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CHAPTER 4
CH7 PROBLEM SOLVING CLASS
R.D. A. BOLINAS
4.9 State whether each of the following
substances is likely to be very soluble in
water. Explain.
(a) Lithium nitrate
(b) Glycine, H2NCH2COOH
(c) Pentane
(d) Ethylene glycol, HOCH2CH2OH
SOLUBLE IONIC COMPOUNDS
• 1. All common compounds of Group 1A(1) ions (Li+, Na+,
• K+, etc.) and ammonium ion (NH4+) are soluble.
• 2. All common nitrates (NO3), acetates (CH3COO or
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C2H3O2), and most perchlorates (ClO4) are soluble.
3. All common chlorides (Cl-), bromides (Br-), and
iodides (I-) are soluble, except those of Ag+, Pb2+ , Cu+,
and Hg2+
4 . All common fluorides (F-) are soluble, except those of
Pb2+ and Group 2A(2).
5. All common sulfates (SO42-) are soluble, except those
of Ca2+ , Sr2+ , Ba2+ , Ag+ , and Pb2+ .
INSOLUBLE IONIC COMPOUNDS
• 1. All common metal hydroxides are insoluble, except
those of Group 1A(1) and the larger members of Group
2A(2) (beginning with Ca2+).
• 2. All common carbonates (CO32- ) and phosphates
(PO43-) are insoluble, except those of Group 1A(1) and
NH4+
• 3. All common sulfides are insoluble except those of
Group1A(1), Group 2A(2), and NH4+
• a) Lithium nitrate, an ionic compound, would be
expected to be soluble in water, and the solubility rules
confirm this.
• 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 not be expected to be soluble 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.
4.11 State whether an aqueous solution of
each of the following substances conducts
an electric current.
Explain your reasoning.
(a) Potassium sulfate
(b) Sucrose, C12H22O11
a) Yes; K2SO4 is an ionic compound that is soluble in
water, producing K+ and SO42– ions.
b) No; sucrose is neither a salt, an acid, nor a base, so it
would be a nonelectrolyte (even though it’s soluble in
water).
4.13 How many total moles of ions are
released when each of the following
samples dissolves completely in water?
(a) 0.734 mol of Na2HPO4
(b) 3.86 g of CuSO4⦁5H2O
(c) 8.66x1020 formula units of NiCl2
4.15 How many moles and numbers of
ions of each type are present in the
following aqueous solutions?
(a) 88 mL of 1.75 M magnesium chloride
(b) 321 mL of a solution containing
0.22 g aluminum sulfate/L
(c) 1.65 L of a solution containing 8.83
1021 formula units of cesium nitrate per
liter
4.17 How many moles of H+ ions are
present in the following aqueous
solutions?
(a) 1.4 mL of 0.75 M hydrobromic acid
(b) 2.47 mL of 1.98 M hydriodic acid
(c) 395 mL of 0.270 M nitric acid
• The acids in this problem are all strong acids, so you
can assume that all acid molecules dissociate
completely to yield H+ ions and associated anions.
One mole of HBr, HI, and HNO3 each produce one mole
of H+ upon dissociation, so moles H+ = moles acid.
Molarity is expressed as moles/L instead of as M.
a) Moles H+ = mol HBr = (1.4 mL) (10–3 L/1mL) (0.75
mol/L) = 1.05 x 10–3 = 1.0 x 10–3 mol H+
• b) Moles H+ = mol HI = (2.47 mL) (10–3L/1 mL) (1.98
mol/L) = 4.8906 x 10–3 = 4.89 x 10–3 mol H+ c) Moles
H+ = mol HNO3 = (395 mL) (10–3 L/1 mL) (0.270 mol/L) =
0.10665 = 0.107 mol H+
4.19 Water “softeners” remove metal ions
such as Ca2+ and Fe3+ by replacing them
with enough Na+ ions to maintain the
same number of positive charges in the
solution. If 1.0 103 L of “hard” water is
0.015 M Ca2 and 0.0010 M Fe3 , how
many moles of Na+ are needed to replace
these ions?
• The moles of the calcium ions and the iron ions are
needed. The moles of each of the ions to be replaced
must be multiplied by the charge to get the total moles of
charge. Since sodium has a +1 charge the total moles of
charge equals the moles of sodium ions.
4.23 The beakers represent the aqueous
reaction of AgNO3 and NaCl. Silver ions
are gray. What colors are used to
represent NO3-, Na+, and Cl-?
Write molecular, total ionic, and net ionic
equations for the reaction.
+

SOLUBLE IONIC COMPOUNDS
• 1. All common compounds of Group 1A(1) ions (Li+, Na+,
• K+, etc.) and ammonium ion (NH4+) are soluble.
• 2. All common nitrates (NO3), acetates (CH3COO or
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C2H3O2), and most perchlorates (ClO4) are soluble.
3. All common chlorides (Cl-), bromides (Br-), and
iodides (I-) are soluble, except those of Ag+, Pb2+ , Cu+,
and Hg2+
4 . All common fluorides (F-) are soluble, except those of
Pb2+ and Group 2A(2).
5. All common sulfates (SO42-) are soluble, except those
of Ca2+ , Sr2+ , Ba2+ , Ag+ , and Pb2+ .
INSOLUBLE IONIC COMPOUNDS
• 1. All common metal hydroxides are insoluble, except
those of Group 1A(1) and the larger members of Group
2A(2) (beginning with Ca2+).
• 2. All common carbonates (CO32- ) and phosphates
(PO43-) are insoluble, except those of Group 1A(1) and
NH4+
• 3. All common sulfides are insoluble except those of
Group1A(1), Group 2A(2), and NH4+
• Plan: Use Table 4.1 to predict the products of this reaction. Ions
not involved in the precipitate are spectator
ions and are not included in the net ionic equation.
•
Solution: Assuming that the left beaker is AgNO3 (because it
has gray Ag+ ion) and the right must be NaCl, then the NO3– is
blue, the Na+ is brown, and the Cl– is green. (Cl– must be
green since it is present with Ag+ in the precipitate in the
beaker on the right.)
• Molecular equation:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Total ionic equation: Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq)
→ AgCl(s) + Na+(aq) + NO3–(aq)
• Net ionic equation: Ag+(aq) + Cl–(aq) → AgCl(s)
4.27 When each of the following pairs of
aqueous solutions is mixed, does a
precipitation reaction occur? If so, write
balanced molecular, total ionic, and net
ionic equations:
(a) Potassium carbonate +
barium hydroxide
(b) Aluminum nitrate + sodium phosphate
• a) Barium carbonate (BaCO3) precipitates; the
potassium hydroxide is a strong base.
• Molecular: K2CO3(aq) + Ba(OH)2(aq) → BaCO3(s) + 2 KOH(aq)
• Total ionic: 2 K+(aq) + CO32–(aq) + Ba2+(aq) + 2 OH–(aq) →
BaCO3(s) + 2 K+(aq) + 2 OH–(aq)
• Net ionic: Ba2+(aq) + CO32–(aq) → BaCO3(s)
• b) Aluminum phosphate (AlPO4) precipitates; the
sodium nitrate is soluble.
• Molecular: Al(NO3)3(aq) + Na3PO4(aq) → AlPO4(s) + 3
NaNO3(aq)
Total ionic: Al3+(aq) + 3 NO3–(aq) + 3 Na+(aq) + PO43–(aq) →
AlPO4(s) + 3 NO3–(aq) + 3 Na+(aq)
4.33 Aluminum sulfate, known as cake alum, has a
wide range of uses, from dyeing leather and cloth to
purifying sewage. In aqueous solution, it reacts with
base to form a white precipitate.
(a) Write balanced total and net ionic equations for its
reaction with aqueous NaOH.
(b) What mass of precipitate forms when185.5 mL of
0.533 M NaOH is added to 627 mL of a solution
that contains 15.8 g of aluminum sulfate per liter?
4.39 Complete the following acid-base
reactions with balanced molecular, total
ionic, and net ionic equations:
(a) Cesium hydroxide(aq) + nitric acid(aq)
(b) Calcium hydroxide(aq) + acetic acid(aq)
• a) Molecular: CsOH(aq) + HNO3(aq) →
CsNO3(aq) + H2O(l)
• Total ionic: Cs+(aq) + OH–(aq) + H+(aq) + NO3–(aq) →
Cs+(aq) + NO3–(aq) + H2O(l)
• Net ionic: OH–(aq) + H+(aq) → H2O(l)
Spectator ions are Cs+ and NO3–.
• b) Molecular: Ca(OH)2(aq) + 2 HC2H3O2(aq) →
Ca(C2H3O2)2(aq) + 2 H2O(l)
• Total ionic: Ca2+(aq) + 2 OH–(aq) + 2 HC2H3O2 (aq) →
Ca2+(aq) + 2 C2H3O2–(aq) + 2 H2O(l)
• Net ionic: OH–(aq) + HC2H3O2(aq) → C2H3O2–(aq) + H2O(l)
Spectator ion is Ca2+.
4.42 If 25.98 mL of a standard
0.1180 M KOH solution reacts with
52.50 mL of CH3COOH solution,
what is the molarity of the acid solution?
• Plan: Write a balanced equation and use the molar ratios
to convert the amount of KOH to the amount of
CH3COOH.
• The reaction is: KOH(aq) + CH3COOH(aq) →
KCH3COO(aq) + H2O(l)
4.43 If 26.25 mL of a standard
0.1850 M NaOH solution is required
to neutralize 25.00 mL of H2SO4,
what is the molarity of the acid
solution?
• The reaction is: 2 NaOH(aq) + H2SO4(aq) →
Na2SO4(aq) + 2 H2O(l)
4.44 An auto mechanic spills 88 mL of
2.6M H2SO4 solution from a rebuilt auto
battery. How many milliliters of 1.6 M
NaHCO3 must be poured on the spill to
react completely with the sulfuric acid?
[Hint: H2O and CO2 are among the
products.]
• The reaction is: 2 NaHCO3(aq) + H2SO4(aq) →
Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)
4.51 Give the oxidation number of sulfur in
the following:
(a) SOCl2
(b) H2S2
(c) H2SO3
(d) Na2S
1. IN GROUP 1A(1), O.N. = +1 in all
compounds
2. IN GROUP 2A(2) O.N. = +2 in all cpds
3. For HYDROGEN, O.N. = +1 with nonmetals, O.N.= -1 with metals and
Boron.
4. FLUORINE: O.N. = -1 in all
compounds
5. OXYGEN: O.N. = -1 in peroxides (-OO-) , O.N. = -2 in all other compounds
6. FOR GROUP 7A(17): O.N. = -1 with
metals, nonmetals except O, and
other halogens lower in the group.
RULES FOR ASSIGNING OXIDATION
NUMBERS:
• 1. Elemental form = 0 O.N. (e.g. O2, Na, Fe…)
• 2. monoatomic ion, O.N. = CHARGE (Na+ = 1)
• 3. in a compound, SUM OF O.N. = 0,
in polyatomic ion, SUM = CHARGE.
1.
2.
3.
4.
5.
6.
IN GROUP 1A(1), O.N. = +1 in all compounds
IN GROUP 2A(2) O.N. = +2 in all cpds
For HYDROGEN, O.N. = +1 with non-metals, O.N.= -1 with metals and Boron.
FLUORINE: O.N. = -1 in all compounds
OXYGEN: O.N. = -1 in peroxides (-O-O-) , O.N. = -2 in all other compounds
FOR GROUP 7A(17): O.N. = -1 with metals, nonmetals except O, and other
halogens lower in the group.
• Consult Table 4.3 for the rules for assigning oxidation
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numbers.
a) +4
b) -1
c) +4
d) -2
4.57 Identify the oxidizing and reducing
agents in the following:
(a) Sn(s) + 2H+(aq) Sn2+(aq) + H2(g)
(b) 2H+(aq) + H2O2 (aq) + 2Fe2+ (aq) 
2Fe3+ (aq) + 2H2O(l)
• LEORA vs. GEROA
• Loses Electrons, OXIDIZED, Reducing Agent
• Gains Electrons, REDUCED, Oxidizing Agent
• a) Oxidizing agent = H+ Reducing agent = Sn
• b) Oxidizing agent = H2O2 Reducing agent = Fe2+
4.59 Identify the oxidizing and reducing
agents in the following:
(a) 8H+ (aq) +Cr2O72-(aq) + 3SO32-(aq) 
2Cr3+(aq) + 3SO42-(aq) + 4H2O(l)
(b) NO3-(aq) + 4Zn(s) + 7OH- (aq) +
6H2O(l)  4Zn(OH)42-(aq) + NH3(aq)
• a) Oxidizing agent = Cr2O72– Reducing agent = SO32–
b) Oxidizing agent = NO3– Reducing agent = Zn
4.66 Balance each of the following redox
reactions and classify it as a combination,
decomposition, or displacement reaction:
(a) Sb(s) + Cl2(g)  SbCl3(s)
(b) AsH3(g)  As(s) + H2(g)
(c) Zn(s) + Fe(NO3)2(aq) 
Zn(NO3)2(aq) +Fe(s)
• Combination: X + Y → Z;
• decomposition: Z → X + Y
• Single displacement: X + YZ → XZ + Y
• Double displacement: WX + YZ → WZ + YX
• a) 2 Sb(s) + 3 Cl2(g) → 2 SbCl3(s) COMBINATION
• b)2AsH3(g)→2As(s)+3H2(g) DECOMPOSITION
c) Zn(s) + Fe(NO3)2(aq) → Zn(NO3)2(aq) + Fe(s) DISPLACEMENT
•
4.85 For the following aqueous reactions, complete and balance
the molecular equation and write a net ionic equation:
(a) Manganese(II) sulfide + hydrobromic acid
(b) Potassium carbonate + strontium nitrate
(c) Potassium nitrite + hydrochloric acid
(d) Calcium hydroxide + nitric acid
(e) Barium acetate + iron(II) sulfate
(f) Barium hydroxide + hydrocyanic acid
(g) Copper(II) nitrate + hydrosulfuric acid
(h) Magnesium hydroxide + chloric acid
(i) Potassium chloride + ammonium phosphate