Chapter 15 Solutions
1.
A homogeneous mixture is a combination of two (or more) pure substances that is
uniform in composition and appearance throughout. Examples of homogeneous mixtures
in the real world include rubbing alcohol (70% isopropyl alcohol, 30% water) and
gasoline (a mixture of hydrocarbons).
2.
solvent, solute
3.
When an ionic solute dissolves in water, a given ion is pulled into solution by the
attractive ion–dipole force exerted by several water molecules. For example, in
dissolving a positive ion, the ion is approached by the negatively charged end of several
water molecules: if the attraction of the water molecules for the positive ion is stronger
than the attraction of the negative ions near it in the crystal, the ion leaves the crystal and
enters solution. After entering solution, the dissolved ion is surrounded completely by
water molecules, which tends to prevent the ion from reentering the crystal.
4.
One substance will mix with and dissolve in another substance if the intermolecular
forces are similar in the two substances, so that when the mixture forms, the forces
between particles in the mixture will be similar to the forces present in the separate
substances. Sugar and ethyl alcohol molecules both contain polar –OH groups, which are
comparable to the polar –OH structure in water. Sugar or ethyl alcohol molecules can
hydrogen-bond with water molecules and intermingle with them freely to form a solution.
Substances like petroleum (whose molecules contain only carbon and hydrogen) are very
nonpolar and cannot form interactions with polar water molecules.
5.
saturated
6.
unsaturated
7.
variable
8.
large
9.
Increasing the surface area of a solid increases the amount of solid that comes in contact
with the solvent. Typically, particles of a solid are broken into smaller pieces by grinding.
This increases the surface-to-volume ratio of each particle and thus increases the amount
of solid in contact with the solvent.
10.
An increase in temperature means an increase in the average kinetic energy. Thus, in a
warmer solution the particles of the liquid solvent are moving more rapidly. Because of
this, there is an increased rate in solute–solvent interaction, which increases the rate of
dissolving.
11.
An increase in temperature means an increase in average kinetic energy. In a warmer
solution, the solvent and solute particles are moving more rapidly. If the solute particles
are gaseous, faster moving particles are more likely to have enough energy to escape
from the liquid.
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154
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155
Solutions
12.
a.
b.
c.
d.
13.
14.
15.
16.
5.00 g CaCl2
1.00 g CaCl2
100 = 5.00% CaCl2
(19.0 g H 2 O + 1.00 g CaCl2 )
15.00 g CaCl2
(285 g H 2 O + 15.00 g CaCl2 )
100 = 5.00% CaCl2
0.00200 g CaCl2
100 = 5.00% CaCl2
(0.0380 g H 2 O + 0.00200 g CaCl2 )
To say that a solution is x% NaCl means that 100 g of the solution would contain x g of
NaCl.
a.
11.5 g solution b.
6.25 g solution c.
54.3 g solution d.
452 g solution 5.34 g KCl
(5.34 g KCl + 152 g H 2 O)
6.25 g NaCl
100 g solution
11.5 g NaCl
100 g solution
0.91 g NaCl
100 g solution
12.3 g NaCl
100 g solution
100 =
67.1 g CaCl2
(67.1 g CaCl2 + 275 g H 2 O)
285 g solution 285 g solution 17.
100 = 5.00% CaCl2
(95.0 g H 2 O + 5.00 g CaCl2 )
100 g solution
7.50 g Na 2 CO3
100.0 g solution
g pentane = 93 g solution = 0.719 g NaCl
= 0.49 g NaCl
= 55.6 g NaCl
5.34 g
157.34 g
100 = 3.39% KCl
100 = 19.6% CaCl2
5.00 g NaCl
g heptane = 93 g solution = 0.719 g NaCl
= 14.3 g NaCl
= 21.4 g Na2CO3
5.2 g heptane
100. g solution
2.9 g pentane
100. g solution
= 4.8 g heptane
= 2.7 g pentane
g hexane = 93 g solution – 4.8 g heptane – 2.7 g pentane = 86 g hexane
18.
0.105
19.
0.221 mol Ca2+; 0.442 mol Cl–
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156
Chapter 15
20.
To say that a solution has a concentration of 5 M means that in 1 L of solution (not
solvent) there would be 5 mol of solute: to prepare such a solution one would place 5 mol
of NaCl in a 1-L flask, and then add whatever amount of water is necessary so that the
total volume would be 1 L after mixing. The NaCl will occupy some space, so the
amount of water to be added will be less than 1.00 L.
21.
Molarity =
a.
moles of solute
liters of solution
250 mL = 0.25 L
M=
b.
22.
M=
Molarity =
a.
0.50 mol KBr
0.50 mol KBr
0.75 L solution
0.50 mol KBr
moles of solute
liters of solution
Molar mass of CuCl2 = 134.45 g
M=
1 mol
134.45 g
0.0316 mol CuCl2
0.125 L solution
125 mL = 0.125 L
= 0.0316 mol CuCl2
= 0.253 M
Molar mass of NaHCO3 = 84.01 g
0.101 g NaHCO3 M=
c.
= 0.67 M
= 0.50 M
1.0 L solution
4.25 g CuCl2 b.
= 1.0 M
0.500 L solution
750 mL = 0.75 L
M=
d.
= 2.0 M
0.25 L solution
500 mL = 0.500 L
M=
c.
0.50 mol KBr
1 mol
84.01 g
0.00120 mol NaHCO3
0.0113 L solution
11.3 mL = 0.0113 L
= 0.00120 mol NaHCO3
= 0.106 M
Molar mass of Na2CO3 = 105.99 g
52.9 g Na2CO3 M=
World of Chemistry
1 mol
105.99 g
0.499 mol Na 2 CO3
1.15 L solution
= 0.499 mol Na2CO3
= 0.434 M
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157
Solutions
d.
Molar mass of KOH = 56.11 g
0.14 mg KOH M=
23.
10 mg
1 mol
0.0015 L solution
56.11 g
1 mol
101.1 g
0.448 mol KNO 3
= 0.448 mol KNO3
molar mass of I2 = 253.8 g
M=
1 mol
253.8 g
225 mL = 0.225 L
= 1.99 M
0.225 L solution
5.15 g I2 = 2.50 10–6 mol KOH
= 1.67 10–3 M = 1.7 10–3 M
molar mass of KNO3 = 101.1 g
M=
25.
3
2.50 x 106 mol KOH
45.3 g KNO3 24.
1g
1.5 mL = 0.0015 L
225 mL = 0.225 L
= 0.0203 mol I2
0.0203 mol I 2
0.225 L solution
= 0.0902 M
molar mass of FeCl3 = 162.2 g
1.01 g FeCl3 1 mol FeCl3
162.2 g FeCl3
= 0.00623 mol FeCl3
10.0 mL = 0.0100 L
M=
0.00623 mol FeCl3
0.0100 L solution
= 0.623 M
Since one mole of FeCl3 contains one mole of Fe3+ and three moles of Cl–, the solution is
0.623 M in Fe3+ and 3(0.623) = 1.87 M in Cl–
26.
molar mass of NaOH = 40.00 g
495 g NaOH =
M=
27.
a.
1 mol
40.00 g
12.4 mol NaOH
20.0 L solution
= 12.4 mol NaOH
= 0.619 M
molar mass of HNO3 = 63.02 g
0.127 L solution 0.105 mol HNO 3
0.0133 mol HNO3 World of Chemistry
1.00 L solution
63.02 g HNO 3
1 mol HNO 3
127 mL = 0.127 L
= 0.0133 mol HNO3
= 0.838 g HNO3
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158
Chapter 15
b.
molar mass of NH3 = 17.03 g
0.155 L solution 2.34 mol NH3 c.
15.1 mol NH 3
1.00 L solution
17.03 g NH 3
molar mass KSCN = 97.19 g
2.01 x 10 3 mol KSCN
0.0122 L solution 0.0299 mol HCl a.
0.0102 L 5.51 L 5.51 L c.
12.2 mL = 0.0122 L
2.45 mol HCl
1.00 L solution
36.46 g HCl
1 mol HCl
0.251 mol NH 4 Cl
1 L solution
53.49 g NH 4 Cl
1 mol NH 4 Cl
= 0.0299 mol HCl
= 1.09 g HCl
450 mL = 0.450 L
= 0.113 mol NH4Cl
= 6.04 g NH4Cl
10.2 mL = 0.0102 L
0.0102 L b.
= 0.491 g KSCN
1 mol KSCN
molar mass of NH4Cl = 53.49 g
0.113 mol NH4Cl 29.
97.19 g KSCN
molar mass of HCl = 36.46 g
0.450 L solution = 5.05 10–3 mol KSCN
1.00 L solution
5.05 10–3 mol KSCN 28.
= 2.34 mol NH3
= 39.9 g NH3
1 mol NH 3
2.51 L solution d.
155 mL = 0.155 L
0.451 mol AlCl3
1.00 L
0.451 mol AlCl3
1.00 L
0.103 mol Na 3 PO 4
1.00 L
0.103 mol Na 3 PO 4
1.00 L
1 mol Al3+
1 mol AlCl3
3 mol Cl
1 mol AlCl3
= 4.60 10–3 mol Al3+
= 1.38 10–2 mol Cl–
3 mol Na +
1 mol Na 3 PO 4
1 mol PO 4 3
1 mol Na 3 PO 4
= 1.70 mol Na+
= 0.568 mol PO43–
1.75 mL = 0.00175 L
0.00175 L 0.00175 L World of Chemistry
1.25 mol CuCl2
1.00 L
1.25 mol CuCl2
1.00 L
1 mol Cu 2+
1 mol CuCl2
2 mol Cl
1 mol CuCl2
= 2.19 10–3 mol Cu2+
= 4.38 10–3 mol Cl–
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159
Solutions
d.
25.2 mL = 0.0252 L
0.00157 mol Ca(OH) 2
0.0252 L 1.00 L
0.00157 mol Ca(OH) 2
0.0252 L 30.
1.00 L
1 mol Ca 2+
1 mol Ca(OH) 2
2 mol OH 1 mol Ca(OH) 2
= 3.96 10–5 mol Ca2+
= 7.91 10–5 mol OH–
250. mL = 0.250 L
0.250 L solution 0.100 mol AgNO3
1.00 L solution
= 0.0250 mol AgNO3
molar mass AgNO3 = 169.9 g
0.0250 mol AgNO3 31.
1 mol AgNO 3
= 4.25 g AgNO3
M1 V1 = M2 V2
a.
M1 = 0.251 M
M2 = ?
V1 = 125 mL
V2 = 250. + 125 = 375 mL
M2 =
b.
c.
(375 mL)
= 0.0837 M
M2 = ?
V1 = 445 mL
V2 = 445 + 250. = 695 mL
(0.499 M ) (445 mL)
(695 mL)
= 0.320 M
M1 = 0.101 M
M2 = ?
V1 = 5.25 L
V2 = 5.25 + 0.250 = 5.50 L
M2 =
d.
(0.251 M ) (125 mL)
M1 = 0.499 M
M2 =
(0.101 M ) (5.25 L)
(5.50 L)
= 0.0964 M
M1 = 14.5 M
M2 = ?
V1 = 11.2 mL
V2 = 11.2 + 250. = 261.2 mL
M2 =
32.
169.9 g AgNO 3
(14.5 M ) (11.2 mL)
(261.2 mL)
= 0.622 M
M1 V1 = M2 V2
HCl:
V2 =
World of Chemistry
M1 = 3.0 M
M2 = 12.1 M
V1 = 225 mL
V2 = ?
(3.0 M ) (225 mL)
(12.1 M )
= 55.8 mL = 56 mL
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160
Chapter 15
HNO3:
V2 =
33.
V2 = ?
= 42.45 mL = 42 mL
M1 = 3.0 M
M2 = 18.0 M
V1 = 225 mL
V2 = ?
(3.0 M ) (225 mL)
(18.0 M )
= 37.5 mL = 38 mL
M1 = 3.0 M
M2 = 17.5 M
V1 = 225 mL
V2 = ?
(3.0 M ) (225 mL)
(17.5 M )
H3PO4:
V2 =
V1 = 225 mL
(15.9 M )
HC2H3O2:
V2 =
M2 = 15.9 M
(3.0 M ) (225 mL)
H2SO4:
V2 =
M1 = 3.0 M
= 38.6 mL = 39 mL
M1 = 3.0 M
M2 = 14.9 M
V1 = 225 mL
V2 = ?
(3.0 M ) (225 mL)
(14.9 M )
= 45.3 mL = 45 mL
M1 V1 = M2 V2
M1 = 3.02 M
M2 = 0.150 M
V1 = ?
V2 = 125 mL = 0.125 L
V1 =
(0.150 M ) (0.125 L)
(3.02 M )
= 0.00621 L = 6.21 mL
The student could prepare her solution by transferring 6.21 mL of the 3.02 M NaOH
solution from a pipet or buret to a 125-mL volumetric flask, and then adding distilled
water to the calibration mark of the flask.
34.
M1 V1 = M2 V2
M1 = 0.200 M
M2 = 0.150 M
V1 = 500. mL = 0.500 L
V2 = ?
V2 =
(0.200 M ) (0.500 L)
(0.150 M )
= 0.667 L = 667 mL
Therefore, 667 – 500. = 167 mL of water must be added.
35.
27.2 mL = 0.0272 L
25.0 mL = 0.0250 L
mol AgNO3 = 0.0272 L solution World of Chemistry
0.104 mol AgNO 3
1.00 L solution
= 0.002829 mol AgNO3
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161
Solutions
1 mol Cl
0.002829 mol AgNO3 M=
36.
0.002829 mol Cl
= 0.002829 mol Cl–
1 mol AgNO 3
= 0.113 M
0.0250 L
Ba(NO3)2(aq) + Na2SO4(aq) BaSO4(s) + 2NaNO3(aq)
12.5 mL = 0.0125 L
moles Ba(NO3)2 = 0.0125 L 0.15 mol Ba(NO 3 ) 2
1.00 L
= 1.88 10–3 mol Ba(NO3)2
From the balanced chemical equation for the reaction, if 1.88 10–3 mol Ba(NO3)2 is to
be precipitated, then 1.88 10–3 mol Na2SO4 will be needed.
1.88 10–3 mol Na2SO4 37.
1 .00 L
0.25 mol Na 2SO 4
36.2 mL = 0.0362 L
= 0.0075 L required = 7.5 mL
37.5 mL = 0.0375 L
Since each formula unit of CaCO3 contains one Ca2+ ion, and since each Na2CO3 formula
unit contains one CO32– ion, we can say that
mol Ca2+ = 0.0362 L CaCl2 0.158 mol CaCl2
mol CO32– = 0.0375 L Na2CO3 1 L CaCl2
= 0.00572 mol Ca2+
0.149 mol Na 2 CO3
1 L Na 2 CO3
= 0.00559 mol CO32–
Since one Ca2+ reacts with one CO32–, Na2CO3 is the limiting reactant. Since 0.00559 mol
CO32– reacts, 0.00559 mol of CaCO3 will form.
molar mass CaCO3 = 100.1 g
0.00559 mol CaCO3 38.
100.1 g CaCO3
1 mol CaCO3
= 0.560 g CaCO3
Pb(NO3)2(aq) + K2CrO4(aq) PbCrO4(s) + 2KNO3(aq)
molar masses: Pb(NO3)2, 331.2 g; PbCrO4, 323.2 g
1.00 g Pb(NO3)2 1 mol Pb(NO3 )2
331.2 g Pb(NO3 )2
= 0.003019 mol Pb(NO3)2
25.0 mL = 0.0250 L
0.0250 L solution 1.00 mol K 2 CrO 4
1.00 L solution
= 0.0250 mol K2CrO4
Pb(NO3)2 is the limiting reactant: 0.003019 mol PbCrO4 will form.
0.003019 mol PbCrO4 World of Chemistry
323.2 g PbCrO 4
1 mol PbCrO 4
= 0.976 g PbCrO4
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162
39.
Chapter 15
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
25.0 mL = 0.0250 L
0.0250 L 0.150 mol NaOH
1.00 L
0.00375 mol NaOH 0.00375 mol HCl 40.
= 0.00375 mol NaOH
1 mol HCl
1 mol NaOH
1 L solution
= 0.00375 mol HCl
= 0.01875 L = 18.8 mL
0.200 mol HCl
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
50.0 mL = 0.0500 L
0.0500 L 0.104 mol HCl
41.
a.
= 0.00520 mol HCl
1.00 L solution
0.00520 mol HCl M=
48.7 mL = 0.0487 L
1 mol NaOH
1 mol HCl
0.00520 mol NaOH
0.0487 L
= 0.00520 mol NaOH
= 0.107 M
NaOH(aq) + HC2H3O2(aq) NaC2H3O2(aq) + H2O(l)
25.0 mL = 0.0250 L
0.0250 L 0.154 mol HC 2 H 3O 2
1.00 L
0.00385 mol HC2H3O2 0.00385 mol NaOH b.
= 0.00385 mol HC2H3O2
1 mol NaOH
1 mol HC 2 H 3O 2
1.00 L
1.00 mol NaOH
= 0.00385 mol NaOH
= 0.00385 L = 3.85 mL NaOH
HF(aq) + NaOH(aq) NaF(aq) + H2O(l)
35.0 mL = 0.0350 L
0.0350 L 0.102 mol HF
1.00 L
0.00357 mol HF 1 mol HF
1 mol NaOH
0.00357 mol NaOH c.
= 0.00357 mol HF
= 0.00357 mol NaOH
1.00 L
1.00 mol NaOH
= 0.00357 L = 3.57 mL
H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l)
10.0 mL = 0.0100 L
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163
Solutions
0.0100 L 0.143 mol H 3 PO 4
0.00143 mol H3PO4 0.00429 mol NaOH d.
= 0.00143 mol H3PO4
1.00 L
3 mol NaOH
= 0.00429 mol NaOH
1 mol H 3 PO 4
1.00 L
1.00 mol NaOH
= 0.00429 L = 4.29 mL
H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)
35.0 mL = 0.0350 L
0.0350 L 0.220 mol H 2SO 4
1.00 L
0.00770 mol H2SO4 2 mol NaOH
= 0.0154 mol NaOH
1 mol H 2SO 4
1.00 L
0.0154 mol NaOH 42.
= 0.00770 mol H2SO4
1.00 mol NaOH
= 0.0154 L = 15.4 mL
When H2SO4 reacts with OH–, the reaction is
H2SO4(aq) + 2OH–(aq) 2H2O(l) + SO42–(aq)
Since each mol of H2SO4 provides two moles of H+ ion, it is only necessary to take half a
mole of H2SO4 to provide one mole of H+ ion. The equivalent weight of H2SO4 is thus
half the molar mass.
43.
1.53 equivalents OH– ion are needed to react with 1.53 equivalents of H+ ion. By
definition, one equivalent of OH– ion exactly neutralizes one equivalent of H+ ion.
44.
N=
a.
number of equivalents of solute
number of liters of solution
equivalent weight NaOH = molar mass NaOH = 40.00 g
0.113 g NaOH 1 equiv NaOH
40.00 g
= 2.83 10–3 equiv NaOH
10.2 mL = 0.0102 L
N=
b.
2.83 x 103 equiv
0.0102 L
= 0.277 N
equivalent weight Ca(OH)2 =
molar mass
2
12.5 mg 1g
3
10 mg
1 equiv
37.05 g
=
74.10 g
= 37.05 g
2
= 3.37 10–4 equiv Ca(OH)2
100. mL = 0.100 L
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164
Chapter 15
N=
c.
3.37 x 103 equiv
0.100 L
= 3.37 10–3 N
equivalent weight H2SO4 =
molar mass
=
98.09 g
2
12.4 g 1 equiv
= 49.05 g
2
= 0.253 equiv H2SO4
49.05 g
155 mL = 0.155 L
N=
45.
46.
0.253 equiv
1 equiv NaOH
a.
0.134 M NaOH b.
0.00521 M Ca(OH)2 c.
4.42 M H3PO4 1 mol NaOH
= 0.134 N NaOH
2 equiv Ca(OH) 2
1 mol Ca(OH) 2
3 equiv H 3 PO 4
1 mol H 3 PO 4
= 0.0104 N Ca(OH)2
= 13.3 N H3PO4
molar mass H3PO4 = 98.0 g
35.2 g H3PO4 M=
1 mol H 3 PO 4
98.0 g H 3 PO 4
0.3592 H 3 PO 4
1.00 L
0.3592 M H3PO4 47.
= 1.63 N
0.155 L
= 0.3592 mol H3PO4
= 0.3592 M = 0.359 M
3 equiv H 3 PO 4
1 mol H 3 PO 4
= 1.08 N
H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)
0.145 M NaOH = 0.145 N NaOH
0.0562 L NaOH 0.145 equiv
1.00 L
56.2 mL = 0.0562 L
= 0.00815 equiv NaOH
0.00815 equiv NaOH requires 0.00815 equiv H2SO4 to react.
0.00815 equiv H2SO4 48.
1.00 L
0.172 equiv
= 0.0474 L = 47.4 mL H2SO4 solution
2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l)
For the 0.125 N H2SO4:
Nacid Vacid = Nbase Vbase
(0.125 N) (24.2 mL) = (0.151 N) (Vbase)
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165
Solutions
Vbase = 20.0 mL of the 0.151 N NaOH solution needed
For the 0.125 M H2SO4:
Since each H2SO4 formula unit produces two H+ ions, the normality of this solution will
be twice its molarity.
0.125 M H2SO4 = 0.250 N H2SO4
Nacid Vacid = Nbase Vbase
(0.250 N) (24.1 mL) = (0.151 N) (Vbase)
Vbase = 39.9 mL of the 0.151 N NaOH solution needed
49.
Colligative properties are properties of a solution that depend only on the number, not the
identity, of the solute particles.
50.
For a solution to boil, bubbles must form in the solution. Solute particles block water
from entering these bubbles. It is not the nature of these particles that matters, but the
number of the particles; thus, it is a colligative property.
51.
Antifreeze solution is a concentrated aqueous solution that has a lower freezing point
than water. It will also have a higher boiling point than water (a solute in water both
lowers the freezing point and raises the boiling point).
52.
millimol CoCl2 = 50.0 mL 0.250 M CoCl2 = 12.5 millimol CoCl2
This contains 12.5 millimol Co2+ and 25.0 millimol Cl–
millimol NiCl2 = 25.0 mL 0.350 M NiCl2 = 8.75 millimol NiCl2
This contains 8.75 millimol Ni2+ and 17.5 millimol Cl–
Total millimol Cl– after mixing = 25.0 + 17.5 = 42.5 millimol Cl–
Total volume after mixing = 50.0 mL + 25.0 mL = 75.0 mL
Mcobalt(II) ion =
Mnickel(II) ion =
Mchloride ion =
53.
12.5 millimol Co2+
75.0 mL
8.75 millimol Ni2+
= 0.117 M
75.0 mL
42.5 millimol Cl
75.0 mL
= 0.167 M
= 0.567 M
AgNO3(s) + NaCl(aq) AgCl(s) + NaNO3(aq)
molar masses: AgNO3, 169.9 g; AgCl, 143.4 g
10.0 g AgNO3 1 mol AgNO 3
169.9 g AgNO 3
= 0.05886 mol AgNO3
50. mL = 0.050 L
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Chapter 15
0.050 L 1.0 x 102 mol NaCl
= 0.00050 mol NaCl
1.00 L
NaCl is the limiting reactant. 0.00050 mol AgCl form.
0.00050 mol AgCl 143.4 g AgNO 3
1 mol
= 0.072 g AgCl (72 mg)
Since 1 mol AgNO3 contains 1 mol Ag+, the mol Ag+ remaining in solution =
0.05886 – 0.00050 = 0.05836 mol AgNO3
0.05836 mol AgNO3 = 0.05836 mol Ag+
Msilver ion =
54.
0.05836 mol Ag +
0.050 L
= 1.167 M = 1.2 M
Ba(NO3)2(aq) + H2SO4(aq) BaSO4(s) + 2HNO3(aq)
37.5 mL = 0.0375 L
0.0375 L 0.221 mol H 2SO 4
1.00 L
= 0.00829 mol H2SO4
Since the coefficients of Ba(NO3)2 and H2SO4 in the balanced chemical equation for the
reaction are both one, then 0.00829 mol of Ba2+ ion will be precipitated from the solution
as BaSO4.
molar mass BaSO4 = 233.4 g
0.00829 mol BaSO4 55.
233.4 g BaSO 4
1 mol BaSO 4
= 1.93 g BaSO4 precipitate
molar mass H2O = 18.0 g
1.0 L water = 1.0 103 mL water 1.0 103 g water
1.0 103 g H2O 56.
1 mol H 2 O
18.0 g H 2 O
= 56 mol H2O
molar mass CaCl2 = 111.0 g
14.2 g CaCl2 1 mol CaCl2
111.0 g CaCl2
= 0.128 mol CaCl2
50.0 mL = 0.0500 L
M=
0.128 mol CaCl2
0.0500 L
World of Chemistry
= 2.56 M
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All rights reserved.
167
Solutions
57.
M1 V1 = M2 V2
a.
M1 = 0.200 M
M2 = ?
V1 = 125 mL
V2 = 125 + 150. = 275 mL
M2 =
b.
V1 = 155 mL
V2 = 155 + 150. = 305 mL
59.
(0.250 M ) (155 mL)
(305 mL)
= 0.127 M
M1 = 0.250 M
M2 = ?
V1 = 0.500 L = 500. mL
V2 = 500. + 150. = 650. mL
(0.250 M ) (500. mL)
(650. mL)
= 0.192 M
M1 = 18.0 M
M2 = ?
V1 = 15 mL
V2 = 15 + 150. = 165 mL
M2 =
58.
= 0.0909 M
M2 = ?
M2 =
d.
(275 mL)
M1 = 0.250 M
M2 =
c.
(0.200 M ) (125 mL)
(18.0 M ) (15 mL)
(165 mL)
= 1.6 M
1 equiv HC 2 H 3O 2
a.
0.50 M HC2H3O2 b.
0.00250 M H2SO4 c.
0.10 M KOH 1 mol HC 2 H 3O 2
2 equiv H 2SO 4
1 mol H 2SO 4
1 equiv KOH
1 mol KOH
= 0.50 N HC2H3O2
= 0.00500 N H2SO4
= 0.10 N KOH
Nacid Vacid = Nbase Vbase
Nacid (10.0 mL) = (3.5 10–2 N)(27.5 mL)
Nacid = 9.6 10–2 N HNO3
60.
[A] =
4 mol
1.0 L
= 4.0 M, [B] =
6 mol
4.0 L
= 1.5 M, [C] =
4 mol
2.0 L
= 2.0 M, [D] =
6 mol
2.0 L
= 3.0 M,
[A] > [D] > [C] > [B]
World of Chemistry
Copyright Houghton Mifflin Company.
All rights reserved.