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CHAPTER 15
ACIDS, BASES, AND SALTS
SOLUTIONS TO REVIEW QUESTIONS
1. The Arrhenius definition is restricted to aqueous solutions, while the Brønsted-Lowry definition is not.
2. By the Arrhenius theory, an acid is a substance that produces hydrogen ions in aqueous solution. A base
is a substance that produces hydroxide ions in aqueous solution.
By the Brønsted-Lowry theory, an acid is a proton donor, while a base accepts protons. Since a proton is
a hydrogen ion, then the two theories are very similar for acids, but not bases. A chloride ion can accept
a proton (producing HCl), so it is a Brønsted-Lowry base, but would not be a base by the Arrhenius
theory, since it does not produce hydroxide ions.
By the Lewis theory, an acid is an electron pair acceptor, and a base is an electron pair donor. Many
individual substances would be similarly classified as bases by Brønsted-Lowry or Lewis theories, since
a substance with an electron pair to donate, can accept a proton. But, the Lewis definition is almost
exclusively applied to reactions where the acid and base combine into a single molecule. The
Brønsted-Lowry definition is usually applied to reactions that involve a transfer of a proton from the
acid to the base. The Arrhenius definition is most often applied to individual substances, not to
reactions. According to the Arrhenius theory, neutralization involves the reaction between a hydrogen
ion and a hydroxide ion to form water.
Neutralization, according to the Brønsted-Lowry theory, involves the transfer of a proton to a negative
ion. The formation of a covalent bond constitutes a Lewis neutralization.
3. Neutralization reactions:
Arrhenius: HCl þ NaOH ! NaCl þ H2 O
ðHþ þ OH ! H2 OÞ
Brønsted-Lowry: HCl þ KCN ! HCN þ KCl
Lewis: AlCl3 þ NaCl ! AlCl4 þ Naþ
ðHþ þ CN ! HCNÞ
–
Cl
Al Cl
Cl
4.
(a)
Br
+
–
Cl
Cl
Cl Al Cl
Cl
–
(b)
–
O H
(c)
–
C
N
These ions are considered to be bases according to the Brønsted-Lowry theory, because they can accept
a proton at any of their unshared pairs of electrons. They are considered to be bases according to the
Lewis acid-base theory, because they can donate an electron pair.
5. Metals that lie above hydrogen in the activity series will form hydrogen gas when they react with an
acid.
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- Chapter 15 6. Carbonates will form carbon dioxide when they react with an acid.
7. NaNO3, sodium nitrate; Ca(NO3)2, calcium nitrate; Al(NO3)3, aluminium nitrate. These are three of the
many possible salts which can be formed from nitric acid.
8. LiCl, lithium chloride; Li2SO4, lithium sulfate; Li3PO4, lithium phosphate. These are three of the many
possible salts which can be formed from lithium hydroxide.
9. An electrolyte must be present in the solution for the bulb to glow.
10. Electrolytes include acids, bases, and salts. (Electrolytes are any compound that conducts electricity in
solution.)
11. First, the orientation of the polar water molecules about the Naþ and Cl is different. The positive end
(hydrogen) of the water molecule is directed towards Cl, while the negative end (oxygen) of the water
molecule is directed towards the Naþ. Second, more water molecules will fit around Cl, since it is
larger than the Naþ ion.
12. The electrolytic compounds are acids, bases, and salts.
13. Names of the compounds in Table 15.3
H2 SO4
sulfuric acid
nitric acid
HNO3
HCl
hydrochloric acid
HBr
hydrobromic acid
perchloric acid
HClO4
NaOH
sodium hydroxide
KOH
potassium hydroxide
calcium hydroxide
CaðOHÞ2
BaðOHÞ2
barium hydroxide
HC2 H3 O2 acetic acid
H2 CO3
carbonic acid
nitrous acid
HNO2
sulfurous acid
H2 SO3
H2 S
hydrosulfuric acid
oxalic acid
H2 C2 O4
boric acid
H3 BO3
HClO
hypochlorous acid
NH3
ammonia
HF
hydrofluoric acid
14. Hydrogen chloride dissolved in water conducts an electric current. HCl reacts with polar water
molecules to produce H3Oþ and Cl ions, which conduct an electric current. Hexane is a nonpolar
solvent, so it cannot pull the HCl molecules apart. Since there are no ions in the hexane solution, it does
not conduct an electric current. HCl does not ionize in hexane.
15. In their crystalline structure, salts exist as positive and negative ions in definite geometric arrangement
to each other, held together by the attraction of the opposite charges. When dissolved in water, the salt
dissociates as the ions are pulled away from each other by the polar water molecules.
16. Testing the electrical conductivity of the solutions shows that CH3OH is a nonelectrolyte, while NaOH is
an electrolyte. This indicates that the OH group in CH3OH must be covalently bonded to the CH3 group.
17. Molten NaCl conducts electricity because the ions are free to move. In the solid state, however, the ions
are immobile and do not conduct electricity.
18. Dissociation is the separation of already existing ions in an ionic compound. Ionization is the formation
of ions from molecules. The dissolving of NaCl is a dissociation, since the ions already exist in the
crystalline compound. The dissolving of HCl in water is an ionization process, because ions are formed
from HCl molecules and H2O.
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- Chapter 15 19. Strong electrolytes are those which are essentially 100% ionized or dissociated in water. Weak
electrolytes are those which are only slightly ionized in water.
20. Ions are hydrated in solution because there is an electrical attraction between the charged ions and the
polar water molecules.
21. The main distinction between water solutions of strong and weak electrolytes is the degree of ionization
of the electrolyte. A solution of an electrolyte contains many more ions than does a solution of a
nonelectrolyte. Strong electrolytes are essentially 100% ionized. Weak electrolytes are only slightly
ionized in water.
22. The HCl molecule is polar and, consequently, is much more soluble in the polar solvent, water, than in
the nonpolar solvent, hexane. There is also a chemical reaction between HCl and H2 O molecules.
! H3 Oþ þ Cl
HCl þ H2 O 23. The pH for a solution with a hydrogen ion concentration of 0.003 M will be between 2 and 3.
24. Tomato juice is more acidic than blood, since its pH is lower.
25. (a)
(b)
(c)
In a neutral solution, the concentration of Hþ and OH are equal.
In an acid solution, the concentration of Hþ is greater than the concentration of OH.
In a basic solution, the concentration of OH is greater than the concentration of Hþ.
26. Pure water is neutral because when it ionizes it produces equal molar concentrations of acid [Hþ] and
base [OH] ions.
27. A neutral solution is one in which the concentration of acid is equal to the concentration of base
½Hþ ¼ ½OH . An acidic solution is one in which the concentration of acid is greater than the
concentration of base ½Hþ > ½OH . A basic solution is one in which the concentration of base is
greater than the concentration of acid ½Hþ < ½OH .
28. A titration is used to determine the concentration of a specific substance (often an acid or a base) in a
sample. A titration determines the volume of a reagent of known concentration that is required to
completely react with a volume of a sample of unknown concentration. An indicator is used to help
visualize the endpoint of a titration. The endpoint is the point at which enough of the reagent of known
concentration has been added to the sample of unknown concentration to completely react with the
unknown solution. An indicator color change is visible when the endpoint has been reached.
29. The net ionic equation for an acid-base reaction in aqueous solutions is:
! H2 O
Hþ þ OH 30. Acid rain is caused by the release of nitrogen and sulfur oxides into the air. When these oxides are
carried through the atmosphere they react with water and form sulfuric acid ðH2 SO4 Þ and nitric acid
ðHNO3 Þ. Precipitation (rain or snow) carries the acids to the ground.
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- Chapter 15 -
SOLUTIONS TO EXERCISES
1. Conjugate acid – base pairs:
(a) NH3 NHþ
4 ; H2 O OH
þ
(b) HC2 H3 O2 C2 H3 O
2 ; H2 O H3 O
2
(c) H2 PO4 HPO4 ; OH H2 O
(d) HCl Cl ; H2 O H3 Oþ
2. Conjugate acid – base pairs:
(a) H2 S HS ; NH3 NH4þ
2
þ
(b) HSO
4 SO4 ; NH3 NH4
(c) HBr Br ; CH3 O CH3 OH
(d) HNO3 NO3 ; H2 O H3 Oþ
3. (a)
(b)
(c)
(d)
(e)
(f)
ZnðsÞ þ 2 HClðaqÞ ! ZnCl2 ðaqÞ þ H2 ðgÞ
AlðOHÞ3 ðsÞ þ 3 H2 SO4 ðaqÞ ! Al2 ðSO4 Þ3 ðaqÞ þ 6 H2 Oðl Þ
Na2 CO3 ðaqÞ þ 2 HC2 H3 O2 ðaqÞ ! 2 NaC2 H3 O2 ðaqÞ þ H2 Oðl Þ þ CO2 ðgÞ
MgOðsÞ þ 2 HIðaqÞ ! MgI2 ðaqÞ þ H2 Oðl Þ
CaðHCO3 Þ2 ðsÞ þ 2 HBrðaqÞ ! CaBr2 ðaqÞ þ 2 H2 Oðl Þ þ 2 CO2 ðgÞ
3 KOHðaqÞ þ H3 PO4 ðaqÞ ! K3 PO4 ðaqÞ þ 3 H2 Oðl Þ
4. Complete and balance these equations:
(a) Fe2 O3 ðsÞ þ 6 HBrðaqÞ ! 2 FeBr3 ðaqÞ þ 3 H2 Oðl Þ
(b) 2 AlðsÞ þ 3 H2 SO4 ðaqÞ ! Al2 ðSO4 Þ3 ðaqÞ þ 3 H2 ðgÞ
(c) 2 NaOHðaqÞ þ H2 CO3 ðaqÞ ! Na2 CO3 ðaqÞ þ 2 H2 Oðl Þ
(d) BaðOHÞ2 ðsÞ þ 2 HClO4 ðaqÞ ! BaðClO4 Þ2 ðaqÞ þ 2 H2 Oðl Þ
(e) MgðsÞ þ 2 HClO4 ðaqÞ ! MgðClO4 Þ2 ðaqÞ þ H2 ðgÞ
(f) K2 OðsÞ þ 2 HIðaqÞ ! 2 KIðaqÞ þ H2 Oðl Þ
5. (a)
Zn þ ð2 Hþ þ 2 Cl Þ ! ðZn2þ þ 2 Cl Þ þ H2
Zn þ 2 Hþ ! Zn2þ þ H2
(b)
2 AlðOHÞ3 þ 6 Hþ þ 3 SO2
! 2 Al3þ þ 3 SO2
þ 6 H2 O
4
4
(c)
AlðOHÞ3 þ 3 Hþ ! Al3þ þ 3 H2 O
2 Naþ þ CO2
þ 2 HC2 H3 O2 ! 2 Naþ þ 2 C2 H3 O
3
2 þ H2 O þ CO2
CO2
3 þ 2 HC2 H3 O2 ! 2 C2 H3 O2 þ H2 O þ CO2
(d)
MgO þ ð2 Hþ þ 2 I Þ ! ðMg2þ þ 2 I Þ þ H2 O
MgO þ 2 Hþ ! Mg2þ þ H2 O
(e)
CaðHCO3 Þ2 þ ð2 Hþ þ 2 Br Þ ! ðCa2þ þ 2 Br Þ þ 2 H2 O þ 2 CO2
CaðHCO3 Þ2 þ 2 Hþ ! Ca2þ þ H2 O þ CO2
(f)
þ 3 H2 O
ð3 Kþ þ 3 OH Þ þ H3 PO4 ! 3 Kþ þ PO3
4
3 OH þ H3 PO4 ! PO3
4 þ 3 H2 O
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- Chapter 15 6. (a)
Fe2 O3 þ ð6 Hþ þ 6 Br Þ ! ð2 Fe3þ þ 6 Br Þ þ 3 H2 O
(b)
Fe2 O3 þ 6 Hþ ! 2 Fe3þ þ 3 H2 O
2 Al þ 6Hþ ! 3 SO2
! 2 Al3þ þ 3 SO2
þ 3 H2
4
4
2 Al þ 6 Hþ ! 2 Al3þ þ 3 H2
(c)
ð2 Naþ þ 2 OH Þ þ H2 CO3 ! 2 Naþ þ CO2
þ 2 H2 O
3
(d)
2 OH þ H2 CO3 ! CO2
3 þ 2 H2 O
þ
2þ
þ 2 ClO
BaðOHÞ2 þ 2 H þ 2 ClO
4 ! Ba
4 þ 2 H2 O
(e)
BaðOHÞ2 þ 2 Hþ ! Ba2þ þ 2 H2 O
2þ
Mg þ 2 Hþ þ 2 ClO
þ 2 ClO
4 ! Mg
4 þ H2
(f)
Mg þ 2 Hþ ! Mg2þ þ H2
K2 O þ ð2 Hþ þ 2 I Þ ! ð2 Kþ þ 2 I Þ þ H2 O
K2 O þ 2 Hþ ! 2 Kþ þ H2 O
7. (a)
(b)
(c)
HNO3 þ NaOH ! H2 O þ NaNO3
2 HC2 H3 O2 þ BaðOHÞ2 ! 2 H2 O þ BaðC2 H3 O2 Þ2
HClO4 þ NH4 OH ! H2 O þ NH4 ClO4
8. (a)
(b)
(c)
2 HBr þ MgðOHÞ2 ! 2 H2 O þ MgBr2
H3 PO4 þ 3 KOH ! 3 H2 O þ K3 PO4
H2 SO4 þ 2 NH4 OH ! 2 H2 O þ ðNH4 Þ2 SO4
9.
LiOH and H2S must be reacted
2 LiOH þ H2 S ! 2 H2 O þ Li2 S
10.
Ca(OH)2 and H2CO3 must be reacted
CaðOHÞ2 þ H2 CO3 ! 2 H2 O þ CaCO3
11. The following compounds are electrolytes:
(a) SO3 , acid in water
(b) K2 CO3 , salt
(e) CuBr2 , salt
(f) HI, acid in water
12. The following compounds are electrolytes:
(b) P2 O5 , acid in water
(c) NaClO, salt
(d) LiOH, base
(f) KMnO4 , salt
13. Molarity of ions.
(a)
1 mol Cu2þ
ð1:25 M CuBr2 Þ
¼ 1:25 M Cu2þ
1 mol CuBr2
2 mol Br
¼ 2:50 M Br
ð1:25 M CuBr2 Þ
1 mol CuBr2
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- Chapter 15 (b)
(c)
(d)
1 mol Naþ
ð0:75 M NaHCO3 Þ
¼ 0:75 M Naþ
1 mol NaHCO3
1 mol HCO
3
¼ 0:75 M HCO
ð0:75 M NaHCO3 Þ
3
1 mol NaHCO3
3 mol Kþ
¼ 10:5 M Kþ
ð3:50 M K3 AsO4 Þ
1 mol K3 AsO4
1 mol AsO3
4
ð3:50 M K3 AsO4 Þ
¼ 3:50 M AsO3
4
1 mol K3 AsO4
2 mol NHþ
4
¼ 1:3 M NHþ
0:65 M ðNH4 Þ2 SO4
4
1 mol ðNH4 Þ2 SO4
1 mol SO2
4
¼ 0:65 M SO2
0:65 M ðNH4 Þ2 SO4
4
1 mol ðNH4 Þ2 SO4
14. Molarity of ions.
(a)
(b)
(c)
(d)
1 mol Fe3þ
¼ 2:25 M Fe3þ
ð2:25 M FeCl3 Þ
1 mol FeCl3
3 mol Cl
¼ 6:75 M Cl
ð2:25 M FeCl3 Þ
1 mol FeCl3
1 mol Mg2þ
¼ 1:20 M Mg2þ
ð1:20 M MgSO4 Þ
1 mol MgSO4
1 mol SO2
4
ð1:20 M MgSO4 Þ
¼ 1:20 M SO2
4
1 mol MgSO4
1 mol Naþ
ð0:75 M NaH2 PO4 Þ
¼ 0:75 M Naþ
1 mol NaH2 PO4
1 mol H2 PO
4
ð0:75 M NaH2 PO4 Þ
¼ 0:75 M H2 PO
4
1 mol NaH2 PO4
1 mol Ca2þ
0:35 M CaðClO3 Þ2
¼ 0:35 M Ca2þ
1 mol CaðClO3 Þ2
2 mol ClO
3
¼ 0:70 M ClO
0:35 M CaðClO3 Þ2
3
1 mol CaðClO3 Þ2
15. We will use the data from No. 13 to solve these problems. 100 mL = 0.100 L
1:25 mol Cu2þ
63:55 g
(a) ð0:100 LÞ
¼ 7:94 g Cu2þ
mol
L
2:50 mol Br 79:90 g
ð0:100 LÞ
¼ 20:0 g Br
L
mol
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- Chapter 15 (b)
(c)
(d)
0:75 mol Naþ
22:99 g
ð0:100 LÞ
¼ 1:7 g Naþ
mol
L
0:75 mol HCO
61:02 g
3
ð0:100 LÞ
¼ 4:6 g HCO
3
mol
L
10:5 mol Kþ 39:10 g
ð0:100 LÞ
¼ 41:1 g Kþ
mol
L
3:50 mol AsO3
138:9 g
4
ð0:100 LÞ
¼ 48:6 g AsO3
4
mol
L
1:3 mol NHþ
18:04 g
4
ð0:100 LÞ
¼ 2:3 g NHþ
4
mol
L
0:65 mol SO2
96:07 g
4
ð0:100 LÞ
¼ 6:2 g SO2
4
1 mol
1L
16. We will use the data from No. 14 to solve these problems 100 mL = 0.100 L
2:25 mol Fe3þ 55:85 g
(a) ð0:100 LÞ
¼ 12:6 g Fe3þ
mol
L
6:75 mol Cl
35:45 g
ð0:100 LÞ
¼ 23:9 g Cl
mol
L
(b)
(c)
(d)
1:20 mol Mg2þ
24:31 g
ð0:100 LÞ
¼ 2:92 g Mg2þ
mol
L
1:20 mol SO2
96:07 g
4
ð0:100 LÞ
¼ 11:5 g SO2
4
mol
L
0:75 mol Naþ
22:99 g
ð0:100 LÞ
¼ 1:7 g Naþ
mol
L
0:75 mol H2 PO
96:99 g H2 PO
4
4
ð0:100 LÞ
¼ 7:3 g H2 PO
4
L
mol
0:35 mol Ca2þ 40:08 g
¼ 1:4 g Ca2þ
ð0:100 LÞ
mol
L
0:70 mol ClO
83:45 g
3
ð0:100 LÞ
¼ 5:8 g ClO
3
mol
L
17. pH ¼ log½Hþ ½Hþ ¼ 10pH
(a)
½Hþ ¼ 1 105
(b)
½Hþ ¼ 2 107
½Hþ ¼ 1 108
(c)
(d)
½Hþ ¼ 2 1010
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- Chapter 15 18. pH ¼ log½Hþ ½Hþ ¼ 10pH
½Hþ ¼ 1 107
(a)
(b)
½Hþ ¼ 5 105
(c)
½Hþ ¼ 2 106
½Hþ ¼ 5 1011
(d)
0:50 mol HCl
¼ 0:028 mol HCl
19. (a) ð55:5 mLÞ
1000 mL
1:25 mol HCl
ð75:0 mLÞ
¼ 0:0938 mol HCl
1000 mL
Total mol HCl ¼ 0:028 mol þ 0:0938 mol ¼ 0:122 mol HCl
Total volume ¼ 0:0555 L þ 0:0750 L ¼ 0:1305 L
0:122 mol HCl
¼ 0:935 M HCl
0:1305 L
1 mol Hþ
¼ 0:935 M Hþ
ð0:935 M HClÞ
1 mol HCl
1 mol Cl
ð0:935 M HClÞ
¼ 0:935 M Cl
1 mol HCl
0:75 mol CaCl2
¼ 0:094 mol CaCl2
(b) ð125 mLÞ
1000 mL
0:25 mol CaCl2
ð125 mLÞ
¼ 0:031 mol CaCl2
1000 mL
Total mol CaCl2 ¼ 0:094 mol þ 0:031 mol ¼ 0:125 mol CaCl2
Total volume ¼ 0:125 L þ 0:125 L ¼ 0:250 L
0:125 mol CaCl2
¼ 0:500 M CaCl2
0:250 L
1 mol Ca2þ
ð0:500 M CaCl2 Þ
¼ 0:500 M Ca2þ
1 mol CaCl2
2 mol Cl
ð0:500 M CaCl2 Þ
¼ 1:00 M Cl
1 mol CaCl2
(c)
NaOH þ HCl ! NaCl þ H2 O
0:333 mol NaOH
ð35:0 mLÞ
¼ 0:0117 mol NaOH
1000 mL
0:250 mol HCl
ð22:5 mLÞ
¼ 0:00563 mol HCl
1000 mL
0.00563 mol HCI reacts with 0.00563 mol NaOH. 0.0061 mol NaOH remains uareacted and
0.00563 mol NaCl is produced. The final volume is 0.0575 L and contains 0.0061 mol NaOH and
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- Chapter 15 0.00563 mol NaCl. Moles of ions are: (0.0061 mol Naþ þ 0.00563 mol Naþ) ¼ 0.0117 mol Naþ,
0.0061 mol OH , and 0.00563 mol Cl . Concentrations of ions are:
0:0177 mol Naþ
¼ 0:203 M Naþ
0:0575 L
0:0061 mol OH
¼ 0:11 M OH
0:0575 L
0:00563 mol Cl
¼ 0:0979 M Cl
0:0575 L
(d)
H2 SO4 þ 2 NaOH ! Na2 SO4 þ 2 H2 O
0:500 mol H2 SO4
¼ 0:00625 mol H2 SO4
ð12:5 mLÞ
1000 mL
0:175 mol NaOH
ð23:5 mLÞ
¼ 0:00411 mol NaOH
1000 mL
1 mol H2 SO4
ð0:00411 mol NaOHÞ
¼ 0:00206 mol H2 SO4 reacted
2 mol NaOH
0.00206 mol H2SO4 reacts with 0.00411 mol NaOH. 0.00419 mol H2SO4 remains unreacted
and 0.00206 mol Na2SO4 is produced. The final volume is 0.0360 L and contains 0.00206 mol
Na2SO4 and 0.00419 mol H2SO4. Moles of ions are 0.00412 mol Naþ, 0.00838 mol Hþ, and
(0.00206 þ 0.00419) ¼ 0.00625 mol SO2
4 . Concentration of ions are:
0:0412 mol Naþ
0:00838 mol Hþ
¼ 0:114 M Naþ
¼ 0:233 M Hþ
0:0360 L
0:0360 L
0:00625 mol SO2
4
¼ 0:174 M SO2
4
0:0360 L
0:10 mol NaCl
20. (a) ð45:5 mLÞ
¼ 0:0046 mol NaCl
1000 mL
0:35 mol NaCl
ð60:5 mLÞ
¼ 0:021 mol NaCl
1000 mL
Total mol NaCl ¼ 0:0046 mol þ 0:021 mol ¼ 0:026 mol NaCl
Total volume ¼ 0:0455 L þ 0:0605 L ¼ 0:1060 L
0:026 mol NaCl
¼ 0:25 M NaCl
0:1060 L
1 mol Naþ
ð0:25 M NaClÞ
¼ 0:25 M Naþ
1 mol NaCl
1 mol Cl
ð0:25 M NaClÞ
¼ 0:25 M Cl
1 mol NaCl
1:25 mol HCl
(b) ð95:5 mLÞ
¼ 0:119 mol HCl
1000 mL
2:50 mol HCl
ð125:5 mLÞ
¼ 0:314 mol HCl
1000 mL
Total mol HCl ¼ 0:119 mol þ 0:314 mol ¼ 0:433 mol HCl
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- Chapter 15 Total volume ¼ 0:0955 L þ 0:1255 L ¼ 0:2210 L
0:433 mol HCl
¼ 1:96 M HCl
0:2210 L 1 mol Hþ
ð1:96 M HClÞ
¼ 1:96 M Hþ
1 mol HCl
1 mol Cl
ð1:96 M HClÞ
¼ 1:96 M Cl
1 mol HCl
0:10 mol BaðNO3 Þ2
¼ 0:0016 M BaðNO3 Þ2
1000 mL
0:20 mol AgNO3
ð10:5 mLÞ
¼ 0:0021 M AgNO3
1000 mL
(c) ð15:5 mLÞ
Number of moles of each substance: 0:0016 mol Ba2þ , 0:0021 mol Agþ , and
ð0:0032 mol þ 0:0021 molÞ ¼ 0:0053 mol NO
3
Total volume ¼ 0:0155 L þ 0:0105 L ¼ 0:0260 L
0:0016 mol Ba2þ
¼ 0:062 M Ba2þ
0:0260 L
0:0021 mol Agþ
¼ 0:081 M Agþ
0:0260 L
(d)
0:0053 mol NO
3
¼ 0:20 M NO
3
0:0260 L
0:25 mol NaCl
ð25:5 mLÞ
¼ 0:0064 mol NaCl
1000 mL
0:15 mol CaðC2 H3 O2 Þ2
ð15:5 mLÞ
¼ 0:0023 mol CaðC2 H3 O2 Þ2
1000 mL
Number of moles of each substance: 0:0064 mol Naþ , 0:0064 mol Cl , 0:0023 mol Ca2þ ,
0:0046 mol C2 H3 O
2.
Total volume ¼ 0:0255 L þ 0:0155 L ¼ 0:0410 L
0:0064 mol Naþ
¼ 0:16 M Naþ
0:0410 L
0:0064 mol Cl
¼ 0:16 M Cl
0:0410 L
0:0023 mol Ca2þ
¼ 0:056 M Ca2þ
0:0410 L
0:0046 mol C2 H3 O
2
¼ 0:11 M C2 H3 O
2
0:0410 L
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- Chapter 15 21. HNO3 ðaqÞ þ H2 OðlÞ ! H3 Oþ ðaqÞ þ NO
3 ðaqÞ
Or
H O
HNO3 ðaqÞ 2! Hþ ðaqÞ þ NO3 ðaqÞ
Because nitric acid ionizes completely it would be both a strong electrolyte and a strong acid.
22. HCNðaqÞ þ H2 OðlÞ Ð H3 Oþ ðaqÞ þ CN ðaqÞ
Or
H2 O
HCNðaqÞ Ð Hþ ðaqÞ þ CN ðaqÞ
HCN is only partially ionized and so it would be a poor electrolyte and a weak acid.
23. The reaction of HCl and NaOH occurs on a 1:1 mole ratio.
HCl þ NaOH ! NaCl þ H2 O
At the endpoint in these titration reactions, equal moles of HCl and NaOH will have reacted.
Moles ¼ (molarity) (volume). At the endpoint, mol HCl ¼ mol NaOH.
Therefore, at the endpoint,
M B VB
MA ¼
M A VA ¼ M B VB
VA
ð37:70 mLÞð0:728 M Þ
(a)
¼ 0:681 M HCl
40:3 mL
(b)
ð33:66 mLÞð0:306 M Þ
¼ 0:542 M HCl
19:00 mL
(c)
ð18:00 mLÞð0:555 M Þ
¼ 0:367 M HCl
27:25 mL
24. The reaction of HCl and NaOH occurs on a 1:1 mole ratio.
HCl þ NaOH ! NaCl þ H2 O
At the endpoint in these titration reactions, equal moles of HCl and NaOH will have reacted.
Moles ¼ (molarity)(volume). At the endpoint, mol HCl ¼ mol NaOH.
Therefore, at the endpoint,
M A VA ¼ M B VB
MB ¼
M A VA
VB
(a)
ð37:19 mLÞð0:126 M Þ
¼ 0:147 M NaOH
31:91 mL
(b)
ð48:04 mLÞð0:482 M Þ
¼ 0:964 M NaOH
24:02 mL
(c)
ð13:13 mLÞð1:425 M Þ
¼ 0:4750 M NaOH
39:39 mL
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- Chapter 15 25. (a)
(b)
(c)
3
2þ
2 PO3
4 ðaqÞ þ 3 Ca ðaqÞ ! Ca3 PO4 2 ðsÞ
2 AlðsÞ þ 6 Hþ ðaqÞ ! 3 H2 ðgÞ þ 2 Al3þ ðaqÞ
þ
CO2
3 ðaqÞ þ 2 H ðaqÞ ! H2 OðaqÞ þ CO2 ðgÞ
26. (a)
(b)
(c)
MgðsÞ þ Cu2þ ðaqÞ ! CuðsÞ þ Mg2þ ðaqÞ
Hþ ðaqÞ þ OH ðaqÞ ! H2 Oðl Þ
þ
SO2
3 ðaqÞ þ 2 H ðaqÞ ! H2 Oðl Þ þ SO2 ðgÞ
27. (a)
1 molar H2SO4 is more acidic. The concentration of Hþ in 1 M H2SO4 is greater than 1 M since
there are two ionizable hydrogens per mole of H2SO4. In HCl the concentration of Hþ will be 1 M,
since there is only one ionizable hydrogen per mole HCl.
1 molar HCl is more acidic. HCl is a strong electrolyte, producing more Hþ than HC2H3O2 which
is a weak electrolyte.
(b)
28. (a)
(b)
2 molar HCl is more acidic. 2 M HCl will yield 2 M Hþ concentration. 1 M HCl will yield 1 M Hþ
concentration.
1 molar H2SO4 is more acidic. Both are strong acids. The concentration of Hþ in 1 M H2SO4 is
greater than in 1 M HNO3 because H2SO4 has two ionizable hydrogens per mole whereas HNO3
has only one ionizable hydrogen per mole.
29. 2 HClO4 ðaqÞ þ CaðOHÞ2 ðsÞ ! CaðClO4 Þ2 ðaqÞ þ 2 H2 Oðl Þ
g CaðOHÞ2 ! mol CaðOHÞ2 ! mol HClO4 ! mL HClO4
mol
2 mol HClO4
1000 mL
50:25 g CaðOHÞ2
¼ 2:58 103 mL HClO4
74:10 g 1 mol CaðOHÞ2
0:525 mol
30. 3 HClðaqÞ þ AlðOHÞ3 ðsÞ ! AlCl3 ðaqÞ þ 3 H2 Oðl Þ
mL HCl ! mol HCl ! mol AlðOHÞ3 ! g AlðOHÞ3
0:125 mol 1 mol AlðOHÞ3
78:00 g
ð275 mL HClÞ
¼ 0:894 g AlðOHÞ3
1000 mL
mol
3 mol HCl
31. NaOH þ HCl ! NaCl þ H2 O
First calculate the grams of NaOH in the sample.
L HCl ! mol HCl ! mol NaOH ! g NaOH
0:2406 mol 1 mol NaOH 40:00 g
ð0:01825 L HClÞ
¼ 0:1756 g NaOH in the sample
L
1 mol HCl
mol
0:1756 g NaOH
ð100Þ ¼ 87:8% NaOH
0:200 g sample
32. NaOH þ HCl ! NaCl þ H2 O
L HCl ! mol HCl ! mol NaOH ! g NaOH
0:466 mol 1 mol NaOH 40:00 g
ð0:04990 L HClÞ
¼ 0:930 g NaOH in the sample
L
1 mol HCl
mol
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- Chapter 15 1:00 g sample 0:930 g NaOH ¼ 0:070 g NaCl in the sample
0:070 g NaCl
ð100Þ ¼ 7:0% NaCl in the sample
1:00 g sample
33. Zn þ 2 HCl ! ZnCl2 þ H2
This is a limiting reactant problem. First find the moles of Zn and HCl from the given data and then
identify the limiting reactant.
1 mol
g Zn ! mol Zn
ð5:00 g ZnÞ
¼ 0:0765 mol Zn
65:39 g
0:350 mol
ð0:100 L HClÞ
¼ 0:0350 mol HCl
L
Therefore Zn is in excess and HCl is the limiting reactant.
1 mol H2
ð0:0350 mol HClÞ
¼ 0:0175 mol H2 produced in the reaction
2 mol HCl
1 atm
P ¼ ð700: torrÞ
T ¼ 27 C ¼ 300: K
¼ 0:921 atm
760 torr
PV ¼ nRT
nRT ð0:0175 mol H2 Þð0:0821 L atm=mol KÞð300: KÞ
¼
¼ 0:468 L H2
P
0:921 atm
34. Zn þ 2 HCl ! ZnCl2 þ H2
V¼
This is a limiting reactant problem. First find moles of Zn and HCl from the given data and then identify
the limiting reactant.
1 mol
g Zn ! mol Zn
ð5:00 g ZnÞ
¼ 0:0765 mol Zn
65:39 g
0:350 mol
ð0:200 L HClÞ
¼ 0:0700 mol HCl
L
Zn is in excess and HCl is the limiting reactant.
1 mol H2
ð0:0700 mol HClÞ
¼ 0:0350 mol H2
2 mol HCl
1 atm
T ¼ 27 C ¼ 300: K
P ¼ ð700: torrÞ
¼ 0:921 atm
760 torr
PV ¼ nRT
V¼
nRT ð0:0350 mol H2 Þð0:0821 L atm=mol KÞð300: KÞ
¼
¼ 0:936 L H2
P
0:921 atm
35. pH ¼ log½Hþ (a)
(b)
(c)
½Hþ ¼ 0:35 M;
½Hþ ¼ 1:75 M;
½Hþ ¼ 2:0 105 M;
pH ¼ logð0:35Þ ¼ 0:46
pH ¼ logð1:75Þ ¼ 0:243
pH ¼ log 2:0 105 ¼ 4:70
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- Chapter 15 36. pH ¼ log½Hþ (a)
(b)
(c)
37. (a)
(b)
(c)
(d)
38. (a)
(b)
(c)
(d)
39. (a)
½Hþ ¼ 0:0020 M;
½Hþ ¼ 7:0 108 M;
½Hþ ¼ 3:0 M;
pH ¼ logð0:0020Þ ¼ 2:70
pH ¼ log 7:0 108 ¼ 7:15
pH ¼ logð3:0Þ ¼ 0:48
Orange juice ¼ 3:7 104 M Hþ
pH ¼ log 3:7 104 ¼ 3:43
Hþ
Vinegar ¼ 2:8 103 M
pH ¼ log 2:8 103 ¼ 2:55
Hþ
shampoo ¼ 2.4 106 M
pH ¼ log 2:4 106 ¼ 5:62
dishwashing detergent ¼ 3.6 108 M Hþ
pH ¼ log 3:6 108 ¼ 7:44
Black coffee ¼ 5:0 105 M Hþ
pH ¼ log 5:0 105 ¼ 4:30
Limewater ¼ 3:4 1011
M Hþ
pH ¼ log 3:4 1011 ¼ 10:47
fruit punch ¼ 2.1 104 M Hþ
pH ¼ log 2:1 104 ¼ 3:68
cranberry apple drink = 1.3
103 M Hþ
pH ¼ log 1:3 103 ¼ 2:89
H2 O
NH3 is a weak base NH3 ðaqÞ Ð NHþ
4 ðaqÞ þ OH ðaqÞ
H2 O
HClðaqÞ ! Hþ ðaqÞ þ Cl ðaqÞ
(b)
HCl is a strong acid
(c)
KOH is a strong base KOH ! Kþ ðaqÞ þ OH ðaqÞ
(d)
HC2 H3 O2 is a weak acid HC2 H3 O2 ðaqÞ Ð Hþ ðaqÞ þ C2 H3 O
2 ðaqÞ
H2 O
H2 O
40. (a)
H2 O
H2 C2 O4 is a weak acid
H2 C2 O4 ðaqÞ Ð Hþ ðaqÞ þ HC2 O
4 ðaqÞ
H2 O
(b)
BaðOHÞ2 is a strong base BaðOHÞ2 ! Ba2þ ðaqÞ þ 2 OH ðaqÞ
(c)
(d)
HClO4 is a strong acid HClO4 ðaqÞ ! Hþ ðaqÞ þ ClO
4 ðaqÞ
HBr is a strong acid HBrðaqÞ ! Hþ ðaqÞ þ Br ðaqÞ
41. (a)
H2 O
CH3 NH2 þHþ ! CH3 NH3 þ
base
conjugate acid
Note that a neutral base forms a positively charged conjugate acid.
(b)
HS þHþ !
base
H2 S
conjugate acid
Note that a negatively charged base forms a neutral acid upon adding a positively charged
hydrogen ion.
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- Chapter 15 42. (a)
HBrO3 Hþ ! BrO3 acid
(b)
conjugate base
þ
þ
NH4 H !
acid
(c)
NH3
conjugate base
H2 PO4 Hþ ! HPO4 2
acid
conjugate base
H3 O2 þ NH3 !
43. HC2acid
base
NH4 þ þ C2 H3 O2 conjugate acid
conjugate base
Note that in any acid base reaction, the original acid and base react to form a new acid and base.
44. S2 ðaqÞ þ H2 OðlÞ ! HS ðaqÞ þ OH ðaqÞ
The sulfide ion is able to act as a Bronsted-Lowry base by accepting a proton from a water molecule.
Note that Bronsted-Lowry bases will often cause the formation of hydroxide ions in aqueous solution.
45. MgðsÞ þ 2 HClðaqÞ ! MgCl2 ðaqÞ þ H2 ðgÞ
46. H2 SO4 ðaqÞ þ CaCO3 ðsÞ ! CaSO4 ðsÞ þ H2 OðlÞ þ CO2 ðgÞ
2
þ
2
47. Na2 SO4 ðaqÞ ! 2 Na ðaqÞ þ SO4 ðaqÞ
H O
48. (a) basic
(d) acidic
(b) acidic
(e) acidic
(c) neutral (f) basic
49. (a) CaCl2 ðsÞ ! Ca2þ ðaqÞ þ 2 Cl ðaqÞ
For each CaCl2 ionic compound, 1 calcium ion and 2 chloride ions result.
Ca2+
(b)
Cl–
Cl–
KFðsÞ ! Kþ ðaqÞ þ F ðaqÞ
For each KF ionic compound, 1 potassium ion and 1 fluoride ion result.
K+
F–
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- Chapter 15 (c)
AlBr3 ðsÞ ! Al3þ ðaqÞ þ 3 Br ðaqÞ
For each AlBr3 ionic compound, 1 aluminum ion and 3 bromide ions result.
Br–
Al3+
Br–
Br–
50. AlBr3 ! Al3þ þ 3 Br
0:142 mol Br
1 mol Al3þ
0:0473 mol Al3þ
¼
¼ 0:0473 M Al3þ
L
3 mol Br
L
51. H2 SO4 þ 2 NaOH ! Na2 SO4 þ 2 H2 O
mol H2 SO4
mL NaOH ! mol NaOH ! mol H2 SO4 ;
¼ M H2 SO4
L
0:313 mol 1 mol H2 SO4
ð35:22 mL NaOHÞ
¼ 0:00551 mol H2 SO4
1000 mL
2 mol NaOH
0:00551 mol H2 SO4
¼ 0:218 M H2 SO4
0:02522 L
52. The acetic acid solution freezes at a lower temperature than the alcohol solution. The acetic acid ionizes
slightly while the alcohol does not. The ionization of the acetic acid increases its particle concentration in
solution above that of the alcohol solution, resulting in a lower freezing point for the acetic acid solution.
53. It is more economical to purchase CH3OH at the same cost per pound as C2H5OH. Because CH3OH has
a lower molar mass than C2H5OH, the CH3OH solution will contain more particles per pound in a given
solution and therefore, have a greater effect on the freezing point of the radiator solution.
Assume 100. g of each compound.
100: g
¼ 2:84 mol
CH3 OH:
34:04 g=mol
CH3 CH2 OH:
100: g
¼ 2:17 mol
46:07 g=mol
54. A hydronium ion is a hydrated hydrogen ion.
Hþ
þ H2 O !
H3 Oþ
ðhydrogen ionÞ
ðhydronium ionÞ
55. Freezing point depression is directly related to the concentration of particles in the solution.
1 mol
1 þ mol
2 mol
3 mol ðparticles in solutionÞ
C12 H22 O11 > HC2 H3 O2 > HCl > CaCl2
Highest freezing point
Lowest freezing point
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Page 213
- Chapter 15 56. (a)
(b)
(c)
100 C pH ¼ log 1 106 ¼ 6:0
25 C pH ¼ log 1 107 ¼ 7:0 pH of H2 O is greater at 25 C
1 106 > 1 107 so, Hþ concentration is higher at 100 C.
The water is neutral at both temperatures, because the H2 O ionizes into equal concentrations of
Hþ and OH at any temperature.
57. As the pH changes by 1 unit, the concentration of Hþ in solution changes by a factor of 10. For
example, the pH of 0.10 M HCl is 1.00, while the pH of 0.0100 M HCl is 2.00.
58. A 1.00 m solution contains 1 mol solute plus 1000 g H2 O. We need to find the total number of moles
and then calculate the mole percent of each component.
1000 g H2 O
¼ 55:49 mol H2 O
18:02 g=mol
55:49 mol H2 O þ 1:00 mol solute ¼ 56:49 total moles
1:00 mol solute
ð100Þ ¼ 1:77% solute
56:49 mol
55:49 mol H2 O
ð100Þ ¼ 98:23% H2 O
56:49 mol
59. Na2 CO3 þ 2 HCl ! 2 NaCl þ CO2 þ H2 O
g Na2 CO3 ! mol Na2 CO3 ! mol HCl ! M HCl
1 mol
2 mol HCl
1
ð0:452 g Na2 CO3 Þ
¼ 0:201 M HCl
106:0 g 1 mol Na2 CO3 0:0424 L
60. H2 SO4 þ 2 KOH ! K2 SO4 þ 2 H2 O
g KOH ! mol KOH ! mol H2 SO4 ! M H2 SO4
1 mol KOH
1 mol H2 SO4
1000 mL
ð6:38 g KOHÞ
¼ 134 mL of 0:4233 M H2 SO4
56:11 g KOH
0:4233 mol H2 SO4
2 mol KOH
61. KOH þ HNO3 ! KNO3 þ H2 O
L HNO3 ! mol HNO3 ! mol KOH ! g KOH
0:240 mol
1 mol KOH
56:11 g
ð0:05000 L HNO3 Þ
¼ 0:673 g KOH
L
1 mol HNO3
mol
62. pH of 1.0 L solution containing 0.1 mL of 1.0 M HCl
1:0 L
1 mol HCl
¼ 1 104 mol HCl added
ð0:1 mLÞ
1000 mL
L
1 104 mol HCl
¼ 1 104 M HCl
1:0 L
1 104 M HCl produces 1 104 M Hþ
pH ¼ log 1 104 ¼ 4:0
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Page 214
- Chapter 15 63. Dilution problem: V1 M 1 ¼ V2 M 2
M1 ¼
V2 M 2
V1
M1 ¼
ð10:0 mLÞð12 M Þ
¼ 0:462 M HCl
ð260:0 mLÞ
64. NaOH þ HCl ! NaCl þ H2 O
1 mol
ð3:0 g NaOHÞ
¼ 0:075 mol NaOH
40:00 g
1L
0:10 mol
ð500: mL HClÞ
¼ 0:050 mol HCl
1000 mL
L
This solution is basic. The NaOH will neutralize the HCl with an excess of 0.025 mol of NaOH
remaining unreacted.
! BaCl2 ðaqÞ þ 2 H2 Oðl Þ
65. BaðOHÞ2 ðaqÞ þ 2 HClðaqÞ 0:38 L BaðOHÞ2
0:35 mol
¼ 0:13 mol BaðOHÞ2
L
0:13 mol BaðOHÞ2 ! 0:26 mol OH
0:65 mol
ð0:500 L HClÞ
¼ 0:33 mol HCl
L
0:33 mol HCl ! 0:33 mol Hþ
0.33 mol Hþ will neutralize 0.26 mol OH and leave 0.07 mol Hþ ð0:33 0:26Þ remaining in solution.
Total volume ¼ 500: mL þ 380 mL ¼ 880 mLð0:88 LÞ
0:07 mol Hþ
¼ 0:08 M Hþ
0:88 L
pH ¼ log½Hþ ¼ log 8 102 ¼ 1:1
½Hþ in solution ¼
The solution is acidic.
0:2000 mol
¼ 0:01000 mol HCl ¼ 0:01000 mol Hþ in 50:00 mL HCl
66. ð0:05000 L HClÞ
L
(a)
(b)
no base added: pH ¼ logð0:2000Þ ¼ 0:700
0:2000 mol
10:00 mL base added: ð0:01000 LÞ
¼ 0:002000 mol NaOH
L
¼ 0:002000 mol OH
ð0:01000 mol Hþ Þ ð0:002000 mol OH Þ ¼ 0:00800 mol Hþ in 60:00 mL solution
0:00800 mol
0:00800
þ
½H ¼
pH ¼ log
¼ 0:880
0:06000 L
0:06000
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Page 215
- Chapter 15 (c)
25:00 mL base added:
0:2000 mol
ð0:02500 LÞ
¼ 0:005000 mol NaOH ¼ mol OH
L
ð0:01000 mol Hþ Þ ð0:005000 mol OH Þ ¼ 0:00500 mol Hþ in 75:00 solution
0:00500 mol
0:00500
þ
½H ¼
pH ¼ log
¼ 1:2
0:07500 L
0:07500
(d)
49.00 mL base added:
0:2000 mol
¼ 0:009800 mol NaOH ¼ mol OH
ð0:04900 LÞ
L
ð0:01000 mol Hþ Þ ð0:009800 mol OH Þ ¼ 0:00020 mol Hþ in 99:00 mL solution
0:00020 mol
0:00020
þ
½H ¼
pH ¼ log
¼ 2:69
0:09900 L
0:09900
(e)
49.90 mL base added:
0:2000 mol
ð0:04990 LÞ
¼ 0:009980 mol NaOH ¼ mol OH
L
ð0:01000 mol Hþ Þ ð0:009800 mol OH Þ ¼ 2 105 mol Hþ in 99:00 mL solution
2 105 mol
2 105
þ
pH ¼ log
½H ¼
¼ 3:7
0:09990 L
0:09990
(f)
(g)
49.99 mL base added:
0:2000 mol
¼ 0:009998 mol NaOH ¼ mol OH
ð0:04999 LÞ
L
ð0:01000 mol Hþ Þ ð0:009998 mol OH Þ ¼ 2 106 mol Hþ in 99:99 mL solution
2 106
2 106
þ
¼ 4:7
½H ¼
pH ¼ log
0:09999 L
9:999 102
50.00 mL of 0.2000 M NaOH neutralizes 50.00 mL of 0.2000 M HCl. No excess acid or base is in
the solution. Therefore, the solution is neutral with a pH ¼ 7.0
8
7
6
5
pH
C15
4
3
2
1
0
0
10
20
30
mL NaOH
40
50
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Page 216
- Chapter 15 67. (a)
(b)
(c)
2 NaOHðaqÞ þ H2 SO4 ðaqÞ ! Na2 SO4 ðaqÞ þ 2 H2 Oðl Þ
mol H2 SO4 ! mol NaOH ! mL NaOH
2 mol NaOH
1000 mL
ð0:0050 mol H2 SO4 Þ
¼ 1:0 102 mL NaOH
1 mol H2 SO4
0:10 mol
1 mol Na2 SO4
142:1 g
ð0:0050 mol H2 SO4 Þ
¼ 0:71 g Na2 SO4
mol
1 mol H2 SO4
68. HNO3 þ KOH ! KNO3 þ H2 O
M A VA ¼ M B VB
ðM A Þð25 mLÞ ¼ ð0:60 M Þð50:0 mLÞ
M A ¼ 1:2 M ðdiluted solutionÞ
Dilution problem M 1 V1 ¼ M 2 V2
ðM A Þð10:0 mLÞ ¼ ð1:2 M Þð100:00 mLÞ
M A ¼ 12 M HNO3 ðoriginal solutionÞ
69. Yes, adding water changes the concentration of the acid, which changes the concentration of the ½Hþ ,
and changes the pH. The pH will rise.
No, the solution theoretically will never reach a pH of 7, but it will approach pH 7 as water is added.
70. H2 SO4 þ 2 NaOH ! Na2 SO4 þ 2 H2 O
0:94 mol H2 SO4
ð0:425 L H2 SO4 Þ
¼ 0:40 mol H2 SO4
L
0:40 mol H2 SO4 ! 0:80 mol Hþ
0:83 mol NaOH
ð0:750 L NaOHÞ
¼ 0:62 mol NaOH
L
0:62 mol NaOH ! 0:62 mol OH
0.80 mol Hþ will neutralize 0.62 mol OH and leave 0.18 mol (0.80 0.62) of Hþ remaining in solution;
so the solution will be acidic.
Total volume ¼ 425 mL þ 750 mL ¼ 1175 mL (1.175 L)
½Hþ in solution ¼
0:18 mol Hþ
¼ 0:15 M Hþ
1:175 L
pH ¼ log ½Hþ ¼ log ð0:15Þ ¼ 0:82
71. (a)
1st determine kind of substance
Copper(II) sulfate is a soluble salt so it will dissociate completely in water
2nd write the dissociation/ionization equation
2
H2 O
2þ
CuSO4 ðaqÞ
! Cu ðaqÞ þ SO4 ðaqÞ
3rd Analyze the equation to determine number of ions formed.
Each CuSO4 will produce 1 Cu2þ ion and 1 SO42 ion, so [Cu2þ] ¼ [SO42] ¼ 1 M
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Page 217
- Chapter 15 (b)
1st determine kind of substance
Nitric acid is a strong acid so it will ionize completely in water
2nd write the dissociation/ionization equation
HNO3 ðaqÞ2! Hþ ðaqÞ þ NO3 ðaqÞ
H O
3rd Analyze the equation to determine number of ions formed.
Each HNO3 will produce 1 Hþ ion and 1 NO3 ion, so [Hþ] ¼ [NO3] ¼ 1 M
(c)
1st determine kind of substance
Sulfuric acid is a strong acid so it will ionize completely in water
2nd write the dissociation/ionization equation
H2 SO4 ðaqÞ2! 2 Hþ ðaqÞ þ SO4 2 ðaqÞ
H O
3rd Analyze the equation to determine number of ions formed.
Each H2SO4 will produce 2 Hþ ions and 1 SO42 ion, so [Hþ] ¼ 2 M and [SO42] ¼ 1 M
(d)
1st determine kind of substance
Calcium sulfide is an insoluble salt so it will not dissociate in water.
2nd write the dissociation/ionization equation
H O
CaSðsÞ2!no reaction
3rd Analyze the equation to determine number of ions formed.
CaS is insoluble so very few of the particles will dissociate and the concentration of all ions will
be close to 0 M.
(e)
1st determine kind of substance
Acetic acid is a weak acid so it will ionize slightly in water
2nd write the dissociation/ionization equation
H2 O
HC2 H3 O2 ðaqÞÐ Hþ ðaqÞ þ C2 H3 O2 ðaqÞ
3rd Analyze the equation to determine number of ions formed.
Each HC2H3O2 will produce some Hþ ions and C2H3O2 ions, so [Hþ] and [C2H3O2] will be
between 0 and 1 M.
72. (a)
(b)
Formula equation
HNO3 ðaqÞ þ LiOHðaqÞ ! H2 OðlÞ þ LiNO3 ðaqÞ
Total ionic equation Hþ ðaqÞ þ NO3 ðaqÞ þ Liþ ðaqÞ þ OH ðaqÞ !
H2 OðlÞ þ Liþ ðaqÞ þ NO3 ðaqÞ
þ
Net ionic equation
H ðaqÞ þ OH ðaqÞ ! H2 OðlÞ
Formula equation
2 HBrðaqÞ þ BaðOHÞ2 ðaqÞ ! 2 H2 OðlÞ þ BaBr2 ðaqÞ
Total ionic equation. 2 Hþ ðaqÞ þ 2Br ðaqÞ þ Ba2þ ðaqÞ þ 2 OH ðaqÞ !
2 H2 OðlÞ þ Ba2þ ðaqÞ þ 2 Br ðaqÞ
þ
Net ionic equation
H ðaqÞ þ OH ðaqÞ ! H2 OðlÞ
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Page 218
- Chapter 15 (c)
Formula equation
HFðaqÞ þ NaOHðaqÞ ! H2 OðlÞ þ NaFðaqÞ
Total ionic equation. HFðaqÞ þ Naþ ðaqÞ þ OH ðaqÞ ! H2 OðlÞ þ Naþ ðaqÞ þ F ðaqÞ
(Note that HF is a weak acid and so it does not ionize to any appreciable extent and is written in its
unionized form.)
Net ionic equation
HFðaqÞ þ OH ðaqÞ ! H2 OðlÞ þ F ðaqÞ
73. 1st write a balanced chemical equation for the reaction of the Hþ in the rain with sodium hydroxide
Hþ ðaqÞ þ NaOHðaqÞ ! Naþ ðaqÞ þ H2 OðlÞ
2nd calculate moles of sodium hydroxide needed to react with all of the Hþ in the rain.
1 L NaOH
0:125 mol NaOH
mol NaOH ¼ ð7:2 mL NaOHÞ
¼ 0:00090 mol NaOH
1000 mL NaOH
1 L NaOH
3rd determine moles of Hþ
1 mol Hþ
mol Hþ ¼ ð0:00090 mol NaOHÞ
¼ 0:00090 mol Hþ
1 mol NaOH
4th determine molarity of Hþ
M Hþ ¼
mol Hþ 0:00090 mol Hþ
¼
¼ 3:6 105 M Hþ
L rain
25 L rain
5th determine pH of rain
pH ¼ log ½Hþ ¼ log 3:6 105 M Hþ ¼ 4:44
74. [Hþ] needs to be 0.00158 M to give the final solution of sodium rhidizonate a pH of 2.800.
V 1M1 ¼ V 2M2
ð500:0 mLÞð0:00158 M Þ ¼ V 2 ð0:60 M Þ
V2 ¼
ð500:0 mLÞð0:00158 M Þ
¼ 1:3 mL 0:60 M HCl
ð0:60 M Þ
To make the solution put 1.3 mL of 0.60 M HCl in a graduated cylinder and fill the cylinder to the
500 mL mark with sodium rhidizonate solution.
75. V 1 M 1 ¼ V 2 M 2
ð3:00 LÞð3:25 M Þ ¼ V 2 ð12:1 M Þ
V2 ¼
ð3:00 LÞð3:25 M Þ
¼ 0:806 L 12:1 M HCl or 806 mL 12:1 M HCl
ð12:1 M Þ
76. CaCO3 ðsÞ þ 2 HClðaqÞ ! CaCl2 ðaqÞ þ H2 OðlÞ þ CO2 ðgÞ
77. First determine the molarity of the two HCl solutions. Take the antilog of the pH value to obtain the ½Hþ .
pH ¼ 0:300;
Hþ ¼ 2:00 M ¼ 2:00 M HCl
pH ¼ 0:150;
Hþ ¼ 1:41 M ¼ 1:41 M HCl
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C15
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Page 219
- Chapter 15 Now treat the calculation as a dilution problem.
V1 M 1
V1 M 1 ¼ V2 M 2
V2 ¼
M2
ð200 mL HClÞð2:00 M Þ
¼ 284 mL solution
1:41 M
284 mL 200 mL ¼ 84 mL H2 O to be added
78. mol acid ¼ mol base
(lactic acid has one acidic H)
1:0 g acid
0:65 mol NaOH
¼ ð0:017 LÞ
¼ 0:011 mol NaOH
molar mass
L
mol acid ¼ mol base
mol HC3 H5 O3 ¼ 0:011 mol
1:0 g
¼ 91 g=mol ¼ molar mass
0:011 mol
molar mass (91 g/mol) = mass of empirical formula (90.08 g/mol)
Therefore the molecular formula, HC3 H5 O3 , is the same as the empirical formula.
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