SOLUTIONS TO SELECTED PROBLEMS IN CHAP. 9 & 10
The Arrhenius Theory (Section 9.1)
9.1
Write the dissociation equations for the following that emphasize their behavior
as Arrhenius acids: a) HI
b) HBrO
c) HCN
d) HClO2
Solution
a) HI(aq)  H+(aq) + I -(aq)
c) HCN(aq)  H+(aq) + CN-(aq)
9.3
b) HBrO(aq)  H+(aq) + BrO -(aq)
d) HClO2 (aq)  H+(aq) + ClO-2 (aq)
Each of the following produces a basic solution when dissolved in water. Identify
those that behave as Arrhenius bases and write dissociation equations to
illustrate that behavior. a) CsOH
b) CH3 NH2
c) NH 3
d) Ca(OH)2
Solution
a) CsOH(aq)  Cs+(aq) + OH-(aq)
c) not an Arrhenius base
b) not an Arrhenius base
d) Ca(OH)2 (aq)  Ca2+(aq) + 2OH-(aq)
The BrØnsted Theory (Section 9.2)
9.5
Identify each Br Ønsted acid and base in the following equations. Note that the
reactions are assumed to be reversible.
+
(aq) d) SO23 - (aq)+H2O(l) HSO-3(aq)+OH- (aq)
a) HBr (aq) + -H2 O (l)  H3 O (aq) + Br
b) H2 O (l) + N3 (aq)  NH3 (aq) + OH (aq)
e) HC N ( aq ) + H 2O ( l) H3 O +( aq ) + CN - (a q)
+
c) H2 S(aq) +H2 O(l)  H3 O (aq) +HS (aq)
Solution
a) HBr (aq) + H 2 O (l)  H3 O +(aq) + Br- (aq)
acid
base
acid
base
b) H2 O (l) + N-3 (aq)  NH3 (aq) + OH- (aq)
acid base
acid
base
c) H2 S (aq) + H2 O (l)  H3 O +(aq) + HS- (aq)
acid
base
acid
base
2
d) SO 3 (aq) + H2 O (l) HSO 3 (aq) + OH- (aq)
base
acid
acid
base
+
e) HCN (aq) + H2 O (l)  H3 O (aq) + CN- (aq)
acid
base
acid
base
9.7
Identify each conjugate acid-base pair in the equations for Exercise 9.5
Solution
a)
b)
c)
d)
e)
Acid
HBr
H2O
H2S
H2O
HCN
Conjugate Base
BrOHHSOHCN-
Base
H2O
N-3
H2O
2-
SO3
H2O
Conjugate Acid
H3O +
HN3
H3O +
HSO3
H3O +
9.11
Write a formula for the conjugate base formed when each of the following
behaves as a BrØnsted acid:
a) HSO-3
b) HPO24 c) HClO 3
d) CH3 NH+3
e) H2C2O 4
Solution
a) SO23
9.13
b) PO24 -
c) ClO-3
e) HC2 O-4
d) CH3 NH2
Write a formula for the conjugate acid formed when each of the following
behaves as a BrØnsted base:
a) NH-2
b) CO23 c) OHd) (CH3 )2NH
e) NO-2
Solution
a) NH3
b) HCO-3
d) (CH3 )2 NH+3
c) H2O
e) HNO2
Naming Acids (Section 9.3)
9.21
9.25
Name the following acids. Refer to Table 4.7 as needed.
a) H2Se(aq)
b) HClO3
c) H2SO 4
d) HNO3
Solution
a) hydroselenic acid
d) nitric acid
b) chloric acid
c) sulfuric acid
Refer to Table 4.7 and write the formula for chromic acid.
Solution
H2CrO4
The Self-Ionization of Water (Section 9.4)
9.27
Calculate the molar concentration of OH - in water solutions with the following
H3O + molar concentrations:
a) 1.0 x 10-7
b) 3.2 x 10-3
c) 4.7 x 10-11
d) 1.2
e) 0.043
Solution
1.0 x10 - 14
- 7
- 7 =1.0 x10 M
a)
1.0 x10
1.0 x10- 14
- 12
[OH ] =
M
- 3 = 3.1x10
b)
3.2 x10
1.0 x10 - 14
- 4
[OH- ]=
- 11 = 2.1x10 M
c)
4.7 x10
-
[OH ]=
9.29
d)
e)
-
1.0 x10 - 14
- 15
= 8.3 x10 M
1.2
-
1.0 x10- 14
- 13
= 2.3 x10 M
0.043
[OH ]=
[OH ]=
Calculate the molar concentration of H3 O+ in water solutions with the following
OH- molar concentrations:
a) 1.0 x 10-7
b) 5.2 x 10-4
c) 9.9 x 10-10
d) 0.092
e) 3.7
Solution
+
a)
[H3 O ]=
1.0 x10 - 14
- 7
- 7 =1.0 x10 M
1.0 x10
- 14
[H3O+ ]= 1.0 x10 - 2 = 1.1x10- 13 M
9.2 x10
d)
- 14
+
1.0 x10- 14
- 11
[H
O
]
=
M
3
[H3O+ ] = 1.0 x10 = 2.7 x10- 15 M
- 4 = 1.9 x10
b)
5.2 x10
3.7
e)
- 14
+
1.0 x10
-5
c) [H3O ] = 9.9 x10- 10 =1.0 x10 M
9.31
Classify the solutions represented in Exercises 9.27 and 9.29 as acidic, basic, or
neutral.
Solution
#9.27 a) neutral
#9.29 a) neutral
b) acidic
b) basic
c) basic
c) acidic
d) acidic
d) basic
e) acidic
e) basic
The pH Concept (Section 9.5)
9.33
Classify solutions with the following characteristics as acidic, basic, or neutral.
a) pH=10
b) pH=4
c) pH=7.3
d) pH=6
Solution
a) basic
9.35
b) acidic
c) basic
d) acidic
Determine the pH of water solutions with the following characteristics. Classify
each solution as acidic, basic, or neutral.
a) [H+] = 1.0 x10-5 M
b) [OH -] = 6.0 x10 -3 M
c) [H+] = [OH-]
+
-4
-9
d) [H ] = 9.0 x10 M
e) [OH ] = 3.0 x10 M
Solution
Use the calculator to get pH. pH = -log[H+]
- 14
+
1.0 x10
- 12
b) [H ] = 6.0 x10 - 3 =1.7 x10 M ; pH =12 ; basic
9.37
a) pH = 5.0; acidic
c) pH = 7.0; neutral
d) pH = 3.1; acidic
- 14
[H+ ] = 1.0 x10 - 9 = 3.3 x10- 6 M ; pH= 5.5 ; acidic
3.0 x10
e)
Determine the pH of water solutions with the following characteristics. Classify
each solution as acidic, basic, or neutral.
a) [H+] = 3.7 x10-4 M
b) [H+] = 7.4 x10-8 M
c) [H+] = 1.9 x10 -10 M
d) [OH-] = 1.3 x10 -1 M
e) [OH-] = 6.8 x10 -7 M
Solution
Use the calculator to get pH. pH = -log[H+]
a) pH = 3.4; acidic
b) pH = 7.1; basic
c) pH = 9.7; basic
- 14
[H+ ]= 1.0 x10 - 1 = 7.7 x10 - 14 M ; pH =13 ; basic
d)
1.3 x10
+
1.0 x10 - 14
-8
[H ]=
= 1.5 x10 M ; pH = 7.8 ; basic
7
e)
6, 8 x10
9.39
Determine the [H+] value for solutions with the following characteristics:
a) pH = 4.45
b) pH = 13.12
c) pH = 7.73
Solution
a) [H+] = 3.55 x 10-5 M b) [H+] = 7.586 x 10-14 M
9.41
c) [H+] = 1.86 x 10 -8 M
Convert the following pH values into both [H +] and [OH -] values:
a) pH = 8.00
b) pH = 6.15
c) pH = 1.30
Solution
1.00 x10- 14
-6
- 8 =1.00 x10 M
a)
1.00 x10
+
- 7
1.00
x10 - 14 = 1.41x10- 8 M
[H ]= 7.08 x10 M [OH ] =
b)
7.08 x10- 7
+
-2
1.00 x10- 14
- 13
[H ]= 5.01x10 M [OH ]=
M
- 2 = 2.00 x10
c)
5.01x10
+
-8
[H ]= 1.00 x10 M
9.45
-
[OH ]=
The pH values of specific samples of food items are listed below. Convert each
value to [H +] and classify the sample as acidic, basic, or neutral.
a) milk, pH = 6.39
b) coffee, pH = 5.10
c) orange juice, pH = 4.07
d) vinegar, pH = 2.65
Solution
a) [H+] = 4.07 x10-7 M; acidic
c) [H+] = 8.51 x10-5 M; acidic
b) [H+] = 7.94 x10-6 M; acidic
d) [H+] = 2.24 x10-3 M; acidic
Properties of Acids (Section 9.6)
9.47
Use the information in Table 9.4 and describe how you would prepare each of
the following solutions.
a) about 750 mL of 0.500 M H 2SO4 from dilute sulfuric acid
b) about 200 mL of 0.1 M NaOH from stock sodium hydroxide solution
c) about 1.0 L of 1.0 M acetic acid from glacial acetic acid
Solution
Vd x Cd
Cc
2
or 1 x 10 using question values
Vc =
= 125 mL
Measure 125 mL of the dilute (3 M) H2SO 4 and
3.00 M
put it in
enough water (about 625 mL) to make 750 mL solution.
Solve equation 7.9 for Vc .
a)
750 mlx 0.500 M
b)
c)
Vc =
Vc =
200 ml x 0.1M
Take 3 mL of stock NaOH and put it in enough
= 3 mL
6M
water (about 197 mL) to make 200 mL solution.
Vc =
1.0 L x 1.0 M 1000 mL
or 6 rounded for question value
x
= 56 mL
18 M
1L
Measure 56 mL glacial acetic acid and
put it in enough water (about 944 mL) to make 1.0 L solution.
9.53
Write balanced molecular equations to illustrate five different reactions that
could be used to prepare BaCl2 from hydrochloric acid (HCl) and other
appropriate substances:
Solution
a) BaO(s) + 2HCl(aq)  BaCl2 (aq) + H2 O(l)
b) Ba(OH)2(s) + 2HCl(aq)  BaCl2(aq) + 2H2 O(l)
c) BaCO3(s) + 2HCl(aq)  BaCl 2(aq) + 2H2 O(l) + CO2 (g)
d) Ba(HCO3 )2(s) + 2HCl(aq)  BaCl 2(aq) + 2H2 O(l) + 2CO2(g)
e) Ba(s) + 2HCl(aq)  BaCl2(aq) + H 2(g)
Salts (Section 9.8)
9.61
Identify with ionic formulas the cations and anions of the following salts:
a) LiCI
b) Cu(NO3 )2
c) SrSO4
d) K3 PO4
e) K2 HPO4
f) CaCO3
Solution
a) Li+, Cld) 3K+, PO34 -
9.65
b) Cu2+ , 2NO-3
e) 2K+, HPO24 -
c) Sr2+, SO24 f) Ca 2+, CO23 -
Calculate the mass of water that would be released if the water of hydration
were completely driven off 1.0 mol of: (a) plaster of Paris and (b) gypsum (see
Table 9.6). How would the products of these reactions compare?
Solution
From Table 9.6, plaster of Paris is CaSO4 •H2 O and gypsum is CaSO4•2H 2O.
1mol H2 O
18 g H2 O
1mol CaSO 4 : H2 O x
x
= 18 g H2 O
1mol CaSO4 : H2 O 1molH2 O
a)
2 mol H2 O
18 g H2 O
1mol CaSO 4 : H2 O x
x
= 36 g H2 O
1mol
CaSO
:
H
O
1molH
b)
4
2
2O
The products are water (H2 O) and CaSO4 in both cases. The amount of
water is greater from gypsum.
9.67
Write formulas for the acid and indicated a solid that could be used to prepare
each of the following salts: a) CuCl2 (solid is an oxide); b) MgSO4 (solid is a
carbonate); c) LiBr (solid is a hydroxide)
Solution
a) HCI and CuO
9.69
b) H2SO 4 and MgCO3
c) HBr and LiOH
Write balanced full equations to illustrate each salt preparation described in
Exercise 9.67.
Solution
a) CuO + 2HCl  CuCl2 + H2O
b) MgCO 3 + H2 SO4  MgSO4 + H2O + CO2
c) LiOH + HBr  LiBr + H 2O
9.73
Determine the number of equivalents and milliequivalents in each of the
following:
a) 0.10 mol KI b) 0.25 mol MgCl2 c) 4.73 x 10-2 mol AgNO 3
Solution
a) 0.10 molKI x
0.10 eq x
1eq
= 0.10eg
1molKI
1000 meq
=1.0 x102 meq
eq
c) 4.73 x10- 2 mol AgNO 3 x
4.73 x10- 2 eq x
9.75
b) 0.25 molMgCl2 x
0.50 eq x
1eq
= 0.50 eg
1molMgCl2
1000 meq
= 5.0 x102 meq
eq
1eq
= 4.73 x10- 2 eg
1mol AgNO 3
1000 meq
= 47.3 meq
eq
Determine the number of equivalents and milliequivalents in 5.00 g of each of
the following salts. Include any waters of hydration given in the salt formula
when you calculate salt formula weights.
a) NaCl
b) NaNO3
c) Na 3PO 4
d) MgSO4 • 7H2O
Solution
a)
b)
c)
5.00 g NaCl x
1mol NaCl
1eq
x
= 8.55 x10- 2 eq = 85.8 meq
58.5 gNaCl 1mol NaCl
5.00 g NaNO3 x
1mol NaNO 3
1eq
x
= 5.88 x10- 2 eq = 58.8 meq
85.0 g NaNO3 1molNaNO3
5.00 g Na3 PO4 x
1mol Na 3PO4
1eq
x
= 9.15 x10- 2 eq = 91.5 meq
164 gNa3 PO4 1molNa3 PO4
d) 5.00 g MgSO4 H2 O (MW = 24.3 + 32.1 + 64.0 + 7(18) = 246.4)
1molMgSO4 :H2O
2eq
5.00gMgSO4 :H2Ox
x
=4.06x10- 2 eq=40.6meq
246.4 gMgSO4 :H2O 1molMgSO4 :H2O
9.77
A sample of intracellular fluid contains 45.1 meq/L of Mg2+ ion. Assume the Mg2+
comes from dissolved MgCl2 and calculate the number of moles and number of
grams of MgCl2 that would be found in 250 mL of the intracellular fluid.
Solution
45.1meq
1eq
1mol MgCl2
-3
x 0.250 L soln x
x
= 5.64 x10 mol MgCl2
L soln
1000 meq
2 eq
95.2 gMgCl2
5.64 x10- 3 mol MgCl2 x
= 0.537 g MgCl2
1mol MgCl2
Strengths of Acids and Bases (Section 9.9)
9.79
Illustrate the difference between weak, moderately strong, and strong acids by
writing dissociation reactions for the hypothetical acid HB using arrows of
various lengths.
Solution
9.81
Arrange the four acids classified as weak in Table 9.7 in order of increasing
strength (weakest first, strongest last).
Solution
Use the % dissociation (smallest = weakest) or Ka (smallest = weakest).
H3BO 3, H2 CO3, HC2 H3 O2 , and HNO2
9.83
Write dissociation reactions and K a expressions for the following weak acids:
a) hypobromous acid, HBrO
b) sulfurous acid, H 2SO3 (1st H only)
c) hydrogen sulfite ion, HSO3
d) hydroselenic acid; H2 Se (1st H only)
e) arsenic acid, H 3AsO4 (1st H only)
Solution
+
a) HBrO H + BrO
 +
b) H2 SO3 H +HSO3
c)
HSO-3 H+ + SO23 -
 +
d) H2 Se H + HSe
+
e) H3 AsO 4  H +H2 AsO4
9.85
[H+ ][BrO- ]
Ka = [HBrO]
[H+ ] [HSO-3 ]
Ka =
[H2 SO3 ]
[H+ ][SO23 - ]
Ka =
[HSO-3 ]
[H+ ][HSe- ]
Ka = [H Se]
2
[H+ ][H2 AsO-4 ]
Ka = [H AsO ]
3
4
Equal molar solutions are made of three monoprotic acids: HA, HB, and HC.
The pHs of the solution are respectively 4.82, 3.16, and 5.47. Rank the acids in
order of increasing acid strength and explain your reasoning.
Solution
The strongest acid will have the highest [H3O+] and lowest pH. HC, HA, HB
9.87
Arsenic acid (H3 AsO4 ) is a moderately weak, triprotic acid. Write equations
showing its stepwise dissociation. Which of the three anions formed in these
reactions will be the strongest BrØnsted base? Which will be the weakest
BrØnsted base? Explain your answers.
Solution
2+
3+
+
21) H3 AsO4 H +H2 AsO4 2) H2 AsO4 H + HAsO4 3) HAsO 4 H + AsO4
H3AsO 4 is the strongest acid; H2 AsO-4 is the weakest base.
2-
HAsO4 is the weakest acid;
3-
AsO4 , is the strongest base.
Analysis of Acids and Bases (Section 9.10)
9.89
Describe the difference between the information obtained by measuring the pH
of an acid solution and by titrating the solution, using and indicator, with base.
Solution
The pH of a solution measures the [H+] actually present in the solution at that
instant. The titration determines the total amount of [H+] ultimately available
from the acid for reaction with the base.
9.91
Determine the number of moles of NaOH that could be neutralized by each of
the following:
a) 1.00 L of 0.200 M HCl
b) 500 mL of 0.150 M HNO3
Solution
0.200 mol HCl 1mol NaOH
x
= 0.200 mol NaOH
a)
1L soln
1mol HCl
0.150 mol HNO3 1mol NaOH
0.500 L soln x
x
= 0.0750 mol NaOH
1L soln
1molHNO3
b)
1.00 L soln x
Titration Calculations (Section 9.11)
9.95
Write a balanced molecular equation to represent the neutralization reaction
between HCl and each of the following bases:
a) Cd(OH)2
b) Cr(OH) 3
c) Fe(OH)2
Solution
a) Cd(OH)2 + 2HCl  CdCl2 + 2H2O
c) Fe(OH)2 + 2HCl  FeCl2 + 2H2O
9.97
b) Cr(OH) 3 + 3HCl  CrCl3 + 3H2 O
A 25.00 mL sample of gastric juice is titrated with 0.0210 M NaOH solution. The
titration to the equivalence point requires 29.8 mL of NaOH solution. If the
equation for the reaction is HCl(aq) + NaOH(aq)  NaCl(aq) + H 2O(l)
what is the molarity of HCl in the gastric juice?
Solution
moles HCl =29.8mL x
1L
0.0210molNaOH 1molHCl
x
x
=6.26 x10- 4 molHCl
1000 ml
1L
1molNaOH
-4
1000 mL
MofHCl = 6.26X10 molHCl x
=0.0250MHCl
1L
25.00 mL
9.99
A 20.00 mL sample of each of the following acid solution is to be titrated to the
equivalence point using 0.120 M NaOH solution. Determine the milliliters of
NaOH solution that will be needed for each acid sample.
a) 0.180 M HCl
b) 0.180 M H 2SO4
c) 0.100 M HCl
d) 10.00 g of H 3PO4 in 250 mL solution
e) 0.150 mol H2MoO4 in 500 mL solution f) 0.215 mol H2 MoO4 in 700 mL
solution
Solution
At the equivalence point, the mol H+ = mol OH- or meq acid = meq base.
Va Ma = V bMb , where the molarity of the acid and base must be the molarity of
H+ or OH-.
For the 0.120 M NaOH, the solution is 0.120 M in OH-.
a) For 0.180 M HCI, the solution is 0.180 M in H+.
0.180 M
M
Vb = Va x a = 20.00 mL x
= 30.0 mL NaOH
Mb
0.120 M
b) For 0.180 M H2SO 4, there are 2 H+ per H2 SO4; so it is 0.360 M in H +.
0.360 M
M
Vb = Va x a = 20.00 mL x
= 60.0 mL NaOH
Mb
0.120 M
c) For 0.100 M HCI, the solution is 0.100 M in H+.
0.100 M
M
Vb = Va x a = 20.00 mL x
= 16.7 mL NaOH
Mb
0.120 M
1molH3 PO4
molH 3PO4 = 10.00 gH3 PO4 x
= 0.1021mol H3PO 4
97.99 g H3 PO4
d)
0.1021molH3 PO4
M=
= 0.408 MH3PO4
0.250 L soln
For 0.408 M H3 PO4, there are 3 H+ per H3 PO4 ; so it is 1.23 M H+ .
1.23 M
M
Vb = Va x a = 20.00 mL x
= 205 mL NaOH
Mb
0.120 M
0.150 molH2 MoO4
= 0.300 M H2MoO 4
e) M ofH 2MoO 4 =
0.500 L
+
Since there are 2H per H2 MoO4, the solution is 0.600 M H+
Vb = Va x
0.600 M
Ma
= 20.00 mL x
= 100 mL NaOH
Mb
0.120 M
0.215 mol H2MoO4
= 0.307 MH2 MoO 4
f) M of H2 MoO 4 =
0.700 L
+
Since there are 2H per H 2MoO4 , the solution is 0.614 M H+
0.614 M
M
Vb = Va x a = 20.00 mL x
= 102 mL NaOH
Mb
0.120 M
9.101 The following acid solutions were titrated to the equivalence point with the base
listed. Use the titration data to calculate the molarity of each acid solution.
a) 25.00 mL of HI solution required 27.15 mL of 0.250 M NaOH solution
b) 20.00 mL of H 2SO4 solution required 11.12 mL of 0.109 M KOH solution
c) 25.00 mL of gastric juice (HCl) required 18.40 mL of 0.0250 M NaOH
solution
Solution
a) molesHI= 27.15 mL x
1L
0.250molNaOH
x
x 1molHI = 6.79 x10- 3 molHI
1000 mL
1L
1molNaOH
-3
1000 mL
=0.272 MHI
1L
1L
0.109molKOH 1molH2SO4
b) molesH2 SO4 =11.12mL x
x
x
=6.06x10- 4 molH2 SO4
1000mL
1L
2molKOH
MHI= 6.79 x10 molHI x
25mL
MH2SO4 =
6.06x10- 4 molH2SO4 1000mL
x
=0.303MH2SO4
20mL
1L
c) molesHCl=18.40mL x
1L
0.250 molNaOH 1molHCl
x
x
= 4.60 x10- 3 molHCl
1000 mL
1L
1molNaOH
-3
MHCl= 4.60 x10 molHCl x
25mL
1000 mL
=0.184MHCl
1L
9.103 A 20.00 mL sample of diprotic oxalic acid (H2 C2 O4 ) solution is titrated with 0.250
M NaOH solution. A total of 27.86 mL of NaOH is required. Calculate:
a) the number of moles of oxalic acid in the 20.00 mL sample
b) the molarity of the oxalic acid solution
c) the number of grams of oxalic acid in the 20.00 mL sample
Solution
1L
0.250molNaOH 1molH2C2O4
x
x
=3.48x10- 3 molH2C2O4
a)
1000mL
1L
2molNaOH
3.48 x10- 3 molH2 C2O4 1000 mL
M=
x
= 0.174 MH2C2O 4
1L
b)
20 mL
90.0 g H2 C2 O4
3.48 x10- 3 mol H2 C2 O4 x
= 0.313 g H2 C2 O 4
1mol H2 C 2 O4
c)
molH2 C2O4 =27.86mL
Hydrolysis Reactions of Salts (9.12)
9.105 A solution of solid NH4Cl in pure water is acidic (the pH is less than 7). Explain.
Solution
The NH+4 is a BrØnsted acid, establishing an equilibrium with H 2O:
NH+4 + H2 O  H3 O + + NH3
The H3O + formed makes the solution acidic (pH is less than 7).
9.107 Predict the relative pH (greater than 7, less than 7, etc) for water solutions of the
following salts. Table 9.9 may be useful. For each solution in which the pH is
greater or less than 7, explain why and write a net ionic equation to justify your
answer. a) potassium sulfite, K2 SO3 b) lithium nitrite, LiNO2 c) sodium
carbonate, Na 2CO3; d) methylammonium chloride, CH 3NH3 Cl (CH3NH2 is a
weak base)
Solution
a) Basic. SO 23 - is a weak base. SO 23 - + H2 O  HSO -3 + O Hb) Basic. NO -2 is a weak base. NO -2 + H2 O  HNO 2 + OH-
c) Basic. CO 23 - is a weak base. CO 23 - + H2 O  HCO -3 + OHd) Acidic. CH3 NH+3 is a weak acid. CH3 NH+3 + H2 O  H3 O + + CH3 NH2
9.109 A chemist has 20.00 mL samples of 0.100 M acid A and 0.100 M acid B in
separate flasks. Both acids are monoprotic. Unfortunately, the flasks are not
labeled, so the scientist does not know which sample is in which flask. But
fortunately, it is known that acid A is strong and acid B is weak. Before thinking
about the problem, the chemist adds 20.00 mL of 0.100 M NaOH to each flask.
Explain how the chemist could use a pH meter (or pH paper) to determine which
flask originally contained which acid.
Solution
The 20.00 mL of NaOH would react with all of the acid in both cases, giving
essentially a solution of the sodium salt of the acid. The salt of the strong
acid would not hydrolyze and would have a pH = 7. The salt of the weak acid
would hydrolyze, giving a basic solution (pH greater than 7).
9.111 How would the pH of equal molar solutions of the following salts compare
(highest, lowest, etc)? NaH2 PO4, Na2 HPO4, and Na 3PO4
Solution
The Na 3PO4 would have the highest pH, definitely basic, because the PO34 is the strongest base of the three anions. The NaH2 PO4 would be the lowest
pH, definitely acidic, because H2PO-4 is the strongest acid of the three
anions. The pH of the Na 2HPO4 would be higher than NaH2PO4 and lower
than Na3 PO4.
Buffers (Section 9.13)
9.113 Could a mixture of ammonia (NH 3), a weak base, and ammonium chloride
(NH4 Cl) behave as a buffer when dissolved in water? Use equations to justify
your answer.
Solution
Yes. The NH3 could react with added acid, and the NH+4 could react with
added base. The equations are:
NH3 + H+ " NH+4 and NH+4 + OH- " NH3 + H2 O
9.115 Calculate the pH of a buffer made by dissolving 1 mol formic acid (HCOOH) and
1 mol sodium formate (HCOONa) in 1 L of solution (see Table 9.9).
Solution
From Table 9.9, pKa = 3.74
[B-] = [HB] = 1.0
pH = pKa + log
Using Equation 9.54,
pH = 3.74 + log 1 = 3.74
[B- ]
[HB]
9.117 Which of the following weak acids and its conjugate base would you use to
make a buffer with a pH of 5.00? Explain your reasons. Acetic acid, nitrous acid,
or formic acid.
Solution
The best buffer capacity is when the pH equals the pK a. The closer the pH Is
to pKa , the higher the buffer capacity. Since pKa for acetic acid is the closest
to 5.00, a mixture of acetic acid and acetic ion would be best.
9.119 Calculate the pH of the buffers with the acid and conjugate base concentrations
listed below.
23a) [HPO 4 ]= 0.33 M, [PO4 ] = 0.52 M
b) [HNO2 ]= 0.029M, [NO2 ]= 0.065 M
2c) [HCO3 ] = 0.50 M, [CO3 ]= 0.15 M
Solution
2a) pKa forHPO4 = 12.66
pH = 12.66 + log 0.52 = 12.86
0.33
b) pKa forHNO2 = 3.33
pH = 3.33 + log 0.065
0.029 = 3.68
c) pKa for HCO-3 = 10.25
pH = 10.25 + log 0.15 = 9.73
0.50
9.121 A citric acid-citrate buffer has pH = 3.20. You want to increase the pH to a value
of 3.35. Would you add citric acid or sodium citrate to the solution? Explain.
Solution
Add sodium citrate. There are at least two viable explanations: (1) to
increase pH, add the BrØnsted base, citrate; or (2) in Equation 9.54, if the
[B-] increases, the value of the log term increases, which increases pH.
ADDITIONAL EXERCISES
9.123 Consider the following dissociation reaction of a weak acid, HA:
It was determined that 2.63% of the acid in a 0.150 M solution of HA in water
was dissociated. Calculate the pH of the 0.150 M solution.
Solution
9.125 When sodium metal, Na, reacts with water, hydrogen gas, H 2, and sodium
hydroxide, in solution are produced. Write a balanced equation for the reaction,
and explain how a basic solution is produced.
Solution
In this single replacement
reaction, sodium and hydrogen exchange places. The resulting sodium
hydroxide is a strong base.
SOLUTIONS TO CHEMISTRY FOR THOUGHT
9.132 In an early industrial method, H2 SO4 was manufactured in lead-lined chambers.
Propose an explanation for this.
Solution
Pb reacts with H2 SO4 to make PbSO4 and H2. The PbSO4 is not soluble and
coats the surface of the lead, preventing further reaction between the acid
and the metal.
9.133 A saturated solution of solid Ca(OH)2 in water has a [OH-] of only 2.50 x 10 -2 M,
and yet Ca(OH)2 is a strong base. Explain this apparent contradiction.
Solution
Ca(OH)2 is only slightly soluble in water, but the limited amount of Ca(OH) 2
which dissolves is completely ionized (completely dissociated). Hence, it is a
strong base because it ionizes completely, but the concentration of [OH-] is
limited by the solubility.
9.134 Imagine a solution of weak acid is being titrated with a strong base. Describe the
substances present in the solution being titrated when it has been titrated
halfway to the equivalence point. How is the pH of this half-titrated solution
related to pKa for the weak acid?
Solution
At that point in the titration of HB, one half of the original HB has reacted to
form B -. Thus, the remaining concentration of HB equals the concentration of
[B- ]
log
=0
B . Using Equation 9.54, the pH = pKa because
[HB]
9.135 Calculate K a for the following weak acids based on the equilibrium
concentrations and pH values given.
a) Benzoic acid, represented by HBz:
pH = 2.61, [Bz-] = 2.48 x 10 -3 , [HBz] = 0.0975
b) Abietic acid, represented by HAb:
pH = 4.16, [Ab-] = 6.93 x 10-5, [HAb] = 0.200
c) Cacodylic acid, represented by HCc:
pH = 3.55, [Cc-] = 0.000284, [HCc] = 0.150
Solution
Rearranging Equation 9.54 gives
-3
[2.48 x10 ]
pKa = 2.61- log
= 4.20
a)
[0.0975]
pK a = pH- log
[B- ]
[HB]
Ka = 6.93 x10
-5
[6.93 x10- 5 ]
b) pKa = 4.16 - log [0.200] = 7.62
[0.000284]
pKa = 3.55 - log
= 6.27
c)
[0.150]
Ka = 2.40 x10- 8
Ka = 5.37 x10 - 7
9.136 Bottles of ketchup are routinely left on the counters of cafes, yet the ketchup
does not spoil. Why not?
Solution
The vinegar in the ketchup makes the pH low enough (acidity high) that the
ketchup is like pickles (see Chemistry Around Us 9.2). The microorganisms
that makes many foods spoil cannot survive in the high acidity.
9.137 Refer to Figure 9.2 and answer the question. What implications does this
reaction have for the long-term durability of marble structures when exposed to
acid rain?
Solution
The CaCO 3 in the marble reacts with the acid to produce CO2 gas. The acid
rain will react with marble, eroding the surface. Marble structures will
ultimately erode away under long-term exposure to acid rain.
9.138 Refer to Figure 9.3 and answer the question. Do you think the results would be
the same if sulfuric acid (H2 SO4 ) were substituted for the HCl? Explain.
Solution
Yes. The most active metal reacts most vigorously. The reaction would be
the same using H 2SO4 , instead of HCI, for the three metals tested. If either
the chloride salt or the sulfate salt of the metal were insoluble, there would
then be a detectable difference in the reactions using HCI or H 2SO4 .
(See 9.1)
9.140 Refer to Figure 9.6 and explain why the pH reading of the meter might not be
7.00 at the equivalence point in a titration. Would the pH at the equivalence
point be greater or less than 7 if hydrochloric acid was titrated with aqueous
ammonia?
Solution
A titration of a strong acid with a strong base would have pH = 7.00 at the
equivalence point. If a weak acid is titrated with a strong base, hydrolysis of
the B- makes the pH greater than 7 at the equivalence point. If a weak base
(NH3 ) is titrated with a strong acid (HCI), the pH will be less than 7 at the
equivalence point.
Radioactive Nuclei (Section 10.1)
10.3
Group the common nuclear radiations (Table 10.1) into the following categories:
a) those with a negative charge b) those with a mass number greater than 0
c) those that consist of particles
Solution
a) The beta particle has a negative charge.
b) The alpha and neutron have mass numbers greater than 0.
c) The alpha, beta, neutron, and positron radiations consist of particles.
10.5
Discuss how the charge and mass of particles that radiation comprises influence
the range or ability of the radiation to penetrate matter.
Solution
Particles having a high mass do not penetrate very far. The alpha particles
with a high charge and high mass are extremely limited in penetration.
Equations for Nuclear Reactions (Section 10.2)
10.7
Write appropriate symbols for the following particles using the, ZX symbolism.
a) a tin-117 nucleus
b) a nucleus of the chromium (Cr) isotope containing 26 neutrons
c) a nucleus of element number 20 that contains 24 neutrons
Solution
a) 117
50 Sn
10.9
b)
50
24 Cr
c)
44
20Ca
Complete the following equations, using appropriate notations and formulas:
Solution
a) 104Be  - 01+ 105B
b)
210
 42 + 206
83Bi
81Ti
c)
15
 01+ 157N
8O
0
44
0
46
d) 44
e) 84Be  42+ 42He
f) 46
22Ti + - 1e  21Sc
23V  1+ 22 Ti
10.11 Write balanced equations to represent decay reactions of the following isotopes.
The decay process or daughter isotope is given in parentheses.
Solution:
a)
c)
e)
66
0
66
29Cu  - 1+ 30Zn
19  0
+ 199F
8O
- 1
108
0
108
50 Sn + - 1e  49In
Isotope Half-Life (Section 10.3)
b)
d)
f)
22
0
+ 22
11Na  1
10Ne
192  4
188
78Pt 2+ 76Os
67  0
67
28Ni
- 1+ 29Cu
10.13 What is meant by a half-life?
Solution
A half-life is the time required for one half of the radioactive isotopes present
at any time to undergo radioactive decay.
10.15 An isotope of lead, 194
82Pb , has a half-life of 11 minutes. What fraction of the lead194 atoms in a sample would remain after 44 minutes had elapsed?
Solution
The time period, 44 minutes, is four half-life
periods. The fraction remaining is
1 x 1 x 1 x 1 x = ( 1)4 = 1
2 2 2 2
2
16
10.17 An archaeologist sometime in the future analyzes the iron used in an old
building. The iron contains tiny amounts of nickel-63, with a half-life of 92 years.
On the basis of the amount of nickel-63 and its decay products found, it is
estimated that about 0.78% (1/128) of the original nickel-63 remains. If the
building was constructed in 1980, in what year did the archeologist make the
discovery?
Solution
Since1/128 of the original is left,( 1 )n = 1 . n= 7 half - lives
2
128
7 half - lives x 92 yrs/half - life =644 yrs after 1980.The year is about 2624
Health Effects of Radiation (Section 10.4)
10.23 A radiologist found that in a 30 minute period the dose from a radioactive source
was 75 units at a distance of 12.0 m from the source. What would the dose be in
the same amount of time at a distance of 2.00 m from the source?
Solution
2
Ix dy
= 2; where Ix =?; Iy = 75units; dx = 2.00m; dy =12.0 m
Iy dx
Ix =Iy x
d2y
(12.0 m)2
=
75
units
x
= 2.7 x 103 units
d2x
(2.00 m)2
Measurement Units for Radiation (Section 10.5)
10.25 Explain why the rem is the best unit to use when evaluating the radiation
received by a person working in an area where exposure to several types of
radiation is possible.
Solution
The rem measures both the intensity of the radiation and the extent of
damage by the different types of radiation.
10.27 Describe how scintillation counters and Geiger-Muller counters detect radiation.
Solution
A scintillation counter counts the flashes of light as radiation strikes a
phosphor. A Geiger counter measures the pulses of electrical current as the
radiation creates a path for electricity to flow through the gas in the detector.
Medical Uses of Radioisotopes (Section 10.6)
10.29 Explain what a diagnostic tracer is and list the ideal characteristics one should
have.
Solution
A diagnostic tracer is a radioactive isotope that concentrates in a specific
organ. It should have a short, but reasonable, half-life, have a stable
daughter isotope, be non-toxic, and give off radiation that can penetrate out
of the tissue and be detected.
10.31 List the ideal characteristics of a radioisotope that is to be administered
internally for therapeutic use.
Solution
A therapeutic radioisotope administered internally should be an alpha or beta
emitter; have a half-life long enough to do the desired radiation; be non-toxic
(and have non-toxic, non-radioactive daughter isotopes); and be
concentrated by the target tissue.
10.33 Gold-198 is a emitter used to treat leukemia. It has a half-life of 2.7 days. The
dosage is 1.0 mCi/kg body weight. How long would it take for a 70 mCi dose to
decay to an activity of 2.2 mCi?
Solution
The fraction left is 2.2 mCi = about 1 or( 1 )5. n = 5 half - lives
32
2
70 mCi
2.7 days x 5 = 13.5 days or about 14 days
Nonmedical Uses of Radioisotope (Section 10.7)
10.35 Suppose you are planning a TV commercial for motor oil. Describe how you
would set it up to show your oil prevents engine wear better than a competing
brand.
Solution
Use a radioisotope to make the metal surfaces on two identical engines. Use
a Geiger counter to measure the radioactivity present in the oil after running
the engines for the same length of time. The one with the higher radioactivity
in the oil had more wear on the metal parts.
10.37 Explain why carbon-14 with a half-life of about 5600 years would not be a good
radioisotope to use if you want to determine the age of a coal bed thought to be
several million years old.
Solution
After 10 half-lives, the amount of the radioisotope left is less than 1/1000 of
the original amount. This would almost be undetectable. Ten half-lives for
14C = 56,000 years. The amount of 14C left after several million years would
be essentially zero.
Induced Nuclear Reactions (Section 10.8)
10.43 To make indium-111 for diagnostic work, silver-109 is bombarded with

particles. What isotope forms when a silver-109 captures an particle? Write
a balanced equation to represent the process. What two particles must be
emitted by the isotope to form indium-111?
Solution
1 09
47 Ag
1
+ 42  111
49In + 2 0n
Two neutrons are emitted as a result of the bombardment.
Nuclear Energy (Section 10.9)
10.45 Write one balanced equation that illustrates nuclear fission.
Solution
235
1
95
1
 139
92U + 0n
54 Xe + 38 Sr + 2 0n
10.47 Explain why critical mass is an important concept in neutron-induced fission
chain reactions.
Solution
A critical mass is when there is enough fissionable material present to
support a self-sustained fission reaction. This process is used for
commercial power generation.
10.49 Complete the following equations, which represent two additional ways uranium235 can undergo nuclear fission.
Solution
ADDITIONAL EXERCISES
10.51 Polonium-210 decays by alpha emission and has a half-life of 138 days.
Suppose a sample of Po-210 contains 6.02x1020 Po-210 atoms and undergoes
decay for 138 days. What was the mass in grams of the original sample?
Assume all alpha particles formed by the decay escape, and the daughter nuclei
remains in the sample as nonradioactive isotopes, then calculate the mass of
the sample in grams after 138 days.
Solution
Original mass is 6.02x10 20 atoms/6.02x1023 atoms/mol*210g/mol = 0.210 g.
At 138 days, 50% of the Po-210 has decayed to daughter product by alpha
emission. 50% of the Po-210 remains (0.105g) and the daughter has lost 4
out of 210 amu per decay (1.90%). The mass of the daughter would be
0.103 g. Total final mass at 138 days would be 0.208 grams.
10.55 At one time, nuclear bombs were tested by exploding them above ground. The
fallout from such tests contained some Sr-90, a radioactive isotope of strontium.
If Sr-90 gets into the food supply, it can be incorporated into the bones of
humans and other animals. Explain how this can happen though strontium is
not a normal component of bones.
Solution
Strontium is not normally available in the food chain but, if introduced in any
manner, can behave as described. This is because it is chemically similar to
calcium.
ALLIED HEALTH EXAM CONNECTION
10.57 Rank the following from highest penetrating power to lowest penetrating power.
a) alpha particles; b) beta particles; c) gamma rays; d) X rays
Solution
Gamma rays = X rays > beta particles > alpha particles
10.59 If a sample of a radioactive element with a half-life of 100 years has a mass of
31.5 kg remaining after 400 years, what was the mass in kg of the original
sample.
Solution
After each half-life, 50% of the material remains. After 4 half-lives have
passed, only 6.25% would remain. The original amount would be found by:
31.5 kg / 0.0625 = 504 kg.
10.61 Which of the following statements best expresses the fundamental principle of
the mass-energy equivalency represented by the Einstein equation, E=mC2 ?
a) small mass = much energy
b) small mass = little energy
c) little energy = great mass
Solution
Item a comes closest. Any mass will equate to a large amount of energy.
SOLUTIONS TO CHEMISTRY FOR THOUGHT
10.62 One (unrealized) goal of ancient alchemists was to change one element into
another. Do such changes occur naturally? Explain your reasoning.
Solution
Yes. In naturally occurring 
and decay, the radioisotope changes from one
kind of atom into a different kind of atom.
10.63 Refer to Figure 10.2. Some isotopes used as positron emitters in PET scans are
fluorine-18, oxygen-15, and nitrogen-13. What element results in each case
when a positron is emitted?
Solution
10.64 Consider the concept of half-life and decide if, in principle, a radioactive isotope
ever completely disappears by radioactive decay. Explain your reasoning.
Solution
In principle, no. Half of the remaining radioisotope decays each half-life.
Eventually, a condition will be reached where there is only a single atom left,
which would finally decay.
10.66 Nuclear wastes typically have to be stored for at least 20 half-lives before they
are considered safe. This can be a time of hundreds of years for some isotopes.
With that in mind, would you consider sending such wastes into outer space a
responsible solution to the nuclear waste disposal problems? Explain your
answer.
Solution
Sending nuclear wastes into space, ultimately to reach the sun, is a possible
solution. But the risk of launch failure has to be small enough to be an
acceptable risk. A launch failure would spread the radioactivity into the
environment of the Earth.
10.68 Uranium-238 is the most abundant naturally occurring isotope of uranium. It
undergoes radioactive decay to form other isotopes. In the first three steps of
this decay, uranium-238 is converted to thorium-234, which is converted to
protactinium-234, which is converted to uranium-234. Assume only one particle
in addition to the daughter is produced in each step, and use equations for the
decay processes to determine the type of radiation emitted during each step.
Solution
Step 1. U-238  Th-234 + 
Step 2. Th-234  Pa-234 + -
Step 3. Pa-234  U-234 + -