HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 207
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 Na+ 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 Na+. Second, more water
molecules will fit around Cl-, since it is larger than the Na+ 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 BrønstedLowry 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.
- 207 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 208
- Chapter 15 8.
Neutralization reactions:
Arrhenius: HCl + NaOH ¡ NaCl + H 2O
(H + + OH - ¡ H 2O)
Brønsted-Lowry: HCl + KCN ¡ HCN + KCl
(H + + CN - ¡ HCN)
+
Lewis: AlCl 3 + NaCl ¡ AlCl 4 + Na
–
Cl
Al Cl
Cl
+
–
Cl
Cl
Cl Al Cl
Cl
–
9.
(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.
10.
The electrolytic compounds are acids, bases, and salts.
11.
Names of the compounds in Table 15.3
sulfuric acid
H 2SO4
HNO3
nitric acid
HCl
hydrochloric acid
HBr
hydrobromic acid
HClO4
perchloric acid
NaOH
sodium hydroxide
KOH
potassium hydroxide
Ca(OH)2 calcium hydroxide
Ba(OH)2 barium hydroxide
HC2H 3O2
H 2CO3
HNO2
H 2SO3
H 2S
H 2C2O4
H 3BO3
HClO
NH 3
HF
acetic acid
carbonic acid
nitrous acid
sulfurous acid
hydrosulfuric acid
oxalic acid
boric acid
hypochlorous acid
ammonia
hydrofluoric acid
12.
Hydrogen chloride dissolved in water conducts an electric current. HCl reacts with polar
water molecules to produce H 3O + 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.
- 208 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 209
- Chapter 15 14.
Testing the electrical conductivity of the solutions shows that CH 3OH is a nonelectrolyte,
while NaOH is an electrolyte. This indicates that the OH group in CH 3OH must be
covalently bonded to the CH 3 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 H 2O.
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 H + and OH - are equal.
In an acid solution, the concentration of H + is greater than the concentration
of OH -.
In a basic solution, the concentration of OH - is greater than the concentration
of H +.
21.
The net ionic equation for an acid-base reaction in aqueous solutions is:
H + + OH - ¡ H 2O.
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 H 2O molecules. HCl + H 2O ¡ H 3O + + Cl-
23.
Pure water is neutral because when it ionizes it produces equal molar concentrations of
acid [H +] 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.
- 209 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 210
- Chapter 15 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 ([H +] = [OH -]). An acidic solution is one in which the concentration of acid is
greater than the concentration of base ([H +] 7 [OH -]). A basic solution is one in which
the concentration of base is greater than the concentration of acid ([H +] 6 [OH -]).
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 (H 2SO4) and nitric acid (HNO3). 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.
- 210 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 211
CHAPTER 15
SOLUTIONS TO EXERCISES
1.
(a)
(b)
(c)
(d)
H 2SO4 - HSO4 - ; H 2C2H 3O2 + - HC2H 3O2
step 1: H 2SO4 - HSO4 - ; H 3O + - H 2O
step 2: HSO4 - - SO4 2- ; H 3O + - H 2O
HClO4 - ClO4 - ; H 3O + - H 2O
H 3O + - H 2O; CH 3OH - CH 3O -
2.
Conjugate acid-base pairs:
(a)
HCl - Cl-; NH 4 + - NH 3
(b) HCO3 - - CO3 2- ; H 2O - OH (c)
H 3O + - H 2O; H 2CO3 - HCO3 (d) HC2H 3O2 - C2H 3O2 - ; H 3O + - H 2O
3.
(a)
(b)
(c)
(d)
(e)
(f)
4.
Complete and balance these equations:
(a) Fe2O3(s) + 6 HBr(aq) : 2 FeBr3(aq) + 3 H2O(l)
(b) 2 Al(s) + 3 H2SO4(aq) : Al2(SO4)3(aq) + 3 H2(g)
(c) 2 NaOH(aq) + H2CO3(aq) : Na2CO3(aq) + 2 H2O(l)
(d) Ba(OH)2(s) + 2 HClO4(aq) : Ba(ClO4)2(aq) + 2 H2O(l)
(e) Mg(s) + 2 HClO4(aq) : Mg(ClO4)2(aq) + H2(g)
(f) K2O(s) + 2 HI(aq) : 2 KI(aq) + H2O(l)
5.
(a)
(b)
Zn(s) + 2 HCI(aq) : ZnCl2(aq) + H2(g)
2 Al(OH)3(s) + 3 H2SO4(aq) : Al2(SO4)3(aq) + 6 H2O(l)
Na2CO3(aq) + 2 HC2H3O2(aq) : 2 NaC2H3O2(aq) + H2O(l) + CO2(g)
MgO(s) + 2 HI(aq) : MgI2(aq) + H2O(l)
Ca(HCO3)2(s) + 2 HBr(aq) : CaBr2(aq) + 2H2O(l) + 2CO2(g)
3 KOH(aq) + H3PO4(aq) : K3PO4(aq) + 3 H2O(l)
Zn + (2 H + + 2 Cl - ) : (Zn2 + + 2 Cl - ) + H2
Zn + 2 H + : Zn2 + + H2
2 Al(OH)3 + (6 H + + 3 SO24 - ) : (2 Al3 + + 3 SO24 - ) + 6 H2O
Al(OH)3 + 3 H + : Al3 + + 3 H2O
- 211 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 212
- Chapter 15 -
(c)
(d)
(e)
(f)
6.
(a)
(b)
(c)
(d)
(e)
(f)
7.
8.
9.
(2 Na + + CO23 - ) + 2 HC2H3O2 : (2 Na + + 2 C2H3O2- ) + H2O + CO2
CO23 - + 2 HC2H3O2 : 2 C2H3O2- + H2O + CO2
MgO + (2 H + + 2 I - ) : (Mg2 + + 2 I - ) + H2O
MgO + 2 H + : Mg2 + + H2O
Ca(HCO3)2 + (2 H + + 2 Br - ) : (Ca2 + + 2 Br - ) + 2 H2O + 2 CO2
Ca(HCO3)2 + 2 H + : Ca2 + + H2O + CO2
(3 K + + 3 OH - ) + H3PO4 : (3 K + + PO34 - ) + 3 H2O
3 OH - + H3PO4 : PO34 - + 3 H2O
Fe2O3 + (6 H + + 6 Br - ) : (2 Fe3 + + 6 Br - ) + 3 H2O
Fe2O3 + 6 H + : 2 Fe3 + + 3 H2O
2 Al + (6 H + + 3 SO24 - ) : (2 Al3 + + 3 SO24 - ) + 3 H2
2 Al + 6 H + : 2 Al3 + + 3 H2
(2 Na + + 2 OH - ) + H2CO3 : (2 Na + + CO23 - ) + 2 H2O
2 OH - + H2CO3 : CO23 - + 2 H2O
Ba(OH)2 + (2 H + + 2 ClO4- ) : (Ba2 + + 2 ClO4- ) + 2 H2O
Ba(OH)2 + 2 H + : Ba2 + + 2 H2O
Mg + (2 H + + 2 ClO4- ) : (Mg2 + + 2 ClO4- ) + H2
Mg + 2H + : Mg2 + + H2
K2O + (2 H + + 2 I - ) : (2 K + + 2 I - ) + H2O
K2O + 2 H + : 2 K + + H2O
The following compounds are electrolytes:
(a) HCl, acid in water
(c)
(b) CO2, acid in water
CaCl 2, salt
The following compounds are electrolytes:
NaHCO3, salt, base in water
(a)
(e)
AgNO3, salt
(c)
(f)
(d) HCOOH, acid
RbOH, base
K 2CrO4, salt
Calculation of molarity of ions.
(a)
(0.015 M NaCl) ¢
1 mol Na+
≤ = 0.015 M Na+
1 mol NaCl
(0.015 M NaCl) ¢
1 mol Cl≤ = 0.015 M Cl1 mol NaCl
- 212 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 213
- Chapter 15 -
(b)
(c)
(d)
10.
(a)
(4.25 M NaKSO4) ¢
1 mol Na+
≤ = 4.25 M Na+
1 mol NaKSO4
(4.25 M NaKSO4) ¢
1 mol K +
≤ = 4.25 M K +
1 mol NaKSO4
(4.25 M NaKSO4) ¢
1 mol SO4 2≤ = 4.25 M SO4 21 mol NaKSO4
(0.20 M CaCl 2) ¢
1 mol Ca2+
≤ = 0.20 M Ca2+
1 mol CaCl 2
(0.20 M CaCl 2) ¢
2 mol Cl≤ = 0.40 M Cl1 mol CaCl 2
a
22.0 g KI
1 mol
1000 mL
ba
ba
b = 0.265 M KI
500. mL
166.0 g
L
(0.265 M KI) ¢
1 mol K +
≤ = 0.265 M K +
1 mol KI
(0.265 M KI) ¢
1 mol I ≤ = 0.265 M I 1 mol KI
(0.75 M ZnBr2) ¢
1 mol Zn2+
≤ = 0.75 M Zn2+
1 mol ZnBr2
(0.75 M ZnBr2) ¢
(b)
(c)
2 mol Br ≤ = 1.5 M Br 1 mol ZnBr2
(1.65 M Al 2(SO4)3) ¢
3 mol SO4 2≤ = 4.95 M SO4 21 mol Al 2(SO4)3
(1.65 M Al 2(SO4)3) ¢
2 mol Al3+
≤ = 3.30 M Al3+
1 mol Al 2(SO4)3
¢
900. g(NH 4)2SO4
1 mol
b = 0.340 M (NH 4)2SO4
≤a
20.0 L
132.2 g
- 213 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 214
- Chapter 15 -
(d)
(0.340 M (NH 4)2SO4) ¢
2 mol NH 4 +
≤ = 0.680 M NH 4 +
1 mol (NH 4)2SO4
(0.340 M (NH 4)2SO4) ¢
1 mol SO4 2≤ = 0.340 M SO4 21 mol (NH 4)2SO4
¢
0.0120 g Mg(ClO3)2
1 mol
b = 0.0628 M Mg(ClO3)2
≤a
0.00100 L
191.2 g
(0.0628 M Mg(ClO3)2) ¢
1 mol Mg 2+
≤ = 0.0628 M Mg 2+
1 mol Mg(ClO3)2
2 mol ClO3 (0.0628 M Mg(ClO3)2) ¢
≤ = 0.126 M ClO3 1 mol Mg(ClO3)2
11.
The molarity of each ion, as calculated in Exercise 9 will be used to calculate the mass of
each ion present in 100. mL of solution. 100. mL = 0.100 L
(a)
(b)
(0.100 L) ¢
22.99 g
0.015 mol Na+
b = 0.034 g Na+
≤a
L
mol
(0.100 L) ¢
35.45 g
0.015 mol Clb = 0.053 g Cl≤a
L
mol
(0.100 L) ¢
22.99 g
4.25 mol Na+
b = 9.77 g Na+
≤a
L
mol
(0.100 L) ¢
39.10 g
4.25 mol K +
b = 16.6 g K +
≤a
L
mol
96.07 g
4.25 mol SO4 2(0.100 L) ¢
b = 40.8 g SO4 2≤a
L
mol
(c)
(0.100 L) ¢
40.08 g
0.20 mol Ca2+
b = 0.80 g Ca2+
≤a
L
mol
(0.100 L) ¢
35.45 g
0.40 mol Clb = 1.4 g Cl≤a
L
mol
- 214 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 215
- Chapter 15 -
(d)
12.
(0.100 L) ¢
39.10 g
0.265 mol K +
b = 1.04 g K +
≤a
L
mol
(0.100 L) ¢
126.9 g
0.265 mol I b = 3.36 g I ≤a
L
mol
The molarity of each ion, as calculated in Exercise 10, will be used to calculate the mass
of each ion present in 100 mL of solution. 100. mL = 0.100 L
(a)
(b)
(c)
(d)
(0.100 L) ¢
65.39 g
0.75 mol Zn2+
b = 4.9 g Zn2+
≤a
L
mol
(0.100 L) ¢
79.90 g
1.5 mol Br b = 12 g Br ≤a
L
mol
(0.100 L) ¢
26.98 g
3.30 mol Al3+
b = 8.90 g Al3+
≤a
L
mol
(0.100 L) ¢
96.07 g
4.95 mol SO4 2b = 47.6 g SO4 2≤a
L
mol
(0.100 L) ¢
18.04 g
0.680 mol NH 4 +
b = 1.23 g NH 4 +
≤a
L
mol
(0.100 L) ¢
96.07 g
0.340 mol SO4 2b = 3.27 g SO4 2≤a
L
mol
0.0628 mol Mg 2+
24.31 g
(0.100 L) ¢
b = 0.153 g Mg 2+
≤a
L
mol
(0.100 L) ¢
13.
pH = -log[H+] [H+] = 10-pH
(a)
(b)
(c)
14.
83.45 g
0.126 mol ClO3 b = 1.05 g ClO3 ≤a
L
mol
[H+] = 3.16 * 10-9
[H+] = 1.0 * 10-7
[H+] = 3.16 * 10-3
pH = -log[H+] [H+] = 10-pH
(a)
[H+] = 3.98 * 10-3
- 215 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 216
- Chapter 15 (b)
(c)
15.
(a)
[H+] = 1.0 * 10-10
[H+] = tap water with a pH = 3.98 * 10 - 7
1.0 mol NaCl
b = 0.030 mol NaCl
1000 mL
1.0 mol NaCl
(40.0 mL)a
b = 0.040 mol NaCl
1000 mL
Total mol NaCl = 0.030 mol + 0.040 mol = 0.070 mol NaCl
0.070 mol NaCl
= 1.0 M NaCl
0.070 L
1.0 mol Na+
(1.0 M NaCl) ¢
≤ = 1.0 M Na+
1.0 mol NaCl
(30.0 mL)a
(1.0 M NaCl) ¢
1.0 mol Cl≤ = 1.0 M Cl1.0 mol NaCl
(b)
HCl + NaOH ¡ NaCl + H 2O
1L
1.0 mol
(30.0 mL HCl)a
ba
b = 0.030 mol HCl
1000 mL
L
1L
1.0 mol
(30.0 mL NaOH)a
ba
b = 0.030 mol NaOH
1000 mL
L
0.030 mol HCl reacts with 0.030 mol NaOH and produces 0.030 mol NaCl. The
final volume is 0.060 L. 0.030 mol NaCl>0.060 L = 0.50 M NaCl. Since there is
one mole each of sodium and chloride ions per mole of NaCl, the molar
concentration of Na+ and Cl- will be 0.50 M Na+ and 0.50 M Cl-.
(c)
KOH + HCl ¡ KCl + H 2O
1L
0.40 mol KOH
ba
b = 0.040 mol KOH
1000 mL
L
0.80 mol HCl
1L
ba
b = 0.080 mol HCl
(100.0 mL)a
1000 mL
L
0.040 mol KOH reacts with 0.040 mol HCl. 0.040 mol HCl remains unreacted and
0.040 mol KCl is produced. The final volume is 200.0 mL and contains 0.040 mol
HCl and 0.040 mol KCl. Moles of ions are: 0.040 mol H +, 0.040 mol K +, and
0.080 mol Cl-. Concentrations of ions are:
0.040 mol H +
= 0.20 M H +
molarity K + = molarity H +
0.200 L
0.080 mol Cl= 0.40 M Cl0.200 L
(100.0 mL)a
- 216 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 217
- Chapter 15 16.
(a)
100.0 mL of 2.0 M KCl and 100.0 mL of 1.0 M CaCl 2 are mixed, giving a final
volume of 200.0 mL and concentrations of 1.0 M KCl and 0.50 M CaCl 2 . The
concentration of K + will be 1.0 M and the concentration of Ca2+ will be 0.50 M.
The chloride ion concentration will be 2.0 M (1.0 M from the KCl and 2(0.50 M)
from the CaCl 2).
(b)
(35.0 mL)a
0.20 mol Ba(OH)2
1L
b¢
≤ = 0.0070 mol Ba(OH)2
1000 mL
L
0.20 mol H 2SO4
1L
b¢
≤ = 0.0070 mol H 2SO4
1000 mL
L
Final volume = 35.0 mL + 35.0 mL = 70.0 mL
(35.0 mL)a
H 2SO4
+ Ba(OH)2 ¡ BaSO4(s) + 2 H 2O
0.0070 mol
0.0070 mol
0.0070 mol
0.014 mol
The H 2SO4 and the Ba(OH)2 react completely producing insoluble BaSO4 and
H 2O. No ions are present in solution.
(c)
2.0 mol
b = 1.0 mol NaCl
L
1.00 mol
(1.00 L AgNO3)a
b = 1.0 mol AgNO3
L
NaCl(aq) + AgNO3(aq) ¡ AgCl(s) + NaNO3(aq)
1.0 mol
1.0 mol
1.0 mol
1.0 mol
(0.500 L NaCl)a
The AgCl is insoluble and produces no ions. The 1.0 mol NaNO3 will produce
1.0 mol Na+ ions and 1.0 mol NO 3- ions. The final volume of the solution is 1.5 L.
The concentration of the ions are:
1.0 mol Na+
= 0.67 M Na+
1.5 L
17.
1.0 mol NO3= 0.67 M NO31.5 L
The reaction of HCl and NaOH occurs on a 1 : 1 mole ratio.
HCl + NaOH ¡ NaCl + H 2O
At the endpoint in these titration reactions, equal moles of HCl and NaOH will have
reacted. Moles = (molarity)(volume). At the endpoint, mol HCl = mol NaOH.
Therefore, at the endpoint,
M BVB
MA =
M AVA = M BVB
VA
- 217 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 218
- Chapter 15 (a)
18.
19.
20.
21.
(37.70 mL)(0.728 M) = (40.13 mL)(M HCl)
(37.70 mL)(0.728 M)
= 0.684 M HCl
M HCl =
40.13 mL
(b)
(33.66 mL)(0.306 M)
= 0.542 M HCl
19.00 mL
(c)
(18.00 mL)(0.555 M)
= 0.367 M HCl
27.25 mL
The reaction of HCl and NaOH occurs on a 1 : 1 mole ratio.
HCl + NaOH ¡ NaCl + H 2O
At the endpoint in these titration reactions, equal moles of HCl and NaOH will have
reacted. Moles = (molarity)(volume). At the endpoint, mol HCl = mol NaOH.
Therefore, at the endpoint,
M AVA
M AVA = M BVB
MB =
VB
(a)
(37.19 mL)(0.126 M)
= 0.147 M NaOH
31.91 mL
(b)
(48.04 mL)(0.482 M)
= 0.964 M NaOH
24.02 mL
(c)
(13.13 mL)(1.425 M)
= 0.4750 M NaOH
39.39 mL
(a)
SO4 2-(aq) + Ba2+(aq) ¡ BaSO4(s)
(b)
CaCO3(s) + 2 H +(aq) ¡ Ca2+(aq) + CO2(g) + H 2O(l)
(c)
Mg(s) + 2 HC2H 3O2(aq) ¡ Mg 2+(aq) + 2 C2H 3O2 -(aq) + H 2(g)
(a)
H 2S(g) + Cd2+(aq) ¡ CdS(s) + 2 H +(aq)
(b)
Zn(s) + 2 H +(aq) ¡ Zn2+(aq) + H 2(g)
(c)
Al3+(aq) + PO4 3-(aq) ¡ AlPO4(s)
The more acidic solution is listed followed by an explanation.
(a)
(b)
1 molar H 2SO4 . The concentration of H + in 1 M H 2SO4 is greater than 1 M since
there are two ionizable hydrogens per mole of H 2SO4 . In HCl the concentration of
H + will be 1 M, since there is only one ionizable hydrogen per mole HCl.
1 molar HCl. HCl is a strong electrolyte, producing more H + than HC2H 3O2 which
is a weak electrolyte.
- 218 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 219
- Chapter 15 22.
The more acidic solution is listed followed by an explanation.
(a)
(b)
23.
2 molar HCl. 2 M HCl will yield 2 M H + concentration. 1 M HCl will yield 1 M H +
concentration.
1 molar H 2SO4 . Both are strong acids. The concentration of H + in 1 M H 2SO4 is
greater than in 1 M HNO3 because H 2SO4 has two ionizable hydrogens per mole
whereas HNO3 has only one ionizable hydrogen per mole.
3 HCl + Al(OH)3 ¡ AlCl 3 + 3 H 2O
g Al(OH)3 ¡ mol Al(OH)3 ¡ mol HCl ¡ mL HCl
0.245 M HCl contains 0.245 mol HCl>1000 mL
(10.0 g Al(OH)3)a
1 mol
3 mol HCl
1000 mL
b¢
b
≤a
78.00 g 1 mol Al(OH)3 0.245 mol
= 1.57 * 103 mL of 0.245 M HCl
24.
2 HCl + Ca(OH)2 ¡ CaCl 2 + 2 H 2O
M Ca(OH)2 ¡ mol Ca(OH)2 ¡ mol HCl ¡ mL HCl
0.245 M HCl contains 0.245 mol HCl>1000 mL
(0.0500 L Ca(OH)2)a
0.100 mol
2 mol HCl
1000 mL
b¢
b
≤a
L
1 mol Ca(OH)2 0.245 mol
= 40.8 mL of 0.245 M HCl
25.
NaOH + HCl ¡ NaCl + H 2O
First calculate the grams of NaOH in the sample.
L HCl ¡ mol HCl ¡ mol NaOH ¡ g NaOH
(0.01825 L HCl)a
a
26.
40.00 g
0.2406 mol 1 mol NaOH
ba
ba
b = 0.1756 g NaOH
L
1 mol HCl
mol
in the sample
0.1756 g NaOH
b(100) = 87.8% NaOH
0.200 g sample
NaOH + HCl ¡ NaCl H 2O
L HCl ¡ mol HCl ¡ mol NaOH ¡ g NaOH
(0.04990 L HCl)a
40.00 g
0.466 mol 1 mol NaOH
ba
ba
b = 0.930 g NaOH in the sample
L
1 mol HCl
mol
- 219 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 220
- Chapter 15 -
1.00 g sample - 0.930 g NaOH = 0.070 g NaCl in the sample
a
27.
0.070 g NaCl
b(100) = 7.0% NaCl in the sample
1.00 g sample
Zn + 2 HCl ¡ ZnCl 2 + H 2
This is a limiting reactant problem. First find the moles of Zn and HCl from the given
data and then identify the limiting reactant.
1 mol
g Zn ¡ mol Zn
b = 0.0765 mol Zn
(5.00 g Zn)a
65.39 g
(0.100 L HCl)a
0.350 mol
b = 0.0350 mol HCl
L
Therefore Zn is in excess and HCl is the limiting reactant.
(0.0350 mol HCl) ¢
1 mol H 2
≤ = 0.0175 mol H 2 produced in the reaction
2 mol HCl
T = 27°C = 300. K
P = (700. torr) a
1 atm
b = 0.921 atm
760 torr
PV = nRT
V =
28.
(0.0175 mol H 2)(0.0821 L atm>mol K)(300. K)
nRT
=
= 0.468 L H 2
P
0.921 atm
Zn + 2 HCl ¡ ZnCl 2 + H 2
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
(5.00 g Zn)a
b = 0.0765 mol Zn
g Zn ¡ mol Zn
65.39 g
(0.200 L HCl)a
0.350 mol
b = 0.0700 mol HCl
L
Zn is in excess and HCl is the limiting reactant.
(0.0700 mol HCl) ¢
1 mol H 2
≤ = 0.0350 mol H 2
2 mol HCl
T = 27°C = 300. K
P = (700. torr) a
1 atm
b = 0.921 atm
760 torr
PV = nRT
(0.0350 mol H 2)(0.0821 L atm>mol K)(300. K)
nRT
=
= 0.936 L H 2
V =
P
0.921 atm
- 220 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 221
- Chapter 15 29.
30.
31.
32.
33.
34.
35.
Calculation of the pH solutions:
pH = -log11 * 10-22 = 2.0
(a)
H + = 0.01 M = 1 * 10-2 M;
(b)
H + = 1.0 M;
(c)
H + = 6.5 * 10-9 M;
(a)
H + = 1 * 10-7 M;
(b)
H + = 0.50 M;
(c)
H + = 0.00010 M = 1.0 * 10-4 M;
(a)
Orange juice = 3.7 * 10-4 M H +
pH = -log13.7 * 10-42 = 3.43
(b)
Vinegar = 2.8 * 10-3 M H +
pH = -log12.8 * 10-32 = 2.55
(a)
Black coffee = 5.0 * 10-5 M H +
pH = -log15.0 * 10-52 = 4.30
(b)
Limewater = 3.4 * 10-11 M H +
pH = -log13.4 * 10-112 = 10.47
(a)
NH3 is a weak base
NH3(aq) ∆ NH4+(aq) + H2O(aq)
(b)
HCl is a strong acid
HCl(aq) ¡ H + (aq) + Cl - (aq)
(c)
KOH is a strong base
(d)
HC2H3O2 is a weak acid
(a)
H2C2O4 is a weak acid
(b)
Ba(OH)2 is a strong base
(c)
HClO4 is a strong acid
(d)
HBr is a strong acid
(a)
acidic; pH = 4
(d)
basic; pH = 9
(b)
basic
(e)
neutral; pH = 7
(c)
basic
(f)
acidic
pH = -log 1.0 = 0
pH = -log16.5 * 10-92 = 8.19
pH = -log11 * 10-72 = 7.0
pH = -log15.0 * 10-12 = 0.30
pH = -log11.0 * 10-42 = 4.00
H2O
H2O
H2O
KOH ¡ K + (aq) + OH - (aq)
H2O
HC2H3O2(aq) ∆ H + (aq) + C2H3O2- (aq)
H2O
H2C2O4(aq) ∆ 2 H + (aq) + C2O24 - (aq)
H2O
Ba(OH)2 ¡ Ba2 + (aq) + 2OH - (aq)
H2O
HClO4(aq) ¡ H + (aq) + ClO4- (aq)
HBr(aq) ¡ H + (aq) + Br - (aq)
- 221 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 222
- Chapter 15 36.
(a)
CaCl 2(s) ¡ Ca2+(aq) + 2 Cl-(aq)
For each CaCl 2 ionic compound, 1 calcium ion and 2 chloride ions result.
Ca2+
(b)
Cl–
Cl–
KF(s) ¡ K +(aq) + F -(aq)
For each KF ionic compound, 1 potassium ion and 1 fluoride ion result.
K+
(c)
F–
AlBr3(s) ¡ Al3+(aq) + 3 Br -(aq)
For each AlBr3 ionic compound, 1 aluminum ion and 3 bromide ions result.
Al3+
37.
Br –
CaI 2 ¡ Ca2+ + 2 I -
¢
38.
Br –
0.520 mol I 1 mol Ca2+
0.260 mol Ca2+
≤¢
≤ = ¢
≤ = 0.260 M Ca2+
L
2 mol I
L
Ba(OH)2 + 2 HCl ¡ BaCl 2 + 2 H 2O
M HCl ¡ mol HCl ¡ mol Ba(OH)2 ¡ M Ba(OH)2
- 222 -
Br –
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 223
- Chapter 15 -
a
0.430 mol HCl
1L
ba
b (29.26 mL) = 0.0126 mol HCl
L
1000 mL
(0.0126 mol HCl)a
1 mol Ba(OH)2
b = 0.00630 mol Ba(OH)2
2 mol HCl
0.00630 mol Ba(OH)2
= 0.309 M Ba(OH)2
0.02040 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 CH 3OH at the same cost per pound as C2H 5OH.
Because CH 3OH has a lower molar mass than C2H 5OH, the CH 3OH 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.
41.
CH 3OH:
100. g
= 2.84 mol
34.04 g>mol
CH 3CH 2OH:
100. g
= 2.17 mol
46.07 g>mol
A hydronium ion is a hydrated hydrogen ion.
H+
+
(hydrogen ion)
H 2O
¡
H 3O +
(hydronium ion)
42.
Freezing point depression is directly related to the concentration of particles in the
solution.
HCl
C12H 22O11 7
HC2H 3O2
7
7
CaCl 2
Highest freezing point
Lowest freezing point
1 mol
2 mol
3 mol (particles in solution)
1+ mol
43.
(a)
100°C
25°C
pH = -log11 * 10-62 = 6.0
pH = -log11 * 10-72 = 7.0
pH of H 2O is greater at 25°C
(b)
1 * 10-6 7 1 * 10-7 so, H + concentration is higher at 100°C.
(c)
The water is neutral at both temperatures, because the H 2O ionizes into equal
concentrations of H + and OH - at any temperature.
- 223 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 224
- Chapter 15 44.
As the pH changes by 1 unit, the concentration of H + 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 H 2O. We need to find the total number
of moles and then calculate the mole percent of each component.
a
1000 g H 2O
b = 55.49 mol H 2O
18.02 g>mol
55.49 mol H 2O + 1.00 mol solute = 56.49 total moles
a
1.00 mol solute
b (100) = 1.77% solute
56.49 mol
55.49 mol H 2O
¢
≤ (100) = 98.22% H 2O
56.49 mol
46.
Na 2CO3 + 2 HCl ¡ 2 NaCl + CO2 + H 2O
g Na 2CO3 ¡ mol Na 2CO3 ¡ mol HCl ¡ M HCl
(0.452 g Na 2CO3)a
47.
1 mol
2 mol HCl
1
b¢
b = 0.201 M HCl
≤a
106.0 g 1 mol Na 2CO3 0.0424 L
2 HCl + Ca(OH)2 ¡ CaCl 2 + 2 H 2O
g Ca(OH)2 ¡ mol Ca(OH)2 ¡ mol HCl ¡ mL HCl
(2.00 g Ca(OH)2)a
48.
1 mol
2 mol HCl
1000 mL
b¢
b = 437 mL of 0.1234 M HCl
≤a
74.10 g 1 mol Ca(OH)2 0.1234 mol
KOH + HNO3 ¡ KNO3 + H 2O
L HNO3 ¡ mol HNO3 ¡ mol KOH ¡ g KOH
(0.05000 L HNO3)a
49.
56.11 g
0.240 mol
1 mol KOH
b¢
b = 0.673 g KOH
≤a
L
1 mol HNO3
mol
pH of 1.0 L solution containing 0.1 mL of 1.0 M HCl
1.0 L
1 mol HCl
ba
b = 1 * 10-4 mol HCl added
(0.1 mL)a
1000 mL
L
1 * 10-4 mol HCl
= 1 * 10-4 M HCl
1.0 L
- 224 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 225
- Chapter 15 -
1 * 10-4 M HCl produces 1 * 10-4 M H +
pH = -log11 * 10-42 = 4.0
50.
Dilution problem
V1M 1 = V2M 2
51.
V1 =
V2M 2
(50.0 L)(5.00 M)
=
= 13.9 L of 18.0 M H 2SO4
M1
18.0 M
NaOH + HCl ¡ NaCl + H 2O
1 mol
(3.0 g NaOH)a
b = 0.075 mol NaOH
40.00 g
1L
0.10 mol
ba
b = 0.050 mol HCl
(500. mL HCl)a
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.
Ba(OH)2(aq) + 2 HCl(aq) ¡ BaCl 2(aq) + 2 H 2O(l)
0.35 mol
(0.380 L Ba(OH)2)a
b = 0.13 mol Ba(OH)2
L
0.13 mol Ba(OH)2 ¡ 0.26 mol OH 0.65 mol
(0.5000 L HCl)a
b = 0.33 mol HCl
L
0.33 mol HCl ¡ 0.33 mol H +
0.33 mol H + will neutralize 0.26 mol OH - and leave 0.07 mol H +10.33 - 0.262
remaining in solution.
Total volume = 500.0 mL + 380 mL = 880 mL (0.88 L)
[H +] in solution =
0.07 mol H +
= 0.08 M H +
0.88 L
pH = -log[H +] = -log18 * 10-22 = 1.1
53.
(0.05000 L HCl)a
(a)
0.2000 mol
b = 0.01000 mol HCl = 0.01000 mol H + in 50.00 mL HCl
L
no base added: pH = -log(0.2000) = 0.700
- 225 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 226
- Chapter 15 -
(b)
10.00 mL base added: (0.01000 L)a
0.2000 mol
b = 0.002000 mol NaOH
L
= 0.002000 mol OH (0.01000 mol H +) - (0.002000 mol OH -) = 0.00800 mol H + in 60.00 mL solution
[H +] =
(c)
0.00800 mol
0.06000 L
pH = -log a
0.00800
b = 0.880
0.06000
25.00 mL base added:
0.2000 mol
b = 0.005000 mol NaOH = mol OH (0.02500 L)a
L
(0.01000 mol H +) - (0.005000 mol OH -) = 0.00500 mol H + in 75.00 mL solution
[H +] =
(d)
0.00500 mol
0.07500 L
pH = -log a
0.00500
b = 1.2
0.07500
49.00 mL base added:
(0.04900 L)a
0.2000 mol
b = 0.009800 mol NaOH = mol OH L
(0.01000 mol H +) - (0.009800 mol OH -) = 0.00020 mol H + in 99.00 mL solution
[H +] =
(e)
0.00020 mol
0.09900 L
pH = -log a
0.00020
b = 2.69
0.09900
49.90 mL base added:
0.2000 mol
(0.04990 L)a
b = 0.009980 mol NaOH = mol OH L
(0.01000 mol H +) - (0.009980 mol OH -) = 2 * 10-5 mol H + in 99.90 mL
solution
[H +] =
(f)
2 * 10-5 mol
0.09990 L
pH = -log ¢
2 * 10-5
≤ = 3.7
0.09990
49.99 mL base added:
0.2000 mol
(0.04999 L)a
b = 0.009998 mol NaOH = mol OH L
(0.01000 mol H +) - (0.009998 mol OH -) = 2 * 10-6 mol H + in 99.99 mL
solution
- 226 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 227
- Chapter 15 -
[H +] =
(g)
2 * 10-6 mol
0.09999 L
pH = -log ¢
2 * 10-6
≤ = 4.7
9.999 * 10-2
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 = 7.0
8
7
6
pH
5
4
3
2
1
0
54.
10
20
30
mL NaOH
40
50
(a)
2 NaOH(aq) + H 2SO4(aq) ¡ Na 2SO4(aq) + 2 H 2O(l)
(b)
mol H 2SO4 ¡ mol NaOH ¡ mL NaOH
(c)
55.
0
(0.0050 mol H 2SO4) ¢
2 mol NaOH
1000 mL
b = 1.0 * 102 mL NaOH
≤a
1 mol H 2SO4
0.10 mol
(0.0050 mol H 2SO4) ¢
1 mol Na 2SO4 142.1 g
b = 0.71 g Na 2SO4
≤a
1 mol H 2SO4
mol
HNO3 + KOH ¡ KNO3 + H 2O
M AVA = M BVB
(M A)(25 mL) = (0.60 M)(50.0 mL)
M A = 1.2 M (diluted solution)
Dilution problem M 1V1 = M 2V2
(M 1)(10.0 mL) = (1.2 M)(100.00 mL)
M 1 = 12 M HNO3 (original solution)
56.
Yes, adding water changes the concentration of the acid, which changes the concentration
of the [H +], 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.
- 227 -
HEINS15-207-228v4.qxd
12/30/06
2:40 PM
Page 228
- Chapter 15 57.
First determine the molarity of the two HCl solutions. Take the antilog of the pH value to
obtain the [H +].
pH = - 0.300;
H+ = 2.00 M = 2.00 M HCl
pH = - 0.150;
H+ = 1.41 M = 1.41 M HCl
Now treat the calculation as a dilution problem.
V1M1
V1M1 = V2M2 V2 =
M2
(200 mL HCl)(2.00 M)
= 284 mL solution
1.41 M
284 mL - 200 mL = 84 mL H 2O to be added
58.
(lactic acid has one acidic H)
mol acid = mol base
1.0 g acid
0.65 mol
= (0.017 L)a
b = 0.01105 mol
molar mass
L
1.0 g
= 0.01105 mol
molar mass
1.0 g
= 90.17 g/mol
0.01105 mol
mass of empirical formula (HC3H 5O3) = 90.17 g>mol
molar mass = mass of empirical formula
Therefore the molecular formula is HC3H 5O3
- 228 -