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C15 09/16/2013 13:25:32 Page 197
C H A P T E R 1 5
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: HClþ NaOH�����! NaClþ H2O Hþ þ OH� �����! H2Oð ÞBrønsted-Lowry: HClþ KCN�����! HCNþ KCl Hþ þ CN� �����! HCNð ÞLewis: AlCl3 þ NaCl�����! AlCl �
4 þ Naþ
Cl–
ClAl Cl +
–Cl
Cl
ClAl ClCl
4. (a)–
Br (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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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 watermolecule is directed towards the Naþ. Second, more water molecules will fit around Cl�, since it islarger than the Naþ ion.
12. The electrolytic compounds are acids, bases, and salts.
13. Names of the compounds in Table 15.3
H2SO4 sulfuric acid HC2H3O2 acetic acid
HNO3 nitric acid H2CO3 carbonic acid
HCl hydrochloric acid HNO2 nitrous acid
HBr hydrobromic acid H2SO3 sulfurous acid
HClO4 perchloric acid H2S hydrosulfuric acid
NaOH sodium hydroxide H2C2O4 oxalic acid
KOH potassium hydroxide H3BO3 boric acid
Ca OHð Þ2 calcium hydroxide HClO hypochlorous acid
Ba OHð Þ2 barium hydroxide 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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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 H2O molecules.
HClþ H2O�����! H3Oþ þ Cl�
23. The pH for a solution with a hydrogen ion concentration of 0.003M will be between 2 and 3.
24. Tomato juice is more acidic than blood, since its pH is lower.
25. (a) In a neutral solution, the concentration of Hþ and OH� are equal.
(b) In an acid solution, the concentration of Hþ is greater than the concentration of OH�.(c) 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] andbase [OH�] ions.
27. A neutral solution is one in which the concentration of acid is equal to the concentration of base
Hþ½ � ¼ OH�½ �. An acidic solution is one in which the concentration of acid is greater than theconcentration of base Hþ½ � > OH�½ �. A basic solution is one in which the concentration of base is
greater than the concentration of acid Hþ½ � < OH�½ �.
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:
Hþ þ OH� �����! H2O
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 H2SO4ð Þ and nitric acidHNO3ð Þ. Precipitation (rain or snow) carries the acids to the ground.
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SOLUTIONS TO EXERCISES
1. Conjugate acid – base pairs:
(a) NH3 � NHþ4 ; H2O� OH�
(b) HC2H3O2 � C2H3O�2 ;H2O� H3O
þ
(c) H2PO�4 � HPO2�
4 ;OH� � H2O
(d) HCl� Cl�;H2O� H3Oþ
2. Conjugate acid – base pairs:
(a) H2S� HS�;NH3 � NH þ4
(b) HSO�4 � SO2�
4 ;NH3 � NH þ4
(c) HBr� Br�; CH3O� � CH3OH
(d) HNO3 � NO �3 ;H2O� H3O
þ
3. (a) Zn sð Þ þ 2 HCl aqð Þ ! ZnCl2 aqð Þ þ H2 gð Þ(b) Al OHð Þ3 sð Þ þ 3 H2SO4 aqð Þ ! Al2 SO4ð Þ3 aqð Þ þ 6 H2O lð Þ(c) Na2CO3 aqð Þ þ 2 HC2H3O2 aqð Þ ! 2 NaC2H3O2 aqð Þ þ H2O lð Þ þ CO2 gð Þ(d) MgO sð Þ þ 2 HI aqð Þ ! MgI2 aqð Þ þ H2O lð Þ(e) Ca HCO3ð Þ2 sð Þ þ 2 HBr aqð Þ ! CaBr2 aqð Þ þ 2 H2O lð Þ þ 2 CO2 gð Þ(f) 3 KOH aqð Þ þ H3PO4 aqð Þ ! K3PO4 aqð Þ þ 3 H2O lð Þ
4. Complete and balance these equations:
(a) Fe2O3 sð Þ þ 6 HBr aqð Þ ! 2 FeBr3 aqð Þ þ 3 H2O lð Þ(b) 2 Al sð Þ þ 3 H2SO4 aqð Þ ! Al2 SO4ð Þ3 aqð Þ þ 3 H2 gð Þ(c) 2 NaOH aqð Þ þ H2CO3 aqð Þ ! Na2CO3 aqð Þ þ 2 H2O lð Þ(d) Ba OHð Þ2 sð Þ þ 2 HClO4 aqð Þ ! Ba ClO4ð Þ2 aqð Þ þ 2 H2O lð Þ(e) Mg sð Þ þ 2 HClO4 aqð Þ ! Mg ClO4ð Þ2 aqð Þ þ H2 gð Þ(f) K2O sð Þ þ 2 HI aqð Þ ! 2 KI aqð Þ þ H2O lð Þ
5. (a) Znþ 2 Hþ þ 2 Cl�ð Þ ! Zn2þ þ 2 Cl�ð Þ þ H2
Znþ 2 Hþ ! Zn2þ þ H2
(b) 2 Al OHð Þ3 þ 6 Hþ þ 3 SO2�4
� � ! 2 Al3þ þ 3 SO2�4
� �þ 6 H2O
Al OHð Þ3 þ 3 Hþ ! Al3þ þ 3 H2O
(c) 2 Naþ þ CO2�3
� �þ 2 HC2H3O2 ! 2 Naþ þ 2 C2H3O�2
� �þ H2Oþ CO2
CO2�3 þ 2 HC2H3O2 ! 2 C2H3O
�2 þ H2Oþ CO2
(d) MgOþ 2 Hþ þ 2 I�ð Þ ! Mg2þ þ 2 I�ð Þ þ H2O
MgOþ 2 Hþ ! Mg2þ þ H2O
(e) Ca HCO3ð Þ2 þ 2 Hþ þ 2 Br�ð Þ ! Ca2þ þ 2 Br�ð Þ þ 2 H2Oþ 2 CO2
Ca HCO3ð Þ2 þ 2 Hþ ! Ca2þ þ H2Oþ CO2
(f) 3 Kþ þ 3 OH�ð Þ þ H3PO4 ! 3 Kþ þ PO3�4
� �þ 3 H2O
3 OH� þ H3PO4 ! PO3�4 þ 3 H2O
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6. (a) Fe2O3 þ 6 Hþ þ 6 Br�ð Þ ! 2 Fe3þ þ 6 Br�ð Þ þ 3 H2O
Fe2O3 þ 6 Hþ ! 2 Fe3þ þ 3 H2O
(b) 2 Alþ 6Hþ ! 3 SO2�4
� � ! 2 Al3þ þ 3 SO2�4
� �þ 3 H2
2 Alþ 6 Hþ ! 2 Al3þ þ 3 H2
(c) 2 Naþ þ 2 OH�ð Þ þ H2CO3 ! 2 Naþ þ CO2�3
� �þ 2 H2O
2 OH� þ H2CO3 ! CO2�3 þ 2 H2O
(d) Ba OHð Þ2 þ 2 Hþ þ 2 ClO�4
� � ! Ba2þ þ 2 ClO�4
� �þ 2 H2O
Ba OHð Þ2 þ 2 Hþ ! Ba2þ þ 2 H2O
(e) Mgþ 2 Hþ þ 2 ClO�4
� � ! Mg2þ þ 2 ClO�4
� �þ H2
Mgþ 2 Hþ ! Mg2þ þ H2
(f) K2Oþ 2 Hþ þ 2 I�ð Þ ! 2 Kþ þ 2 I�ð Þ þ H2O
K2Oþ 2 Hþ ! 2 Kþ þ H2O
7. (a) HNO3 þ NaOH ! H2Oþ NaNO3
(b) 2 HC2H3O2 þ BaðOHÞ2 ! 2 H2Oþ BaðC2H3O2Þ2(c) HClO4 þ NH4OH ! H2Oþ NH4ClO4
8. (a) 2 HBrþMgðOHÞ2 ! 2 H2OþMgBr2(b) H3PO4 þ 3 KOH ! 3 H2Oþ K3PO4
(c) H2SO4 þ 2 NH4OH ! 2 H2Oþ ðNH4Þ2SO4
9. LiOH and H2S must be reacted
2 LiOHþ H2S ! 2 H2Oþ Li2S
10. Ca(OH)2 and H2CO3 must be reacted
CaðOHÞ2 þ H2CO3 ! 2 H2Oþ CaCO3
11. The following compounds are electrolytes:
(a) SO3, acid in water (e) CuBr2, salt
(b) K2CO3, salt (f) HI, acid in water
12. The following compounds are electrolytes:
(b) P2O5, acid in water (d) LiOH, base
(c) NaClO, salt (f) KMnO4, salt
13. Molarity of ions.
(a) 1:25M CuBr2ð Þ 1 mol Cu2þ
1 mol CuBr2
� �¼ 1:25M Cu2þ
1:25M CuBr2ð Þ 2 mol Br�
1 mol CuBr2
� �¼ 2:50M Br�
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(b) 0:75M NaHCO3ð Þ 1 mol Naþ
1 mol NaHCO3
� �¼ 0:75M Naþ
0:75M NaHCO3ð Þ 1 mol HCO�3
1 mol NaHCO3
� �¼ 0:75M HCO�
3
(c) 3:50M K3AsO4ð Þ 3 mol Kþ
1 mol K3AsO4
� �¼ 10:5M Kþ
3:50M K3AsO4ð Þ 1 mol AsO3�4
1 mol K3AsO4
� �¼ 3:50M AsO3�
4
(d) 0:65M NH4ð Þ2SO4
� � 2 mol NHþ4
1 mol NH4ð Þ2SO4
� �¼ 1:3M NHþ
4
0:65M NH4ð Þ2SO4
� � 1 mol SO2�4
1 mol NH4ð Þ2SO4
� �¼ 0:65M SO2�
4
14. Molarity of ions.
(a) 2:25M FeCl3ð Þ 1 mol Fe3þ
1 mol FeCl3
� �¼ 2:25M Fe3þ
2:25M FeCl3ð Þ 3 mol Cl�
1 mol FeCl3
� �¼ 6:75M Cl�
(b) 1:20MMgSO4ð Þ 1 molMg2þ
1 mol MgSO4
� �¼ 1:20MMg2þ
1:20MMgSO4ð Þ 1 mol SO2�4
1 mol MgSO4
� �¼ 1:20M SO2�
4
(c) 0:75M NaH2PO4ð Þ 1 mol Naþ
1 mol NaH2PO4
� �¼ 0:75M Naþ
0:75M NaH2PO4ð Þ 1 mol H2PO�4
1 mol NaH2PO4
� �¼ 0:75M H2PO
�4
(d) 0:35M Ca ClO3ð Þ2� � 1 mol Ca2þ
1 mol Ca ClO3ð Þ2
� �¼ 0:35M Ca2þ
0:35M Ca ClO3ð Þ2� � 2 mol ClO�
3
1 mol Ca ClO3ð Þ2
� �¼ 0:70M ClO�
3
15. We will use the data from No. 13 to solve these problems. 100 mL = 0.100 L
(a) ð0:100 LÞ 1:25 mol Cu2þ
L
� �63:55 g
mol
� �¼ 7:94 g Cu2þ
ð0:100 LÞ 2:50 mol Br�
L
� �79:90 g
mol
� �¼ 20:0 g Br�
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(b) ð0:100 LÞ 0:75 mol Naþ
L
� �22:99 g
mol
� �¼ 1:7 g Naþ
ð0:100 LÞ 0:75 mol HCO�3
L
� �61:02 g
mol
� �¼ 4:6 g HCO�
3
(c) ð0:100 LÞ 10:5 mol Kþ
L
� �39:10 g
mol
� �¼ 41:1 g Kþ
ð0:100 LÞ 3:50 mol AsO3�4
L
� �138:9 g
mol
� �¼ 48:6 g AsO3�
4
(d) ð0:100 LÞ 1:3 mol NHþ4
L
� �18:04 g
mol
� �¼ 2:3 g NHþ
4
0:100 Lð Þ 0:65 mol SO2�4
1 L
� �96:07 g
1 mol
� �¼ 6:2 g SO2�
4
16. We will use the data from No. 14 to solve these problems 100 mL = 0.100 L
(a) ð0:100 LÞ 2:25 mol Fe3þ
L
� �55:85 g
mol
� �¼ 12:6 g Fe3þ
ð0:100 LÞ 6:75 mol Cl�
L
� �35:45 g
mol
� �¼ 23:9 g Cl�
(b) ð0:100 LÞ 1:20 mol Mg2þ
L
� �24:31 g
mol
� �¼ 2:92 gMg2þ
ð0:100 LÞ 1:20 mol SO2�4
L
� �96:07 g
mol
� �¼ 11:5 g SO2�
4
(c) ð0:100 LÞ 0:75 mol Naþ
L
� �22:99 g
mol
� �¼ 1:7 g Naþ
ð0:100 LÞ 0:75 mol H2PO�4
L
� �96:99 g H2PO
�4
mol
� �¼ 7:3 g H2PO
�4
(d) ð0:100 LÞ 0:35 mol Ca2þ
L
� �40:08 g
mol
� �¼ 1:4 g Ca2þ
ð0:100 LÞ 0:70 mol ClO�3
L
� �83:45 g
mol
� �¼ 5:8 g ClO�
3
17. pH ¼ �log Hþ½ � Hþ½ � ¼ 10�pH
(a) Hþ½ � ¼ 1� 10�5
(b) Hþ½ � ¼ 2� 10�7
(c) Hþ½ � ¼ 1� 10�8
(d) Hþ½ � ¼ 2� 10�10
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18. pH ¼ �log Hþ½ � Hþ½ � ¼ 10�pH
(a) Hþ½ � ¼ 1� 10�7
(b) Hþ½ � ¼ 5� 10�5
(c) Hþ½ � ¼ 2� 10�6
(d) Hþ½ � ¼ 5� 10�11
19. (a) 55:5 mLð Þ 0:50 mol HCl
1000 mL
� �¼ 0:028 mol HCl
75:0 mLð Þ 1:25 mol HCl
1000 mL
� �¼ 0:0938 mol HCl
Total mol HCl ¼ 0:028 molþ 0:0938 mol ¼ 0:122 mol HCl
Total volume ¼ 0:0555 Lþ 0:0750 L ¼ 0:1305 L
0:122 mol HCl
0:1305 L¼ 0:935M HCl
0:935M HClð Þ 1 mol Hþ
1 mol HCl
� �¼ 0:935M Hþ
0:935M HClð Þ 1 mol Cl�
1 mol HCl
� �¼ 0:935M Cl�
(b) 125 mLð Þ 0:75 mol CaCl2
1000 mL
� �¼ 0:094 mol CaCl2
125 mLð Þ 0:25 mol CaCl2
1000 mL
� �¼ 0:031 mol CaCl2
Total mol CaCl2 ¼ 0:094 molþ 0:031 mol ¼ 0:125 mol CaCl2
Total volume ¼ 0:125 Lþ 0:125 L ¼ 0:250 L
0:125 mol CaCl2
0:250 L¼ 0:500M CaCl2
0:500M CaCl2ð Þ 1 mol Ca2þ
1 mol CaCl2
� �¼ 0:500M Ca2þ
0:500M CaCl2ð Þ 2 mol Cl�
1 mol CaCl2
� �¼ 1:00M Cl�
(c) NaOHþ HCl ! NaClþ H2O
35:0 mLð Þ 0:333 mol NaOH
1000mL
� �¼ 0:0117 mol NaOH
22:5 mLð Þ 0:250 mol HCl
1000 mL
� �¼ 0:00563 mol HCl
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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0.00563 mol NaCl. Moles of ions are: (0.0061 mol Naþþ 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
0:0575 L¼ 0:203M Naþ
0:0061 mol OH�
0:0575 L¼ 0:11M OH�
0:00563 mol Cl�
0:0575 L¼ 0:0979M Cl�
(d) H2SO4 þ 2 NaOH ! Na2SO4 þ 2 H2O
12:5 mLð Þ 0:500 mol H2SO4
1000 mL
� �¼ 0:00625 mol H2SO4
23:5 mLð Þ 0:175 mol NaOH
1000mL
� �¼ 0:00411 mol NaOH
0:00411 mol NaOHð Þ 1 mol H2SO4
2 mol NaOH
� �¼ 0:00206 mol H2SO4 reacted
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(0.00206þ 0.00419)¼ 0.00625 mol SO2�
4 . Concentration of ions are:
0:0412 mol Naþ
0:0360 L¼ 0:114M Naþ
0:00838 mol Hþ
0:0360 L¼ 0:233M Hþ
0:00625 mol SO2�4
0:0360 L¼ 0:174M SO2�
4
20. (a) 45:5 mLð Þ 0:10 mol NaCl
1000 mL
� �¼ 0:0046 mol NaCl
60:5 mLð Þ 0:35 mol NaCl
1000 mL
� �¼ 0:021 mol NaCl
Total mol NaCl ¼ 0:0046 molþ 0:021 mol ¼ 0:026 mol NaCl
Total volume ¼ 0:0455 Lþ 0:0605 L ¼ 0:1060 L
0:026 mol NaCl
0:1060 L¼ 0:25M NaCl
0:25M NaClð Þ 1 mol Naþ
1 mol NaCl
� �¼ 0:25M Naþ
0:25M NaClð Þ 1 mol Cl�
1 mol NaCl
� �¼ 0:25M Cl�
(b) 95:5 mLð Þ 1:25 mol HCl
1000 mL
� �¼ 0:119 mol HCl
125:5 mLð Þ 2:50 mol HCl
1000 mL
� �¼ 0:314 mol HCl
Total mol HCl ¼ 0:119 molþ 0:314 mol ¼ 0:433 mol HCl
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Total volume ¼ 0:0955 Lþ 0:1255 L ¼ 0:2210 L
0:433 mol HCl
0:2210 L¼ 1:96M HCl
1:96M HClð Þ 1 mol Hþ
1 mol HCl
� �¼ 1:96M Hþ
1:96M HClð Þ 1 mol Cl�
1 mol HCl
� �¼ 1:96M Cl�
(c) 15:5 mLð Þ 0:10 mol Ba NO3ð Þ21000 mL
� �¼ 0:0016M Ba NO3ð Þ2
10:5 mLð Þ 0:20 mol AgNO3
1000 mL
� �¼ 0:0021M AgNO3
Number of moles of each substance: 0:0016 mol Ba2þ, 0:0021 mol Agþ, and0:0032 molþ 0:0021 molð Þ ¼ 0:0053 mol NO�
3
Total volume ¼ 0:0155 Lþ 0:0105 L ¼ 0:0260 L
0:0016 mol Ba2þ
0:0260 L¼ 0:062M Ba2þ
0:0021 mol Agþ
0:0260 L¼ 0:081M Agþ
0:0053 mol NO�3
0:0260 L¼ 0:20M NO�
3
(d) 25:5 mLð Þ 0:25 mol NaCl
1000 mL
� �¼ 0:0064 mol NaCl
15:5 mLð Þ 0:15 mol Ca C2H3O2ð Þ21000 mL
� �¼ 0:0023 mol Ca C2H3O2ð Þ2
Number of moles of each substance: 0:0064 mol Naþ, 0:0064 mol Cl�, 0:0023 mol Ca2þ,0:0046 mol C2H3O
�2 .
Total volume ¼ 0:0255 Lþ 0:0155 L ¼ 0:0410 L
0:0064 mol Naþ
0:0410 L¼ 0:16M Naþ
0:0064 mol Cl�
0:0410 L¼ 0:16M Cl�
0:0023 mol Ca2þ
0:0410 L¼ 0:056M Ca2þ
0:0046 mol C2H3O�2
0:0410 L¼ 0:11M C2H3O
�2
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21. HNO3ðaqÞ þ H2OðlÞ ! H3OþðaqÞ þ NO3
�ðaqÞOr
HNO3ðaqÞ ��!H2O HþðaqÞ þ NO3�ðaqÞ
Because nitric acid ionizes completely it would be both a strong electrolyte and a strong acid.
22. HCNðaqÞ þ H2OðlÞÐH3OþðaqÞ þ CN�ðaqÞ
Or
HCNðaqÞÐH2O
HþðaqÞ þ CN�ðaqÞHCN 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.
HClþ NaOH�����! NaClþ H2O
At the endpoint in these titration reactions, equal moles of HCl and NaOH will have reacted.
Moles¼ (molarity) (volume). At the endpoint, mol HCl¼mol NaOH.
Therefore, at the endpoint,
MAVA ¼ MBVB MA ¼ MBVB
VA
(a)37:70 mLð Þ 0:728Mð Þ
40:3 mL¼ 0:681M HCl
(b)33:66 mLð Þ 0:306Mð Þ
19:00 mL¼ 0:542M HCl
(c)18:00 mLð Þ 0:555Mð Þ
27:25 mL¼ 0:367M HCl
24. The reaction of HCl and NaOH occurs on a 1:1 mole ratio.
HClþ NaOH ! NaClþ H2O
At the endpoint in these titration reactions, equal moles of HCl and NaOH will have reacted.
Moles¼ (molarity)(volume). At the endpoint, mol HCl¼mol NaOH.
Therefore, at the endpoint,
MAVA ¼ MBVB MB ¼ MAVA
VB
(a)37:19 mLð Þ 0:126Mð Þ
31:91 mL¼ 0:147M NaOH
(b)48:04 mLð Þ 0:482Mð Þ
24:02 mL¼ 0:964M NaOH
(c)13:13 mLð Þ 1:425Mð Þ
39:39 mL¼ 0:4750M NaOH
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25. (a) 2 PO3�4 aqð Þ þ 3 Ca2þ aqð Þ ! Ca3 PO3�
4
� �2sð Þ
(b) 2 Al sð Þ þ 6 Hþ aqð Þ ! 3 H2 gð Þ þ 2 Al3þ aqð Þ(c) CO2�
3 aqð Þ þ 2 Hþ aqð Þ ! H2O aqð Þ þ CO2 gð Þ
26. (a) Mg sð Þ þ Cu2þ aqð Þ ! Cu sð Þ þMg2þ aqð Þ(b) Hþ aqð Þ þ OH� aqð Þ ! H2O lð Þ(c) SO2�
3 aqð Þ þ 2 Hþ aqð Þ ! H2O lð Þ þ SO2 gð Þ27. (a) 1 molar H2SO4 is more acidic. The concentration of Hþ in 1M H2SO4 is greater than 1M since
there are two ionizable hydrogens per mole of H2SO4. In HCl the concentration of Hþ will be 1M,
since there is only one ionizable hydrogen per mole HCl.
(b) 1 molar HCl is more acidic. HCl is a strong electrolyte, producing more Hþ than HC2H3O2 which
is a weak electrolyte.
28. (a) 2 molar HCl is more acidic. 2M HCl will yield 2M Hþ concentration. 1M HCl will yield 1M Hþ
concentration.
(b) 1 molar H2SO4 is more acidic. Both are strong acids. The concentration of Hþ in 1M H2SO4 is
greater than in 1M HNO3 because H2SO4 has two ionizable hydrogens per mole whereas HNO3
has only one ionizable hydrogen per mole.
29. 2 HClO4 aqð Þ þ Ca OHð Þ2 sð Þ ! Ca ClO4ð Þ2 aqð Þ þ 2 H2O lð Þg Ca OHð Þ2 ! mol Ca OHð Þ2 ! mol HClO4 ! mLHClO4
50:25 g Ca OHð Þ2� � mol
74:10 g
� �2 mol HClO4
1 mol Ca OHð Þ2
� �1000 mL
0:525 mol
� �¼ 2:58� 103 mL HClO4
30. 3 HCl aqð Þ þ Al OHð Þ3 sð Þ ! AlCl3 aqð Þ þ 3 H2O lð ÞmLHCl ! mol HCl ! mol Al OHð Þ3 ! g Al OHð Þ3275 mLHClð Þ 0:125 mol
1000 mL
� �1 mol Al OHð Þ33 mol HCl
� �78:00 g
mol
� �¼ 0:894 g Al OHð Þ3
31. NaOHþ HCl ! NaClþ H2O
First calculate the grams of NaOH in the sample.
L HCl ! mol HCl ! mol NaOH ! g NaOH
0:01825 L HClð Þ 0:2406 mol
L
� �1 mol NaOH
1mol HCl
� �40:00 g
mol
� �¼ 0:1756 g NaOH in the sample
0:1756 g NaOH
0:200 g sample
� �100ð Þ ¼ 87:8%NaOH
32. NaOHþ HCl ! NaClþ H2O
LHCl ! mol HCl ! mol NaOH ! g NaOH
0:04990 L HClð Þ 0:466 mol
L
� �1 mol NaOH
1mol HCl
� �40:00 g
mol
� �¼ 0:930 g NaOH in the sample
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1:00 g sample� 0:930 g NaOH ¼ 0:070 g NaCl in the sample
0:070 g NaCl
1:00 g sample
� �100ð Þ ¼ 7:0%NaCl in the sample
33. Znþ 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.
g Zn ! mol Zn 5:00 g Znð Þ 1 mol
65:39 g
� �¼ 0:0765 mol Zn
0:100 L HClð Þ 0:350 mol
L
� �¼ 0:0350 mol HCl
Therefore Zn is in excess and HCl is the limiting reactant.
0:0350 mol HClð Þ 1 mol H2
2 mol HCl
� �¼ 0:0175 mol H2 produced in the reaction
T ¼ 27�C ¼ 300:K P ¼ 700: torrð Þ 1 atm
760 torr
� �¼ 0:921 atm
PV ¼ nRT
V ¼ nRT
P¼ 0:0175 mol H2ð Þ 0:0821 L atm=mol Kð Þ 300:Kð Þ
0:921 atm¼ 0:468 L H2
34. Znþ 2 HCl ! ZnCl2 þ H2
This is a limiting reactant problem. First find moles of Zn and HCl from the given data and then identify
the limiting reactant.
g Zn ! mol Zn 5:00 g Znð Þ 1 mol
65:39 g
� �¼ 0:0765 mol Zn
0:200 L HClð Þ 0:350 mol
L
� �¼ 0:0700 mol HCl
Zn is in excess and HCl is the limiting reactant.
0:0700 mol HClð Þ 1 mol H2
2 mol HCl
� �¼ 0:0350 mol H2
T ¼ 27�C ¼ 300:K P ¼ 700: torrð Þ 1 atm
760 torr
� �¼ 0:921 atm
PV ¼ nRT
V ¼ nRT
P¼ 0:0350 mol H2ð Þ 0:0821 L atm=mol Kð Þ 300:Kð Þ
0:921 atm¼ 0:936 L H2
35. pH ¼ �log Hþ½ �(a) Hþ½ � ¼ 0:35M; pH ¼ �log 0:35ð Þ ¼ 0:46(b) Hþ½ � ¼ 1:75M; pH ¼ �log 1:75ð Þ ¼ �0:243(c) Hþ½ � ¼ 2:0� 10�5 M; pH ¼ �log 2:0� 10�5
� � ¼ 4:70
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36. pH ¼ �log Hþ½ �(a) Hþ½ � ¼ 0:0020M; pH ¼ �log 0:0020ð Þ ¼ 2:70(b) Hþ½ � ¼ 7:0� 10�8 M; pH ¼ �log 7:0� 10�8
� � ¼ 7:15
(c) Hþ½ � ¼ 3:0M; pH ¼ �log 3:0ð Þ ¼ �0:48
37. (a) Orange juice ¼ 3:7� 10�4 M Hþ
pH ¼ �log 3:7� 10�4� � ¼ 3:43
(b) Vinegar ¼ 2:8� 10�3 M Hþ
pH ¼ �log 2:8� 10�3� � ¼ 2:55
(c) shampoo ¼ 2.4 � 10�6 M Hþ
pH ¼ �log 2:4� 10�6� � ¼ 5:62
(d) dishwashing detergent ¼ 3.6 � 10�8 M Hþ
pH ¼ �log 3:6� 10�8� � ¼ 7:44
38. (a) Black coffee ¼ 5:0� 10�5 M Hþ
pH ¼ �log 5:0� 10�5� � ¼ 4:30
(b) Limewater ¼ 3:4� 10�11 M Hþ
pH ¼ �log 3:4� 10�11� � ¼ 10:47
(c) fruit punch ¼ 2.1 � 10�4 M Hþ
pH ¼ �log 2:1� 10�4� � ¼ 3:68
(d) cranberry apple drink = 1.3 � 10�3 M Hþ
pH ¼ �log 1:3� 10�3� � ¼ 2:89
39. (a) NH3 is a weak base NH3 aqð Þ ÐH2 O
NHþ4 aqð Þ þ OH� aqð Þ
(b) HCl is a strong acid HCl aqð Þ�!H2 OHþ aqð Þ þ Cl� aqð Þ
(c) KOH is a strong base KOH�!H2 OKþ aqð Þ þ OH� aqð Þ
(d) HC2H3O2 is a weak acid HC2H3O2 aqð Þ ÐH2 O
Hþ aqð Þ þ C2H3O�2 aqð Þ
40. (a) H2C2O4 is a weak acid H2C2O4 aqð Þ ÐH2 O
Hþ aqð Þ þ HC2O�4 aqð Þ
(b) Ba OHð Þ2 is a strong base Ba OHð Þ2 �!H2 O
Ba2þ aqð Þ þ 2 OH� aqð Þ(c) HClO4 is a strong acid HClO4 aqð Þ�!H2 O
Hþ aqð Þ þ ClO�4 aqð Þ
(d) HBr is a strong acid HBr aqð Þ ! Hþ aqð Þ þ Br� aqð Þ
41. (a) CH3NH2base
þHþ ! CH3NH3þ
conjugate acid
Note that a neutral base forms a positively charged conjugate acid.
(b) HS�base
þHþ ! H2Sconjugate acid
Note that a negatively charged base forms a neutral acid upon adding a positively charged
hydrogen ion.
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42. (a) HBrO3acid
�Hþ ! BrO3�
conjugate base
(b) NH4þ
acid�Hþ ! NH3
conjugate base
(c) H2PO4�
acid�Hþ ! HPO4
2�conjugate base
43. HC2H3O2acid
þNH3base
! NH4þ
conjugate acidþ C2H3O2
�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 H2OðlÞ ! HS�ðaqÞ þ OH�ðaqÞThe 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 ! MgCl2ðaqÞ þ H2ðgÞ46. H2SO4ðaqÞ þ CaCO3ðsÞ ! CaSO4ðsÞ þ H2OðlÞ þ CO2ðgÞ47. Na2SO4ðaqÞ ��!H2O 2 NaþðaqÞ þ SO4
2�ðaqÞ48. (a) basic (d) acidic
(b) acidic (e) acidic
(c) neutral (f) basic
49. (a) CaCl2 sð Þ �����! Ca2þ aqð Þ þ 2 Cl� aqð Þ
For each CaCl2 ionic compound, 1 calcium ion and 2 chloride ions result.
Ca2+ Cl– Cl–
(b) KF sð Þ �����! Kþ aqð Þ þ F� aqð ÞFor each KF ionic compound, 1 potassium ion and 1 fluoride ion result.
K+ F–
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(c) AlBr3 sð Þ ! Al3þ aqð Þ þ 3 Br� aqð ÞFor each AlBr3 ionic compound, 1 aluminum ion and 3 bromide ions result.
Al3+ Br– Br– Br–
50. AlBr3 ! Al3þ þ 3 Br�
0:142 mol Br�
L
� �1 mol Al3þ
3 mol Br�
� �¼ 0:0473 mol Al3þ
L
� �¼ 0:0473M Al3þ
51. H2SO4 þ 2 NaOH ! Na2SO4 þ 2 H2O
mLNaOH ! mol NaOH ! mol H2SO4;mol H2SO4
L
� �¼ M H2SO4
35:22 mLNaOHð Þ 0:313 mol
1000 mL
� �1 mol H2SO4
2 mol NaOH
� �¼ 0:00551 mol H2SO4
0:00551 mol H2SO4
0:02522 L
� �¼ 0:218M H2SO4
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.
CH3OH:100: g
34:04 g=mol¼ 2:84 mol
CH3CH2OH:100: g
46:07 g=mol¼ 2:17 mol
54. A hydronium ion is a hydrated hydrogen ion.
Hþ þ H2O �����! H3Oþ
hydrogen ionð Þ hydronium ionð Þ55. Freezing point depression is directly related to the concentration of particles in the solution.
1 mol 1þ mol 2 mol 3 mol particles in solutionð ÞC12H22O11 > HC2H3O2 > HCl > CaCl2
Highest freezing point Lowest freezing point
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56. (a) 100�C pH ¼ �log 1� 10�6� � ¼ 6:0
25�C pH ¼ �log 1� 10�7� � ¼ 7:0 pH of H2O is greater at 25�C
(b) 1� 10�6 > 1� 10�7 so, Hþ concentration is higher at 100�C.(c) The water is neutral at both temperatures, because the H2O 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 0.10M HCl is 1.00, while the pH of 0.0100M HCl is 2.00.
58. A 1.00 m solution contains 1 mol solute plus 1000 g H2O. We need to find the total number of moles
and then calculate the mole percent of each component.
1000 g H2O
18:02 g=mol
� �¼ 55:49 mol H2O
55:49 mol H2Oþ 1:00 mol solute ¼ 56:49 total moles
1:00 mol solute
56:49 mol
� �100ð Þ ¼ 1:77% solute
55:49 mol H2O
56:49 mol
� �100ð Þ ¼ 98:23%H2O
59. Na2CO3 þ 2 HCl�����! 2 NaClþ CO2 þ H2O
g Na2CO3 �����! mol Na2CO3 �����! mol HCl�����! M HCl
0:452 g Na2CO3ð Þ 1 mol
106:0 g
� �2 mol HCl
1 mol Na2CO3
� �1
0:0424 L
� �¼ 0:201M HCl
60. H2SO4 þ 2 KOH ! K2SO4 þ 2 H2O
g KOH ! mol KOH ! mol H2SO4 ! M H2SO4
6:38 g KOHð Þ 1 mol KOH
56:11 g KOH
� �1 mol H2SO4
2 mol KOH
� �1000 mL
0:4233 mol H2SO4
� �¼ 134 mL of 0:4233M H2SO4
61. KOHþ HNO3 �����! KNO3 þ H2O
LHNO3 �����! mol HNO3 �����! mol KOH�����! g KOH
0:05000 L HNO3ð Þ 0:240 mol
L
� �1 mol KOH
1mol HNO3
� �56:11 g
mol
� �¼ 0:673 g KOH
62. pH of 1.0 L solution containing 0.1 mL of 1.0M HCl
0:1 mLð Þ 1:0 L
1000 mL
� �1 mol HCl
L
� �¼ 1� 10�4 mol HCl added
1� 10�4 mol HCl
1:0 L¼ 1� 10�4 M HCl
1� 10�4 M HCl produces 1� 10�4 M Hþ
pH ¼ �log 1� 10�4� � ¼ 4:0
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63. Dilution problem: V1M1 ¼ V2M2
M1 ¼ V2M2
V1
M1 ¼ 10:0 mLð Þ 12Mð Þ260:0 mLð Þ ¼ 0:462M HCl
64. NaOHþ HCl�����! NaClþ H2O
3:0 g NaOHð Þ 1 mol
40:00 g
� �¼ 0:075 mol NaOH
500:mLHClð Þ 1 L
1000 mL
� �0:10 mol
L
� �¼ 0:050 mol HCl
This solution is basic. The NaOH will neutralize the HCl with an excess of 0.025 mol of NaOH
remaining unreacted.
65. Ba OHð Þ2 aqð Þ þ 2 HCl aqð Þ �����! BaCl2 aqð Þ þ 2 H2O lð Þ
0:38 L Ba OHð Þ2� � 0:35 mol
L
� �¼ 0:13 mol Ba OHð Þ2
0:13 mol Ba OHð Þ2 �����! 0:26 mol OH�
0:500 L HClð Þ 0:65 mol
L
� �¼ 0:33 mol HCl
0:33 mol HCl�����! 0:33 mol Hþ
0.33 mol Hþ will neutralize 0.26 mol OH� and leave 0.07 mol Hþ 0:33� 0:26ð Þ remaining in solution.
Total volume ¼ 500:mLþ 380 mL ¼ 880 mL 0:88 Lð Þ
Hþ½ � in solution ¼ 0:07 mol Hþ
0:88 L¼ 0:08M Hþ
pH ¼ �log Hþ½ � ¼ �log 8� 10�2� � ¼ 1:1
The solution is acidic.
66. 0:05000 L HClð Þ 0:2000 mol
L
� �¼ 0:01000 mol HCl ¼ 0:01000 mol Hþ in 50:00 mLHCl
(a) no base added: pH ¼ �log 0:2000ð Þ ¼ 0:700
(b) 10:00 mL base added: 0:01000 Lð Þ 0:2000 mol
L
� �¼ 0:002000mol NaOH
¼ 0:002000mol OH�
0:01000 mol Hþð Þ � 0:002000 mol OH�ð Þ ¼ 0:00800 mol Hþin 60:00 mL solution
Hþ½ � ¼ 0:00800 mol
0:06000 LpH ¼ �log
0:00800
0:06000
� �¼ 0:880
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(c) 25:00 mL base added:
0:02500 Lð Þ 0:2000 mol
L
� �¼ 0:005000 mol NaOH ¼ mol OH�
0:01000 mol Hþð Þ � 0:005000mol OH�ð Þ ¼ 0:00500 mol Hþ in 75:00 solution
Hþ½ � ¼ 0:00500 mol
0:07500 LpH ¼ �log
0:00500
0:07500
� �¼ 1:2
(d) 49.00 mL base added:
0:04900 Lð Þ 0:2000 mol
L
� �¼ 0:009800 mol NaOH ¼ mol OH�
0:01000 mol Hþð Þ � 0:009800mol OH�ð Þ ¼ 0:00020 mol Hþ in 99:00 mL solution
Hþ½ � ¼ 0:00020 mol
0:09900 LpH ¼ �log
0:00020
0:09900
� �¼ 2:69
(e) 49.90 mL base added:
0:04990 Lð Þ 0:2000 mol
L
� �¼ 0:009980 mol NaOH ¼ mol OH�
0:01000 mol Hþð Þ � 0:009800mol OH�ð Þ ¼ 2� 10�5 mol Hþ in 99:00 mL solution
Hþ½ � ¼ 2� 10�5 mol
0:09990 LpH ¼ �log
2� 10�5
0:09990
� �¼ 3:7
(f) 49.99 mL base added:
0:04999 Lð Þ 0:2000 mol
L
� �¼ 0:009998 mol NaOH ¼ mol OH�
0:01000 mol Hþð Þ � 0:009998mol OH�ð Þ ¼ 2� 10�6 mol Hþ in 99:99 mL solution
Hþ½ � ¼ 2� 10�6
0:09999 LpH ¼ �log
2� 10�6
9:999� 10�2
� �¼ 4:7
(g) 50.00 mL of 0.2000M NaOH neutralizes 50.00 mL of 0.2000M HCl. No excess acid or base is in
the solution. Therefore, the solution is neutral with a pH¼ 7.08
7
6
5
4pH
3
2
1
00 10 20 30
mL NaOH40 50
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67. (a) 2 NaOH aqð Þ þ H2SO4 aqð Þ �����! Na2SO4 aqð Þ þ 2 H2O lð Þ(b) mol H2SO4 �����! mol NaOH�����! mLNaOH
0:0050 mol H2SO4ð Þ 2 mol NaOH
1mol H2SO4
� �1000 mL
0:10 mol
� �¼ 1:0� 102 mLNaOH
(c) 0:0050 mol H2SO4ð Þ 1 mol Na2SO4
1 mol H2SO4
� �142:1 g
mol
� �¼ 0:71 g Na2SO4
68. HNO3 þ KOH�����! KNO3 þ H2O
MAVA ¼ MBVB
MAð Þ 25 mLð Þ ¼ 0:60Mð Þ 50:0 mLð ÞMA ¼ 1:2M diluted solutionð Þ
Dilution problemM1V1 ¼ M2V2
MAð Þ 10:0 mLð Þ ¼ 1:2Mð Þ 100:00 mLð ÞMA ¼ 12M HNO3 original solutionð Þ
69. Yes, adding water changes the concentration of the acid, which changes the concentration of the Hþ½ �,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. H2SO4 þ 2 NaOH ! Na2SO4 þ 2 H2O
0:425 L H2SO4ð Þ 0:94 mol H2SO4
L
� �¼ 0:40 mol H2SO4
0:40 mol H2SO4 ! 0:80 mol Hþ
0:750 L NaOHð Þ 0:83 mol NaOH
L
� �¼ 0:62 mol NaOH
0:62 mol NaOH ! 0:62 mol OH�
0.80mol Hþ will neutralize 0.62mol OH� and leave 0.18mol (0.80� 0.62) of Hþ remaining in solution;
so the solution will be acidic.
Total volume ¼ 425mL þ 750mL ¼ 1175mL (1.175L)
Hþ½ � in solution ¼ 0:18 mol Hþ
1:175 L¼ 0:15M Hþ
pH ¼ �log Hþ½ � ¼ �log 0:15ð Þ 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
CuSO4ðaqÞ��!H2O Cu2þðaqÞ þ SO42�ðaqÞ
3rd Analyze the equation to determine number of ions formed.
Each CuSO4 will produce 1 Cu2þ ion and 1 SO4
2� ion, so [Cu2þ] ¼ [SO42�] ¼ 1 M
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(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
HNO3ðaqÞ��!H2O HþðaqÞ þ NO3�ðaqÞ
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 1M
(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
H2SO4ðaqÞ��!H2O 2 HþðaqÞ þ SO42�ðaqÞ
3rd Analyze the equation to determine number of ions formed.
Each H2SO4 will produce 2 Hþ ions and 1 SO4
2� ion, so [Hþ] ¼ 2M and [SO42�] ¼ 1M
(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
CaSðsÞ��!H2O 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 0M.
(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
HC2H3O2ðaqÞÐH2O
HþðaqÞ þ C2H3O2�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 bebetween 0 and 1M.
72. (a) Formula equation HNO3ðaqÞ þ LiOHðaqÞ ! H2OðlÞ þ LiNO3ðaqÞTotal ionic equation HþðaqÞ þ NO3
�ðaqÞ þ LiþðaqÞ þ OH�ðaqÞ !H2OðlÞ þ LiþðaqÞ þ NO3
�ðaqÞNet ionic equation HþðaqÞ þ OH�ðaqÞ ! H2OðlÞ
(b) Formula equation 2 HBrðaqÞ þ BaðOHÞ2ðaqÞ ! 2 H2OðlÞ þ BaBr2ðaqÞTotal ionic equation. 2 HþðaqÞ þ 2Br�ðaqÞ þ Ba2þðaqÞ þ 2 OH�ðaqÞ !
2 H2OðlÞ þ Ba2þðaqÞ þ 2 Br�ðaqÞNet ionic equation HþðaqÞ þ OH�ðaqÞ ! H2OðlÞ
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(c) Formula equation HFðaqÞ þ NaOHðaqÞ ! H2OðlÞ þ NaFðaqÞTotal ionic equation. HFðaqÞ þ NaþðaqÞ þ OH�ðaqÞ ! H2OðlÞ þ NaþðaqÞ 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 ! H2OðlÞ þ F�ðaqÞ73. 1st write a balanced chemical equation for the reaction of the Hþ in the rain with sodium hydroxide
HþðaqÞ þ NaOHðaqÞ ! NaþðaqÞ þ H2OðlÞ2nd calculate moles of sodium hydroxide needed to react with all of the Hþ in the rain.
mol NaOH ¼ 7:2 mLNaOHð Þ 1 L NaOH
1000 mLNaOH
� �0:125 mol NaOH
1 L NaOH
� �¼ 0:00090 mol NaOH
3rd determine moles of Hþ
mol Hþ ¼ 0:00090 mol NaOHð Þ 1 mol Hþ
1 mol NaOH
� �¼ 0:00090 mol Hþ
4th determine molarity of Hþ
M Hþ ¼ mol Hþ
L rain¼ 0:00090 mol Hþ
25 L rain¼ 3:6� 10�5 M Hþ
5th determine pH of rain
pH ¼ �log Hþ½ � ¼ �log 3:6� 10�5 M Hþ� � ¼ 4:44
74. [Hþ] needs to be 0.00158M to give the final solution of sodium rhidizonate a pH of 2.800.
V1M1 ¼ V2M2
500:0 mLð Þ 0:00158Mð Þ ¼ V2 0:60Mð Þ
V2 ¼ 500:0 mLð Þ 0:00158Mð Þ0:60Mð Þ ¼ 1:3 mL 0:60M HCl
To make the solution put 1.3mL of 0.60M HCl in a graduated cylinder and fill the cylinder to the
500mL mark with sodium rhidizonate solution.
75. V1M1 ¼ V2M2
3:00 Lð Þ 3:25Mð Þ ¼ V2 12:1Mð Þ
V2 ¼ 3:00 Lð Þ 3:25Mð Þ12:1Mð Þ ¼ 0:806 L 12:1M HCl or 806 mL 12:1M HCl
76. CaCO3ðsÞ þ 2 HClðaqÞ ! CaCl2ðaqÞ þ H2OðlÞ þ CO2ðgÞ
77. First determine the molarity of the two HCl solutions. Take the antilog of the pH value to obtain the Hþ½ �.pH ¼ 0:300; Hþ ¼ 2:00M ¼ 2:00M HCl
pH ¼ 0:150; Hþ ¼ 1:41M ¼ 1:41M HCl
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Now treat the calculation as a dilution problem.
V1M1 ¼ V2M2 V2 ¼ V1M1
M2
200 mLHClð Þ 2:00Mð Þ1:41M
¼ 284 mL solution
284 mL� 200 mL ¼ 84 mLH2O to be added
78. mol acid¼mol base (lactic acid has one acidic H)
1:0 g acid
molar mass¼ 0:017 Lð Þ 0:65 mol NaOH
L
� �¼ 0:011 mol NaOH
mol acid¼mol base
mol HC3H5O3 ¼ 0:011 mol
1:0 g
0:011 mol¼ 91 g=mol ¼ molar mass
molar mass (91 g/mol) = mass of empirical formula (90.08 g/mol)
Therefore the molecular formula, HC3H5O3, is the same as the empirical formula.
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