCl
mmol
 80.10 mg / tablet
4 tablets
10 mg CH 5 N 3
4.536  10 1 kg 1 tablet
 100 lb 

 5.68 tablets or 6 tablets
kg
lb
80.10 mg
Proceeding in the same way as part (a), we find the results for parts (b) and (c) in the
no. tablets 
spreadsheet that follows.
Fundamentals of Analytical Chemistry: 8th ed.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
A
Problem 16-34
B
C
Conc. HCl, M
Vol. HCl, mL
Conc. NaOH, M
Vol. NaOH, mL
No. Tablets
0.1750
100.0
0.1080
11.37
4
mg CH5N3/tablet
80.0991169
(a)
(b)
(c)
Chapter 16
D
E
Patient Wt, lb Proper dose No. tablets
100 5.67546831
6
150 8.51320247
9
275 15.6075379
16
Spreadsheet Documentation
B9 = (((B3*B4)-(B5*B6))*(1/3)*59.07)/B7
C12 = 10*B12*0.4546*(1/$B$9)
D12 = ROUND(C12,0)
16-35
 0.1224 mmol HCl
 1 mmol N 14.007 g N
 22.66 mL HCl  


mL
mmol HCl 1000 mmol


%N 
 100%  3.92% N
0.992 g sample
16-36 Multiplication factor for meat is 6.25 protein/N
6.25 protein
 24.48% protein
N
6.50 oz tuna 28.3 g 24.48 g protein


 45.0 g protein / can
can
oz
100 g tuna
3.916% N 
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-37
A
B
C
1
16-37 Nitrogen in a plant food preparation
2
Weight sample, g
3
Volume of HCl, mL
4
Conc. HCl, M
5
Vol. NaOH, mL
6
Conc. NaoH, M
E
F
0.5843
50.00
0.1062
11.89
0.0925 In Cell B8, the mmol of HCl/g sample is calculated by subtracting
7
8
D
the mmol of NaOH used from the total mmol of HCl added and
mmol HCl/g sample
9
7.20550 dividing by the sample weight.
Molar masses
Percentages
10
(a) N
11
(b) urea
12
(c) (NH4)2SO4
132.141
47.61 % (NH4)2SO4
13
(d) (NH4)3PO4
149.09
35.81 % (NH4)3PO4
14
Spreadsheet Documentation
15
B8=($B$3*$B$4-$B$5*$B$6)/$B$2
16
C10=$B$8*1*B10/1000*100
The percentages are calculated in Cells C10:C13 from the no. of
17
C11=$B$8*1/2*B11/1000*100
mmol of HCl/g sample times the no. of mmol compound/mmol HCl
18
C12=$B$8*1/2*B12/1000*100
times the molar mass of the compound divided by 1000 (mmolar
19
C13=$B$8*1/3*B13/1000*100
mass).
14.007
10.09 %N
60.06
21.64 % urea
16-38
 0.05063 mmol
  0.04917 mmol

mmol HCl consumed  
 50.00 mL   
 7.46 mL   2.165 mmol
mL
mL

 

1 mmol N
14.007 g
2.165 mmol HCl 

1 mmol HCl 1000 mmol
%N 
 100%  3.335% N
0.9092 g
% protein  3.335%  5.7  19.0% protein
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-39 In the first titration,
 0.08421 mmol HCl

mmol HCl consumed  
 30.00 mL  
mL


 0.08802 mmol NaOH

 10.17 mL   1.63114 mmol HCl

mL


and
1.63114 mmol HCl  mmol NH 4 NO 3  2  mmol ( NH 4 ) 2 SO4 
The amounts of the two species in the original sample are
mmol NH 4 NO 3  2  mmol ( NH 4 ) 2 SO 4   1.63114 mmol 
200 mL
 6.5246 mmol (1)
50 mL
In the second titration,
 0.08421 mmol HCl

mmol HCl consumed  
 30.00 mL  
mL


 0.08802 mmol NaOH

 14.16 mL   1.27994 mmol HCl

mL


and
1.27994 mmol HCl  (2  mmol NH 4 NO3 )  2  mmol ( NH 4 ) 2 SO4 
The amounts of the two species in the original sample are
( 2  mmol NH 4 NO 3 )  2  mmol ( NH 4 ) 2 SO 4   1.27994 mmol 
200 mL
 10.2395 mmol
25 mL
(2)
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
Subtracting equation (1) from equation (2) gives
mmol NH 4 NO 3  10.2395 mmol  6.52455 mmol  3.7149 mmol NH 4 NO 3
10.2395 mmol  ( 2  3.7149 mmol)
 1.4048 mmol ( NH 4 ) 2 SO 4
2
80.04 g NH 4 NO 3
3.7149 mmol NH 4 NO 3 
1000 mmol
% NH 4 NO 3 
 100%  24.39% NH 4 NO 3
1.219 g sample
132.14 g ( NH 4 ) 2 SO 4
1.4048 mmol ( NH 4 ) 2 SO 4 
1000 mmol
% ( NH 4 ) 2 SO 4 
 100%
1.219 g sample
 15.23% ( NH 4 ) 2 SO 4
mmol ( NH 4 ) 2 SO 4 
16-40 For the first aliquot,
mmol HCl consumed  mmol NaOH  mmol KOH  ( 2  mmol K 2 CO 3 )
 0.05304 mmol HCL

mmol KOH  ( 2  mmol K 2 CO 3 )  
 40.00 mL HCl  
mL


 0.04983 mmol NaOH

 4.74 mL NaOH   1.8854 mmol

mL


For the second aliquot,
0.05304 mmol HCl
 28.56 mL HCl  1.5148 mmol HCl( KOH )
mL
1.8854 mmol  1.5148
mmol K 2 CO 3 
 0.1853 mmol K 2 CO 3
2
mmol HCl  mmol KOH 
56.11 g KOH
1000 mmol
 100%  69.84% KOH

50 mL 
1.217 g 

500 mL 

1.5148 mmol KOH 
% KOH 
138.21 g K 2 CO 3
1000 mmol
 100%  21.04% K 2 CO 3

50 mL 
1.217 g 

500 mL 

0.18530 mmol K 2 CO 3 
% K 2 CO 3 
100%  (69.84%  21.04%)  9.12% H 2 O
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-41 For the first aliquot,
mmol HCl  mmol NaOH  mmol NaHCO 3  ( 2  mmol Na 2 CO 3 )
 0.01255 mmol HCl

mmol NaHCO 3  ( 2  mmol Na 2 CO 3 )  
 50.00 mL HCl  
mL


 0.01063 mmol NaOH

 2.34 mL NaOH   0.6026 mmol

mL


For the second aliquot,
mmol NaHCO 3  mmol NaOH  mmol HCl
 0.01063 mmol NaOH
  0.01255 mmol HCl


 25.00 mL NaOH   
 7.63 mL HCl 
mL
mL

 

 0.1700 mmol NaHCO 3
mmol Na 2 CO3 
0.6026 mmol  0.1700 mmol
 0.2163 mmol Na 2 CO3
2
84.01 g NaHCO 3
1000 mmol
 100%  28.56% NaHCO 3

25.00 g 
 0.5000 g 

250.0 g 

0.1700 mmol NaHCO 3 
% NaHCO 3 
105.99 g Na 2 CO 3
1000 mmol
 100%  45.85% Na 2 CO 3

25.00 mL 
 0.5000 g 

250.0 mL 

0.2163 mmol Na 2 CO 3 
% Na 2 CO 3 
100%  ( 28.56%  45.85%)  25.59% H 2 O
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-42
1
2
A
B
C
16-42 Titrations with 0.06122 M HCl
M HCl
0.06122
D
E
3
M Na3PO4
4
5
6
7
8
(a)
Add one proton
to thymolphthalein
endpoint
mL Na3PO4 mmol base mL HCl
10.00
0.55550
9.07
15.00
0.83325
13.61
25.00
1.38875
22.68
40.00
2.22200
36.30
9
10
11
12
13
14
15
(b)
Add two protons
to bromocresol
green endpoint
mL Na3PO4 mmol base mL HCl
10.00
1.11100
18.15
15.00
1.66650
27.22
20.00
2.22200
36.30
25.00
2.77750
45.37
(c)
mL solution mmol base mL HCl
16
M Na3PO4
0.02102
20.00
1.17180
19.14
17
18
19
20
21
M Na2HPO4
0.01655
Add two protons
to phosphate and
one to monohydrogen phosphate
25.00
30.00
40.00
1.46475
1.75770
2.34360
23.93
28.71
38.28
22
23
24
25
26
27
28
29
30
31
32
33
34
35
(d)
M NaOH
Add one proton
0.05555
0.01655 mL Na3PO4 mmol base mL HCl
15.00
0.56355
9.21
20.00
0.75140
12.27
35.00
1.31495
21.48
40.00
1.50280
24.55
Spreadsheet Documentation
D5=C5*$B$3
D10=C10*$B$3*2
D16=$B$15*2*C16+$B$16*C16
D23=C23*$B$16+C23*$B$22
E5=D5/$B$2
E10=D10/$B$2
E16=D16/$B$2
E23=D23/$B$2
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-43
1
2
3
4
A
B
C
16-43 Titrations with 0.07731 M NaOH
M NaOH
0.07731
(a) and (b)
M HCl
0.03000
5
6
7
8
9
10
11
12
13
M H3PO4
(a) React with one
proton to bromocresol green
endpoint
(b) React with two
protons to thymolphthalein end point
14
(c) NaH2PO4
15
16
17
18
19
20
M NaH2PO4
One proton reacts
to thymol-phthalein
endpoint
(d) Mixture
M H3PO4
0.06407
21
M NaH2PO4
0.03000
D
E
0.01000
mL solution mmol acid mL NaOH
25.00
1.0000
12.93
mL solution mmol acid mL NaOH
25.00
1.2500
16.17
mL solution mmol acid
mL NaOH
10.00
0.64070
8.29
20.00
1.28140
16.57
30.00
1.92210
24.86
40.00
2.56280
33.15
mL solution mmol acid mL NaOH
0.02000
20.00
1.40000
18.11
22
23
(d) React with two
24
25
protons from H3PO4
and one from
26
27
28
29
30
31
32
33
34
35
NaH2PO4
Spreadsheet Documentation
D7=C7*$B$4+C7*$B$5
E6=D6/$B$2
D11=C6*$B$4+2*C6*$B$5
E11=D11/$B$2
D15=$B$15*C15
E15=D15/$B$2
D20=2*$B$20*C20+$B$21*C20
E20=D20/$B$2
25.00
1.75000
22.64
30.00
2.10000
27.16
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-44
A
B
C
D
E
F
1
16-44 Titrations of carbonate mixtures
2
M HCl
0.1202
(a)
22.42
22.44
3
Volume, mL
25.00
(b)
15.67
42.13
4
M NaOH
40.00
(c)
29.64
36.42
5
M Na2CO3
105.99
(d)
16.12
32.23
6
M NaHCO3
84.01
(e)
0.00
33.33
7
Table 14-2 gives the volume relationships in titrations of these mixtures.
Vol. to phenol., mL Vol. to BCG, mL
8
9
(a) Since essentially the same volume is used for each endpoint, there is only NaOH
10
present. We use the average volume to calculate the no. of mg NaOH/mL
11
mmol NaOH
12
mg NaOH/mL
2.6961
4.314
13
(b) Since Vphth< ½Vbcg, only carbonate and bicarbonate are present.
14
mmol carbonate
15
mmol total
1.8835
mmol bicarbonate
5.0640
16
(c) Now Vphth > ½Vbcg, so we have a mixture of NaOH and Na2CO3
17
mmol carbonate plus NaOH
18
19
mmol NaOH
0.8150
(d) Since Vphth = ½Vbcg, we have only Na2CO3 present
20
mmol carbonate
21
22
mmol carbonate
3.5627
1.2970
mg Na2CO3/ml
1.9376
8.215
(e) Since Vphth = 0, we have only NaHCO3 present which gains one proton.
23
mmol NaHCO3
24
mg NaHCO3/ml
4.0063
13.46
25
Spreadsheet Documentation
26
B12=((D2+E2)/2)*$B$2
D18=B18-C18
27
C12=B12*1*$B$4/$B$3
E18=C18*$B$5/$B$3
28
B15=D3*$B$2
F18=D18*$B$4/$B$3
29
C15=E3*$B$2
B21=$D$5*$B$2
30
D15=C15-2*B15
C21=B21*1*$B$5/$B$3
31
E15=B15*$B$5/$B$3
B24=E6*$B$2
32
F15=D15*$B$6/$B$3
C24=B24*1*$B$6/$B$3
33
B18=D4*$B$2
34
C18=($E$4-$D$4)*$B$2
2.7478
mg Na2CO3/ml
7.985
mg Na2CO3/ml
3.455
mg NaHCO3/ml
4.358
mg NaOH/ml
4.396
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-45
A
1
2
3
4
5
6
7
B
C
D
16-45 Titrations of Mixtures
E
F
Vol. to phenol., mL Vol. to BCG, mL
0.08601
(a)
0.00
18.15
Volume, mL
M NaOH
25.00
(b)
21.00
28.15
40.00
(c)
19.80
39.61
M Na3AsO4
207.89
(d)
18.04
18.03
M Na2HAsO4
185.91
(e)
16.00
37.37
M HCl
We use the method of Problem 14-44. Table 14-2 gives the volume relationships in titrations of similar mixtures .
8
9
10
11
12
13
(a) Since Vphth = 0, we have only Na2HAsO4 present which gains one proton.
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
(c) Since Vphth = ½Vbcg, we have only Na3AsO4 present
mmol NaHCO3
mg Na2HAsO4/ml
1.5611
11.61
(b) Now Vphth > ½Vbcg, so we have a mixture of NaOH and Na3AsO4
mmol Na3AsO4 plus NaOH
mmol Na3AsO4
1.8062
mmol Na3AsO4
mmol NaOH
0.6150
mg Na3AsO4/ml
1.1912
5.114
mg NaOH/ml
1.906
mg Na3AsO4/ml
1.7030
14.16
(d) Since essentially the same volume is used for each endpoint, there is only NaOH
present. We use the average volume to calculate the no. of mg NaOH/mL
mmol NaOH
mg NaOH/mL
1.5512
2.482
(e) Since Vphth< ½Vbcg, only Na3AsO4 and Na2HAsO4 are present.
mmol Na3AsO4
mmol total
1.3762
mmol Na2HAsO4
3.2142
Spreadsheet Documentation
B10=$E2*$B$2
B16=$D$4*$B$2
C10=B10*1*$B$6/$B$3
C16=$E$6*$B$2
B13=D3*$B$2
B23=$D$6*$B$2
C13=($E$3-$D$3)*$B$2
C23=$E$6*$B$2
D13=B13-C13
D23=C23-2*B23
E13=C13*$B$5/$B$3
E23=B23*$B$5/$B$3
F13=D13*$B$4/$B$3
F23=D23*$B$6/$B$3
0.4619
mg Na3AsO4/ml
11.44
mg Na2HAsO4/ml
3.435
Fundamentals of Analytical Chemistry: 8th ed.
Chapter 16
16-46 (a) The equivalent weight of an acid is that weight of the pure material that contains one
mole of titratable protons in a specified reaction.
(b) The equivalent weight of a base is that weight of the pure material that consumes one
mole of protons in a specified reaction.
16-47 (a) With bromocresol green, only one of the two protons in the oxalic acid will react.
Therefore, the equivalent mass is the molar mass, or 126.1 g.
(b) When phenolphthalein is the indicator, two of the protons are consumed. Therefore,
the equivalent mass of oxalic acid is one-half the molar mass, or 63.0 g.
16-48 (a)
1 mmol CH 3COOH
0.1008 mmol NaOH
 45.62 mL NaOH 
mL
mmol NaOH
 0.4598 M CH 3COOH
10.00 mL
(b)
0.4598 mmol CH 3COOH 60.03 mg CH 3COOH
1 mL CH 3COOH
1g



 100%
mL
mmol
1000 mg
1.004 g
 2.75% CH 3COOH