Silberberg7e solution manual ch 04

CHAPTER 4 THREE MAJOR CLASSES OF
CHEMICAL REACTIONS
FOLLOW–UP PROBLEMS
4.1A

Plan: Examine each compound to see what ions, and how many of each, result when the compound is dissolved
in water and match one compound’s ions to those in the beaker. Use the total moles of particles and the molar
ratio in the compound’s formula to find moles and then mass of compound.
Solution:
H O

2
 Li+(aq) + Br–(aq)
a) LiBr(s) 

H O

2
 2Cs+(aq) + CO32–(aq)
Cs2CO3(s) 

H O

2
 Ba2+(aq) + 2Cl–(aq)
BaCl2(s) 
Since the beaker contains +2 ions and twice as many –1 ions, the electrolyte is BaCl2.
 0.05 mol Ba 2  ions   1 mol BaCl2   208.2 g BaCl 2 
b) Mass (g) of BaCl2 = 3 Ba 2  particles 
 1 Ba 2  particle   1 mol Ba 2  ions   1 mol BaCl 
2 




= 31.2 g BaCl2



4.1B



Plan: Write the formula for sodium phosphate and then write a balanced equation showing the ions that result
when sodium phosphate is placed in water. Use the balanced equation to determine the number of ions that result
when 2 formula units of sodium phosphate are placed in water. The molar ratio from the balanced equation gives
the relationship between the moles of sodium phosphate and the moles of ions produced. Use this molar ratio to
calculate the moles of ions produced when 0.40 mol of sodium phosphate is placed in water.
Solution:
a) The formula for sodium phosphate is Na3PO4. When the compound is placed in water, 4 ions are produced for
each formula unit of sodium phosphate: three sodium ions, Na+, and 1 phosphate ion, PO43–.
H O

2
 3Na+(aq) + PO43–(aq)
Na3PO4(s) 
If two formula units of sodium phosphate are placed in water, twice as many ions should be produced:

H O

2
 6Na+(aq) + 2PO43–(aq)

2Na3PO4(s) 
Any drawing should include 2 phosphate ions (each with a 3– charge) and 6 sodium ions (each with a 1+
charge).

b) Moles of ions = 0.40 mol Na3PO4
4.2A

4 mol ions
1 mol Na3 PO4

= 1.6 moles of ions

Plan: Write an equation showing the dissociation of one mole of compound into its ions. Use the given
information to find the moles of compound; use the molar ratio between moles of compound and moles of ions in
the dissociation equation to find moles of ions.
Solution:
a) One mole of KClO4 dissociates to form one mole of potassium ions and one mole of perchlorate ions.
H O

2
 K+(aq) + ClO4–(aq)
KClO4(s) 
Therefore, 2 moles of solid KClO4 produce 2 mol of K+ ions and 2 mol of ClO4– ions.

H O

2
 Mg2+(aq) + 2C2H3O2–(aq)
b)
Mg(C2H3O2)2(s) 

First convert grams of Mg(C2H3O2)2 to moles of Mg(C2H3O2)2 and then use molar ratios to determine the moles of
each ion produced.
 1 mol Mg(C2 H3O2 )2 
Moles of Mg(C2H3O2)2 =  354 g Mg(C2 H3O2 )2  
 = 2.48596 mol
 142.40 g Mg(C2 H3O2 )2 

The dissolution of 2.48596 mol Mg(C2H3O2)2(s) produces 2.49 mol Mg2+ and (2 x 2.48596) = 4.97 mol C2H3O2–
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4-1


H O

2
 2NH4+(aq) + CrO42–(aq)
c)
(NH4)2CrO4(s) 
First convert formula units to moles.
 1 mol (NH 4 ) 2 CrO 4 
Moles of (NH4)2CrO4 = 1.88 x 1024 FU 
= 3.121886 mol
 6.022 x 1023 FU 


The dissolution of 3.121886 mol (NH4)2CrO4(s) produces (2 x 3.121886) = 6.24 mol NH4+ and 3.12 mol CrO42–.



4.2B



Plan: Write an equation showing the dissociation of one mole of compound into its ions. Use the given
information to find the moles of compound; use the molar ratio between moles of compound and moles of ions in
the dissociation equation to find moles of ions.
Solution:
a) One mole of Li2CO3 dissociates to form two moles of lithium ions and one mole of carbonate ions.
H O

2
 2Li+(aq) + CO32–(aq)
Li2CO3(s) 
Therefore, 4 moles of solid Li2CO3 produce 8 mol of Li+ ions and 4 mol of CO32– ions.

H O

2
 2Fe3+(aq) + 3SO42–(aq)
b)
Fe2(SO4)3(s) 
First convert grams of Fe2(SO4)3 to moles of Fe2(SO4)3 and then use molar ratios to determine the moles of each
ion produced.

Moles of Fe2(SO4)3= (112 g Fe2(SO4)3)

1 mol Fe2 (SO4 )3
399.88 g Fe2 (SO4 )3

= 0.2801 = 0.280 mol Fe2(SO4)3

The dissolution of 0.280 mol Fe2(SO4)3(s) produces 0.560 mol Fe3+ (2 x 0.2801) and 0.840 mol SO42–
(3 x 0.2801)
H O

2
 Al3+(aq) + 3NO3–(aq)
c)
Al(NO3)3(s) 
First convert formula units of Al(NO3)3 to moles of Al(NO3)3 and then use molar ratios to determine the moles of
each ion produced.

Moles of Al(NO3)3= (8.09 x 1022 formula units Al(NO3)3)

1 mol Al(NO3 )3

6.022 x 1023 formula units Al(NO3 )3

= 0.1343 mol Al(NO3)3
The dissolution of 0.134 mol Al(NO3)3 (s) produces 0.134 mol Al3+ and 0.403 mol NO3– (3 x 0.1343)
4.3A

Plan: Convert the volume from mL to liters. Convert the mass to moles by dividing by the molar mass of KI.
Divide the moles by the volume in liters to calculate molarity.
Solution:
Amount (moles) of KI = 6.97 g KI

M=

4.3B

0.0420 mol KI

1000 mL

100. mL

1L

1 mol KI
166.0 g KI

= 0.420 M

Plan: Convert the volume from mL to liters. Convert the mass to moles by first converting mg to g and then
dividing by the molar mass of NaNO3. Divide the moles by the volume in liters to calculate molarity.
Solution:
Amount (moles) of NaNO3 = 175 mg NaNO3

M=
4.4A

= 0.0420 mol KI

0.00206 mol NaNO3

1000 mL

15.0 mL

1L

1g

1 mol NaNO3

1000 mg

85.00 g NaNO3

= 0.00206 mol NaNO3

= 0.137 M

Plan: Divide the mass of sucrose by its molar mass to change the grams to moles. Divide the moles of sucrose by
the molarity to obtain the volume of solution.

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4-2


Solution:
Volume (L) of solution =
 1 mol C12 H 22 O11 

1L
135 g C12 H22O11  


 = 0.11951 = 0.120 L
 342.30 g C12 H22 O11  3.30 mol C12 H 22 O11 
Road map:
Mass (g) of sucrose
Divide by M (g/mol)
Amount (moles) of sucrose
Divide by M (mol/L)
Volume (L) of solution
4.4B

Plan: Convert the volume from mL to L; then multiply by the molarity of the solution to obtain the moles of
H2SO4.
Solution:
Amount (mol) of H2SO4 = 40.5 mL H2SO4

1L

0.128 mol H2 SO4

1000 mL

1L

= 0.00518 mol H2SO4

Road map:
Volume (mL) of soln
1000 mL = 1 L
Volume (L) of soln

Multiply by M
(1 L soln = 0.128 mol H2SO4)
Amount (mol) H2SO4
4.5A

Plan: Multiply the volume and molarity to calculate the number of moles of sodium phosphate in the solution.
Write the formula of sodium phosphate. Determine the number of each type of ion that is included in each
formula unit. Use this information to determine the amount of each type of ion in the described solution.
Solution:
Amount (mol) of Na3PO4 = 1.32 L

0.55 mol Na3 PO4
1L

= 0.7260 = 0.73 mol Na3PO4

+

In each formula unit of Na3PO4, there are 3 Na ions and 1 PO43– ion.
Amount (mol) of Na+ = 0.73 mol Na3PO4

3 mol Na+

1 mol Na3 PO4

= 2.2 mol Na+

3-

Amount (mol) of PO43– = 0.73 mol Na3PO4

1 mol PO4

1 mol Na3 PO4

= 0.73 mol PO43–

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4-3


4.5B

Plan: Write the formula of aluminum sulfate. Determine the number of aluminum ions in each formula unit.
Calculate the number of aluminum ions in the sample of aluminum sulfate. Convert the volume from mL to L.
Divide the number of moles of aluminum ion by the volume in L to calculate the molarity of the solution.
Solution:
In each formula unit of Al2(SO4)3, there are 2 Al3+ ions.
Amount (mol) of Al3+ = 1.25 mol Al2(SO4)3
M=

4.6A

1000 mL

875 mL

1L

1 mol Al2 (S04 )3

Mconc Vconc
Mdil

= 2.86 M

=

(4.50 M)(60.0 mL)
(1.25 M)

= 216 mL

Plan: Determine the new concentration from the dilution equation (Mconc)(Vconc) = (Mdil)(Vdil). Convert the molarity
(mol/L) to g/mL in two steps (one step is moles to grams, and the other step is L to mL).
Solution:
M dil 



 7.50 M  25.0 m 3
M concVconc

Vdil
500.m 3



= 0.375 M

 0.375 mol H 2SO4   98.08 g H 2SO4
Concentration (g/mL) = 

1L

 1 mol H 2SO4
–2
= 0.036780 = 3.68x10 g/mL solution
4.7A

= 2.50 mol Al3+

Plan: Determine the new volume from the dilution equation (Mconc)(Vconc) = (Mdil)(Vdil).
Solution:
Vdil =

4.6B

2.50 mol Al3+

2 mol Al3+

  103 L 

 
  1 mL 

Plan: Count the number of particles in each solution per unit volume.

Solution:
Solution A has 6 particles per unit volume while Solution B has 12 particles per unit volume. Solution B is more
concentrated than Solution A. To obtain Solution B, the total volume of Solution A was reduced by half:
 6 particles 1.0 mL 
N V
Vconc  dil dil 
= 0.50 mL
N conc
12 particles 
Solution C has 4 particles and is thus more dilute than Solution A. To obtain Solution C, ½ the volume of solvent
must be added for every volume of Solution A:
 6 particles 1.0 mL 
N V
Vdil  conc conc 
= 1.5 mL
Ndil
 4 particles 

4.7B

Plan: Count the number of particles in each solution per unit volume. Determine the final volume of the solution.
Use the dilution equation, (Nconc)(Vconc) = (Ndil)(Vdil), to determine the number of particles that will be present
when 300. mL of solvent is added to the 100. mL of solution represented in circle A. (Because M is directly
proportional to the number of particles in a given solution, we can replace the molarity terms in the dilution
equation with terms representing the number of particles.)
Solution:
There are 12 particles in circle A, 3 particles in circle B, 4 particles in circle C, and 6 particles in circle D.
The concentrated solution (circle A) has a volume of 100. mL. 300. mL of solvent are added, so the volume of the
diluted solution is 400. mL.
Ndil =

Nconc Vconc
Vdil

=

(12particles)(100.0 mL)
(400. mL)

= 3 particles

There are 3 particles in circle B, so circle B represents the diluted solution.
4.8A

Plan: Determine the ions present in each substance on the reactant side and write new cation-anion combinations.
Use Table 4.1 to determine if either of the combinations of ions is not soluble. If a precipitate forms there will be

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4-4


a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the
substances and balancing. In the total ionic equation, all soluble substances are written as separate ions. The net
ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form
(spectator ions) on each side of the reaction arrow.
Solution:
a) The resulting ion combinations that are possible are iron(III) phosphate and cesium chloride. According to
Table 4.1, iron(III) phosphate is insoluble, so a reaction occurs. We see that cesium chloride is soluble.

Total ionic equation:
Fe3+(aq) + 3Cl–(aq) + 3Cs+(aq) + PO43–(aq)  FePO4(s) + 3Cl–(aq) + 3Cs+(aq)
Net ionic equation:
Fe3+(aq) + PO43–(aq)  FePO4(s)
b) The resulting ion combinations that are possible are sodium nitrate (soluble) and cadmium hydroxide
(insoluble). A reaction occurs.
Total ionic equation:
2Na+(aq) + 2OH–(aq) + Cd2+(aq) + 2NO3–(aq)  Cd(OH)2(s) + 2Na+(aq) + 2NO3–(aq)
Note: The coefficients for Na+ and OH– are necessary to balance the reaction and must be included.
Net ionic equation:
Cd2+(aq) + 2OH–(aq)  Cd(OH)2(s)
c) The resulting ion combinations that are possible are magnesium acetate (soluble) and potassium bromide
(soluble). No reaction occurs.
4.8B

Plan: Determine the ions present in each substance on the reactant side and write new cation-anion combinations.
Use Table 4.1 to determine if either of the combinations of ions is not soluble. If a precipitate forms there will be
a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the
substances and balancing. In the total ionic equation, all soluble substances are written as separate ions. The net
ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form
(spectator ions) on each side of the reaction arrow.
Solution:
a) The resulting ion combinations that are possible are silver chloride (insoluble, an exception) and barium nitrate
(soluble). A reaction occurs.
Total ionic equation:
2Ag+(aq) + 2NO3–(aq) + Ba2+(aq) + 2Cl–(aq)  2AgCl(s) + Ba2+(aq) + 2NO3–(aq)
Net ionic equation:
Ag+(aq) + Cl–(aq)  AgCl(s)
b) The resulting ion combinations that are possible are ammonium sulfide (soluble) and potassium carbonate
(soluble). No reaction occurs.

c) The resulting ion combinations that are possible are lead(II) sulfate (insoluble, an exception) and nickel(II)
nitrate (soluble). A reaction occurs.
Total ionic equation:
Ni2+(aq) + SO42–(aq) + Pb2+(aq) + 2NO3–(aq)  PbSO4(s) + Ni2+(aq) + 2NO3–(aq)
Net ionic equation:
Pb2+(aq) + SO42–(aq)  PbSO4 (s)

4.9A

Plan: Look at the ions (number and charge) produced when each of the given compounds dissolves in water and
find the match to the ions shown in the beaker. Once the ions in each beaker are known, write new cation-anion
combinations and use Table 4.1 to determine if any of the combination of ions is not soluble. If a precipitate forms
there will be a reaction and chemical equations may be written. The molecular equation simply includes the
formulas of the substances and must be balanced. In the total ionic equation, all soluble substances are written as
separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing
in identical form (spectator ions) on each side of the reaction arrow.
Solution:
a) Beaker A has four ions with a +2 charge and eight ions with a –1 charge. The beaker contains dissolved
Zn(NO3)2 which dissolves to produce Zn2+ and NO3– ions in a 1:2 ratio. The compound PbCl2 also has a +2 ion
and –1 ion in a 1:2 ratio but PbCl2 is insoluble so ions would not result from this compound.

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4-5


b) Beaker B has three ions with a +2 charge and six ions with a –1 charge. The beaker contains dissolved
Ba(OH)2 which dissolves to produce Ba2+ and OH– ions in a 1:2 ratio. Cd(OH)2 also has a +2 ion and a –1 ion in
a 1:2 ratio but Cd(OH)2 is insoluble so ions would not result from this compound.

c) The resulting ion combinations that are possible are zinc hydroxide (insoluble) and barium nitrate (soluble).
The precipitate formed is Zn(OH)2. The spectator ions are Ba2+ and NO3–.
Balanced molecular equation: Zn(NO3)2(aq) + Ba(OH)2(aq)  Zn(OH)2(s) + Ba(NO3)2(aq)
Total ionic equation:
Zn2+(aq) + 2NO3–(aq) + Ba2+(aq) + 2OH– (aq)  Zn(OH)2(s) + Ba2+(aq) + 2NO3– (aq)
Net ionic equation:
Zn2+(aq) + 2OH–(aq)  Zn(OH)2(s)
d) Since there are only six OH– ions and four Zn2+ ions, the OH– is the limiting reactant.
Mass of Zn(OH)2= (6 OH– ions)

0.050 molOH

1 mol Zn(OH

99.40 g Zn(OH

1OH particle

2 molOH

1 mol Zn(OH

= 14.9100 = 15 g Zn(OH)2
4.9B

Plan: Look at the ions (number and charge) produced when each of the given compounds dissolves in water and
find the match to the ions shown in the beaker. Once the ions in each beaker are known, write new cation-anion
combinations and use Table 4.1 to determine if any of the combination of ions is not soluble. If a precipitate forms
there will be a reaction and chemical equations may be written. The molecular equation simply includes the
formulas of the substances and must be balanced. In the total ionic equation, all soluble substances are written as

separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing
in identical form (spectator ions) on each side of the reaction arrow.
Solution:
a) Beaker A has eight ions with a +1 charge and four ions with a –2 charge. The beaker contains dissolved
Li2CO3 which dissolves to produce Li+ and CO32–ions in a 2:1 ratio. The compound Ag2SO4 also has a +1 ion
and –2 ion in a 2:1 ratio but Ag2SO4 is insoluble so ions would not result from this compound.
b) Beaker B has three ions with a +2 charge and six ions with a –1 charge. The beaker contains dissolved CaCl2
which dissolves to produce Ca2+ and Cl– ions in a 1:2 ratio. Ni(OH)2 also has a +2 ion and a –1 ion in a 1:2 ratio
but Cd(OH)2 is insoluble so ions would not result from this compound.
c) The resulting ion combinations that are possible are calcium carbonate (insoluble) and lithium chloride
(soluble). The precipitate formed is CaCO3. The spectator ions are Li+ and Cl–.
Balanced molecular equation: Li2CO3(aq) + CaCl2(aq)  CaCO3(s) + 2LiCl(aq)
Total ionic equation:
2Li+(aq) + CO32– (aq) + Ca2+(aq) + 2Cl– (aq)  CaCO3(s) + 2Li+(aq) + 2Cl– (aq)
Net ionic equation:
Ca2+(aq) + CO32–(aq)  CaCO3 (s)
d) Since there are only four CO32– ions and three Ca2+ ions, the Ca2+ is the limiting reactant.
Mass of CaCO3= (3 Ca2+ ions)

4.10A

0.20 molCa2+

1Ca

2+

particle

1 mol CaCO3

1 molCa

2+

100.09 g CaCO3
1 mol CaCO3

= 60.0540 = 60. g CaCO3

Plan: We are given the molarity and volume of calcium chloride solution, and we must find the volume of sodium
phosphate solution that will react with this amount of calcium chloride. After writing the balanced equation, we
find the amount (mol) of calcium chloride from its molarity and volume and use the molar ratio to find the amount
(mol) of sodium phosphate required to react with the calcium chloride. Finally, we use the molarity of the sodium
phosphate solution to convert the amount (mol) of sodium phosphate to volume (L).
Solution:
The balanced equation is: 3CaCl2(aq) + 2Na3PO4(aq)  Ca3(PO4)2(s) + 6NaCl(aq)
Finding the volume (L) of Na3PO4 needed to react with the CaCl2:
Volume (L) of Na3PO4 = 0.300 L CaCl2

0.175 moles CaCl2

2 mol Na3 PO4

1L

1L

3 mol CaCl2

0.260 mol Na3 PO4

= 0.1346 = 0.135 L Na3PO4
4.10B

Plan: We are given the mass of silver chloride produced in the reaction of silver nitrate and sodium chloride, and
we must find the molarity of the silver nitrate solution. After writing the balanced equation, we find the amount

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4-6


(mol) of silver nitrate that produces the precipitate by dividing the mass of silver chloride produced by its molar
mass and then using the molar ratios from the balanced equation. Then we calculate the molarity by dividing the
moles of silver nitrate by the volume of the solution (in L).
The balanced equation is: AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Amount (mol) of AgNO3 = 0.148 g AgCl
Molarity (M) of AgNO3 =
4.11A

1 mol AgCl

1 mol AgNO3

143.4 g AgCl

1 mol AgCl

1.03 x 10–3 mol AgNO3

1000 mL

45.0 mL

1L

= 0.001032 = 1.03x10–3 mol AgNO3

= 2.29 x 10–2 M

Plan: Multiply the volume in liters of each solution by its molarity to obtain the moles of each reactant. Write a
balanced equation. Use molar ratios from the balanced equation to determine the moles of lead(II) chloride that
may be produced from each reactant. The reactant that generates the smaller number of moles is limiting. Change
the moles of lead(II) chloride from the limiting reactant to the grams of product using the molar mass of lead(II)
chloride.
Solution:
(a)The balanced equation is:
Pb(C2H3O2)2(aq) + 2NaCl(aq)  PbCl2(s) + 2NaC2H3O2(aq)
 103 L   1.50 mol Pb(C2 H3O2 )2 
Moles of Pb(C2H3O2)2 =  268 mL  
 
 = 0.402 mol Pb(C2H3O2)2
1L

 1 mL  

 103 L   3.40 mol NaCl 
Moles of NaCl = 130 mL  
 

 = 0.442 mol NaCl
1L

 1 mL  


1 mol PbCl2
Moles of PbCl2 from Pb(C2H3O2)2 =  0.402 mol Pb(C2 H3O2 )2  

 1 mol Pb(C2 H3O2 )2 
= 0.402 mol PbCl2
 1 mol PbCl2 
Moles of PbCl2 from NaCl =  0.442 mol NaCl  
 = 0.221 mol PbCl2
 2 mol NaCl 
The NaCl is limiting. The mass of PbCl2 may now be determined using the molar mass.
 278.1 g PbCl2 
Mass (g) of PbCl2 =  0.221 mol PbCl2  
 = 61.4601 = 61.5 g PbCl2
 1 mol PbCl2 
(b) Ac is used to represent C2H3O2:
Amount (mol)
Pb(Ac)2 + 2NaCl
→
PbCl2 + 2NaAc
Initial
0.402
0.442
0
0

Change
– 0.221
– 0.442
+0.221
+0.442
Final
0.181
0
0.221
0.442
4.11B

Plan: Write a balanced equation. Multiply the volume in liters of each solution by its molarity to obtain the
moles of each reactant. Use molar ratios from the balanced equation and the molar mass of iron(III) hydroxide to
determine the mass (g) of iron(III) hydroxide that may be produced from each reactant. The smaller mass is the
amount of product actually made.
Solution:
(a)The balanced equation is: FeCl3(aq) + 3NaOH(aq)  3NaCl(aq) + Fe(OH)3(s)
Mass(g) of Fe(OH)3 produced from FeCl3 =
155 mL FeCl3

1L

0.250 mol FeCl3

1 mol mol Fe(OH)3

106.87 g Fe(OH)3

1000 mL

1L

1 mol FeCl3

1 mol mol Fe(OH)3

= 4.14 g Fe(OH)3

Mass(g) of Fe(OH)3 produced from NaOH =
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4-7


215 mL NaOH

1L

0.300 mol NaOH

1 mol mol Fe(OH)3

106.87 g Fe(OH)3

1000 mL

1L

3 mol NaOH

1 mol mol Fe(OH)3

= 2.30 g Fe(OH)3

NaOH produces the smaller amount of product, so it is the limiting reagent and 2.30 g of Fe(OH)3 are produced.
(b) To complete the reaction table, we need the moles of FeCl3 and NaOH that react. Remember that NaOH is the
limiting reagent and will be consumed in this reaction.
Amount (mol) of FeCl3 =
155 mL FeCl3

1L

0.250 mol FeCl3

1000 mL

1L

= 0.0388 mol FeCl3

Amount (mol) of NaOH =
215 mL NaOH

1L

0.300 mol NaOH

1000 mL

1L

Amount (mol)
Initial
Change
Final
4.12A

FeCl3 + 3NaOH
0.0388
0.0645
– 0.0215
–0.0645
0.0173
0

= 0.0645 mol NaOH
→

3NaCl +
0
+0.0645
0.0645

Fe(OH)3
0
+0.0215
0.0215

Plan: Convert the given volume from mL to L and multiply by the molarity (mol/L) to find moles of KOH. Each
mole of the strong base KOH will produce one mole of hydroxide ions. Finally, multiply the amount (mol) of
hydroxide ions by Avogadro’s number to calculate the number of hydroxide ions produced.
Solution:
H O

2
 K+(aq) + OH–(aq)
KOH(s) 

No. of OH– ions produced = 451 mL NaOH

1L

1.20 mole KOH

1 mol OH-

6.022 x 10 23 OH- ions

1000 mL

1L

1 mole KOH

1 mole OH-

= 3.2591 x 10
4.12B

23

= 3.26 x 1023 OH– ions

Plan: Convert the given volume from mL to L and multiply by the molarity (mol/L) to find moles of HCl. Each
mole of the strong acid HCl will produce one mole of hydrogen ions. Finally, multiply the amount (mol) of
hydrogen ions by Avogadro’s number to calculate the number of hydrogen ions produced.
Solution:
H O

2
 H+(aq) + Cl–(aq)
HCl(g) 

No. of H+ ions produced = 65.5 mL HCl

1L

0.722 mole HCl

1 mol H+

1000 mL

1L

1 mole HCl

= 2.8479 x 10

6.022 x 10 23 H+ ions
1 mole H+
22

= 2.85 x 1022 H+ ions

4.13A

Plan: According to Table 4.2, both reactants are strong and therefore completely dissociate in water. Thus, the key
reaction is the formation of water. The other product of the reaction is soluble.
Solution:
Molecular equation: 2HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2H2O(l)
Total ionic equation:
2H+(aq) + 2NO3–(aq) + Ca2+(aq) + 2OH–(aq)  Ca2+(aq) + 2NO3–(aq) + 2H2O(l)
Net ionic equation: 2H+(aq) + 2OH–(aq)  2H2O(l) which simplifies to H+(aq) + OH–(aq)  H2O(l)

4.13B

Plan: According to Table 4.2, both reactants are strong and therefore completely dissociate in water. Thus, the key
reaction is the formation of water. The other product of the reaction is soluble.
Solution:
Molecular equation: HI(aq) + LiOH(aq)  LiI(aq) + H2O(l)
Total ionic equation: H+(aq) + I–(aq) + Li+(aq) + OH–(aq)  Li+(aq) + I–(aq) + H2O(l)
Net ionic equation: H+(aq) + OH–(aq)  H2O(l)

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4-8

4.14A

Plan: The reactants are a weak acid and a strong base. The acidic species is H+ and a proton is transferred to OH–
from the acid. The only spectator ion is the cation of the base.
Solution:
Molecular equation:
2HNO2(aq) + Sr(OH)2(aq)  Sr(NO2)2(aq) + 2H2O(l)
Ionic equation:
2HNO2(aq) + Sr2+(aq) + 2OH–(aq)  Sr2+(aq) + 2NO2– (aq) + 2H2O(l)
Net ionic equation:
2HNO2(aq) + 2OH–(aq)  2NO2– (aq) + 2H2O(l) or
HNO2(aq) + OH–(aq)  NO2– (aq) + H2O(l)

The salt is Sr(NO2)2, strontium nitrite, and the spectator ion is Sr2+.
4.14B

Plan: The reactants are a strong acid and the salt of a weak base. The acidic species is H3O+ and a proton is
transferred to the weak base HCO3– to form H2CO3, which then decomposes to form CO2 and water.
Solution:
Molecular equation:
2HBr(aq) + Ca(HCO3)2(aq)  CaBr2(aq) + 2H2CO3(aq)
Ionic equation:
2H3O+(aq) + 2Br–(aq) + Ca2+(aq) + 2HCO3–(aq)  2CO2(g) + 4H2O(l) + 2Br–(aq) + Ca2+(aq)
Net ionic equation:
H3O+(aq) + HCO3–(aq)  CO2(g) + 2H2O(l)
2H3O+(aq) + 2HCO3–(aq)  2CO2(g) + 4H2O(l) or

The salt is CaBr2, calcium bromide.

4.15A

Plan: Write a balanced equation. Determine the moles of HCl by multiplying its molarity by its volume, and,
through the balanced chemical equation and the molar mass of aluminum hydroxide, determine the mass of
aluminum hydroxide required for the reaction.
Solution:
The balanced equation is: Al(OH)3(s) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)
Mass(g) of Al(OH)3 = 3.4 x 10–2 L HCl

0.10 mol HCl

1 mol Al(OH)3

78.00 g Al(OH)3

1L

3 mol HCl

1 mol Al(OH)3

= 0.08840 = 0.088 g Al(OH)3

4.15B

Plan: Write a balanced equation. Determine the moles of NaOH by multiplying its molarity by its volume, and,
through the balanced chemical equation and the molar mass of acetylsalicylic acid, determine the mass of
acetylsalicylic acid in the tablet.
Solution:
The balanced equation is: HC9H7O4(aq) + NaOH(aq)  NaC9H7O4(aq) + H2O(l)

Mass(g) of HC9H7O4= 14.10 mL NaOH

1L

0.128 mol NaOH

1 mol HC9 H7 O4

180.15 g HC9 H7 O4

1000 mL

1L

1 mol NaOH

1 mol HC9 H7 O4

= 0.3251 = 0.325 g HC9H7O4
4.16A

Plan: Write a balanced chemical equation. Determine the moles of HCl by multiplying its molarity by its volume,
and, through the balanced chemical equation, determine the moles of Ba(OH)2 required for the reaction. The
amount of base in moles is divided by its molarity to find the volume.
Solution:
The molarity of the HCl solution is 0.1016 M. However, the molar ratio is not 1:1 as in the example problem.
According to the balanced equation, the ratio is 2 moles of acid per 1 mole of base:
2HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2H2O(l)

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4-9


 10 3 L   0.1016 mol HCl   1 mol Ba(OH) 2  

L
Volume (L) of Ba(OH)2 =  50.00 mL  
  2 mol HCl   0.1292 mol Ba(OH) 
 1 mL  
L


2 


= 0.0196594 = 0.01966 L Ba(OH)2 solution

4.16B

Plan: Write a balanced chemical equation. Determine the moles of H2SO4 by multiplying its molarity by its
volume, and, through the balanced chemical equation, determine the moles of KOH required for the reaction. The
amount of base in moles is divided by its volume (in L) to find the molarity.
Solution:
The balanced equation is: 2KOH(aq) + H2SO4(aq)  K2SO4(aq) + 2H2O(l)
Amount (mol) of KOH = 20.00 mL H2SO4
Molarity =

0.009808 mol KOH

1000 mL

18.15 mL

1L

1L

0.2452 mol H2 SO4

2 mol KOH

1000 mL

1L

1 mol H2 SO4

= 0.009808 mol KOH

= 0.5404 M

4.17A

Plan: Apply Table 4.4 to the compounds. Do not forget that the sum of the O.N.’s (oxidation numbers) for a
compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion.
Solution:
a) Sc = +3 O = –2 In most compounds, oxygen has a –2 O.N., so oxygen is often a good starting point. If each
oxygen atom has a –2 O.N., then each scandium must have a +3 oxidation state so that the sum of O.N.’s equals

zero: 2(+3) + 3(–2) = 0.
b) Ga = +3 Cl = –1 In most compounds, chlorine has a –1 O.N., so chlorine is a good starting point. If each
chlorine atom has a –1 O.N., then the gallium must have a +3 oxidation state so that the sum of O.N.’s equals
zero: 1(+3) + 3(–1) = 0.
c) H = +1 P = +5 O = –2 The hydrogen phosphate ion is HPO42–. Again, oxygen has a –2 O.N. Hydrogen has a +1
O.N. because it is combined with nonmetals. The sum of the O.N.’s must equal the ionic charge, so the following
algebraic equation can be solved for P: 1(+1) + 1(P) + 4(–2) = –2; O.N. for P = +5.
d) I = +3 F = –1 The formula of iodine trifluoride is IF3. In all compounds, fluorine has a –1 O.N., so fluorine is
often a good starting point. If each fluorine atom has a –1 O.N., then the iodine must have a +3 oxidation state so
that the sum of O.N.’s equal zero: 1(+3) + 3(–1) = 0.

4.17B

Plan: Apply Table 4.4 to the compounds. Do not forget that the sum of the O.N.’s (oxidation numbers) for a
compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion.
Solution:
a) K = +1 C = +4 O = –2 In all compounds, potassium has a +1 O.N, and in most compounds, oxygen has a –2
O.N. If each potassium has a +1 O.N. and each oxygen has a –2 O.N., carbon must have a +4 oxidation state so
that the sum of the O.N.’s equals zero: 2(+1) + 1(+4) + 3(–2) = 0.
b) N = –3 H = +1 When it is combined with a nonmetal, like N, hydrogen has a +1 O.N. If hydrogen has a +1
O.N., nitrogen must have a –3 O.N. so the sum of the O.N.’s equals +1, the charge on the polyatomic ion: 1(–3) +
4(+1) = +1.
c) Ca = +2 P = –3 Calcium, a group 2A metal, has a +2 O.N. in all compounds. If calcium has a +2 O.N., the
phosphorus must have a –3 O.N. so the sum of the O.N.’s equals zero: 3(+2) + 2(–3) = 0.
d) S = +4 Cl = –1 Chlorine, a group 7A nonmetal, has a –1 O.N. when it is in combination with any nonmetal
except O or other halogens lower in the group. If chlorine has an O.N. of –1, the sulfur must have an O.N. of +4
so that the sum of the O.N.’s equals zero: 1(+4) + 4(–1) = 0.

4.18A

Plan: Apply Table 4.4 to determine the oxidation numbers for all the compounds in the reaction. Do not forget
that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the
sum must equal the charge on the ion. After determining oxidation numbers for all atoms in the reaction, identify
the atoms for which the oxidation numbers have changed from the left hand side of the equation to the right hand
side of the equation. If the oxidation number of a particular atom increases, that atom has been oxidized, and the
compound, element, or ion containing that atom on the reactant side of the equation is the reducing agent. If the
oxidation number of a particular atom decreases, that atom has been reduced, and the compound, element, or ion
containing that atom on the reactant side of the equation is the oxidizing agent.

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4-10


Solution:
a) Oxidation numbers in NCl3: N = +3, Cl = –1
Oxidation numbers in H2O: H = +1, O = –2
Oxidation numbers in NH3: N = –3, H = +1
Oxidation numbers in HOCl: H = +1, O = –2, Cl = +1
The oxidation number of nitrogen decreases from +3 to –3, so N is reduced and NCl3 is the oxidizing agent. The
oxidation number of chlorine increases from –1 to +1, so Cl is oxidized and NCl3 is also the reducing agent.
b) Oxidation numbers in AgNO3: N = +1, N = +5, O = –2
Oxidation numbers in NH4I: N = –3, H = +1, I = –1
Oxidation numbers in AgI: Ag = +1, I = –1
Oxidation numbers in NH4NO3: N (in NH4+) = –3, H = +1, N (in NO3–) = +5, O = –2
None of the oxidation numbers change, so this is not an oxidation-reduction reaction.
c) Oxidation numbers in H2S: H = +1, S = –2
Oxidation numbers in O2: O = 0
Oxidation numbers in SO2: S = +4, O = –2

Oxidation numbers in H2O: H = +1, O = –2
The oxidation number of oxygen decreases from 0 to –2, so O is reduced and O2 is the oxidizing agent. The
oxidation number of sulfur increases from –2 to +4, so S is oxidized and H2S is the reducing agent.
4.18B

Plan: Apply Table 4.4 to determine the oxidation numbers for all the compounds in the reaction. Do not forget
that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the
sum must equal the charge on the ion. After determining oxidation numbers for all atoms in the reaction, identify
the atoms for which the oxidation numbers have changed from the left hand side of the equation to the right hand
side of the equation. If the oxidation number of a particular atom increases, that atom has been oxidized, and the
compound, element, or ion containing that atom on the reactant side of the equation is the reducing agent. If the
oxidation number of a particular atom decreases, that atom has been reduced, and the compound, element, or ion
containing that atom on the reactant side of the equation is the oxidizing agent.
Solution:
a) Oxidation numbers in SiO2: Si = +4, O = –2
Oxidation numbers in HF: H = +1, F = –1
Oxidation numbers in SiF4: Si = +4, F = –1
Oxidation numbers in H2O: H = +1, O = –2
None of the oxidation numbers change, so this is not an oxidation-reduction reaction.
b) Oxidation numbers in C2H6: C = –3, H = +1
Oxidation numbers in O2: O = 0
Oxidation numbers in CO2: C = +4, O = –2
Oxidation numbers in H2O: H = +1, O = –2
The oxidation number of oxygen decreases from 0 to –2, so O is reduced and O2 is the oxidizing agent. The
oxidation number of carbon increases from –3 to +4, so C is oxidized and C2H6 is the reducing agent.
c) Oxidation numbers in CO: C = +2, O = –2
Oxidation numbers in I2O5: I = +5, O = –2
Oxidation numbers in I2: I = 0
Oxidation numbers in CO2: C = +4, O = –2
The oxidation number of iodine decreases from +5 to 0, so I is reduced and I2O5 is the oxidizing agent. The

oxidation number of carbon increases from +2 to +4, so C is oxidized and CO is the reducing agent.

4.19A

Plan: Write a balanced reaction. Find moles of KMnO4 by multiplying its volume and molarity and determine
the moles of Ca2+ in the sample using the molar ratio from the reaction. Divide moles of Ca2+ by the volume of
the milk sample to obtain molarity. In part (b), moles of Ca2+ will be converted to grams.
Solution:
2KMnO4(aq) + 5CaC2O4(s) + H2SO4(aq) 
2MnSO4(aq) + K2SO4(aq) + 5CaSO4(s) + 10CO2(g) + 8H2O(l)
 10 3 L   4.56 x 103 mol KMnO 4 
Moles of KMnO4 =  6.53 mL  

 1 mL  
1L




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4-11


= 2.97768 x 10–5 mol KMnO4
 5 mol CaC 2 O 4   1 mol Ca 2+  
1
  1 mL 
Molarity of Ca2+ = 2.97768 x 105 mol KMnO 4 

 
 
  3 
2
mol
KMnO
1
mol
CaC
O
2.50
mL
  10 L 
4 
2 4 

–2
–2
–2
2+
= 2.97768 x 10 = 2.98 x 10 mol/L = 2.98 x 10 M Ca
 2.97768 x 10 2 mol Ca 2+   40.08 g Ca 2+ 
b) Concentration (g/L) = 
= 1.193454 = 1.19 g/L
 
2+ 


L


  1 mol Ca

This concentration is consistent with the typical value in milk.



4.19B



Plan: Find moles of K2Cr2O7 by multiplying its volume and molarity, and determine the moles of Fe2+ in the
sample using the molar ratio from the reaction provided in the problem statement (remember that the Fe2+ is part
of the FeSO4 compound and that there is one Fe2+ ion for every formula unit of FeSO4). Divide moles of Fe2+ by
the volume of the FeSO4 solution to obtain molarity. In part (b), convert the moles of Fe2+ to grams. Then divide
the mass of iron ion by the total mass of the ore and multiply by 100% to find the mass percent of iron in the ore.
Solution:
a) The balanced equation (provided in the problem statement) is:
6FeSO4(aq) + K2Cr2O7(aq) + 7H2SO4(aq)  3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) + 7H2O(l) + K2SO4(aq)
Amount (mol) of Fe2+ = 21.85 mL K2Cr2O7

1L

0.250 mol K2 Cr2 O7

6 mole FeSO4

1 mol Fe2+

1000 mL

1L

1 mol K2 Cr2 O7

1 mol FeSO4

= 0.0328 mol Fe2+
Molarity of Fe2+ =

0.0328 mol Fe

2+

30.0 mL

b) Mass (g) of iron = 0.0328 mol Fe2+
Mass % of iron in sample =
4.20A

1.83 g Fe
2.58 g ore

1000 mL
1L
1 mol Fe
1 mol Fe

= 1.0933 = 1.09 M

2+

55.85 g Fe
1 mol Fe

= 1.8319 = 1.83 g Fe

(100%) = 70.9302 = 70.9%

Plan: To classify a reaction, compare the number of reactants used versus the number of products formed. Also
examine the changes, if any, in the oxidation numbers. Recall the definitions of each type of reaction:
Combination: X + Y  Z;
Decomposition: Z  X + Y
Single displacement: X + YZ  XZ + Y
Solution:
a) Combination; S8(s) + 16F2(g)  8SF4(g)
O.N.:
S=0
F=0
S = +4
F = –1
Sulfur changes from 0 to +4 oxidation state; it is oxidized and S8 is the reducing agent.
Fluorine changes from 0 to –1 oxidation state; it is reduced and F2 is the oxidizing agent.
b) Displacement; 2CsI(aq) + Cl2(aq)  2CsCl(aq) + I2(aq)
O.N.:
Cs = +1 Cl = 0
Cs = +1 I = 0
I = –1
Cl = –1
Total ionic eqn: 2Cs+(aq) + 2I–(aq) + Cl2(aq)  2Cs+(aq) + 2Cl–(aq) + I2(aq)
Net ionic eqn: 2I–(aq) + Cl2(aq)  2Cl–(aq) + I2(aq)

Iodine changes from –1 to 0 oxidation state; it is oxidized and CsI is the reducing agent. Chlorine
changes from 0 to –1 oxidation state; it is reduced and Cl2 is the oxidizing agent.
c) Displacement; 3Ni(NO3)2(aq) + 2Cr(s)  2Cr(NO3)3(aq) + 3Ni(s)
O.N.:
Ni = +2
Cr = 0
Cr = +3
Ni = 0
N = +5
N = +5
O = –2
O = –2
Total ionic eqn: 3Ni+2(aq) + 6NO3–(aq) + 2Cr(s)  2Cr+3(aq) + 6NO3–(aq) + 3Ni(s)
Net ionic eqn: 3Ni+2(aq) + 2Cr(s)  2Cr+3(aq) + 3Ni(s)
Nickel changes from +2 to 0 oxidation state; it is reduced and Ni(NO3)2 is the oxidizing agent.
Chromium changes from 0 to +3 oxidation state; it is oxidized and Cr is the reducing agent.

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4-12


4.20B

Plan: To classify a reaction, compare the number of reactants used versus the number of products formed. Also
examine the changes, if any, in the oxidation numbers. Recall the definitions of each type of reaction:
Combination: X + Y  Z;
Decomposition: Z  X + Y
Single displacement: X + YZ  XZ + Y

Solution:
a) Displacement; Co(s) + 2HCl(aq)
 CoCl2(aq) + H2(g)
(molecular equation)
O.N.:
Co = 0 H = +1, Cl = –1
Co = +2, Cl = –1 H = 0
(total ionic equation)
Co(s) + 2H+(aq) + 2Cl–(aq)  Co2+(aq) + 2Cl–(aq) + H2(g)
Co(s) + 2H+(aq)  Co2+(aq) + H2(g)
(net ionic equation)
Cobalt changes from 0 to +2 oxidation state; it is oxidized and Co is the reducing agent.
Hydrogen changes from +1 to 0 oxidation state; it is reduced and HCl is the oxidizing agent.
b) Combination; 2CO (g) +
O2(g)  2CO2(g)
O.N.:
C = +2, O = –2
O=0
C = +4, O = –2
Carbon changes from +2 to +4 oxidation state; it is oxidized and CO is the reducing agent. Oxygen
changes from 0 to –2 oxidation state; it is reduced and O2 is the oxidizing agent.
c) Decomposition; 2N2O5(s)  4NO2(g) + O2(g)
O.N.:
N = +5
N = +4 O = 0
O = –2
O = –2
Nitrogen changes from +5 to +4 oxidation state; it is reduced and N2O5 is the oxidizing agent. Oxygen
changes from –2 to 0 oxidation state; it is oxidized and N2O5 is also the reducing agent.

END–OF–CHAPTER PROBLEMS
4.1

Plan: Review the discussion on the polar nature of water.
Solution:
Water is polar because the distribution of its bonding electrons is unequal, resulting in polar bonds, and the shape
of the molecule (bent) is unsymmetrical.

4.2

Plan: Review the discussion on water soluble compounds.
Solution:
Ionic and polar covalent compounds are most likely to be soluble in water. Because water is polar, the partial
charges in its molecules are able to interact with the charges, either ionic or dipole-induced, in other substances.

4.3

Plan: Solutions that conduct an electric current contain electrolytes.
Solution:
Ions must be present in an aqueous solution for it to conduct an electric current. Ions come from ionic compounds
or from other electrolytes such as acids and bases.

4.4

Plan: Review the discussion on ionic compounds in water.
Solution:
The ions on the surface of the solid attract the water molecules (cations attract the “negative” ends and anions attract the
“positive” ends of the water molecules). The interaction of the solvent with the ions overcomes the attraction of the
oppositely charged ions for one another, and they are released into the solution.

4.5

Plan: Recall that ionic compounds dissociate into their ions when dissolved in water. Examine the charges of the
ions in each scene and the ratio of cations to anions.
Solution:
a) CaCl2 dissociates to produce one Ca2+ ion for every two Cl– ions. Scene B contains four 2+ ions
and twice that number of 1– ions.
b) Li2SO4 dissociates to produce two Li+ ions for every one SO42– ion. Scene C contains eight 1+ ions
and half as many 2– ions.
c) NH4Br dissociates to produce one NH4+ ion for every one Br– ion. Scene A contains equal numbers of 1+

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4-13


4.6

and 1– ions.
Plan: Write the formula for magnesium nitrate and note the ratio of magnesium ions to nitrate ions.
Solution:
Upon dissolving the salt in water, magnesium nitrate, Mg(NO3)2, would dissociate to form one Mg2+ ion for every
two NO3– ions, thus forming twice as many nitrate ions. Scene B best represents a volume of magnesium nitrate
solution. Only Scene B has twice as many nitrate ions (red circles) as magnesium ions (blue circles).

4.7

Plan: Review the discussion of ionic compounds in water.
Solution:

In some ionic compounds, the force of the attraction between the ions is so strong that it cannot be overcome by
the interaction of the ions with the water molecules. These compounds will be insoluble in water.

4.8

Plan: Review the discussion of covalent compounds in water.
Solution:
The interaction with water depends on the structure of the molecule. If the interaction is strong, the substance will
be soluble; otherwise, the substance will not be very soluble. Covalent compounds that contain polar
groups interact well with the polar solvent water and therefore dissolve in water. Covalent compounds that do
not contain polar bonds are not soluble in water.

4.9

Plan: Review the discussion of covalent compounds in water.
Solution:
Some covalent compounds that contain the hydrogen atom dissociate into ions when dissolved in water. These
compounds form acidic solutions in water; three examples are HCl, HNO3, and HBr.

4.10

Plan: Count the total number of spheres in each box. The number in box A divided by the volume change in each
part will give the number we are looking for and allow us to match boxes.
Solution:
The number in each box is: A = 12, B = 6, C = 4, and D = 3.
a) When the volume is tripled, there should be 12/3 = 4 spheres in a box. This is box C.
b) When the volume is doubled, there should be 12/2 = 6 spheres in a box. This is box B.
c) When the volume is quadrupled, there should be 12/4 = 3 spheres in a box. This is box D.

4.11

Plan: Recall that molarity = moles of solute/volume (L) of solution.
Solution:
a) Mdil = molarity of the diluted solution
Mconc = molarity of the concentrated solution
Vdil = volume of the diluted solution
Vconc = volume of the concentrated solution
The equation works because the quantity (moles) of solute remains the same when a solution
is diluted; only the amount of solvent changes.
M x V = moles;
Mdil x Vdil = Mconc x Vconc
molesdil = molesconc
moles solute
b) Molarity =
liters solution
Moles CaCl2 = molarity x liters of solution; Mass CaCl2 = molarity x liters of solution x molar mass of CaCl2.

4.12

Plan: Recall that molarity = moles of solute/volume (L) of solution. Here you can use the number of particles in
place of moles of solute.
Solution:
a) Solution B has the highest molarity as it has the largest number of particles, 12, in a volume of 50 mL.
b) Solutions A and F both have 8 particles in a volume of 50 mL and thus the same molarity. Solutions C, D,
and E all have 4 particles in a volume of 50 mL and thus have the same molarity.
c) Mixing Solutions A and C results in 8 + 4 = 12 particles in a volume of 100 mL. That is a lower molarity than
that of Solution B which has 12 particles in a volume of 50 mL or 24 particles in a volume of 100 mL.
d) Adding 50 mL to Solution D would result in 4 particles in a total volume of 100 mL; adding 75 mL to Solution
F would result in 4 particles in a volume of 100 mL. The molarity of each solution would be the same.

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4-14


e) Solution A has 8 particles in a volume of 50 mL while Solution E has the equivalent of 4 particles in a volume
of 50 mL. The molarity of Solution E is half that of Solution A. Therefore half of the volume, 12.5 mL, of
Solution E must be evaporated. When 12.5 mL of solvent is evaporated from Solution E, the result will be 2
particles in 12.5 mL or 8 particles in 50 mL as in Solution A.
4.13

Plan: Remember that molarity is moles of solute/volume of solution.
Solution:
Volumes may not be additive when two different solutions are mixed, so the final volume may be slightly
different from 1000.0 mL. The correct method would state, “Take 100.0 mL of the 10.0 M solution and add water
until the total volume is 1000 mL.”

4.14

Plan: Compounds that are soluble in water tend to be ionic compounds or covalent compounds that have polar
bonds. Many ionic compounds are soluble in water because the attractive force between the oppositely charged
ions in an ionic compound are replaced with an attractive force between the polar water molecule and the ions
when the compound is dissolved in water. Covalent compounds with polar bonds are often soluble in water since
the polar bonds of the covalent compound interact with those in water.
Solution:
a) Benzene, a covalent compound, is likely to be insoluble in water because it is nonpolar and water is polar.
b) Sodium hydroxide (NaOH) is an ionic compound and is therefore likely to be soluble in water.
c) Ethanol (CH3CH2OH) will likely be soluble in water because it contains a polar –OH bond like water.
d) Potassium acetate (KC2H3O2) is an ionic compound and will likely be soluble in water.

4.15

Plan: Compounds that are soluble in water tend to be ionic compounds or covalent compounds that have polar
bonds. Many ionic compounds are soluble in water because the attractive force between the oppositely charged
ions in an ionic compound are replaced with an attractive force between the polar water molecule and the ions
when the compound is dissolved in water. Covalent compounds with polar bonds are often soluble in water since
the polar bonds of the covalent compound interact with those in water.
Solution:
a) Lithium nitrate is an ionic compound and is expected to be soluble in water.
b) Glycine (H2NCH2COOH) is a covalent compound, but it contains polar N–H and O–H bonds. This would make
the molecule interact well with polar water molecules, and make it likely that it would be soluble.
c) Pentane (C5H12) has no bonds of significant polarity, so it would be expected to be insoluble in the polar solvent
water.
d) Ethylene glycol (HOCH2CH2OH) molecules contain polar O–H bonds, similar to water, so it would be expected
to be soluble.

4.16

Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds,
acids, and bases.
Solution:
a) Cesium bromide, CsBr, is a soluble ionic compound, and a solution of this salt in water contains Cs+ and Br–
ions. Its solution conducts an electric current.
b) HI is a strong acid that dissociates completely in water. Its aqueous solution contains H+ and I– ions, so it
conducts an electric current.

4.17

Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds,

acids, and bases.
Solution:
a) Potassium sulfate, K2SO4, is an ionic compound that is soluble in water, producing K+ and SO42– ions. Its
solution conducts an electric current.
b) Sucrose is neither an ionic compound, an acid, nor a base, so it would be a nonelectrolyte (even though it’s
soluble in water). Its solution does not conduct an electric current.

4.18

Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into
ions with the correct molar ratios. Convert mass and formula units to moles of compound and use the molar ratio
to convert moles of compound to moles of ions.

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4-15


Solution:
a) Each mole of NH4Cl dissolves in water to form 1 mole of NH4+ ions and 1 mole of Cl– ions, or a total of 2
moles of ions: NH4Cl(s) → NH4+(aq) + Cl–(aq).
 2 mol ions 
Moles of ions =  0.32 mol NH4 Cl  
 = 0.64 mol of ions
 1 mol NH4 Cl 
b) Each mole of Ba(OH)2•8H2O forms 1 mole of Ba2+ ions and 2 moles of OH– ions, or a total of 3 moles of ions:
Ba(OH)2•8H2O(s) → Ba2+(aq) + 2OH–(aq). The waters of hydration become part of the larger bulk of water.
Convert mass to moles using the molar mass.
 1 mol Ba(OH)2 •8H 2O  


3 mol ions
Moles of ions =  25.4 g Ba(OH)2 •8H 2O  


 315.4 g Ba(OH)2 •8H 2O   1 mol Ba(OH)2 •8H 2O 
= 0.2415980 = 0.242 mol of ions
c) Each mole of LiCl produces 2 moles of ions (1 mole of Li+ ions and 1 mole of Cl– ions):
LiCl(s) → Li+(aq) + Cl–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of LiCl contains
6.022x1023 units of LiCl, more easily expressed as formula units.

  2 mol ions 
1 mol LiCl
Moles of ions = 3.55x1019 FU LiCl 


23
 6.022 x10 FU LiCl   1 mol LiCl 





= 1.17901x10–4 = 1.18x10–4 mol of ions

4.19

Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into
ions with the correct molar ratios. Convert mass and formula units to moles of compound and use the molar ratio
to convert moles of compound to moles of ions.

Solution:
a) Each mole of Rb2SO4 dissolves in water to form 2 moles of Rb+ ions and 1 mole of SO42– ions, or a total of
3 moles of ions: Rb2SO4(s) → 2Rb+(aq) + SO42–(aq).

 3 mol ions 
Moles of ions =  0.805 mol Rb2SO4  
 = 2.415 = 2.42 mol of ions
 1 mol Rb2SO4 
b) Each mole of Ca(NO3)2 forms 1 mole of Ca2+ ions and 2 moles of NO3– ions, or a total of
3 moles of ions: Ca(NO3)2(s) → Ca2+(aq) + 2NO3–(aq). Convert mass to moles using molar mass.
 1 mol Ca(NO3 )2   3 mol ions 
Moles of ions = 3.85x103 g Ca(NO3 )2 


 164.10 g Ca(NO3 )2   1 mol Ca(NO3 )2 
= 7.03839x10–5 = 7.04x10–5 mol of ions
c) Each mole of Sr(HCO3)2 produces 3 moles of ions (1 mole of Sr2+ ions and 2 moles of HCO3– ions):
Sr(HCO3)2(s) → Sr2+(aq) + 2HCO3–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of Sr(HCO3)2
contains 6.022x1023 units of Sr(HCO3)2, more easily expressed as formula units.



1 mol Sr(HCO3 ) 2
3 mol ions
Moles of ions = 4.03x1019 FU Sr(HCO 3 ) 2 
 6.022 x1023 FU Sr(HCO )   1 mol Sr(HCO ) 
3 2 
3 2 

= 2.0076x10–4 = 2.01x10–4 mol of ions





4.20





Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into
ions with the correct molar ratios. Convert mass and formula units to moles of compound and use the molar ratio
to convert moles of compound to moles of ions.
Solution:
a) Each mole of K3PO4 forms 3 moles of K+ ions and 1 mole of PO43– ions, or a total of 4 moles of ions:
K3PO4(s)  3K+(aq) + PO43–(aq)
 4 mol ions 
Moles of ions =  0.75 mol K 3 PO4  
 = 3.0 mol of ions.
 1 mol K3 PO4 
b) Each mole of NiBr2•3H2O forms 1mole of Ni2+ ions and 2 moles of Br– ions, or a total of 3 moles of ions:
NiBr2•3H2O(s)  Ni2+(aq) + 2Br –(aq). The waters of hydration become part of the larger bulk of water.

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4-16

Convert mass to moles using the molar mass.

 1 mol NiBr2 •3H 2O  

3 mol ions
Moles of ions = 6.88 x 103 g NiBr2 •3H 2O 


 272.54 g NiBr2 •3H 2O   1 mol NiBr2 •3H 2O 
= 7.5732x10–5 = 7.57x10–5 mol of ions
c) Each mole of FeCl3 forms 1mole of Fe3+ ions and 3 moles of Cl– ions, or a total of 4 moles of ions:
FeCl3(s)  Fe3+(aq) + 3Cl–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of FeCl3 contains
6.022x1023 units of FeCl3, more easily expressed as formula units.

  4 mol ions 
1 mol FeCl 3
Moles of ions = 2.23x1022 FU FeCl 3 
 6.022x1023 FU FeCl   1 mol FeCl 
3
3 

= 0.148124 = 0.148 mol of ions





4.21



Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into
ions with the correct molar ratios. Convert mass and formula units to moles of compound and use the molar ratio
to convert moles of compound to moles of ions.
Solution:
a) Each mole of Na2HPO4 forms 2 moles of Na+ ions and 1 mole of HPO42– ions, or a total of 3 moles of ions:
Na2HPO4(s)  2Na+(aq) + HPO42–(aq).
 3 mol ions 
Moles of ions =  0.734 mol Na 2 HPO4  
 = 2.202 = 2.20 mol of ions
 1 mol Na 2 HPO4 
b) Each mole of CuSO4•5H2O forms 1 mole of Cu2+ ions and 1 mole of SO42– ions, or a total of 2 moles of ions:
CuSO4•5H2O(s)  Cu+2(aq) + SO42–(aq). The waters of hydration become part of the larger bulk of water.
Convert mass to moles using the molar mass.
 1 mol CuSO4 •5H 2 O  

2 mol ions
Moles of ions =  3.86 g CuSO4 •5H 2 O  
 

 249.69 g CuSO4 •5H 2 O   1 mol CuSO4 •5H 2 O 
= 3.0918x10–2 = 3.09x10–2 mol of ions
c) Each mole of NiCl2 forms 1mole of Ni2+ ions and 2 moles of Cl– ions, or a total of 3 moles of ions:
NiCl2(s)  Ni2+(aq) + 2Cl–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of NiCl2 contains
6.022x1023 units of NiCl2, more easily expressed as formula units.

  3 mol ions 
1 mol NiCl 2
Moles of ions = 8.66x1020 FU NiCl 2 
 6.022x1023 FU NiCl   1 mol NiCl 

2 
2 

= 4.31418x10–3 = 4.31x10–3 mol of ions



4.22





moles solute 

Plan: In all cases, use the known quantities and the definition of molarity  M 
to find the
L of solution 

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The
chemical formulas must be written to determine the molar mass. (a) You will need to convert milliliters to liters,
multiply by the molarity to find moles, and convert moles to mass in grams. (b) Convert mass of solute to moles
and volume from mL to liters. Divide the moles by the volume. (c) Multiply the molarity by the volume.
Solution:
a) Calculating moles of solute in solution:
 103 L   0.267 mol Ca(C 2 H 3O 2 ) 2 
Moles of Ca(C2H3O2)2 = 185.8 mL 
 = 0.0496086 mol Ca(C2H3O2)2
 1 mL  
1L




Converting from moles of solute to grams:
 158.17 g Ca(C2 H3O2 )2 
Mass (g) of Ca(C2H3O2)2 = 0.0496086 mol Ca(C2 H3O2 )2 

 1 mol Ca(C2 H3O2 )2 

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4-17


= 7.84659 = 7.85 g Ca(C2H3O2)2
b) Converting grams of solute to moles:

 1 mol KI 
 = 0.127108 moles KI
 166.0 g KI 

Moles of KI =  21.1 g KI  

 10 3 L 
Volume (L) = 500. mL 
= 0.500 L
 1 mL 



Molarity of KI =

0 .127108 mol KI
= 0.254216 = 0.254 M KI
0.500 L

 0.850 mol NaCN 
c) Moles of NaCN = 145.6 L 
 = 123.76 = 124 mol NaCN
1L


4.23

moles solute 

Plan: In all cases, use the known quantities and the definition of molarity  M 
to find the
L of solution 

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The
chemical formulas must be written to determine the molar mass. (a) You will need to convert mass of solute to
moles and divide by the molarity to obtain volume in liters, which is then converted to milliliters. (b) Multiply
the volume by the molarity to obtain moles of solute. Use Avogadro’s number to determine the number of ions
present. (c) Divide mmoles by milliliters; molarity may not only be expressed as moles/L, but also as
mmoles/mL.
Solution:
a) Converting mass of solute to moles:
 1 mol KOH 

Moles of KOH = 8.42 g KOH 
 = 0.15006 mol KOH
 56.11 g KOH 
 1L 
Volume (L) of KOH solution = 0.15006 mol KOH 
 = 0.066398 L KOH solution
 2.26 mol 
 1L 
Volume (mL) of KOH solution = 0.066398 L KOH  3
 = 66.39823 = 66.4 mL KOH solution
 10 mL 
 2.3 mol CuCl 2 
b) Moles of CuCl2 = 52 L 
 = 119.6 mol CuCl2
L



 1 mol Cu 2  
Moles of Cu2+ ions = 119.6 mol CuCl2 
= 119.6 mol Cu2+ ions
 1 mol CuCl 
2


Converting moles of ions to number of ions:
 6.022 x1023 Cu 2  ions 
Number of Cu2+ ions = 119.6 mol Cu 2  ions 
= 7.2023x1025 = 7.2x1025 Cu2+ ions
 1 mol Cu 2  ions 








 135 mmol glucose 
c) M glucose = 
 = 0.490909 = 0.491 M glucose
275 mL


Note: Since 1 mmol is 10–3 mol and 1 mL is 10–3 L, we can use these units instead of converting to mol and L
since molarity is a ratio of mol/L. Molarity may not only be expressed as moles/L, but also as mmoles/mL.
4.24

moles solute 

Plan: In all cases, use the known quantities and the definition of molarity  M 
to find the
L
of solution 

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The
chemical formulas must be written to determine the molar mass. (a) Convert volume in milliliters to liters,

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4-18


multiply the volume by the molarity to obtain moles of solute, and convert moles to mass in grams. (b) The
simplest way will be to convert the milligrams to millimoles. Molarity may not only be expressed as moles/L, but
also as mmoles/mL. (c) Convert the milliliters to liters and find the moles of solute and moles of ions by
multiplying the volume and molarity. Use Avogadro’s number to determine the number of ions present.
Solution:
a) Calculating moles of solute in solution:
 10 3 L   5.62 x10 2 mol K 2SO 4 
Moles of K2SO4 =  475 mL  
 = 0.026695 mol K2SO4
 1 mL  
L



Converting moles of solute to mass:
 174.26 g K 2SO4 
Mass (g) of K2SO4 =  0.026695 mol K 2SO4  
 = 4.6519 = 4.65 g K2SO4
 1 mol K 2SO4 
b) Calculating mmoles of solute:
 7.25 mg CaCl2   1 mmol CaCl2 
Mmoles of CaCl2 = 
 = 0.065327 mmoles CaCl2

1 mL

 110.98 mg CaCl2 

Calculating molarity:
 0.065327 mmol CaCl2 
Molarity of CaCl2 = 
 = 0.065327 = 0.0653 M CaCl2
1 mL


If you believe that molarity must be moles/liters then the calculation becomes:
 7.25 mg CaCl 2   10 3 g   1 mL   1 mol CaCl 2 
Molarity of CaCl2 = 
 = 0.065327 = 0.0653 M CaCl2
 
  3  
1 mL

  1 mg   10 L   110.98 g CaCl 2 
Notice that the two central terms cancel each other.
c) Converting volume in L to mL:
 103 L 
Volume (L) = 1 mL 
= 0.001 L
 1 mL 


Calculating moles of solute and moles of ions:
 0.184 mol MgBr2 
–4
Moles of MgBr2 = 0.001 L 
 = 1.84x10 mol MgBr2
1

L


 1 mol Mg 2 
Moles of Mg2+ ions = 1.84 x 104 mol MgBr2 
 1 mol MgBr
2







–4
2+
 = 1.84x10 mol Mg ions


 6.022 x1023 Mg 2  ions 
Number of Mg2+ ions = 1.84x10 4 Mg 2  ions 
 1 mol Mg 2  ions 


= 1.1080x1020 = 1.11x1020 Mg2+ ions



4.25



moles solute 

to find the
Plan: In all cases, use the known quantities and the definition of molarity  M 
L of solution 

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The
chemical formulas must be written to determine the molar mass. (a) Convert mass of solute to moles and volume
from mL to liters. Divide the moles by the volume. (b) You will need to convert mass of solute to moles and
divide by the molarity to obtain volume in liters. (c) Divide the mmoles of solute by the molarity to obtain
volume in mL.
Solution:
a) Calculating moles of solute:
 1 mol AgNO3 
Moles of AgNO3 = 46.0 g AgNO3 
 = 0.2707475 mol AgNO3
 169.9 g AgNO3 

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4-19


 10 3 L 
= 0.335 L
Volume (L) = 335 mL 
 1 mL 




Calculating the molarity:
 0.2707574 mol AgNO 3 
Molarity of AgNO3 = 
 = 0.80823 = 0.808 M AgNO3
0.335 L


b) Calculating moles of solute:
 1 mol MnSO4 
Moles of MnSO4 =  63.0 g MnSO4  
 = 0.417218 mol MnSO4
 151.00 g MnSO4 
Calculating volume of solution:


1L
Volume (L) of solution =  0.417218 mol MnSO4  
 = 1.08368 = 1.08 L MnSO4 solution
 0.385 mol MnSO4 



1mL
c) Volume (mL) of ATP solution = 1.68 mmol ATP  

2
 6.44 x10 mmol ATP 

= 26.087 = 26.1 mL ATP solution
4.26

Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into
ions with the correct molar ratios. Convert the information given to moles of compound and use the molar ratio
to convert moles of compound to moles of ions. Avogadro’s number is used to convert moles of ions to numbers
of ions.
Solution:
a) Each mole of AlCl3 forms 1mole of Al3+ ions and 3 moles of Cl– ions: AlCl3(s)  Al3+(aq) + 3Cl–(aq).
Molarity and volume must be converted to moles of AlCl3.
 10 3 L   0.45 mol AlCl3 
Moles of AlCl3 = 130. mL  
 = 0.0585 mol AlCl3
 1 mL  
L



 1 mol Al3 
Moles of Al3+ =  0.0585 mol AlCl3  
 1 mol AlCl
3



3+
 = 0.0585 = 0.058 mol Al


 6.022x1023 Al 3 

Number of Al3+ ions = 0.0585 mol Al 3  
 1 mol Al 3 






 3 mol Cl
Moles of Cl– =  0.0585 mol AlCl3  
 1 mol AlCl
3



–
 = 0.1755 = 0.18 mol Cl


 6.022x1023 Cl 
Number of Cl– ions = 0.1755 mol Cl  
 1 mol Cl 







22

22
3+
 = 3.52287x 10 = 3.5x10 Al ions



23
23
–
 = 1.05686x10 = 1.1x10 Cl ions


b) Each mole of Li2SO4 forms 2 moles of Li+ ions and 1 mole of SO42– ions: Li2SO4(s)  2Li+(aq) + SO42–(aq).
 10 3 L   2.59 g Li 2SO 4   1 mol Li 2SO 4 
–4
Moles of Li2SO4 =  9.80 mL  
 = 2.3087x10 mol Li2SO4

 1 mL  
1
L
109.94
g
Li
SO

2
4 



 2 mol Li 
Moles of Li+ = 2.3087x10 4 mol Li 2SO 4 
 1 mol Li SO
2
4







–4
–4
+
 = 4.6174x10 = 4.62x10 mol Li


 6.022 x 1023 Li 
Number of Li+ ions = 4.6174x104 mol Li  

1 mol Li 







20

20
+
 = 2.7806x10 = 2.78x10 Li ions


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4-20


 1 mol SO 4 2 
Moles of SO42– = 2.3087x10 4 mol Li 2SO 4 
 1 mol Li SO
2
4







–4
–4
2–
 = 2.3087x10 = 2.31x10 mol SO4


 6.022 x 1023 SO 4 2  

Number of SO42– ions = 2.3087x104 mol SO 4 2  


1 mol SO 4 2 


= 1.39030x1020 = 1.39x1020 SO42– ions
c) Each mole of KBr forms 1 mole of K+ ions and 1 mole of Br– ions: KBr(s)  K+(aq) + Br–(aq).
 10 3 L   3.68 x1022 FU KBr  

1 mol KBr
Moles of KBr =  245 mL  
 
 = 0.01497 mol KBr
23
 1 mL  
L


  6.022x10 FU KBr 





 1 mol K  
Moles of K+ =  0.01497 mol KBr  
= 0.01497 = 1.50x10–2 mol K+
 1 mol KBr 



 6.022 x1023 K 
Number of K+ ions = 0.01497 mol K  
 1 mol K 







21
21 +
 = 9.016x10 = 9.02x10 K ions


 1 mol Br  
Moles of Br– =  0.01497 mol KBr  
= 0.01497 = 1.50x10–2 mol Br–
 1 mol KBr 


 6.022 x1023 Br 
Number of Br– ions = 0.01497 mol Br  
 1 mol Br 




4.27




21
21
–
 = 9.016x10 = 9.02x10 Br ions


Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into
ions with the correct molar ratios. Convert the information given to moles of compound and use the molar ratio
to convert moles of compound to moles of ions. Avogadro’s number is used to convert moles of ions to numbers
of ions.
Solution:
a) Each mole of MgCl2 forms 1 mole of Mg2+ ions and 2 moles of Cl– ions: MgCl2(s)  Mg2+(aq) + 2Cl–(aq).
 103 L   1.75 mol MgCl2 
Moles of MgCl2 =  88.mL  
 = 0.154 mol MgCl2
 1 mL  
L



 1 mol Mg 2 
Moles of Mg2+ =  0.154 mol MgCl2  
 1 mol MgCl
2



2+
 = 0.154 = 0.15 mol Mg


 6.022x1023 Mg 2  
Number of Mg2+ ions = 0.154 mol Mg 2  
= 9.27388x1022 = 9.3x1022 Mg2+ ions
 1 mol Mg 2  







 2 mol Cl
Moles of Cl– =  0.154 mol MgCl2  
 1 mol MgCl
2



–
 = 0.308 = 0.31 mol Cl


 6.022x1023 Cl  
Number of Cl– ions = 0.308 mol Cl  
= 1.854776x1023 = 1.9x1023 Cl– ions

 1 mol Cl  


b) Each mole of Al2(SO4)3 forms 2 moles of Al3+ ions and 3 moles of SO42– ions:
Al2(SO4)3(s)  2Al3+(aq) + 3SO42–(aq).
 103 L   0.22 g Al2 (SO 4 )3   1 mol Al 2 (SO 4 )3 
Moles of Al2(SO4)3 =  321 mL  
 



1L
  342.14 g Al2 (SO 4 )3 
 1 mL  





= 2.06406x10–4 mol Al2(SO4)3
 2 mol Al3 
Moles of Al3+ = 2.06406x10 4 mol Al 2 (SO 4 )3 
 1 mol Al (SO )
2
4 3







–4
–4
3+
 = 4.12812x10 = 4.1x10 mol Al


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4-21


 6.022x1023 Al3 
Number of Al3+ ions = 4.12812x104 mol Al3  
 1 mol Al3 







20
20
3+
 = 2.4860x10 = 2.5x10 Al ions


 3 mol SO4 2  
Moles of SO42– = 2.06406x104 mol Al2 (SO 4 )3 
= 6.19218x10–4 = 6.2x10–4 mol SO42
 1 mol Al (SO ) 
2
4 3 

23
 6.022 x10 SO 4 2  
Number of SO42– ions = 6.19218 x104 mol SO 4 2  

 1 mol SO 2 
4


= 3.7289x1020 = 3.7x1020 SO42– ions
c) Each mole of CsNO3 forms 1 mole of Cs+ ions and 1 mole of NO3– ions: CsNO3(s)  Cs+(aq) + NO3–(aq)
 8.83x1021 FU CsNO3  

1 mol CsNO3
= 0.024194 mol CsNO3
Moles of CsNO3 = 1.65 L  


23

  6.022x10 FU CsNO 
L
3 












 1 mol Cs 
Moles of Cs+ =  0.024194 mol CsNO3  
 1 mol CsNO
3



+
 = 0.024194 = 0.0242 mol Cs


 6.022x1023 Cs 
Number of Cs+ ions = 0.024194 mol Cs  
 1 mol Cs 




Moles of NO3– =



 1 mol NO3
 1 mol CsNO3

 0.024194 mol CsNO3  


22
22
+
 = 1.45695x10 = 1.46x10 Cs ions



–
 = 0.024194 = 0.0242 mol NO3


 6.022x1023 NO3
Number of NO3– ions = 0.024194 mol NO3 
 1 mol NO 
3




4.28




22
22
–
 = 1.45695x10 = 1.46x10 NO3 ions


Plan: These are dilution problems. Dilution problems can be solved by converting to moles and using the new
volume; however, it is much easier to use M1V1 = M2V2. The dilution equation does not require a volume in
liters; it only requires that the volume units match. In part c), it is necessary to find the moles of sodium ions in
each separate solution, add these two mole amounts, and divide by the total volume of the two solutions.
Solution:
a) M1 = 0.250 M KCl
V1 = 37.00 mL
M2 = ?
V2 = 150.00 mL
M1V1= M2V2
0.250 M 37.00 mL 
M x V1
=
= 0.061667 = 0.0617 M KCl
M2 = 1
150.0 mL
V2
V1 = 25.71 mL M2 = ?
V2 = 500.00 mL
b) M1 = 0.0706 M (NH4)2SO4
M1V1= M2V2
0.0706 M 25.71 mL 

M x V1
M2 = 1
=
= 0.003630 = 0.00363 M (NH4)2SO4
500.0 mL
V2
 10 3 L   0.348 mol NaCl   1 mol Na  
c) Moles of Na+ from NaCl solution = 3.58 mL 

 
 1 mL  
1L
  1 mol NaCl 


= 0.00124584 mol Na+
 10 3 L   6.81x10 2 mol Na 2SO 4   2 mol Na  
Moles of Na+ from Na2SO4 solution =  500. mL  
 

 1 mL  
1L


  1 mol Na 2SO 4 
= 0.0681 mol Na+
+
Total moles of Na ions = 0.00124584 mol Na+ ions + 0.0681 mol Na+ ions = 0.06934584 mol Na+ ions
Total volume = 3.58 mL + 500. mL = 503.58 mL = 0.50358 L
total moles Na  ions

0.06934584 mol Na  ions
=
= 0.1377057 = 0.138 M Na+ ions
Molarity of Na+ =
total volume
0.50358 L

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4-22


4.29

Plan: These are dilution problems. Dilution problems can be solved by converting to moles and using the new
volume; however, it is much easier to use M1V1 = M2V2. The dilution equation does not require a volume in
liters; it only requires that the volume units match.
Solution:
a) M1 = 2.050 M Cu(NO3)2
V1 = ?
M2 = 0.8543 M Cu(NO3)2
V2 = 750.0 mL
M1V1= M2V2
0.8543 M 750.0 mL 
M 2 x V2
=
= 312.5488 = 312.5 mL
V1 =
2.050 M

M1
 1.63 mol CaCl2   2 mol Cl  
–
M1 Cl– = 
 = 3.26 M Cl ions
 
1L

  1 mol CaCl 2 
V1 = ?
M2 = 2.86x10–2 M Cl– ions
V2 = 350. mL

b) M1 = 1.63 M CaCl2
M1 = 3.26 M Cl–
M1V1= M2V2





2.86x102 M  350. mL 
M 2 x V2
=
= 3.07055= 3.07 mL
V1 =
3.26 M
M1
V1 = 18.0 mL
M2 = 0.0700 M Li2CO3

c) M1 = 0.155 M Li2CO3
M1V1= M2V2
0.155 M 18.0 mL 
M x V1
=
= 39.8571 = 39.9 mL
V2 = 1
M2
0.0700 M 
4.30

V2 = ?

Plan: Use the density of the solution to find the mass of 1 L of solution. Volume in liters must be converted to
volume in mL. The 70.0% by mass translates to 70.0 g solute/100 g solution and is used to find the mass of
HNO3 in 1 L of solution. Convert mass of HNO3 to moles to obtain moles/L, molarity.
Solution:
 1 mL   1.41 g solution 
a) Mass (g) of 1 L of solution = 1 L solution  3  
 = 1410 g solution
 10 L 
1 mL




 70.0 g HNO3 
Mass (g) of HNO3 in 1 L of solution = 1410 g solution 
 = 987 g HNO3/L
 100 g solution 

 1 mol HNO3 
b) Moles of HNO3 = 987 g HNO3 
 = 15.6617 mol HNO3
 63.02 g HNO3 
 15.6617 mol HNO 3 
Molarity of HNO3 = 
 = 15.6617 = 15.7 M HNO3
1 L solution


4.31

Plan: Use the molarity of the solution to find the moles of H2SO4 in 1 mL. Convert moles of H2SO4 to mass of
H2SO4, divide that mass by the mass of 1 mL of solution, and multiply by 100 for mass percent. Use the density
of the solution to find the mass of 1 mL of solution.
Solution:
 18.3 mol H 2SO 4   10 3 L 
–2
a) Moles of H2SO4 in 1 mL = 
 = 1.83x10 mol H2SO4/mL
 
1
L
1
mL




 98.08 g H2SO4 

b) Mass of H2SO4 in 1 mL = 1.83x102 mol H 2SO4 
 = 1.79486 g H2SO4
 1 mol H 2SO4 
 1.84 g 
Mass of 1 mL of solution = 1 mL 
 = 1.84 g solution
 1 mL 



Mass percent =



1.79486 g H 2SO4
mass of H 2 SO 4
100  = 97.5467 = 97.5% H2SO4 by mass
100  =
1.84 g solution
mass of solution

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4-23


4.32

moles solute 


. Convert mass of solute to moles and volume from
Plan: Recall the definition of molarity  M 
L of solution 

mL to liters. Divide the moles by the volume.
Solution:
 1 mol NaClO 
Moles of NaClO = 20.5 g NaClO 
 = 0.2785896 mol NaClO
 74.44 g NaClO 

 103 L 
Volume (L) = 375 mL 
= 0.375 L
 1 mL 



 0.2753896 mol NaClO 
Molarity of NaClO = 
 = 0.73437 = 0.734 M NaClO
0.375 L


4.33

Plan: The first part of the problem is a simple dilution problem (M1V1 = M2V2). The volume in units of gallons can
be used. In part b), convert mass of HCl to moles and use the molarity to find the volume that contains that
number of moles.

Solution:
a) M1 = 11.7 M
V1 = ?
M2 = 3.5 M
V2 = 3.0 gal
3.5
M
3.0
gal



M 2 x V2
=
= 0.897436 gal
V1 =
11.7 M
M1
Instructions: Be sure to wear goggles to protect your eyes! Pour approximately 2.0 gal of water into the
container. Add slowly and with mixing 0.90 gal of 11.7 M HCl into the water. Dilute to 3.0 gal with water.
b) Converting from mass of HCl to moles of HCl:
 1 mol HCl 
Moles of HCl = 9.66 g HCl 
 = 0.264948 mol HCl
 36.46 g HCl 
Converting from moles of HCl to volume:

  1 mL 
1L
Volume (mL) of solution = 0.264948 mol HCl 

  3 
11.7
mol
HCl

  10 L 
= 22.64513 = 22.6 mL muriatic acid solution

4.34

Plan: Convert the mass of the seawater in kg to g and use the density to convert the mass of the seawater to
volume in L. Convert mass of each compound to moles of compound and then use the molar ratio in the
dissociation of the compound to find the moles of each ion. The molarity of each ion is the moles of ion divided
by the volume of the seawater. To find the total molarity of the alkali metal ions [Group 1A(1)], add the moles of
the alkali metal ions and divide by the volume of the seawater. Perform the same calculation to find the total
molarity of the alkaline earth metal ions [Group 2A(2)] and the anions (the negatively charged ions).
Solution:
a) The volume of the seawater is needed.
 103 g   cm 3   1 mL   103 L 
Volume (L) of seawater = 1.00 kg  
= 0.97560976 L
 1 kg   1.025 g   1 cm 3   1 mL 





The moles of each ion are needed. If an ion comes from more than one source, the total moles are needed.
NaCl:
Each mole of NaCl forms 1 mole of Na+ ions and 1 mole of Cl– ions: NaCl(s)  Na+(aq) + Cl–(aq)

 1 mol NaCl 
Moles of NaCl =  26.5 g NaCl  
 = 0.4534565 mol NaCl
 58.44 g NaCl 
 1 mol Na  
Moles of Na+ =  0.4534565 mol NaCl  
= 0.4534565 mol Na+
 1 mol NaCl 



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4-24


 1 mol Cl 
= 0.4534565 mol Cl–
Moles of Cl– =  0.4534565 mol NaCl  
 1 mol NaCl 


MgCl2:
Each mole of MgCl2 forms 1 mole of Mg2+ ions and 2 moles of Cl– ions: MgCl2(s)  Mg2+(aq) + 2Cl–(aq)
 1 mol MgCl2 
Moles of MgCl2 =  2.40 g MgCl2  
 = 0.025207 mol MgCl2
 95.21 g MgCl2 
 1 mol Mg 2 

Moles of Mg2+ =  0.025207 mol MgCl 2  
 1 mol MgCl
2



2+
 = 0.025207 mol Mg


 2 mol Cl  
Moles of Cl– =  0.025207 mol MgCl 2  
= 0.050415 mol Cl–
 1 mol MgCl 
2 

MgSO4:
Each mole of MgSO4 forms 1 mole of Mg2+ ions and 1 mole of SO42– ions: MgSO4(s)  Mg2+(aq) + SO42–(aq)

 1 mol MgSO4 
Moles of MgSO4 =  3.35 g MgSO4  
 = 0.0278308 mol MgSO4
 120.37 g MgSO4 
 1 mol Mg 2 
Moles of Mg2+ =  0.0278308 mol MgSO 4  
 1 mol MgSO
4




2+
 = 0.0278308 mol Mg


 1 mol SO 4 2  
Moles of SO42– =  0.0278308 mol MgSO 4  
= 0.0278308 mol SO42–
 1 mol MgSO 
4 

CaCl2:
Each mole of CaCl2 forms 1 mole of Ca2+ ions and 2 moles of Cl– ions: CaCl2(s)  Ca2+(aq) + 2Cl–(aq)
 1 mol CaCl2
Moles of CaCl2 = 1.20 g CaCl 2  
 110.98 g CaCl 2

  1 mol Ca 2 
 
  1 mol CaCl 2

 1 mol Ca 2 
Moles of Ca2+ =  0.0108128 mol CaCl2  
 1 mol CaCl
2



 = 0.0108128 mol CaCl2



2+
 = 0.0108128 mol Ca


 2 mol Cl 
Moles of Cl– =  0.0108128 mol CaCl 2  
= 0.0216255 mol Cl–
 1 mol CaCl 
2 

KCl:
Each mole of KCl forms 1 mole of K+ ions and 1 mole of Cl– ions: KCl(s)  K+(aq) + Cl–(aq)

 1 mol KCl 
Moles of KCl = 1.05 g KCl  
 = 0.0140845 mol KCl
 74.55 g KCl 
 1 mol K  
Moles of K+ =  0.0140845 mol KCl  
= 0.0140845 mol K+
 1 mol KCl 


 1 mol Cl  
Moles of Cl– =  0.0140845 mol KCl  
= 0.0140845 mol Cl–
 1 mol KCl 



NaHCO3:
Each mole of NaHCO3 forms 1 mole of Na+ ions and 1 mole of HCO3– ions: NaHCO3(s)  Na+(aq) + HCO3–(aq)
 1 mol NaHCO3 
Moles of NaHCO3 =  0.315 g NaHCO3  
 = 0.00374955 mol NaHCO3
 84.01 g NaHCO3 
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