Chapter 1
1-19 How many grams of perchloric acid, HClO4, are contained in 37.6 g of 70.5 wt% aqueous
perchloric acid? How many grams of water are in the same solution?
wt%  Weight percent 
mass of subs tan ce
x ( 100 )
mass of total solution or total sample

g HClO4 
 0.705
37.6 g solution   26.5 g HClO4
g
solution 

37.6 g solution  26.5 g HClO4  11.1 g H 2 O
1-30 What is the maximum volume of 0.25M sodium hypochlorite solution (NaOCl, laundry
bleach) that can be prepared by dilution of 1.00 L of 0.80 M NaOCl?
McVc = MdVd
mol 
mol 


 0.80
1.00 L   0.25
Vd
L 
L 


Vd  0.80 / 0.25  3.2 L
Chapter 2
2-9 The densities (g/ml) of several substances are:
acetic acid 1.05
CCl4 1.59
Sulfur 2.07
lithium 0.53
mercury 13.5
PbO2 9.4
lead 11.4
iridium 22.5
From figure 2.5, predict which substance will have the smallest percentage buoyancy correction
and which will have the greatest.
PbO2: lowest correct density closest to 8.0 g/ml
Lithium: largest, lowest density (0.53 g/ml)
0.0012g/ml
)
8.0 g / ml
m
0.0012g/ml
(1 
)
d
m' ( 1 
Chapter 3
3-16 Find the absolute and percent relative uncertainty and express each answer with a
reasonable number of significant figures:
(c) [4.97 ± 0.05 – 1.86 ± 0.01]/21.1 ± 0.2 =
Error for subtraction:
0.0510 
0.05 2  0.012
= [3.11 ± 0.0510]/21.1 ± 0.2 both 4.97 & 1.86 have two numbers to the right
of decimal point
Error for division, convert to relative uncertainty:
= [3.11 ± 1.64%]/21.1 ± 0.95% 1.64% = 0.051/3.11 & 0.95% = 0.2/21.1
1.90 
1.64 2  0.95 2
= 0.147 ± 1.90%
both 3.11 and 21.1 have 3 significant figures
=0.147 ± 0.003
[1.90% x 0.147 = 0.0027 round up to 0.003]
Chapter 27
What is the %KCl in a solid if 5.1367 g of solid gives rise to 0.8246 g AgCl?
Cl-
+
Ag+ 
AgCl(s)
 1 mol AgCl   1 mol KCl   74.55 g KCl 
 
 
  0.4287 g
gKCl  0.8246 g AgCl  
 143 .4 g AgCl   1 mol AgCl   mol KCl 
 0.4287 
%KCl  
  100  8.346 %
 5.1367 
Note: 4 significant figures
27-35 A mixture weighing 7.290 mg contained only cyclohexane, C6H12 (FM 84.159), and
oxirane, C2H4O (FM 44.053). When the mixture was analyzed by combustion analysis, 21.999
mg of CO2 (FM 44.010) was produced. Find the weight percent of oxirane in the mixture.
FM
C6H12 +
84.159
C2H4O 
44,053
CO2 + H2O
44.010
Let x = mg of C6H12 and y = mg of C2H4O
x + y = 7.290
Also:
CO2 = 6 (moles of C6H12) + 2(moles of C2H4O)
Conserve number of carbon atoms:
xl 
 yl  21.999

6

  2
 84.159 l 
 44.053  44.010
Make substitution x = 7.290 –y and solve for y
y = 0.767 mg = 0.767 mg / 7.290 mg = 10.5 wt%
27-21. A mixture containing only Al2O3 (FM 101.96) and Fe2O3 (FM 159.69) weighs 2.019 g.
When heated under a stream of H2, Al2O3 is unchanged, but Fe2O3 is converted into metallic Fe
plus H2O (g). If the residue weighs 1.774 g, what is the weight percent of Fe2O3 in the original
mixture?
Fe2O3
+
Al2O3
2.019 g
heat

H2
Fe
+
Al2O3
1.774 g
Mass of oxygen lost: 2.019 g – 1.774 g = 0.245 g
Moles of oxygen atoms lost: (0.245 g )(1 mole / 15.9994 g) = 0.01531 moles
Fe2O3 : 3 moles of oxygen = 1 mole of Fe2O3
Moles of Fe2O3 = 1/3(0.01531) = 0.005105
Mass of Fe2O3 = (0.005105 moles)(159.69 g /mole) = 0.815 g
wt% = (0.815 g / 2.019 g)x100 = 40.4%
Chapter 4
4-A(i) For the following bowling scores 116.0, 97.9, 114.2, 106.8 and 108.3, find the mean,
median, range and standard deviation.
Mean ( x ) 
116.0  97.9  114.2  106.8  108.3
 108.6
5
Median = 97.9, 106.8, 108.3, 114.2, 116.0  108.3 (middle)
Range = 116.0 – 97.9 = 18.1
S tan dard Deviation ( s ) 
116.0  108.6 2  97.9  108.6 2  114.2  108.6 2  106.8  108.6 2  ( 108.3  108.6 )2
5  1
S tan dard Deviation ( s ) 
54.76  114 .49  31.36  3.24  0.09
54.76  114 .49  31.36  3.24  0.09
203 .94


4
4
4
S tan dard Deviation ( s )  50.985  7.1
4-A(ii) A bowler has a mean score of 108.6 and a standard deviation of 7.1. What fraction of the
bowler’s scores will be less than 90.2?
Determine how many standard deviations the value 80.2 is from the mean.
z
xx
s

108.6  80.2
7.1
 4.00
From Gaussian table:
Area below 2.60 standard deviation is 0.5000 - 0.499968 = 0.000032 = 3.2x10-3%
Therefore, the bowler only has a 3.2x10-3% chance of bowling a game below 80.2
4-A(iii) For the following bowling scores 116.0, 97.9, 114.2, 106.8 and 108.3, a bowler has a
mean score of 108.6 and a standard deviation of 7.1. What is the 90% confidence interval for the
mean?
2.132 7.1  108.6  6.8
ts
x
 108.6 
n
 5
Degrees of freedom 5-1 =4, 90% confidence from student’s t table = 2.132
95% confident range contains “true” mean : (true mean between 12.0 & 13.0)
4-A(iv) For the following bowling scores 116.0, 97.9, 114.2, 106.8 and 108.3, a bowler has a
mean score of 108.6 and a standard deviation of 7.1. Using the Q test, decide whether the
number 97.9 should be discarded.
97.9, 106.8, 108.3, 114.2, 116.0
Q
Gap
106 .8  97.9 8.9


 0.49  Q table  0.64
Range 116 .0  97.9 18.1
Therefore, 97.9 should be retained.
Chapter 5
Ex: The amount of protein in a sample is measured by the samples absorbance of light at a given
wavelength. Using standards, a best fit line of absorbance vs. mg protein gave the following
parameters:
m = 0.01630 sm = 0.00022
b = 0.1040
sb = 0.0026
An unknown sample has an absorbance of 0.246 ± 0.0059. What is the amount of protein in the
sample?
x
x
y  b 0.246  0.1040

 8.71 g
m
0.01630
y  b 0.246 ( 0.0059 )  0.1040 ( 0.0026 )

m
0.01630 ( 0.0002 2 )
x  8.71 g  0.4 g ( or  4.7 %)
5-19. Low concentrations of Ni-EDTA near the detection limit gave the following counts in a
mass spectral measurement: 175, 104, 164, 193, 131, 189, 155, 133, 151, 176. Ten measurements
of a blank had a mean of 45 counts. A sample containing 1.00 mM Ni-EDTA gave 1,797 counts.
Estimate the detection limit for Ni-EDTA
Standard deviation for the 10 measurements: 28.2
Detection limit:
y dl  45  3 ( 28.2 )  129 .6 counts
Convert counts to molarity:
m
y sampl e  y blank
sample concentration

1797  45
counts
 1.752 x10 9
1.00 M
M
Minimum detectable concentration:
c
( 3 )( 28.2 )
3s

 4.8 x10 8 M
m 1.752 x10 9 counts / M
5-24 Tooth enamel consists mainly of the mineral calcium hydroxyapatite, Ca10(PO4)6(OH)2.
Trace elements in teeth of archaeological specimens provide anthropologists with clues about
diet and disease of ancient people. Students at Hamline University measured strontium in enamel
from extracted wisdom teeth by atomic absorption spectroscopy. Solutions with a constant total
volume of 10.0 mL contained 0.750 mg of dissolved tooth enamel plus variable concentrations of
added Sr. Find the concentration of Sr.
Added Sr (ng/mL = ppb)
Signal (arbitrary units)
0
28.0
2.50
34.3
5.00
42.8
7.50
51.5
10.00
58.6
y = 3.136x + 27.36
y-intercept = -8.72 ng/mL = ppb  concentration of unknown in the 10 mL sample
5.29 A solution containing 3.47 mM X (analyte) and 1.72 mM S (standard) gave peak areas of
3,473 and 10,222, respectively, in a chromatographic analysis. Then 1.00 mL of 8.47 mM S was
added to 5.00 mL of unknown X, and the mixture was diluted to 10.0 mL. The solution gave
peak areas of 5,428 and 4,431 for X and S, respectively
(a) Calculate the response factor for the analyte
(b) Find the concentration of S (mM) in the 10.0 mL of mixed solution.
(c) Find the concentration of X (mM) in the 10.0 mL of mixed solution.
(d) Find the concnetration of X in the original unknown.
(a)
Ax
As 3473
 10222 
F

 F
  F  0.1684
[X]
[ S ] 3.47
 1.72 
(b) Simple dilution
 1.00 mL 
[ S ]  ( 8.47 )
  0.847
 10.00 mL 
(c) Use answers to a and b
Ax
As 5428
 4431 
F

 0.1684 
  [ X ]  6.16 M
[X]
[S] [X ]
 0.847 
(d) Simple dilution
 10.00 mL 
[ x ]  ( 6.16 )
  12.3 M
 5.00 mL 
Chapter 6
6-16: Find [Cu2+] in a solution saturated with Cu4(OH)6(SO4) if [OH-] is fixed at 1.0x10-6M.
Note that Cu4(OH)6(SO4) gives 1 mol of SO42- for 4 mol of Cu2+?
K sp  2.3  10 69
[Ksp table  appendix F on page AP9]
Let x = [Cu2+], then [SO42-]=1/4x
1
K sp  [Cu  ]4 [OH- ]6 [SO 24- ]  ( x )4 ( 1.0 x10 6 )6 ( x )  2.3 x10 69
4
 ( x )5 ( 2.5 x10 37 )  2.3 x10 69  x 5  9.2 x10 33  x  3.9 x10 7 M
6-16 (B). Find [Cu2+] in a solution saturated with Cu4(OH)6(SO4) if [OH-] is fixed at 1.0x10-6M
and 0.10M Na2SO4 is added to the solution.
Initial
Concentration
Final
concentration
Cu4(OH)6(SO4)
solid
solid
Cu+
0
OH1.0x10-6
SO4-2
0.10M
x
1.0x10-6
1/4x + 0.10M
Let x = [Cu2+], then [SO42-]=1/4x
Assume 1/4x << 0.10M
K sp  [Cu  ] 4 [OH- ]6 [SO 24- ]  ( x ) 4 ( 1.0 x10 6 )6 ( 0.10 )  2.3 x10 69
 ( x ) 4 ( 1.0 x10 37 )  2.3 x10 69  x 4  2.3 x10 32  x  1.2 x10 8 M
Check assumption: ( ¼)1.2x10-8<< 0.10M
3.1x10-9 << 0.10M  true
1.2x10-8M < 3.9x10-7M solubility of Cu2+ is reduced
6-25 Given the following equilibria, calculate the concentration of each zinc-containing species in
a solution saturated with Zn(OH)2(s) and containing [OH-] at a fixed concentration of 3.2x107
M.
Ksp = 3.0x10-16
1 = 2.5 x104
3 = 7.2x1015
4 = 2.8x1015
Zn(OH)2 (s)
Zn(OH+)
Zn(OH)3Zn(OH)42-
K sp  [Zn 2 ][OH - ] 2  3.0  10 16  [Zn 2 ] 

 1  [Zn (OH) ]


[Zn 2  ][OH - ]
K sp
- 2
[OH ]
 3.0  10
16
( 3.2 x10
7 2
)
 2.9  10 3 M
 2.5  10 4  [Zn (OH) ]   1 [Zn 2  ][OH - ]


 2.5  10 4 2.9  10 3 3.2  10 7  2.3  10 5 M
3 

[Zn (OH)3  ]

[Zn 2  ][OH - ]3
 7.2  10 15  [Zn (OH)3  ]   3 [Zn 2  ][OH - ]3


3
 7.2  10 15 2.9  10  3 3.2  10 7
 4  [Zn(OH)4


2
]
[Zn
2
- 4
][OH ]

 6.8  10 7 M
 2.8  10 15  [Zn(OH)4  2 ]   4 [Zn 2  ][OH - ] 4
 2.8  10 15 2.9  10  3 3.2  10 7

4
 8.5  10 14 M
pH Ex: (a) What is the pH of a solution containing 1x10-6 M H+?
pH   log[ H  ]   log( 1  10 6 M )  6.0
(b) What is the [OH-] of a solution containing 1x10-6 M H+?
K sp  [Zn 2 ][OH - ] 2  3.0  10 16  [Zn 2 ] 
K sp
- 2
[OH ]
 3.0  10
16
( 3.2 x10
7 2
)
 2.9  10 3 M
6-49. Write the Kb reaction of CN-. Given that the Ka value for HCN is 6.2x10-10, calculate Kb for
CN-.
CN- + H2O <--> HCN + OHKw  Ka  Kb  Kb  Kw / Ka
 Kb 
( 1.0  10 14 )
( 6.2  10
10
)
 1.6  10 5
Chapter 7
7-A (a) Suppose 29.41 mL of I3- solution is required to react with 0.1970 g of pure ascorbic acid,
what is the molarity of the I3- solution?
(0.1970g)(1 mole/176.124 g) = 1.1185x10-3 mol (1.1185 mmol) of ascorbic acid
1 mole ascorbic acid = 1 mole I3-  1.1185 mmol I3Molarity of I3- : 1.1185 mmol/29.41 mL = 0.03803 M
(b) A vitamin C tablet containing ascorbic acid plus an inert binder was ground to a powder, and
0.4242g was titrated by 31.63 mL of I3-. Find the weight percent of ascorbic acid in the
tablet.
(31.63 mL)(0.03803M) = 1.203 mmol of I31 mole ascorbic acid = 1 mole I3-  1.203 mmol ascorbic acid
(1.203x10-3 mol)(176.124 g/mol) = 0.2119g ascorbic acid
(0.2119g)/(0.4242g)x100 =49.94%
Chapter 8
8-3. What is the ionic strength of a 0.0087 M KOH and 0.0002 M La(IO3)3 solution? Assume
complete dissociation and no formation of LaOH2+

1
 c i z i2
2 i
[K+]=[OH-]=0.0087
3x[La+3] =[IO3-]
K+
OHLa+3
IO32
2
2
½[0.0087x1 +0.0087x(-1) + 0.0002x3 +0.0006x(-1)2] = 0.0099 M
8-11. What is the pH of a solution containing 0.010M HCl plus 0.040 M KClO4?
First determine the ionic strength of the solution, since the ion charges are all 1:
 = 0.010M (HCl) + 0.040M (KClO4) = 0.050 M
Using table, H+ = 0.86
[H+] = 0.010M
pH   log AH    log[ H  ]  H    log[( 0.010 )( 0.86 )]   log[ 8.6  10 3 ]  2.07
Ignoring difference between activity and concentration:
pH   log AH    log[ H  ]  H    log[ 0.010 ]  2.00
8-9 (a) What is the [Hg22+] in a saturated solution of Hg2Br2 with 0.00100M KCl, where
and KCl acts as an “inert salt”?
First determine the ionic strength,  = 0.00100M (KCl) negligible contribution from
Hg2Br2
Using table, Hg2+ = 0.867, Br- = 0.965
[Hg2+2] = x, [Br-] = 2x
K sp  A
A2
Hg2 2 
Br

 [ Hg2 2  ]
K sp  5.6  10  23  4 x 3 
x3
Hg2 2 
2
Hg2 2  Br 
[ Br  ] 2  2
Br 
 ( x )
Hg 2 2 
( 2 x )2  2
Br 
 ( 4 )( 0.867 )( 0.964 )2 x 3  3.223 x 3
5.6  10 23
 2.6  10 8 M
3.223
(b) What is the [Hg22+] in a saturated solution of Hg2Br2 with 0.00100M KBr?
First determine the ionic strength, m = 0.00100M (KBr) negligible contribution from
Hg2Br2
Using table, gHg2+ = 0.867, gBr- = 0.965
[Br-] = 0.00100M (KBr), negligible contribution from Hg2Br2
K sp  A
Hg2 2 
[ Hg2 2  ] 
A2
Br 
 [ Hg2 2  ]
K sp
 Hg
2
[ Br  ] 2  2
2
[ Hg2 2  ]  7.0 x10 17 M
Br 
Hg2 2 

[ Br  ] 2  2
Br 
5.6 x10 23
( 0.867 )( 0.00100 M )2 ( 0.964 )2
8-24 Write a mass balance for a solution of Fe2(SO4)3, if the species are Fe3+, Fe(OH)2+,
Fe(OH)2+, Fe2(OH)24+, FeSO4+, SO42- and HSO4-.
Fe2(SO4)3  3(total Fe) = 2(total SO4)
(3){[Fe3+] + [Fe(OH)2+] + [Fe(OH)2+] + 2[Fe2(OH)24+] +[ FeSO4+] } =
(2){[ FeSO4+] + [SO42-] + [HSO4-]}
2 in front of Fe2(OH)24+ because it contains 2 Fe.
Chapter 9
9-11. (a) A 0.0450 M solution of benzoic acid has a pH of 2.78. Calculate pKa for this acid
Concentrations:
A10-2.78
HA
F-10-2.78
H+
10-2.78
F=0.0450M; pH =-log[H+]; [H+] = 10-pH = [A-]
Ka 
( 10 2.78 )2
[ H  ][ A  ]

 6.35  10 5
 2.78
[ HA ]
( 0.0450  10
)
pKa   log[ 6.35 x10 5 ]  4.20
(b) What is the percent fraction dissociation?

x 10 2.78 M 1.66 x10 3


 0.0369  3.69%
F 0.0450 M
0.0450
9-40. (a) Calculate how many milliters of 0.626 M KOH should be added to 5.00 g of MOBS
(FW: 223.29) to give a pH of 7.40?
Initial moles:
Final moles:
HA
0.0224
0.0224-x
OHx
-
Ax
 [A - ] 

pH  7.40  pK a  log 
 [HA] 


x
7.40  7.48  log
0.0224  x
x
 0.08  log
0.0224  x
x
1.86 x10 2
 1.86 x10 2  1.832 x  x 
 0.01017 mol
0.0224  x
1.832
0.01017 mol
volume 
 16.2 mL
0.626 M
0.832 
(b) What is the pH if an additional 5 mL of the KOH solution is added?
Total moles of KOH = (21.2 ml)(0.626M)=0.01327 mol
 [A - ] 

pH  7.40  pK a  log 
 [HA] 


0.01327
pH  7.48  log
 7.48  log( 1.453 )  7.64
0.0224  0.01327
Chapter 10
10-11 How many grams of Na2CO3 (FM 105.99) should be mixed with 5.00 g of NaHCO3 (FM
84.01) to produce 100 mL of buffer with pH 10.00?
pK a1  6.351
pK a 2  10.329
We know, [CO3-2] and [HCO3-], so use pKa2
pKa from Appendix G acid dissociation constants page AP12
 [CO 2- ] 
( xg ) /( 105 .99 g / mol )
3 
pH  pK a 2  log 
 10.00  10.329  log
 [HCO- ] 
( 5.00 g ) /( 84.0 g / mol )
3 

x
 0.329  log
6.3089
0.4688  x / 6.3089
x  2.96 g
Note: volume not used since it simply cancels.
10-12 How many milliliters of 0.202 M NaOH should be added to 25.0 mL of 0.0233 M of
salicylic acid (2-hydroxybenzoic acid) to adjust the pH to 3.50?
pK a1  2.972
pK a 2  13.7
Treat as monoprotic acid
At pH 3, mixture of H2A and HAMoles of salicylic acid (H2A) = (25.0 mL)(0.0233 M) = 0.5825 mmol
Initial moles:
Final moles:
3.50  2.972  log
H2 A
0.5825
0.5825-x
OHx
-
HAx
x
0.5825  x
x
0.5825  x
3.373  x / 0.5825  x
0.528  log
1.965  3.373 x  x
1.965  4.373 x
x  0.4493 mmol  ( 0.4493 mmol ) /( 0.202 M )  2.223 mL NaOH
10-20 How many milliters of 1.00 M KOH should be added to 100 mL of solution containing
10.0 g of histidine hydrochloride (His.HCl FM 191.62) to get a pH of 9.30?
Treat as monoprotic acid.
histidine hydrochloride is the intermediate form (H2His+) between pK1 & pK2.
1) Must add enough KOH (1:1 molar ratio) to convert all H2His+ to HHis
2) Must added more KOH to obtain mixture of HHis and His- to obtain pH of 9.30
Initial moles of H2His+ = 10.0g/(191.62 g/mol) = 0.05219 mol
Require 0.05219 mol of KOH plus:
Initial moles:
Final moles:
HHis
0.05219
0.05219-x
OHx
-
Hisx
[ His  ]
x
 9.30  9.28  log
[ HHis ]
0.05219  x
x
0.02  log
0.05219  x
1.047  x / 0.05219  x
pH  pK 3  log
0.0546496  1.047 x  x
0.0546496  2.047 x
x  0.02670 mol
Total KOH moles  0.02670  0.05219  0.0789 mol  ( 0.0789 mol ) / 1.00 MKOH )  78.9 mL
Chaper 11
11-8 a) What is the pH at the equivalence point when 0.100 M hydroxyacetic acid is titrated with
0.0500 M KOH?
Equivalence point  exactly enough KOH to consume hydroxyacetic acid (HA)
Twice the volume of KOH (0.0500) is required to titrate hydroxyacetic acid (0.100)
Formal concentration of A- = (volume of HA/(volume of HA + volume KOH))(0.100M)
= V/(V+2V)(0.100M) set V = 1
= 1/(1+3)(0.100M)
= 0.0333M
The solution only contains A-  weak base (Ka = 1.48x10-4, Appendix G, AP14)
F-x
x
x
K
K
x2
x2
1.0 x10 14
 Kb  w 
 Kb  w 
 6.757 x10 11
Fx
K a 0.0333  x
K a 1.48 x10  4
0  x 2  6.757 x10 11 x  2.25 x10 12
solve quadratic  x  1.50 x10 6 M  pH   log( 1.50 x10 6 )  8.18
b) What indicator would be a good choice to monitor the endpoint?
Cresol red to phenolphtelen or any number of inidcators that change color around
pH 8.18 (see table)
Chapter 12
ex: What is the concentration of free Fe3+ in a solution of 0.10 M Fe(EDTA)- at pH 8.00?
Kf = 1025.1=1.3x1025 from table 12-2
𝒂Y4- at pH 8.0 = 4.2x10-3 from table 12-1
Fe3+
0
x
Initial conc:
Final conc:
K 'f 
[ Fe( EDTA )- ]
[ Fe
3
][ EDTA ]
EDTA Fe(EDTA)0
0.10
x
0.10-x

( 0.10  x )
 5.46  10 22
( x )( x )
Solve quadratic for x:
 x  [ Fe 3  ]  [ EDTA ]  1.4  10 12
Chapter 14
14-25 (a): Calculate Eo and Go for the following reaction:
Identify the half-reactions, look for atoms that with a change charge (ionic) state:
Standard reaction potentials are listed in appendix H.
E+o = 1.92V
E-o = 1.229V
Eo = 0.69V
E cell  E   E   1.92  1.229  0.69V
G   nFE  ( 4 mol )( 9.649 x10 4
C
J
)( 0.69 )  2.7 x10 5 J
mol
C
14-19 (a) : Calculate the cell voltage if the concentration of NaF and KCl were each 0.10 M in
the following cell:
(anode, E-) Pb(s) | PbF2(s) | F- (aq) || Cl- (aq) | AgCl(s) | Ag(s) (cathode, E+)
Identify the half-reactions, look for atoms that with a change charge (ionic) state:
Standard reaction potentials are listed in appendix H.
Eo = 0.222
Eo = -0.350
Solve the Nernst equation for each half-reaction:
right half  cell : E   0 .222 
0 .05916
0 .05916
log[ Cl  ] 2  0 .222 
log[ 0.10 M ] 2
2
2
E   0.222  0.0592  0.2812
0 .05916
0 .05916
log[ F  ] 2  0 .222 
log[ 0.10 M ] 2
2
2
E   0.350  0.0592  2.908
left half  cell : E   0.350 
E cell  E   E   0.2812  ( 2.908 )  0.572V
14-25 (b): Calculate K for the following reaction:
Identify the half-reactions, look for atoms that with a change charge (ionic) state:
Standard reaction potentials are listed in appendix H.
E+o = 0.017V
E-o = 0.356V
Eo = -0.339V
E cell  E   E   0.017  0.356  0.339V
K  K sp  10
nE o
0.05916
 10
( 1 )( 0.339 )
0.05916
 1.9 x10 6
14-40: If the voltage for the following cell is 0.512V, find Ksp for Cu(IO3)2:
Identify the half-reactions, look for atoms that with a change charge (ionic) state:
Standard reaction potentials are listed in appendix H.
E+o = 0.339V
E-o = -0.236V
Eo = 0.575V
E  0.512V  E o 
0 .05916
[ Ni 2  ]
0 .05916
[ 0.0025 M ]
log
 0.575 
log
2
2
[ Cu 2  ]
[ Cu 2  ]
( 0.512V  0.575V )
[ 0.0025 M ]
 log
0 .05916
[ Cu 2  ]
2
[ 0.0025 M ]
2.1298  log
[ Cu 2  ]
[ 0.0025 M ]
1.348 x10 2 
[ Cu 2  ]
[ 0.0025 M ]
[ Cu 2  ] 
 1.85 x10 5
2
1.348 x10
K sp  [ Cu 2  ][ IO3  ] 2  ( 1.85 x10 5 )( 0.10 )2  1.85 x10 7
Chapter 15
15-8: A 10.0 mL solution of 0.0500 M AgNO3 was titrated with 0.0250M NaBr in the cell:
S.C.E. || titration solution | Ag(s)
Find the cell voltage for 10.0 mL of titrant.
The equivelence volume (Ve) = 20.0 mL [20.0 mL x 0.250M = 10.0 mL x 0.0500M]
Titration reaction relative to S.C.E (E- = 0.241V):
E+o = 0.799V
E   0 .799 
0 .05916
1
log
1
[ Ag  ]
[Ag+] = (10.0/20.0)(0.0500M)(10.0/20.0)=0.0125M
Fraction
Dilution
Remaining
Factor (original volume/final volume)
(original volume-added volume)/total volume
E cell  0 .799 
E cel l  0.445V
0 .05916
1
log
 0.241
1
[ 0.0125 M ]
Chapter 16
16-17: A 50.00 mL sample containing La3+ was titrated with sodium oxalate to precipitate
La2(C2O4)3, which was washed, dissolved in acid, and titrated with 18.0 mL of 0.006363
M KMnO4. Calculate the molarity of La3+ in the unknown.
Need to identify the titration reaction, first determine the two ½ reactions.
Oxidation with Potassium permanganate:
Reduction:
Oxidation:
Then write a balanced reaction:
18.04 mL of 0.006363 M KMnO4 = 0.1148 mmol of MnO4Reacts with (5/2)(0.1148) = 0.2870 moles of H2C2O4
which came from (2/3)(0.2870) = 0.1913 mmol of La3+ [La2(C2O4)3]
[La3+] = 0.1913 mmol/50.00 mL = 3.826 mM
Chapter 18
18-B: A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a
1.000 cm cuvet. A blank had an absorbance of 0.029. The absorbance of an unknown solution
of compound A was 0.375. Find the concentration of A in the unknown.
First, find the molar absorptivity of compound A:

A
0.624  0.029

 1.50 x10 3 M 1 cm 1
cb ( 3.96 x10  4 M )( 1.000 cm )
(Correct absorbance for blank)
Use molar absorptivity to calculate concentration of unknown:
c
A
0.375  0.029

 2.31 x10  4 M
b ( 1.50 x10 3 M 1 cm 1 )( 1.000 cm )
18-20: In formaldehyde, the transition n p*(T1) occurs at 397 nm, and the np*(S1) transition
comes at 355 nm. What is the difference in energy (kJ/mol) between the S1 and T1 states?
n *(T1)
E  h  h
c

 6.6261 x10  34 Js
2.9979 x10 8 s 1
397 x10
9
 5.00 x10 19 J
m
Convert to J/mol, multiply by Avogadro’s number
5.00 x10 19 J / molecule  6.022 x10 23 molecules / mol  301kJ / mol
n p*(S1)
E  h  h
c

 6.6261 x10  34 Js
2.9979 x10 8 s 1
355 x10
9
 5.60 x10 19 J
m
Convert to J/mol, multiply by Avogadro’s number
5.60 x10 19 J / molecule  6.022 x10 23 molecules / mol  337 kJ / mol
The difference between the T1 and S1 statest is 337-301 = 36 kJ/mol
Chapter 23
23-11: Butanoic acid has a partition coefficient of 3.0 (favoring benzene) when distributed
between water and benzene. Find the formal concentration of butanoic acid in each phase
when 100 mL of 0.10 M aqueous butanoic acid is extracted with 25 mL of benzene at pH
4.00 and pH 10.00.
Ka for butanoic acid = 1.52x10-5 from appendix G AP12
At pH 4.00:
D
K[H ]
([ H  ]  K a )

( 3 )( 10 4.00 )
( 10  4.00  1.52 x10 5 )
 2.60
Fraction remaining in water:
q
V1
100 mL

 0.606
( V1  DV2 ) ( 100 mL  2.60  25 mL )
Molarity in water:
[ bu tan oic acid ]  ( 0.606 )( 0.10 M )  0.0606 M
Molarity in benzene:
Total moles in system = ( 0.100 L )( 0.10 M )  0.010 mol
Fraction in benzene = (1-0.606) = 0.394
Molarity in benzene = (0.394)(0.010 mol)/(0.025 L) = 0.16 M
At pH 10.00:
D
K[H ]

([ H ]  K a )
( 3 )( 10 10.00 )

( 10
10.00
 1.52 x10
5
 1.97 x10 5
)
Fraction remaining in water:
q
V1
100 mL

 0.9999951
( V1  DV2 ) ( 100 mL  1.97 x10 5  25 mL )
Molarity in water:
[ bu tan oic acid ]  (~ 1 )( 0.10 M )  0.10 M
Molarity in benzene:
Total moles in system = ( 0.100 L )( 0.10 M )  0.010 mol
Fraction in benzene = (1-0.9999951) = 4.9x10-4
Molarity in benzene = (4.9x10-4)(0.010 mol)/(0.025 L) = 2x10-6 M
23-24: The retention volume of a solute is 76.2 mL for a column with V m = 16.6 mL and Vs =
12.7 mL. Calculate the capacity factor and the partition coefficient for this solute.
Vm – volume of mobile phase
Vs – volume of stationary phase
Capacity factor:
k' 
Vr' Vr  Vm 76.2  16.6


 3.59
Vm
Vm
16.6
Partition coefficient:
V
16.6
K  k' m  ( 3.59 )
 4.69
Vs
12.7
23-42: Two compounds with partition coefficients of 15 and 18 are to be separated on a column
with Vm/Vs = 3.0 and tm = 1.0 min. Calculate the number of theoretical plates needed to
produce a resolution of 1.5
Want Rs = 1.5,
Rs  1.5 
N
  1
4
Need t2 and t1
k2 '  K 2
Vs
1
 18
 6.0
Vm
3.0
k1'  K 1
Vs
1
 15
 5.0
Vm
3.0
t  tm
k1'  1
 t1  t m ( k1' 1 )  ( 1.0 min)( 5.0  1 )  6.0 min
tm
t  tm
k2 '  2
 t 2  t m ( k 2 ' 1 )  ( 1.0 min)( 6.0  1 )  7.0 min
tm
Determine 


t 2 7.0

 1.167
t1 6.0
Determine N
N
1.167  1
4
( 1.5 )( 4 )
N 
 36
0.167
R s  1 .5 
N  1.3 x10 3 plates