CHAPTER 4 | Solution Chemistry and the Hydrosphere
4.70. Collect and Organize
We use the solubility rules to determine which sulfides would be insoluble and therefore contribute to the
black “smoke” from underwater hydrothermal vents.
Analyze
The solubility rules that would apply to these sulfides would be the following:
(1) compounds with cations of group 1 ions and NH 4+ are soluble and
(2) all other compounds are insoluble as sulfides
Solve
Of the list, Na+ and Li+ would be the only soluble sulfides. All the other cations would form insoluble sulfides
with formulas of MnS, FeS, CaS, MgS, ZnS, PbS, and CuS.
Think about It
Several insoluble sulfides may contribute to the black “smoke” in the hydrothermal vents under the oceans.
4.72. Collect and Organize
If Cr3+ and Cd2+ can be removed from wastewater by treatment with NaOH(aq), they must form insoluble salts
and, therefore, can be filtered off.
Analyze
Referring to Table 4.5 we see that all hydroxides, except those of Na + (and other alkali metals and NH4+) and
those of Ba2+, Ca2+, Sr2+, are insoluble so Cr3+ and Cd2+ form precipitates in a hydroxide solution. Using the
rules to write formulas for the salts learned in Chapter 2, cadmium hydroxide is Cd(OH) 2 and chromium(III)
hydroxide is Cr(OH)3. The net ionic equations would describe the formation of the insoluble compounds
without the spectator ions. In both equations, Na + will not be involved in the reaction and will not appear in the
net ionic equation.
Solve
The net ionic equations are
Cd2+(aq) + 2 OH–(aq)  Cd(OH)2(s)
Cr3+(aq) + 3 OH–(aq)  Cr(OH)3(s)
Think about It
It is important in the net ionic equation to specify the precipitate as (s). If (aq) is used after the formula for a
salt, it implies that the salt is soluble.
4.74. Collect and Organize
Silver chloride is an insoluble salt so we can assume that 100% of it precipitates from mixing 10 mL of a
solution of 5  10–3 M Cl– with 0.500 M AgNO3. We have to calculate the moles of Cl– and Ag+ in the solution
to determine if either is the limiting reactant before we can determine the mass of AgCl formed.
Analyze
Because we know the volume and concentration of the Cl– solution, we can calculate the moles of Cl–
contained in that solution. We are given an unspecified volume of AgNO 3 solution, so this must be available in
excess. The relevant net ionic equation describes the precipitation of the insoluble AgCl:
Ag+(aq) + Cl–(aq)  AgCl(s)
155
156 | Chapter 4
Therefore, the stoichiometric ratio of Cl– to AgCl formed is 1:1.
Solve
 5  10–3 mol 
mol Cl–  10 mL  
 5  10–5 mol Cl–
 1000 mL 
Because 1 mol of Cl– reacts with 1 mol of Ag+ to form 1 mol of AgCl, 5  10–5 mol of AgCl forms in the
reaction. The mass of AgCl formed is
143.32 g
5  10–5 mol 
 7  10 –3 g AgCl
1 mol
Think about It
Precipitation reactions involving insoluble salts can give us the amount of a cation or anion present in a
solution.
4.75. Collect and Organize
To determine how much MgCO3 precipitates in this reaction, we have to determine whether either Na 2CO3 or
Mg(NO3)2 is the limiting reactant. The net ionic reaction for the reaction is
Mg2+(aq) + CO32–(aq)  MgCO3(s)
Analyze
From the given volumes of each reactant and its concentration, we first need to calculate the moles of Mg 2+
and CO32– present in the mixed solution. These react in a 1:1 molar ratio to form MgCO 3. By comparison of
the moles of Mg2+ and CO32– we are able to determine the limiting reactant. Because 1 mol of MgCO 3 will
form from 1 mol of either Mg2+ or CO32–, the moles of the limiting reactant must equal the moles of MgCO 3
formed. From the moles of MgCO3 formed, we can calculate the mass formed using the molar mass of MgCO 3
(84.31 g/mol).
Solve
mol CO32–  10.0 mL Na 2 CO3 
mol Mg 2+  5.00 mL Mg  NO3  2 
0.200 mol 1 mol CO32–

 2.00  10 –3 mol
1000 mL 1 mol Na 2 CO3
0.0500 mol
1 mol Mg 2+

 2.50  10–4 mol
1000 mL 1 mol Mg  NO3  2
The limiting reactant is Mg(NO3)2 and 2.50  10–4 MgCO3 will form.
The mass of MgCO3 produced is
84.31 g
2.50  10–4 mol 
 2.11  10 –2 g MgCO3
1 mol
Think about It
In every stoichiometric equation, the moles of the reactants are important. For species in solution, the moles
can be found by multiplying the volume of the solution by the concentration, just as finding moles from a mass
of substance involves dividing the mass of substance by the molar mass.
Solution Chemistry and the Hydrosphere | 157
4.77. Collect and Organize
From the balanced equation, 1 mol of O2 is required to react with 4 mol of Fe(OH) +, the Fe(II) species.
Knowing the volume and concentration of Fe(II), we are asked to find the grams of O2 needed to form the
insoluble Fe(OH)3 product.
Analyze
The moles of Fe(OH)+ in solution can be found by multiplying the volume of the Fe(II) solution (75 mL) by its
concentration (0.090 M). The number of moles of O2 required in the reaction is one-fourth of the moles of
Fe(OH)+ present. From moles of O2 we can use the molar mass of O2 (32.00 g/mol) to calculate the grams of
O2 needed.
Solve
0.090 mol
 6.75  10–3 mol
1000 mL
1 mol O2
32.00 g O2
Mass of O2  6.75  10–3 mol Fe(OH)+ 

 5.4  10–2 g
+
1 mol
4 mol Fe(OH)
mol Fe(OH)+  75 mL 
Think about It
Because the molar ratio of Fe(OH)+ to O2 is 1: 4, we require fewer moles of O2 in this reaction than we have of
the Fe(II) species.
4.94. Collect and Organize
The oxidation number for nitrogen in these species varies depending on the number of hydrogens bonded to
nitrogen and the overall charge on the species.
Analyze
The oxidation number for hydrogen in these species is +1. The oxidation number for nitrogen will be negative
and can be determined by:
Oxidation number on N = charge on species – (number of H atoms)  (+1)
Solve
(a) N2: oxidation number on N = 0 – 0(+1) = 0
(b) N2H4: oxidation number on N = 0 – 4(+1) = –4. This oxidation number is for two N atoms, so the oxidation
number on each N is –2.
(c) NH4+: oxidation number on N = +1 – 4(+1) = –3
Think about It
The oxidation number for nitrogen, N2, is also clear following the rule that oxidation numbers for pure
elements are equal to zero.
4.96. Collect and Organize
Each of the reactions is either an oxidation or reduction half-reaction. We need to balance each by adding the
correct number of electrons to either the reactant side or the product side of the equation to balance the charge.
Analyze
A reduction reaction must have electrons added to a reactant so the electrons, after balancing, appear on the
left-hand side of the equation. An oxidation reaction must lose electrons so the electrons, after balancing,
appear on the right side of the equation.
Solve
(a) As written, the reactant side has a charge of 2+ and the product side has a charge of 3+. We need to add 1
electron to the product side to balance the charge.
Fe2+(aq)  Fe3+(aq) + 1 e–
158 | Chapter 4
This reaction is an oxidation.
(b) As written, the reactant side has a charge of 0 and the product side has a charge of 1–. We need to add 1
electron to the reactant side to balance the charge.
e– + AgI(s)  Ag(s) + I–(aq)
This reaction is a reduction.
(c) As written, the reactant side has a charge of 3+ and the product side has a charge of 2+. We need to add 1
electron to the reactant side to balance the charge.
e– + VO2+(aq) + 2 H+(aq)  VO2+(aq) + H2O ( )
This reaction is a reduction.
(d) As written, the reactant side has a charge of 0 and the product side has a charge of 10+. We need to add 10
electrons to the product side to balance the charge.
I2(s) + 6 H2O ( )  2 IO3–(aq) + 12 H+(aq) + 10 e–
This reaction is an oxidation.
Think about It
Remember to compute total charge on each side of the arrow. The charge of 2 mol of H + is 2+ not 1+.
4.98. Collect and Organize
To write the balanced half-reaction, we have to identify the reactants and products, balance the atoms, and then
balance the charge for the equation.
Analyze
We are given the formulas for the reactant, MnCO3, and the product, MnO2. Because the reaction occurs under
acidic conditions, we can use H2O and H+ to balance the atoms for these reactions. We are also told that this is
an oxidation reaction so we expect to add electrons to the product side in order to balance charge. The
carbonate species that may also be present is HCO3–.
Solve
Write reactants and products in an unbalanced equation:
MnCO3  MnO2 + HCO3–
The manganese and carbon atoms are already balanced. Balance oxygen atoms by adding water to the side
deficient in oxygen:
2 H2O + MnCO3  MnO2 + HCO3–
Balance hydrogen atoms by adding H+ to the side deficient in hydrogen:
2 H2O + MnCO3  MnO2 + HCO3– + 3 H+
Balance charge by adding 2 electrons to the product side (and add the state of each chemical species for
completeness):
2 H2O ( ) + MnCO3(s)  MnO2(s) + HCO3–(aq) + 3 H+(aq) + 2 e–
Think about It
This reaction is indeed an oxidation as electrons are produced.
4.100. Collect and Organize
For every species we are asked to find the oxidation number for each atom. From those oxidation numbers, we
can see which species is oxidized and which species is reduced.
Analyze
In these reactions, we can use the typical oxidation numbers of –2 for oxygen and +1 for hydrogen. Any
species, such as O2, that is an element has an oxidation number of 0. The oxidation numbers for the atoms
must add up to the charge on the species.
Solve
(a) Reactants
SiO2: Si = +4, O = –2
Products
H4SiO4: H = +1, Si = +4, O = –2
Solution Chemistry and the Hydrosphere | 159
H2O: H = +1, O = –2
No atoms change oxidation number. This is not a redox reaction.
(b) Reactants
Products
MnCO3: Mn = +2, C = +4, O = –2
MnO2: Mn = +4, O = –2
O 2: O = 0
CO2: C = +4, O = –2
Manganese is oxidized (Mn2+ to Mn4+) and oxygen is reduced (O2 to O2–).
(c) Reactants
Products
NO2: N = +4, O = –2
NO3–: N = +5, O = –2
H2O: H = +1, O = –2
NO: N = +2, O = –2
H+: H = +1
Nitrogen is both oxidized (N4+ to N5+) and reduced (N4+ to N2+) in this reaction.
Think about It
A reaction (as in c) in which a single reactant is both oxidized and reduced is called a disproportionation
reaction.
4.106. Collect and Organize
For each reaction, we can split the reaction into reduction and oxidation half-reactions, then balance them
under the specified conditions for hydrogen and oxygen, then for charge. Finally, we add the reduction and
oxidation half-reactions to obtain the overall reaction.
Analyze
The difference between balancing redox half-reactions in acid versus base is how we balance for hydrogen
(H+ in acid, OH– in base). We use the strategy that balances the reaction under acidic conditions first, then add
sufficient OH– to neutralize the acid on both sides of the equation to give the equation in basic solution.
Solve
(a) The half-reactions are
Mn 2+  MnO2
O2  H 2 O
(We can decide that O2 is reduced to H2O because Mn2+ is oxidized.) Balancing the oxygen and hydrogen in
acid solution first:
2 H 2 O + Mn 2+  MnO2 + 4 H +
4 H + + O2  2 H 2 O
Balancing for charge with electrons:
2 H 2 O + Mn 2+  MnO2 + 4 H + + 2 e –
4 e – + 4 H + + O2  2 H 2O
Adding together:
 2 H O  Mn
2
2+

 MnO2  4 H   2 e –  2
4 e –  4 H +  O2  2 H 2 O
4 H 2 O + 2 Mn 2+ + 4 H + + O2  2 MnO2 + 8 H+ + 2 H2 O
Simplifying:
2 H 2 O( )  2 Mn 2+ (aq)  O 2 ( g )  2 MnO 2 ( s)  4 H + ( aq)
By adding 4 OH– to both sides of the equation, we can place the reaction in basic solution:
4 OH –  2 H 2 O  2 Mn 2+  O2  2 MnO2  4 H +  4 OH –
4 OH –  2 H 2 O  2 Mn 2+  O2  2 MnO2  4 H 2 O
4 OH – (aq)  2 Mn 2+ (aq)  O 2 ( g )  2 MnO 2 ( s)  2 H 2O( )
(b) The half-reactions with balanced atoms (except H and O) are
160 | Chapter 4
MnO 2  Mn 2+
2 I–  I2
Balancing oxygen and hydrogen with H2O and H+ gives
4 H + + MnO2  Mn 2+ + 2 H 2 O
2 I –  I2
Balancing charge with electrons:
2 e – + 4 H + + MnO2  Mn 2+ + 2 H 2 O
2 I–  I2 + 2 e–
Adding the half-reactions:
2 e–  4 H+  MnO2  Mn 2+  2 H 2 O
2 I –  I2  2 e –
4 H (aq)  MnO2 ( s)  2 I (aq)  Mn 2+ (aq)  2 H 2 O( )  I 2 ( s)
(c) The half-reactions with balanced atoms (except H and O) are
I2  2 I –
+
–
2 S2 O32–  S4 O6 2–
The oxygen atoms are also balanced now! Balancing charge with electrons:
2 e – + I2  2 I –
2 S2 O32–  S4 O6 2– + 2 e –
Adding the half-reactions:
2 e–  I2  2 I –
2 S2 O32–  S4 O 6 2–  2 e –
I 2 ( s )  2 S2 O32– (aq)  2 I – (aq)  S4 O 6 2– ( aq)
Think about It
To balance reactions in either basic or neutral solution, first balance in acid. If you need to place the reaction
in basic solution, remember to add the same amount of OH– to both sides of the equation.