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C15 09/16/2013 13:25:32 97

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


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


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


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


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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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


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 .


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


HNO3ðaqÞ��!H2O HþðaqÞ þ NO3�ðaqÞ


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


H2SO4ðaqÞ��!H2O 2 HþðaqÞ þ SO42�ðaqÞ


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.


CaSðsÞ��!H2O no reaction


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


HC2H3O2ðaqÞÐH2O

HþðaqÞ þ C2H3O2�ðaqÞ


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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C15 09/16/2013 13:25:36 Page 219

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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- Chapter 15 -

acids, bases, and salts - chemeketa …faculty.chemeketa.edu/lemme/heinsolutions/ch15.pdf · acids,...

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