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. An electrolyte must be present in the solution for the bulb to glow.
3. Electrolytes include acids, bases, and salts.
4. 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 Clþ. Second, more water molecules will fit around Cl, since it is
larger than the Naþ ion.
5. The pH for a solution with a hydrogen ion concentration of 0.003 M will be between 2 and 3.
6. Tomato juice is more acidic than blood, since its pH is lower.
7. 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.
- 181 -
- Chapter 15 8. Neutralization reactions:
Arrhenius: HCl þ NaOH ! NaCl þ H2 O
ðHþ þ OH ! H2 OÞ
Brønsted-Lowry: HCl þ KCN ! HCN þ KCl
! AlCl4 þ Naþ
Lewis: AlCl3 þ NaCl Cl
Cl Cl
Cl
9.
(a)
Br
+
–
Cl
(b)
Cl
Cl Al Cl
Cl
–
O H
–
(c)
ðHþ þ CN ! HCNÞ
–
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
10. The electrolytic compounds are acids, bases, and salts.
11. Names of the compounds in Table 15.3
sulfuric acid
H2 SO4
nitric acid
HNO3
HCl
hydrochloric acid
HBr
hydrobromic acid
HClO4
perchloric acid
NaOH
sodium hydroxide
KOH
potassium hydroxide
CaðOHÞ2
calcium hydroxide
barium hydroxide
BaðOHÞ2
HC2 H3 O2 acetic acid
H2 CO3
carbonic acid
HNO2
nitrous acid
sulfurous acid
H2 SO3
H2 S
hydrosulfuric acid
oxalic acid
H 2 C 2 O4
H3 BO3
boric acid
HClO
hypochlorous acid
NH3
ammonia
HF
hydrofluoric acid
12. Hydrogen chloride dissolved in water conducts an electric current. HCl reacts with polar water
molecules to produce H3Oþ and Cl ions, which conduct 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.
13. 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.
14. 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.
15. 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.
16. 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.
- 182 -
- Chapter 15 17. Strong electrolytes are those which are essentially 100% ionized or dissociated in water. Weak
electrolytes are those which are only slightly ionized in water.
18. Ions are hydrated in solution because there is an electrical attraction between the charged ions and the
polar water molecules.
19. 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.
20. (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.
21. The net ionic equation for an acid-base reaction in aqueous solutions is:
Hþ þ OH ! H2 O
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.
HCl þ H2 O ! H3 Oþ þ Cl
23. Pure water is neutral because when it ionizes it produces equal molar concentrations of acid [Hþ] and
base [OH] ions.
24. The fundamental difference between a colloidal dispersion and a true solution lies in the size of the
particles. In a true solution particles are usually ions or hydrated molecules and are less than 1 nm in
size. In colloidals the particles are aggregates of ions or molecules, ranging in size from 1-1000 nm.
25. Dialysis is the process of removing dissolved solutes from a colloidal dispersion by use of a dialyzing
membrane. The dissolved solutes pass through the membrane leaving the colloidal dispersion behind.
Dialysis is used in artificial kidneys to remove soluble waste products from the blood.
26. A neutral solution is one in which the concentration of acid is equal to the concentration of base
ð½Hþ ¼ ½OH Þ. An acidic solution is one in which the concentration of acid is greater than the
concentration of base ð½Hþ > ½OH Þ. A basic solution is one in which the concentration of base is
greater than the concentration of acid ð½Hþ < ½OH Þ.
27. 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.
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.
- 183 -
- Chapter 15 -
SOLUTIONS TO EXERCISES
1. Conjugate acid – base pairs:
(a) NH3 NHþ
4 ; H2 O OH
(b) HC2 H3 O2 C2 H3 O2 ; H2 O H3 Oþ
2
(c) H2 PO
4 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
(b)
Zn þ 2 Hþ ! Zn2þ þ H2
! 2 Al3þ þ 3 SO2
þ 6 H2 O
2 AlðOHÞ3 þ 6 Hþ þ 3 SO2
4
4
(c)
AlðOHÞ3 þ 3 Hþ ! Al3þ þ 3 H2 O
þ 2 HC2 H3 O2 ! 2 Naþ þ 2 C2 H3 O
2 Naþ þ CO2
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
(f)
CaðHCO3 Þ2 þ 2 Hþ ! Ca2þ þ H2 O þ CO2
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
- 184 -
- 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 6 Hþ ! 3 SO2
! 2 Al3þ þ 3 SO2
þ 3 H2
4
4
2 Al þ 6 Hþ ! 2 Al3þ þ 3 H2
(c)
þ 2 H2 O
ð2 Naþ þ 2 OH Þ þ H2 CO3 ! 2 Naþ þ CO2
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
þ 2 ClO
Mg þ 2 Hþ þ 2 ClO
4 ! Mg
4 þ H2
Mg þ 2 Hþ ! Mg2þ þ H2
(f)
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. The following compounds are electrolytes:
(a) SO3 , acid in water
(b) K2 CO3 , salt
(e) CuBr2 , salt
(f) HI, acid in water
8. The following compounds are electrolytes:
(b) P2 O5 , acid in water
(c) NaClO, salt
(d) LiOH, base
(f) KMnO4 , salt
9. Molarity of ions.
(a)
(b)
(c)
(d)
1 mol Cu2þ
¼ 1:25 M Cu2þ
ð1:25 M CuBr2 Þ
1 mol CuBr2
2 mol Br
¼ 2:50 M Br
ð1:25 M CuBr2 Þ
1 mol CuBr2
1 mol Naþ
¼ 0:75 M Naþ
ð0:75 M NaHCO2 Þ
1 mol NaHCO2
1 mol HCO
3
ð0:75 M NaHCO3 Þ
¼ 0:75 M HCO
3
1 mol NaHCO3
3 mol Kþ
¼ 10:5 M Kþ
ð3:50 M K3 AsO4 Þ
1 mol K3 AsO2
1 mol AsO3
4
¼ 3: 50 M AsO 3
ð3:50 M K3 AsO4 Þ
4
1 mol K3 AsO4
2 mol NHþ
4
0:65 M ðNH4 Þ2 SO4
¼ 1: 3 M NH þ
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
- 185 -
- Chapter 15 10. Molarity of ions.
(a)
(b)
(c)
(d)
1 mol Fe3þ
¼ 2:25 M Fe3þ
1 mol FeCl3
3 mol Cl
ð2:25 M FeCl3 Þ
¼ 6:75 M Cl
1 mol FeCl3
1 mol Mg2þ
¼ 1:20 M Mg2þ
ð1:20 M MgSO4 Þ
1 mol MgSO4
1 mol SO2
4
¼ 1:20 M SO2
ð1:20 M MgO4 Þ
4
1 mol MgSO4
1 mol Naþ
¼ 0:75 M Naþ
ð0:75 M NaH2 PO4 Þ
1 mol NaH2 PO4
1 mol H2 PO
4
¼ 0:75 M H2 PO
ð0:75 M NaH2 PO4 Þ
4
1 mol NaH2 PO4
1 mol Ca2þ
¼ 0:35 M Ca2þ
0:35 M CaðClO3 Þ2
1 mol CaðClO3 Þ2
2 mol ClO
3
¼ 0:70 M ClO
0:35 M CaðClO3 Þ2
3
1 mol CaðClO3 Þ2
ð2:25 M FeCl3 Þ
11. We will use the data from No. 9 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
L
mol
2:50 mol Br 79:90 g
ð0:100 LÞ
¼ 20:0 g Br
L
mol
0:75 mol Naþ 22:99 g
¼ 1:7 g Naþ
(b) ð0:100 LÞ
L
mol
0:75 mol HCO
61:02 g
3
ð0:100 LÞ
¼ 4:6 g HCO
3
L
mol
10:5 mol Kþ
39:10 g
¼ 41:1 g kþ
(c) ð0:100 LÞ
L
mol
3:50 mol AsO3
138:9 g
4
ð0:100 LÞ
¼ 48:6 g AsO3
4
L
mol
1:3 mol NHþ
18:04 g
4
(d) ð0:100 LÞ
¼ 2:3 g NHþ
4
L
mol
12. We will use the data from No. 10 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
L
mol
6:75 mol Cl
35:45 g
ð0:100 LÞ
¼ 23:9 g Cl
L
mol
- 186 -
- Chapter 15 (b)
(c)
(d)
1:20 mol Mg2þ
24:31 g
ð0:100 LÞ
¼ 2:92 g Mg2þ
L
mol
1:20 mol SO2
96:07 g
4
ð0:100 LÞ
¼ 11:5 g SO2
4
L
mol
0:75 mol Naþ
22:99 g
ð0:100 LÞ
¼ 1:7 g Naþ
L
mol
0:75 mol H2 PO
96:99 g H2 PO
4
4
¼ 7:3 g H2 PO2
ð0:100 LÞ
4
L
mol
0:35 mol Ca2þ 40:08 g
ð0:100 LÞ
¼ 1:4 g Ca2þ
L
mol
0:70 mol ClO
83:45 g
3
ð0:100 LÞ
¼ 5:8 g ClO
3
L
mol
13. pH ¼ log½Hþ ½Hþ ¼ 10pH
(a)
½Hþ ¼ 1 105
(b)
½Hþ ¼ 3 109
(c)
½Hþ ¼ 6 102
14. pH ¼ log½Hþ ½Hþ ¼ 10pH
(a)
½Hþ ¼ 1 107
(b)
½Hþ ¼ 2 104
(c)
½Hþ ¼ 3 1012
0:50 mol HCl
¼ 0:028 mol HCl
15. (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
¼ 0:031 mol CaCl2
ð125 mLÞ
1000 mL
- 187 -
- Chapter 15 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
¼ 1:00 M Cl
ð0:500 M CaCl2 Þ
1 mol CaCl2
(c)
NaOH þ HCl ! NaCl þ H2 O
0:333 mol NaOH
¼ 0:0117 mol NaOH
ð35:0 mLÞ
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
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
ð12:5 mLÞ
¼ 0:00625 mol H2 SO4
1000 mL
0:175 mol NaOH
ð23:5 mLÞ
¼ 0:00411 mol NaOH
1000 mL
1 mol H2 SO4
¼ 0:00206 mol H2 SO4 reacted
ð0:00411 mol NaOHÞ
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
0:0838 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
- 188 -
- Chapter 15 0:10 mol NaCl
16. (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
¼ 0:25 M Naþ
ð0:25 M NaClÞ
1 mol NaCl
1 mol Cl
¼ 0:25 M Cl
ð0:25 M NaClÞ
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
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
¼ 1:96 M Hþ
ð1:96 M HClÞ
1 mol HCl
1 mol Cl
¼ 1:96 M Cl
ð1:96 M HClÞ
1 mol HCl
0:10 mol BaðNO3 Þ2
(c) ð15:5 mLÞ
¼ 0:0016 M BaðNO3 Þ2
1000 mL
0:20 mol AgNO3
ð10:5 mLÞ
¼ 0:0021 M AgNO3
1000 mL
Number of moles of each substance:
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
0:0053 mol NO
3
¼ 0:20 M NO
3
0:0260 L
- 189 -
0:0016 mol Ba2þ , 0:0021 mol Agþ , and ð0:0032
- Chapter 15 (d)
0:25 mol NaCl
ð25:5 mLÞ
¼ 0:0064 mol NaCl
1000 mL
0:15 mol CaðC2 H3 O2 Þ2
¼ 0:0023 mol CaðC2 H3 O2 Þ2
ð15:5 mLÞ
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
17. 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
M A VA ¼ M B VB
MA ¼
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
18. 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
(a)
MB ¼
M A VA
VB
ð37:19 mLÞð0:126 M Þ
¼ 0:147 M NaOH
31:91 mL
- 190 -
- Chapter 15 -
(b)
ð48:04 mLÞð0:482 M Þ
¼ 0:964 M NaOH
24:02 mL
19. (a)
(b)
(c)
ð13:13 mLÞð1:425 M Þ
¼ 0:4750 M NaOH
39:39 mL
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
20. (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
(c)
21. The more acidic solution is listed followed by an explanation.
(a)
(b)
1 molar H2SO4. 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. HCl is a strong electrolyte, producing more Hþ than HC2H3O2 which is a weak
electrolyte.
22. The more acidic solution is listed followed by an explanation.
(a)
(b)
2 molar HCl. 2 M HCl will yield 2 M Hþ concentration. 1 M HCl will yield 1 M Hþ concentration.
1 molar H2SO4. 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.
23. 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 HClO
1000 mL
50:25 g CaðOHÞ2
¼ 2:58 103 mL
74:10 g 1 mol CaðOHÞ2 0:525 mol
24. 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
3 mol HCl
1000 mL
mol
25. 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
- 191 -
- Chapter 15 26. 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
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
27. 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
ð5:00 g ZnÞ
g Zn ! mol Zn
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
¼ 0:0175 mol H2 produced in the reaction
ð0:0350 mol HClÞ
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
28. 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
¼ 0:0765 mol Zn
g Zn ! mol Zn
ð5:00 g ZnÞ
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
¼ 0:0350 mol H2
ð0:0700 mol HClÞ
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
- 192 -
- Chapter 15 29. 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
30. pH ¼ log½Hþ (a)
(b)
(c)
31. (a)
(b)
32. (a)
(b)
33. (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
4
þ
Orange juice
¼ 3:7 410 M H
¼ 3:43
pH ¼ log 3:7 10
3
þ
Vinegar ¼ 2:8
10 3M
H
¼ 2:55
pH ¼ log 2:8 10
Black coffee ¼ 5:0 105 M Hþ
pH ¼ log 5:0 105 ¼ 4:30
þ
Limewater ¼ 3:4 1011
MH
11
¼ 10:47
pH ¼ log 3:4 10
H2 O
NH3 is a weak base NH3 ðaqÞ Ð NHþ
4 ðaqÞ þ H2 OðaqÞ
H2 O
(b)
HCl is a strong acid HClðaqÞ ! Hþ ðaqÞ þ Cl ðaqÞ
(c)
KOH is a strong base
(d)
HC2 H3 O2 is a weak acid HC2 H3 O2 ðaqÞ Ð Hþ ðaqÞ þ C2 H3 O
2 ðaqÞ
H2 O
KOH ! Kþ ðaqÞ þ OH ðaqÞ
H2 O
H2 O
34. (a)
H2 C2 O4 is a weak acid H2 C2 O4 ðaqÞ Ð Hþ ðaqÞ þ HC2 O
4 ðaqÞ
(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
þ
HBr is a strong acid HBrðaqÞ ! H ðaqÞ þ Br ðaqÞ
H2 O
H2 O
35. (a) basic
(b) acidic
(c) neutral
36. (a)
(d) acidic
(e) acidic
(f) basic
! Ca2þ ðaqÞ þ 2 Cl ðaqÞ
CaCl2 ðsÞ For each CaCl2 ionic compound, 1 calcium ion and 2 chloride ions result.
Ca2+
Cl–
Cl–
- 193 -
- Chapter 15 (b)
KFðsÞ ! Kþ ðaqÞ þ F ðaqÞ
For each KF ionic compound, 1 potassium ion and 1 fluoride ion result.
K+
(c)
F–
AlBr3 ðsÞ ! Al3þ ðaqÞ þ 3 Br ðaqÞ
For each AlBr3 ionic compound, 1 aluminum ion and 3 bromide ions result.
Al3+
37.
Br–
Br–
Br–
! Ca2þ þ 2 I
CaI2 0:520 mol I
1 mol Ca2þ
0:260 mol Ca2þ
¼
¼ 0:260 M Ca2þ
L
2 mol I
L
38. H2 SO4 þ 2 NaOH ! Na2 SO4 þ 2 H2 O mol H2 SO4
¼ M H2 SO4
mL NaOH ! mol NaOH ! mol H2 SO4 ;
L
0:313 mol 1 mol H2 SO4
ð35:22 mL NaOHÞ
¼ 0:00551 mol H2 SO4
2 mol NaOH
1000 mL
0:00551 mol H2 SO4
¼ 0:218 M H2 SO4
0:02522 L
39. 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.
40. 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
CH3 OH:
¼ 2:84 mol
34:04 g=mol
CH3 CH2 OH:
100: g
¼ 2:17 mol
46:07 g=mol
- 194 -
- Chapter 15 41. A hydronium ion is a hydrated hydrogen ion.
þ H2 O !
H 3 Oþ
Hþ
ðhydrogen ionÞ
ðhydronium ionÞ
42. Freezing point depression is directly related to the concentration of particles in the solution.
C12 H22 O11 > HC2 H3 O2 > HCl > CaCl2
Highest freezing point
2 mol 3 mol ðparticles in solutionÞ
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.
1 mol
43. (a)
(b)
(c)
Lowest freezing point
1 þ mol
44. 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.
45. 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
46. Na2 CO3 þ 2 HCl ! 2 NaCl þ CO2 þ H2 O
! mol Na2 CO3 ! mol HCl ! M HCl
g Na2 CO3 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
! CaCl2 þ 2 H2 O
47. 2 HCl þ CaðOHÞ2 ! mol CaðOHÞ2 ! mol HCl ! mL HCl
g CaðOHÞ2 1 mol
2 mol HCl
1000 mL
2:00 g CaðOHÞ2
¼ 437 mL of 0:1234 M HCl
74:10 g 1 mol CaðOHÞ2
0:1234 mol
! KNO3 þ H2 O
48. KOH þ HNO3 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
- 195 -
- Chapter 15 49. pH of 1.0 L solution containing 0.1 mL of 1.0 M HCl
1:0 L
1 mol HCl
ð0:1 mLÞ
¼ 1 104 mol HCl added
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
50. Dilution problem: V1 M 1 ¼ V2 M 2
M1 ¼
V1 M 2
V1
M1 ¼
ð10:0 mLÞð12 M Þ
¼ 0:462 M HCl
ð260:0 mLÞ
51. 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.
52. BaðOHÞ2 ðaqÞ þ 2 HClðaqÞ ! BaCl2 ðaqÞ þ 2 H2 Oðl Þ
0:35 mol
0:38 L BaðOHÞ2
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
0:2000 mol
53. ð0:05000 L HClÞ
¼ 0:01000 mol HCl ¼ 0:01000 mol Hþ in 50:00 mL HCl
L
(a)
no base added: pH ¼ logð0:2000Þ ¼ 0:700
- 196 -
- Chapter 15 -
(b)
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
(c)
25:00 mL base added:
0:2000 mol
¼ 0:005000 mol NaOH ¼ mol OH
ð0:02500 LÞ
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
H
½ 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
¼ 0:009980 mol NaOH ¼ mol OH
ð0:04990 LÞ
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
¼ 3:7
½H ¼
pH ¼ log
0:09990
0:09990 L
(f)
49.99 mL base added:
0:2000 mol
ð0:04999 LÞ
¼ 0:009998 mol NaOH ¼ mol OH
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
- 197 -
- Chapter 15 (g)
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
pH
5
4
3
2
1
0
54. (a)
(b)
(c)
0
10
20
30
mL NaOH
40
50
2 NaOHðaqÞ þ H2 SO4 ðaqÞ ! Na2 SO4 ðaqÞ þ 2 H2 Oðl Þ
! mol NaOH ! mL NaOH
mol H2 SO4 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 NaOH
142:1 g
ð0:0050 mol H2 SO4 Þ
¼ 0:71 g Na2 SO4
1 mol H2 SO4
mol
55. HNO3 þ KOH ! KNO3 þ H2 O
M A VA ¼ M B VB
ðM A Þð10:0 mLÞ ¼ ð1:2 M Þð100:00 mLÞ
M A ¼ 1:2 M ðdiluted solutionÞ
Dilution problem M 1 V1 ¼ M 2 V2
ðM 1 Þð25 mLÞ ¼ ð0:60 M Þð50:0 mLÞ
M 1 ¼ 12 M HNO3 ðoriginal solutionÞ
56. 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.
57. 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
Now treat the calculation as a dilution problem.
V1 M 1
V2 ¼
V1 M 1 ¼ V 2 M 2
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
- 198 -
- Chapter 15 58. mol acid ¼ mol base
(lactic acid has one acidic H)
1:0 g acid
0:65 mol
¼ ð0:017 LÞ
¼ 0:01105 mol
molar mass
L
1:0 g
¼ 0:01105 mol
molar mass
1:0 g
¼ 90:17 g=mol
0:01105 mol
mass of empirical formula ðHC3 H5 O3 Þ ¼ 90:17 g=mol
molar mass ¼ mass of empirical formula
Therefore the molecular formula is HC3 H5 O3
- 199 -