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Example 4.1
Calculate the molarity of each ion in an aqueous solution that is 0.00384 M Na 2SO4 and
0.00202 M NaCl. In addition, calculate the total ion concentration of the solution.
Strategy
First, note that there are three types of ions in the solution—Na+, SO42–, and Cl–. Of these
three types, SO42– comes only from the dissolved Na2SO4, Cl– comes only from the NaCl,
and Na+ comes from both solutes. Thus, we can establish [SO42–] just from the molarity of
Na2SO4(aq) and [Cl–] just from the molarity of NaCl(aq). For [Na+], we will have to work
with both molarities. Finally, we sum the molarities of the individual ions to obtain the
total ion concentration.
Solution
Example 4.1 continued
Assessment
As you become more familiar with calculating ion concentrations, you should be able
to work a problem of this sort just by inspecting the formulas of the solutes, thereby
arriving directly at the conclusions; [SO42–] = 0.00384 M, [Cl–] = 0.00202 M, and [Na+] =
(2 x 0.00384) + 0.00202 = 0.00970 M.
Exercise 4.1A
Seawater is essentially 0.438 M NaCl and 0.0512 M MgCl 2, together with several other
minor solutes. What are the molarities of Na+, Mg2+, and Cl– in seawater?
Exercise 4.1B
Each year, oral rehydration therapy (ORT)—feeding a person an electrolyte solution—
saves the lives of a million children worldwide who become dehydrated from diarrhea.
Each liter of the electrolyte solution contains 3.5 g sodium chloride, 1.5 g potassium
chloride, 2.9 g sodium citrate (Na3C6H5O7), and 20.0 g glucose (C6H12O6). Calculate the
molarity of each species present in the solution.
(Hint: Sodium citrate is a strong electrolyte, and glucose is a nonelectrolyte.)
Example 4.2
Barium nitrate, used to produce a green color in fireworks, can be made by the reaction of
nitric acid with barium hydroxide. Write (a) a complete-formula equation, (b) an ionic
equation, and (c) a net ionic equation for this neutralization reaction.
Solution
Example 4.2 continued
Exercise 4.2A
Calcium hydroxide is used to neutralize a waste stream of hydrochloric acid. Write (a) a
complete-formula equation, (b) an ionic equation, and (c) a net ionic equation for this
neutralization reaction.
Exercise 4.2B
Write the (a) complete-formula equation, (b) ionic equation, and (c) net ionic equation for
the following neutralization reaction.
KHSO4(aq) + NaOH(aq)  ?
Example 4.3 A Conceptual Example
Explain the observations illustrated in Figure 4.6.
Analysis and Conclusions
The bulb in Figure 4.6(a) is only dimly lit because acetic acid is a weak acid and therefore
a weak electrolyte [recall Figure 4.3(c)]. The situation in (b) is similar because ammonia
is a weak base and therefore also ionizes only slightly. When the two solutions are mixed,
which is what has been done in (c), the H+ ions from the CH3COOH readily combine with
NH3 molecules to form NH4+ ions:
H+(aq) + NH3(aq)  NH4+(aq)
By removing H+ ions from the solution, this reaction causes more CH3COOH
molecules to ionize, producing more H+ to react with more NH3 and so on. Soon all the
CH3COOH molecules ionize, and the neutralization goes to completion:
The original weak acid and weak base are replaced by an aqueous solution of a salt—
an ionic compound and strong electrolyte. The solution is now a good electrical
conductor, as seen in Figure 4.6(c).
Example 4.3 continued
Exercise 4.3A
In a situation similar to that in Figure 4.6, describe the observations you would expect to
make if the original solutions were CH3NH2(aq) and HNO3(aq).
Exercise 4.3B
Describe the observations you would expect in a situation similar to that in Figure 4.6 if
you start with a slurry of Mg(OH)2(s) in water (milk of magnesia) and add vinegar
[essentially 1 M CH3COOH(aq)] until all the Mg(OH)2(s) dissolves and some vinegar
remains in excess.
Example 4.4
Predict whether a precipitation reaction will occur in each of the following cases. If so,
write a net ionic equation for the reaction.
(a) Na2SO4(aq) + MgCl2 (aq)  ?
(c) K2CO3 (aq) + ZnCl2(aq)  ?
(b) (NH4)2S(aq) + Cu(NO3)2(aq)  ?
Strategy
We must consider the simple compounds that could be formed from the combinations of
ions present. Then we can use the solubility guidelines to determine whether any potential
products are insoluble compounds. If one or more of these potential products are
insoluble, a reaction will occur. If all potential products are soluble, there will be no
reaction.
Example 4.4 continued
Solution
Example 4.4 continued
Exercise 4.4A
Predict whether a reaction will occur in each of the following cases. If so, write a net ionic
equation for the reaction.
(a) MgSO4(aq) + KOH(aq)  ?
(c) Sr(NO3)2(aq) + Na2SO4(aq)  ?
(b) FeCl3(aq) + Na2S(aq)  ?
Exercise 4.4B
In each of the following cases, predict whether a reaction will occur, and, if so, write the
net ionic equation for the reaction.
(a) ZnSO4(aq) + BaS(aq)  ?
(b) Mg(OH)2(s) + NaOH(aq)  ?
(c) NaHCO3(aq) + Ca(OH)2(aq)  ?
Example 4.5 A Conceptual Example
Figure 4.8 shows that the dropwise addition of NH3(aq) to FeCl3(aq) produces a
precipitate. What is the precipitate?
Analysis and Conclusions
The reactants, NH3 and FeCl3, are both soluble in water. That all ammonium compounds
are soluble means the precipitate is not likely to contain NH 4+. It must therefore contain
Fe3+. But what is the anion? Recall that NH3 is a weak base and that it produces OH– ions
in aqueous solution:
Example 4.5 continued
Exercise 4.5A
Suppose that after all the Fe(OH)3(s) is precipitated, a large quantity of HCl(aq) is added
to the beaker in Figure 4.8. Describe what you would expect to see, and write a net ionic
equation for this change.
Exercise 4.5B
Potassium palmitate is a typical water-soluble soap. It is formed by the neutralization of
palmitic acid, CH3(CH2)14COOH, with potassium hydroxide. When calcium chloride is
added to an aqueous solution of potassium palmitate, a gray precipitate is observed. What
is the likely precipitate? Write ionic and net ionic equations for its formation.
Example 4.6
One cup (about 240 g) of a certain clear chicken broth yields 4.302 g AgCl when excess
AgNO3(aq) is added to it. Assuming that all the Cl – is derived from NaCl, what is the
mass of NaCl in the sample of broth?
Strategy
In this problem, chloride ions derived from NaCl are precipitated as AgCl. From the given
mass of AgCl, we can determine the number of moles of chloride ions initially present.
From the moles of Cl–, we can calculate first the moles of NaCl and then the mass of
NaCl.
Example 4.6 continued
Solution
Example 4.6 continued
Assessment
Note that the mass of the broth (about 240 g) does not enter into this calculation. We are
interested only in the mass of NaCl present. On the other hand, if we had been required to
find the percentage of NaCl in the broth, the mass of the broth would have been required
[% NaCl  (1.754 g NaCl/240 g broth) x 100%  0.7% NaCl].
Exercise 4.6A
What is the mass percent NaCl in a mixture of sodium chloride and sodium nitrate if a
0.9056-g sample of the mixture yields 0.9372 g AgCl(s) when allowed to react with
excess AgNO3(aq)?
Exercise 4.6B
Consider the seawater sample described in Exercise 4.1A. How many grams of precipitate
would you expect to get by adding (a) an excess of AgNO3(aq) to 225 mL of the seawater,
(b) an excess of NaOH(aq) to 5.00 L of the seawater? What are the precipitates?
Example 4.7
What are the oxidation numbers assigned to the atoms of each element in
(a) KClO4
(b) Cr2O72–
(c) CaH2
(d) Na2O2
(e) Fe3O4
Strategy
In applying the rules for determining oxidation numbers, we must adhere to the priority
order in which they are listed above. Note, however, that except in the case of
uncombined elements, we cannot apply Rule 1 first.
Solution
(a) The oxidation number of K is +1 (Rule 3). The oxidation number of O is –2 (Rule 6),
and the total for four O atoms is –8. For these two elements, the total is +1 – 8 = –7.
The oxidation number of the Cl atom in this ternary compound must be +7, to give a
total of zero (+1 – 8 + 7 = 0) for all atoms in the formula unit (Rule 1).
(b) The oxidation number of O is –2 (Rule 6), and the total for seven O atoms is –14. The
total of the oxidation numbers in this ion must be –2 (Rule 2). Therefore the total of
the oxidation numbers of two Cr atoms is +12, and that of one Cr atom is +6.
(c) Keeping in mind that the total for the formula unit must be 0 (Rule 1) and that the
oxidation number of Ca is +2 (Rule 3), the oxidation number of H must be –1 rather
than its usual +1 (Rule 5). Thus, for CaH2 the sum of the oxidation numbers is +2 +
(2 x –1) = 0.
Example 4.7 continued
(d) The oxidation number of Na is +1 (Rule 3), and for the two Na atoms, +2. The total
for the formula unit must be 0 (Rule 1). Even though the oxidation number of O is
usually –2 (Rule 6), here it must be –1 so that the total for the two O atoms is –2.
Rule 3 takes priority over Rule 6.
(e) The oxidation number of O is –2 (Rule 6). For four O atoms, the total is –8. The total
for the formula unit must be 0 (Rule 1). The total for three Fe atoms must be +8, and
for each Fe atom, +8/3.
Assessment
Remember that the sum of the oxidation numbers of all the atoms present must add up to
equal the total charge on the molecular or ionic formula. Notice how Rules 1 and 2 were
used at one point or another in each part of this example.
Usually, fractional oxidation numbers, as seen in part (e) signify an average. The
compound Fe3O4 is actually Fe2O3 · FeO. Two of the Fe atoms have oxidation numbers of
+3, and one has an oxidation number of +2. The average is (3 + 3 + 2)/3 = +8/3.
Exercise 4.7A
Assign an oxidation number to each atom in
(a) Al2O3, (b) P4, (c) CH3F, (d) HAsO42–, (e) NaMnO4, (f) ClO2–, (g) CsO2
Exercise 4.7B
Assign an oxidation number to each underlined atom:
(a) HSbF6, (b) CHCl3, (c) P3O105–, (d) S4O62–, (e) C3O2, (f) NO2+, (g) C2O42–
Example 4.8
Explain the difference in what happens when a copper-clad penny is immersed in
(a) hydrochloric acid and (b) nitric acid, as shown in Figure 4.14.
Example 4.8 continued
Analysis and Conclusions
(a) Because copper lies below hydrogen in the activity series of the metals, Cu(s) cannot
reduce H+(aq) to H2(g) and be oxidized to Cu2+(aq). Looking at it the other way, H+ is
not a strong enough oxidizing agent to oxidize Cu(s) to Cu 2+(aq). Chloride ion in
HCl(aq) can only be a reducing agent. As neither H+ nor Cl– can oxidize Cu, we
expect no reaction between Cu(s) and HCl(aq).
(b) In contrast, we clearly observe a reaction between copper and nitric acid. The Cu(s)
is oxidized to Cu2+(aq) (blue color). From the HCl analysis, we know that H+(aq) is
not strong enough as an oxidizing agent to oxidize Cu(s) to Cu 2+(aq), and we know
that NO3–(aq) is an oxidizing agent here because the N atom has its highest possible
oxidation number, +5. Therefore, copper is being oxidized to Cu 2+(aq) by NO3–(aq).
Figure 4.12 suggests that NO3–(aq) might be reduced to any one of several products.
We have no way at this point of predicting what that product might be, but the redbrown gas proves to be nitrogen dioxide, NO2. The oxidation number of nitrogen in
NO2 is the same as in N2O4 that is, +4. Using this information, we can write a net
ionic equation: (not balanced)
Cu(s) + H+(aq) + NO3–(aq)  Cu2+(aq) + NO2(g) + H2O(1) (not balanced)
Finally, the equation is balanced by inspection.
Cu(s) + 4 H+(aq) + NO3–(aq)  Cu2+(aq) + 2 NO2(g) + 2 H2O(1)
Example 4.8 continued
Exercise 4.8A
The oxidizing agent potassium dichromate, K2Cr2O7(s) reacts one way when heated with
hydrochloric acid and another way when heated with nitric acid. With one of the acids, a
gas is evolved and the solution color changes from red-orange to green. With the other
acid, the original red-orange color remains unchanged; that is, no reaction occurs. Explain
this difference in behavior.
Exercise 4.8B
Since 1982, U.S. pennies have been made of zinc with a thin copper coating. If the edge
of a new cent is notched with a knife and dropped in hydrochloric acid overnight, only a
hollow shell of copper remains the next morning. How would the resulting solution differ
from the two solutions in Figure 4.14?
Example 4.9
What volume (mL) of 0.2010 M NaOH is required to neutralize 20.00 mL of 0.1030 M
HCl in an acid–base titration?
Strategy
We need to do four things to solve this problem. In general, other titration problems can
be solved with a similar approach.
Step 1: Write an equation describing the neutralization and obtain a stoichiometric factor
relating moles of NaOH and moles of HCl.
Step 2: Determine how many moles of HCl are to be neutralized.
Step 3: Find the number of moles of NaOH required in the neutralization.
Step 4: Determine the volume of the solution containing this number of moles of NaOH.
Solution
Example 4.9 continued
Assessment
The molarity of the NaOH(aq) is just about twice that of the HCl(aq), and the combining
mole ratio of the acid and base is 1:1. This suggests that the volume of NaOH(aq) required
should be just about one-half that of the HCl(aq), and it is: 10.25 mL of the NaOH(aq)
neutralizes 20.00 mL of the HCl(aq).
Example 4.9 continued
Exercise 4.9A
What volume of 0.01060 M HBr(aq) is required to neutralize 25.00 mL of 0.01580 M
Ba(OH)2
2 HBr(aq) + Ba(OH)2(aq)  BaBr2(aq) + 2 H2O(l)
Exercise 4.9B
A 2.000-g sample of a sulfuric acid solution that is 96.5% H2SO4 by mass is dissolved in a
quantity of water and titrated. What volume of 0.3580 M KOH(aq) is required for the
titration? Assume that at the equivalence point the solution in the flask is K 2SO4(aq).
Example 4.10
A 10.00-mL sample of an aqueous solution of calcium hydroxide is neutralized by 23.30
mL of 0.02000 M HNO3(aq). What is the molarity of the calcium hydroxide solution?
Strategy
To determine the molarity of the Ca(OH)2(aq), we need to determine how many moles of
Ca(OH)2 are consumed in the titration and divide this number by the volume of the
sample (0.01000 L, or 10.00 mL). To determine the number of moles of Ca(OH) 2, we can
proceed through much of the problem as we did in Example 4.9.
Solution
Let’s start with a balanced equation for this reaction:
Example 4.10 continued
Exercise 4.10A
What volume of 0.550 M NaOH(aq) is required to neutralize a 10.00-mL sample of
vinegar that is 4.12% by mass acetic acid, CH3COOH? Assume that the vinegar has a
density of 1.01 g/mL.
Exercise 4.10B
A sample of battery acid is to be analyzed for its sulfuric acid content. A 1.00-mL sample
has a mass of 1.239 g. This 1.00-mL sample is diluted to 250.0 mL with water, and
10.00 mL of the diluted acid is neutralized by 32.44 mL of 0.00986 M NaOH. What is the
mass percent H2SO4 in the battery acid?
Example 4.11
Suppose a 0.4096-g sample from a box of commercial table salt is dissolved in water and
requires 49.57 mL of 0.1387 M AgNO3(aq) to completely precipitate the chloride ion. If
the chloride ion present in solution comes only from the sodium chloride, find the mass of
NaCl in the sample. Is commercial table salt pure sodium chloride?
Strategy
As with other solution stoichiometry problems, we can determine first the number of
moles of NaCl and then the NaCl mass from the volume and molarity of the AgNO 3(aq),
the appropriate stoichiometric factor from the balanced equation for the precipitation
reaction, and the molar mass of NaCl.
Solution
Example 4.11 continued
Assessment
The calculated mass of the NaCl is less than the mass of the sample of table salt, as it
should be because the sample is not pure NaCl. Also, notice that we did not need to use
the mass of table salt in our calculation, although we would need that information if we
wanted to find the mass percent of NaCl in the table salt.
Note that we could have combined the two steps into a single setup in which we
would not have had to record the intermediate result: 6.875 x 10–3 mol AgNO3. The final
answer proves to be the same by either method; 0.4018 g NaCl.
Exercise 4.11A
A solution containing a water-soluble salt of the radioactive element thorium can be
titrated with oxalic acid solution to form insoluble thorium(IV) oxalate. If 25.00 mL of a
solution of thorium(IV) ion requires 19.63 mL of 0.02500 M H2C2O4 for complete
precipitation, find the molar concentration of thorium(IV) ion in the unknown solution.
Exercise 4.11B
A mixture of 0.1015 g of NaCl and 0.1324 g KCl is dissolved in water and titrated with
0.1500 M AgNO3(aq). What volume of the AgNO3(aq) will be needed to reach the
endpoint?
Example 4.12
A 0.2865-g sample of an iron ore is dissolved in acid, and the iron is converted entirely to
Fe2+(aq). To titrate the resulting solution, 0.02645 L of 0.02250 M KMnO4(aq) is required.
What is the mass percent of iron in the ore?
Strategy
We can work this problem in two parts. The first requires finding the number of grams of
Fe in the unknown solution from the titration data. This calculation is similar to ones we
have done for acid–base and precipitation titrations. The second part is a straightforward
calculation of a percentage, using the sample mass of 0.2865 g.
Solution
The purpose of each factor in the calculation involving titration data is indicated in the
following outline:
Example 4.12 continued
Assessment
Because iron is only one component of iron ore, the mass of iron in the ore sample must
be less than 0.2865 g and the percent iron must be less than 100%. These facts can be
used to assess the plausibility of the results obtained in the calculation.
Exercise 4.12A
Suppose the titration in Example 4.12 was carried out with 0.02250 M K 2Cr2O7(aq) rather
than KMnO4(aq). What volume of the K2Cr2O7(aq) would be required? The net ionic
equation is
6 Fe2+(aq) + Cr2O72–(aq) + 14 H+(aq)  6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l)
Exercise 4.12B
A 20.00-mL sample of KMnO4(aq) is required to titrate 0.2378 g sodium oxalate in an
acidic solution. How many milliliters of this same KMnO4(aq) are required to titrate a
25.00-mL sample of 0.1010 M FeSO4 in an acidic solution?
2 MnO4–(aq) + 16 H+(aq) + 5 C2O42–(aq)  2 Mn2+(aq) + 8 H2O(l) + 10 CO2(g)
Cumulative Example
Sodium nitrite is used in the production of fabric dyes, as a meat preservative, as a bleach
for fibers, and in photography. It is prepared by passing nitrogen monoxide gas and
oxygen gas into an aqueous solution of sodium carbonate. Carbon dioxide gas is the other
product of the reaction. (a) Write a balanced equation for the reaction. (b) What mass of
sodium nitrite should be produced in the reaction of 748 g of Na 2CO3, with the other
reactants in excess? (c) In another preparation, the reactants are 225 mL of 1.50 M
Na2CO3(aq), 22.1 g nitrogen monoxide, and excess O2. What mass of sodium nitrite
should be produced if the reaction has a yield of 95.1%?
Strategy
For part (a), we can write the formulas for reactants and products (Sections 2.6 and 2.7),
and then construct and balance the equation (Section 3.7). Using the balanced equation,
we can solve part (b), starting with grams of Na2CO3 and ending with grams of NaNO2
(Section 3.8). Part (c) can be broken down into three steps: Convert volume of Na 2CO3
to moles using molarity (Section 3.11); solve as a limiting reactant problem (Section 3.9)
for the mass of produced from NO and Na2CO3; use that theoretical yield of product
along with percentage yield (Section 3.10) to find the actual yield.
Cumulative Example continued
Solution
Cumulative Example continued
Assessment
The balanced equation has four N atoms, twelve O atoms, four Na atoms, and two C
atoms on each side. Because the molar mass of NaNO2 is roughly half that of Na2CO3
and because the mole ratio between NaNO2 and Na2CO3 is 2:1, the mass of NaNO2 we
get in part (b) should be comparable to the mass of Na2CO3 used. In part (c), because the
yields of product from the two reactants are so similar, it is difficult to determine whether
the limiting reactant is correct except by close inspection of the setup. However, we can
note that the actual yield of NaNO2 is indeed less than the theoretical yield, as it should
be.