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CHEM 111 Chapter 4: Chapter 4Exam

Course Code
CHEM 111
Study Guide

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CHAPTER 4 | Solution Chemistry and the Hydrosphere
4.1. Collect and Organize
This question asks us to differentiate between a strong binary acid and a weak binary acid. A binary acid is an
acid containing hydrogen and another element, such as HCl. A strong acid completely dissociates, meaning that
all binary molecules of HX are present in solution as H+ and X. A weak acid does not completely dissociate,
meaning that some of the HX molecules are in the form of H+ and X and some are present in the form HX.
From Figure P4.1, all of the gray spheres have a + charge so these must represent the H+ in the binary acids.
The green, yellow, and magenta spheres carrying a – charge must represent the Xs in the binary acids HX. If the
X are free in solution and not combined with H+, then that HX must be a strong acid. If, however, HX is found
in the solution as a molecule along with H+ and X, then that HX must be a weak acid.
All the green and magenta spheres are present in solution as ions, not in combination with H+ as HX. Therefore,
HX(green) and HX(magenta) must both be strong acids. HX(yellow), however, is represented as two HX
molecules and three X(yellow) ions. Because only some of the HX(yellow) has dissociated, HX(yellow) is the
weak acid.
Think about It
The extent to which a weak acid may be dissociated can vary. If this were a strong weak acid, perhaps only one
HX(yellow) molecule would be represented in Figure P4.1 along with four X(yellow) anions.
4.2. Collect and Organize
From Figure P4.2 we are asked to identify the substances present when mixing NaCl and AgNO3 as aqueous
solutions. Both of these are soluble salts and upon mixing might form a precipitate.
In aqueous solution NaCl exists as Na+ and Cl and AgNO3 exists as Ag+ and NO3
. In Figure P4.2 we see that
there are 6 bronze anions, 6 magenta cations, and 6 molecules made up of yellow atoms and gray atoms. The
bonded species must represent a precipitate salt that has formed in the solution. From solubility rules we know
that all sodium salts are soluble, so Na+ will not precipitate from the solution in combination with Cl or NO3
Ag+, however, does form an insoluble salt with Cl.
The molecules of gray and yellow must be the precipitate AgCl. The ions left in solution must be the soluble
NaNO3 salt.
(a) Na+ = magenta spheres with positive charge in solution
(b) Cl = yellow spheres in combination with Ag+ as an insoluble salt
(c) NO3
= bronze spheres with negative charge in solution
Think about It
The stoichiometry in Figure P4.2 shows that there are no excess ions in solution; the reactants were mixed in a
1:1 ratio.
4.3. Collect and Organize
We have to consider the possible charges for the highlighted elements in each of the substances. All of the
elements are nonmetals and may have positive or negative charge, depending on what other elements they are
combined with.
In each of the substances, oxygen carries a 2– charge and hydrogen carries a 1+ charge. Knowing that, we can
determine the charge on the X atom in all the compounds. The elements that are shown in Figure P4.3 are

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156 | Chapter 4
green = nitrogen, blue = phosphorus, orange = sulfur, purple = chlorine, and yellow = argon. Nitrogen and
phosphorus could have charges of 3+, 5+, and 3–; sulfur may have charges of 4+, 6+, and 2–; chlorine may
have charges of 1–, 5+, and 7+; and argon is a noble gas that does not lose or gain electrons to become charged
and so it does not combine with other elements, even ones like fluorine and oxygen.
(a) In HX, X has a 1– charge, so HX is HCl (purple).
(b) In H2XO4, X has a charge of 6+, so H2XO4 is H2SO4 (orange).
(c) In HXO3, X has a charge of 5+, which may be either HNO3 or HPO3; nitric acid is a common acid and HPO3
is not very common so HXO3 is HNO3 (green).
(d) In H3XO4, X has a charge of 5+, which may be either H3NO4 or H3PO4; phosphoric acid is a common acid
and H3NO4 is not very common so H3XO4 is H3PO4 (blue).
Think about It
We will find out later, based on electron count, Lewis structures, and formal charge, why H3NO4 is not common
but H3PO4 is.
4.4. Collect and Organize
We are to correlate solubility rules and acid or base character with the groups in the periodic table. The groups
highlighted in Figure P4.4 are green = sodium through cesium in group 1; blue = silver and gold in group 11;
orange = zinc, cadmium, and mercury in group 12; and purple = chlorine through astatine in group 17.
All but one of the groups are metals; these metals will form salts with anionic species (halides and hydroxides).
The nonmetal group will not combine with anions but will form covalent compounds. Using this information
and the information about solubility of salts in Table 4.5, we can now correlate the chemical properties.
(a) Silver in group 11 forms an insoluble halide salt.
(b) Hydroxide compounds of both group 11 (such as AgOH) and group 12 [such as Zn(OH)2] are insoluble.
(c) Group 1 elements form compounds with hydroxide (NaOH, KOH, RbOH, CsOH) that are soluble.
(d) The binary compounds of group 17 form the compounds HCl, HBr, and HI. These are strong acids.
Think about It
This correlation makes remembering some of the strong acids (group 17 HX except for HF) and some of the
strong bases (group 1 MOH) easier to remember.
4.5. Collect and Organize
A solution consists of a solute and a solvent. To decide which is the solvent we have to interpret the definitions
of these terms.
A solvent is something (often a liquid, but not always) in which the solute dissolves. The solute is a substance
that dissolves into a solvent.
When the solute is a salt and the solvent is water, we could define the solvent as the liquid component of the
solution. When the solvent and solute are both liquids or both solids, however, that definition is muddied. The
solvent though is usually in greater quantity (volume) than the solute, so defining the solvent as the component
that is present in the greatest amount may be more broadly applied.
Think about It
A solid–solid solution that involves two metals is called an alloy. An example of an alloy is bronze.
4.6. Collect and Organize
Using the definition of a solvent and a solution, we are asked to determine if a solid could ever act as a solvent.

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Solution Chemistry and the Hydrosphere | 157
A solution is a homogeneous mixture of two substances. If two solids can be mixed homogeneously, then one of
the solids must function as a solvent.
A solid can indeed be a solvent. A mixture of metals (an alloy) is a homogeneous solid solution of one or more
metals in another metal.
Think about It
Solid–solid solutions are ubiquitous from stainless steel to amalgams for dental fillings.
4.7. Collect and Organize
Molarity is expressed in moles per liter (mol/L). We need to convert from 1.00 mmol in 1 mL to mol/L for this
To convert from mmol to mol, we need to divide the mmol by 1000. To convert mL to liters, divide mL by
Because 1.00 mmol is 1.00 u 10–3 mol and 1.00 mL is 1.00 u 10–3 L,
1.00 mmol 1.00 10 mol 1.00 mol/L 1.00
1.00 mL 1.00 10 L
Think about It
Because 1 mmol/mL = 1 mol/L, we can use this relationship to express molarity comfortably in either units.
4.8. Collect and Organize
Does the molarity change when a portion of a solution is removed and placed into a separate container?
A solution is homogeneous throughout, so the concentration (mol/L or even solute/volume) will be the same
throughout the solution.
Because a solution is homogeneous and the solute is evenly distributed throughout the solution, removing any
portion of the solution will not change the molarity.
Think about It
A large container of a solution of the same molarity as that of a small container indeed has the same molarity,
but the number of moles in each container differs. The large container contains more moles of the solute.
4.9. Collect and Organize
The molarity of a solution is the moles of solute in one liter of solution. For each part of this problem, we are
given the moles of solute in a volume (in mL) of solution. Molarity is abbreviated as M (e.g., 2.00 M).
To find the molarity, we need only divide the moles of solute by the volume of solution in liters. To get volume
in liters, we simply divide the milliliters of solution by 1000 or, even more simply, move the decimal three
places to the left (e.g., 100.0 mL = 0.1000 L).
(a) 2
0.56 mol = 5.6 BaCl
0.1000 L M
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