Three Major classes of reaction
Tiga Jenis
Utama Reaksi
Kimia
Introduction
Artists did not have an affordable, stable blue pigment until
1704, when a paint maker accidentally produced Prussian blue
in his laboratory.
The pigment quickly became extremely popular due to its deep
color and because it faded in light much less than other blue
pigments; it has since been used by countless artists, including
van Gogh and Picasso.
This first modern synthetic pigment is produced by a simple
reaction occurring in water and is just one of a myriad of
products formed by aqueous reactions.
In nature, aqueous reactions occur unceasingly in the gigantic
containers we know as oceans. And, in every cell of your body,
thousands of reactions taking place right now enable you to
function. With millions of reactions occurring in and around you,
it would be impossible to describe them all.
Fortunately, it isn’t necessary because when we survey even a
small percentage of reactions, especially those in aqueous
solution, a few major classes emerge.
SOLUTION CONCENTRATION
AND THE ROLE OF WATER AS
A SOLVENT
A solution consists of a smaller quantity of one substance,
the solute, dissolved in a larger quantity of another, the
solvent; in an aqueous solution, water serves as the
solvent.
For any reaction in solution, the solvent plays a key role
that depends on its chemical nature.
Some solvents passively disperse the substances into
individual molecules.
But water is much more active, interacting strongly with
the substances and even reacting with them in some
cases. Let’s focus on how the water molecule interacts
with both ionic and covalent solutes.
The Polar Nature of
Water
On the atomic scale, water’s great solvent power arises from the
uneven distribution of electron charge and its bent molecular shape,
which create a polar molecule:
Uneven charge distribution. The electrons in a covalent bond are
shared between the atoms. In a bond between identical atoms—as in
H2, Cl2, O2—the sharing is equal and electron charge is distributed
evenly between the two nuclei
In covalent bonds between different atoms, the sharing is uneven
because one atom attracts the electron pair more strongly than the
other atom does.
For example, in each OH bond of water, the shared electrons are
closer to the O atom because an O atom attracts electrons more
strongly than an H atom does.
This uneven charge distribution creates a polar bond, one with partially
charged “poles.”
In Figure 4.1B, the asymmetrical shading
shows this distribution, and the symbol
indicates a partial charge.
The O end is partially negative,
represented by red shading and -, and
the H end is partially positive, represented
by blue shading and +. In the ball-andstick model of Figure 4.1C, the polar arrow
points to the negative pole, and the tail,
shaped like a plus sign, marks the positive
pole.
2. Bent molecular shape. The sequence of
the HOH atoms in water is not linear: the
water molecule is bent with a bond angle
of 104.5 (Figure 4.1C).
3. Molecular polarity. The combination of
polar bonds and bent shape makes water
a polar molecule: the region near the O
atom is partially negative, and the region
between the H atoms is partially positive
(Figure 4.1D)
Ionic Compounds in
Water
In an ionic solid, oppositely charged ions are held together by
electrostatic attractions.
Water separates the ions by replacing these attractions with others
between several water molecules and each ion.
Picture a granule of a soluble ionic compound in water: the
negative ends of some water molecules are attracted to the
cations, and the positive ends of other water molecules are
attracted to the anions (Figure 4.2).
Dissolution occurs because the attractions between each type of
ion and several water molecules outweigh the attractions between
the ions.
Gradually, all the ions separate (dissociate), become solvated—
surrounded closely by solvent molecules—and then move randomly
in the solution.
For an ionic compound that doesn’t dissolve in
water, the attraction between ions is greater than
the attraction between the ions and water.
Actually, these so-called insoluble substances do
dissolve to a very small extent, usually several
orders of magnitude less than so-called soluble
substances. For example, NaCl (a “soluble”
compound) is over 4104 times more soluble than
AgCl (an “insoluble” compound):
Solubility of NaCl in H2O at 20 C = 365 g/L
Solubility of AgCl in H2O at 20 C = 0.009 g/L
How Ionic Solutions Behave: Electrolytes and
Electrical Conductivity
Calculating the Number of
Moles of Ions in Solution
From
the formula of the soluble ionic compound, we know
the number of moles of each ion in solution.
For example, the equations for dissolving KBr and CaBr 2
in water to form solvated ions are
KBr(s) K+(aq) + Br-(aq)
CaBr2(s) Ca2+(aq) + 2Br-(aq)
(“H2O” above the arrows means that water is the solvent,
not a reactant.) Note that 1 mol of KBr dissociates into 2
mol of ions—1 mol of K+ and 1 mol of Br-—and 1 mol of
CaBr2 dissociates into 3 mol of ions—1 mol of Ca2+ and 2
mol of Br-.
Sample Problems
What amount (mol) of each ion is in each
solution?
a) 5.0 mol of ammonium sulfate dissolved
in water
b) 78.5 g of cesium bromide dissolved in
water
c) 7.421022 formula units of copper(II)
nitrate dissolved in water
Covalent Compounds in
Water
Water
dissolves many covalent (molecular) compounds also. Table
sugar (sucrose, C12H22O11), beverage (grain) alcohol (ethanol,
CH3CH2OH), and automobile antifreeze (ethylene glycol,
HOCH2CH2OH) are some familiar examples. All contain their own
polar bonds, which interact with the bonds of water.
However, most soluble covalent substances do not separate into
ions, but remain intact molecules. For example,
HOCH2CH2OH(l) HOCH2CH2OH(aq)
As a result, their aqueous solutions do not conduct an electric
current, and these substances are nonelectrolytes.
Many other covalent substances, such as benzene (C6H6) and
octane (C8H18), do not contain polar bonds, and these substances
do not dissolve appreciably in water
Expressing Concentration
in Terms of Molarity
When working quantitatively with any solution, it is
essential to know the concentration— the quantity of
solute dissolved in a given quantity of solution (or of
solvent).
Concentration is an intensive property (like density or
temperature) and thus is independent of the solution
volume: a 50-L tank of a solution has the same
concentration (solute quantity/solution quantity) as a 50mL beaker of the solution.
Molarity (M) is the most common unit of concentration.
It expresses the concentration in units of moles of solute
per liter of solution:
Sample Problems
Glycine (C2H5NO2) has the simplest structure of
the 20 amino acids that make up proteins. What
is the molarity of a solution that contains 53.7 g
of glycine dissolved in 495 mL of solution?
Calculate the molarity of the solution made
when 6.97 g of KI is dissolved in enough water
to give a total volume of 100. mL
Calculate the molarity of a solution that
contains 175 mg of sodium nitrate in a total
volume of 15.0 mL.
Amount-Mass-Number Conversions
Involving Solutions
Sample Problems
1. Biochemists often study reactions in solutions
containing phosphate ion, which is commonly found
in cells. How many grams of solute are in 1.75 L of
0.460 M sodium hydrogen phosphate?
2. In biochemistry laboratories, solutions of sucrose
(table sugar, C12H22O11) are used in high-speed
centrifuges to separate the parts of a biological cell.
How many liters of 3.30 M sucrose contain 135 g of
solute?
3. What amount (mol) of each ion is in 35 mL of 0.84
M zinc chloride?
Preparing and Diluting
Molar Solutions
Notice that the volume term in the denominator
of the molarity expression in Equation 4.1 is the
solution volume, not the solvent volume.
This means that you cannot dissolve 1 mol of
solute in 1 L of solvent to make a 1 M solution.
Because the solute volume adds to the solvent
volume, the total volume (solute 1 solvent)
would be more than 1 L, so the concentration
would be less than 1 M.
Preparing a Solutions
Correctly preparing a solution of a solid solute requires four steps.
Let’s prepare 0.500 L of 0.350 M nickel(II) nitrate hexahydrate
[Ni(NO3)26H2O]:
Weigh the solid. Calculate the mass of solid needed by
converting from volume (L) to amount (mol) and then to mass (g):
Transfer the solid. We need 0.500 L of solution, so we choose a
500-mL volumetric flask (a flask with a fixed volume indicated by
a mark on the neck), add enough distilled water to fully dissolve
the solute (usually about half the final volume, or 250 mL of
distilled water in this case), and transfer the solute. Wash down
any solid clinging to the neck with some solvent
Dissolve the solid. Swirl
the flask until all the solute
is dissolved. If necessary,
wait until the solution is at
room temperature. (As we
discuss in Chapter 13, the
solution process may be
accompanied by heating or
cooling.)
Add solvent to the final
volume. Add distilled water
to bring the solution
volume to the line on the
flask neck; cover and mix
thoroughly again.
Diluting a Solutions
A concentrated solution (higher molarity) is converted to
a dilute solution (lower molarity) by adding solvent, which
means the solution volume increases but the amount
(mol) of solute stays the same.
As a result, the dilute solution contains fewer solute
particles per unit volume and, thus, has a lower
concentration than the concentrated solution (Figure 4.6).
Chemists often prepare a concentrated solution (stock
solution) and store it and dilute it as needed
Solving Dilution
Problems
solve dilution problems, we use the fact that the
To
amount (mol) of solute does not change during the
dilution process.
Therefore, once we calculate the amount of solute
needed to make a particular volume of a dilute solution,
we can then calculate the volume of concentrated
solution that contains that amount of solute.
This two-part calculation can be combined into one step
in the following relationship:
where M and V are the molarity and volume of the dilute
(subscript “dil”) and concentrated (subscript “conc”)
solutions
Sample Problems
Isotonic saline is 0.15 M aqueous NaCl. It
simulates the total con centration of ions in
many cellular fluids, and its uses range from
cleaning contact lenses to washing red blood
cells. How would you prepare 0.80 L of isotonic
saline from a 6.0 M stock solution?
A chemist dilutes 60.0 mL of 4.50 M potassium
permanganate to make a 1.25 M solution. What
is the final volume of the diluted solution?
WRITING EQUATIONS FOR
AQUEOUS
IONIC REACTIONS
Chemists use three types of equations to
represent aqueous ionic reactions.
Let’s examine a reaction to see what each type
shows. When solutions of silver nitrate and
sodium chromate are mixed, brick-red, solid
silver chromate (Ag2CrO4) forms
Figure 4.7 depicts the reaction at the
macroscopic level (photos), the atomic level
(blow-up circles), and the symbolic level with the
three types of equations (reacting ions are in red
type):
The molecular equation (top) reveals the least about the species
that are actually in solution because it shows all the reactants and
products as if they were intact, undissociated compounds. Only
the designation for solid, (s), tells us that a change has occurred:
2AgNO3(aq) + Na2CrO4(aq)
Ag2CrO4(s) + 2NaNO3(aq)
• The total ionic equation (middle) is much more accurate
because it shows all the soluble ionic substances as they actually
exist in solution, where they are dissociated into ions. The
Ag2CrO4(s) stands out as the only undissociated substance:
2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + CrO42-(aq)
Ag2CrO4(s) + 2Na+(aq) + 2NO3-(aq)
The charges also balance: four positive and four
negative for a net zero charge on the left side, and
two positive and two negative for a net zero charge
on the right.
Notice that Na+(aq) and NO3-(aq) appear unchanged
on both sides of the equation. These are called
spectator ions (shown with pale colors in the atomic
level scenes).
They are not involved in the actual chemical change
but are present only as part of the reactants; that is,
we can’t add an Ag+ ion without also adding an anion,
in this case, the NO3- ion.
The net ionic equation (bottom) is very useful because it
eliminates the spectator ions and shows only the actual
chemical change:
2Ag+(aq) + CrO42-(aq) Ag2CrO4(s)
The formation of solid silver chromate from silver ions and
chromate ions is the only change.
To make that point clearly, suppose we had mixed solutions of
potassium chromate, K2CrO4(aq), and silver acetate,
AgC2H3O2(aq), instead of sodium chromate and silver nitrate.
The three ionic equations would be
Thus, the same change would have occurred, and only the
spectator ions would differ—K+(aq) and C2H3O2-(aq) instead
of Na+(aq) and NO3-(aq).
PRECIPITATION
REACTIONS
In a precipitation reaction, two soluble ionic compounds react
to form an insoluble product, a precipitate. The reaction you
saw between silver nitrate and sodium chromate is one example.
Precipitates form for the same reason that some ionic
compounds don’t dissolve: the electrostatic attraction between
the ions outweighs the tendency of the ions to remain solvated
and move throughout the solution.
When the two solutions are mixed, the ions collide and stay
together, and a solid product “comes out of solution.” Thus, the
key event in a precipitation reaction is the formation of an
insoluble product through the net removal of ions from solution.
Figure 4.8 (on the next page) shows the process for calcium
fluoride.
Predicting Whether a
Precipitate Will Form
To
predict whether a precipitate will form when
we mix two aqueous ionic solutions, we refer to
the short list of solubility rules in Table 4.1.
Let’s see how to apply these rules. When
sodium iodide and potassium nitrate are each
dissolved in water, their solutions consist of
solvated, dispersed ions:
Three steps help us predict if a precipitate forms:
Note the ions in the reactants. The reactant ions are
Na+(aq) + I-(aq) + K+(aq) + NO3-(aq) ?
Consider all possible cation-anion combinations. In
addition to NaI and KNO3, which we know are soluble, the
other cation-anion combinations are NaNO3 and KI.
Decide whether any combination is insoluble. According to
Table 4.1, all compounds of Group lA(l) ions and all nitrate
compounds are soluble. No reaction occurs because all the
cation-anion combinations—NaI, KNO3, NaNO3, and KI—are
soluble:
Na+(aq) + I-(aq) + K+(aq) + NO3-(aq)
Na+(aq) + NO3-(aq) + K+(aq) + I-(aq)
All the ions are spectator ions, so when we eliminate
them, we have no net ionic equation.
Stability Rules
Now, what happens if we substitute a solution of lead(II) nitrate,
Pb(NO3)2, for the KNO3 solution? The reactant ions are Na +, I-, Pb2+,
and NO3-.
In addition to the two soluble reactants, NaI and Pb(NO 3)2, the other
two possible cation-anion combinations are NaNO 3 and PbI2.
According to Table 4.1, NaNO3 is soluble, but PbI2 is not.
The total ionic equation shows the reaction that occurs as Pb 2+ and
I- ions collide and form a precipitate:
2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq)
2Na+(aq) + 2NO3-(aq) + PbI2(s)
And the net ionic equation confirms it:
Pb2+(aq) + 2I-(aq) PbI2(s)
Sample Problem
Does a reaction occur when each of these pairs
of solutions is mixed? If so, write balanced
molecular, total ionic, and net ionic equations,
and identify the spectator ions.
1. Potassium fluoride(aq) + strontium
nitrate(aq)
2. Ammonium perchlorate(aq) + sodium
bromide(aq)
Stoichiometry of
Precipitation Reactions
Solving stoichiometry problems for any reaction that
takes place in solution, such as a precipitation reaction,
requires the additional step of converting the volume of
reactant or product in solution to amount (mol):
1. Balance the equation.
2. Find the amount (mol) of one substance using its molar
mass (for a pure substance) or the volume and
molarity (for a substance in solution).
3. Relate that amount to the stoichiometrically equivalent
amount of another substance.
4. Convert to the desired units.
Summary
Sample Problem
Magnesium is the second most abundant metal in
seawater, after sodium. The first step in its industrial
extraction involves the reaction of the magnesium ion
with calcium hydroxide to precipitate magnesium
hydroxide. What mass of magnesium hydroxide is formed
when 0.180 L of 0.0155 M magnesium chloride reacts
with excess calcium hydroxide?
Solution
Write the balanced equation:
MgCl2(aq) + Ca(OH)2(aq) Mg(OH)2(s) + CaCl2(aq)
Using the molar ratio to convert amount (mol) of MgCl 2 to
amount (mol) of Mg(OH)2
Converting from amount (mol) of Mg(OH)2 to mass (g):
Tugas Mandiri - 4
1. It is desirable to remove calcium ion from hard water to
prevent the formation of precipitates known as boiler scale
that reduce heating efficiency. The calcium ion is reacted
with sodium phosphate to form solid calcium phosphate,
which is easier to remove than boiler scale. What volume of
0.260 M sodium phosphate is needed to react completely
with 0.300 L of 0.175 M calcium chloride?
2. To lift fingerprints from a crime scene, a solution of silver
nitrate is sprayed on a surface to react with the sodium
chloride left behind by perspiration. What is the molarity of
a silver nitrate solution if 45.0 mL of it reacts with excess
sodium chloride to produce 0.148 g of precipitate?
Limiting Reactant Problem
for a Precipitation Reaction
Mercury and its compounds have uses from fillings for
teeth (as a mixture with silver, copper, and tin) to the
production of chlorine. Because of their toxicity,
however, soluble mercury compounds, such as
mercury(II) nitrate, must be removed from industrial
wastewater. One removal method reacts the
wastewater with sodium sulfide solution to produce
solid mercury(II) sulfide and sodium nitrate solution. In
a laboratory simulation, 0.050 L of 0.010 M mercury(II)
nitrate reacts with 0.020 L of 0.10 M sodium sulfide.
a) How many grams of mercury(II) sulfide form?
b) Write a reaction table for this process.
Hg(NO3)2(aq) + Na2S(aq) HgS(s) +
2NaNO3(aq)
Finding the amount (mol) of HgS formed from Hg(NO3)2:
Finding the amount (mol) of HgS from Na2S:
Hg(NO3)2 is the limiting reactant because it forms fewer
moles of HgS.
Converting the amount (mol) of HgS formed from
Hg(NO3)2 to mass (g):
With Hg(NO3)2 as the limiting reactant, the reaction table
is:
A large excess of Na2S remains after the reaction. Note
that the amount of NaNO3 formed is twice the amount of
Hg(NO3)2 consumed, as the balanced equation shows.
Sample Problems
Despite the toxicity of lead, many of its
compounds are still used to make pigments.
a) When 268 mL of 1.50 M lead(II) acetate
reacts with 130. mL of 3.40 M sodium
chloride, how many grams of solid lead(II)
chloride can form?
b) Using the abbreviation “Ac” for the acetate
ion, write a reaction table for the process.
Tugas Mandiri - 5
Iron(III) hydroxide has been used to adsorb
arsenic and heavy metals from contaminated
soil and water. Solid iron(III) hydroxide is
produced by reacting aqueous solutions of
iron(III) chloride and sodium hydroxide.
a) What mass of iron(III) hydroxide is formed
when 155 mL of 0.250 M iron(III) chloride
reacts with 215 mL of 0.300 M sodium
hydroxide?
b) Write a reaction table for this process.
ACID-BASE
REACTIONS
Aqueous acid-base reactions occur in processes as
diverse as the metabolic action of proteins and
carbohydrates, the industrial production of fertilizer, and
the revitalization of lakes damaged by acid rain
These reactions involve water as reactant or product, in
addition to its common role as solvent. Of course, an
acid-base reaction (also called a neutralization
reaction) occurs when an acid reacts with a base, but
the definitions of these terms and the scope of this
reaction class have changed over the years. For our
purposes at this point, we’ll use definitions that apply to
substances found commonly in the lab:
Acids and the Solvated
Proton
Acidic
solutions arise when certain covalent H-containing
molecules dissociate into ions in water. In every case,
these molecules contain a polar bond to H in which the
other atom pulls much more strongly on the electron pair.
A good example is HBr. The Br end of the HBr bond is
partially negative, and the H end is partially positive.
When hydrogen bromide gas dissolves in water, the poles
of H2O molecules are attracted to the oppositely charged
poles of the HBr. The bond breaks, with H becoming the
solvated cation H+(aq) and Br becoming the solvated
anion Br-(aq):
HBr(g) H+(aq) + Br-(aq)
The solvated H+ ion is a very unusual species. The H atom is a proton
surrounded by an electron, so H+ is just a proton. With a full positive
charge concentrated in such a tiny volume, H + attracts the negative
pole of water molecules so strongly that it forms a covalent bond to
one of them. We can show this interaction by writing the solvated H +
ion as an H3O+ ion (hydronium ion) that also is solvated:
The hydronium ion, which we write as H 3O+ [or (H2O)H+], associates
with other water molecules to give species such as H 5O2+ [or (H2O)2H+],
H7O3+ [or (H2O)3H+], H9O4+ [or (H2O)4H+], and so forth. Figure 4.11
shows H7O3+, an H3O+ ion associated (dotted lines) with two H2O
molecules. As a general notation for these various species, we write
H+(aq), but later in this chapter and in much of the text, we’ll use H 3O+
(aq).
Acids and Bases as
Electrolytes
Strong acids and strong bases
dissociate completely into ions.
Therefore, like soluble ionic compounds,
they are strong electrolytes and
conduct a large current, as shown by
the brightly lit bulb (Figure).
Weak acids and weak bases dissociate
very little into ions. Most of their
molecules remain intact. Therefore,
they are weak electrolytes, which
means they conduct a small current.
Because a strong acid (or strong base)
dissociates completely, we can find the
molarity of H+ (or OH-) and the amount
(mol) or number of each ion in solution.
Sample Problem
Nitric acid is a major chemical in the fertilizer and
explosives industries. How many H+(aq) ions are in 25.3
mL of 1.4 M nitric acid?
Proton Transfer in AcidBase Reactions
When we take a closer look (with color) at the reaction
between a strong acid and strong base, as well as several
related reactions, a unifying pattern appears. Let’s examine
three types of reaction to gain insight about this pattern.
Reaction Between a Strong Acid and a Strong Base
When HCl gas dissolves in water, the H + ion ends up bonded
to a water molecule. Thus, hydrochloric acid actually consists
of solvated H3O+ and Cl- ions:
If we add NaOH solution, the total ionic equation shows that
H3O+ transfers a proton to OH- (leaving a water molecule
written as H2O, and forming a water molecule written as
HOH):
Without the spectator ions, the net ionic equation shows
more clearly the transfer of a proton from H3O+ to OH-:
This equation is identical to the one we saw earlier
with the additional H2O molecule left over from the H 3O+.
Figure 4.13 shows this process on the atomic level and
also shows that, if the water is evaporated, the spectator
ions, Cl- and Na+, crystallize as the salt NaCl.
Thus, an acid-base reaction is a proton-transfer
process. In the early 20th century, the chemists
Johannes Brønsted and Thomas Lowry stated:
• An acid is a molecule (or ion) that donates a
proton.
• A base is a molecule (or ion) that accepts a
proton.
Therefore, in an aqueous reaction between strong
acid and strong base, H3O+ ion acts as the acid and
donates a proton to OH- ion, which acts as the base
and accepts it.
Gas-Forming Reactions: Acids
with Carbonates (or Sulfites)
When an ionic carbonate, such as K2CO3, is treated with an acid, such
as HCl, one of the products is carbon dioxide, as the molecular equation
shows:
Square brackets around a species, in this case H 2CO3, mean it is very
unstable: H2CO3 decomposes immediately into water and carbon
dioxide. Combining these two equations cancels [H 2CO3] and gives the
overall equation:
Writing the total ionic equation with H 3O+ ions from the HCl shows the
actual species in solution and the key event in a gas-forming reaction,
removal of ions from solution through formation of both a gas and
water:
Writing the net ionic equation, with the intermediate
formation of H2CO3, eliminates the spectator ions, Cl- and K+,
and makes it easier to see that proton transfer takes place
as each of the two H3O+ ions transfers one proton to the
carbonate ion:
In essence, this is an acid-base reaction with carbonate ion
accepting the protons and, thus, acting as the base.
Ionic sulfites react similarly to form water and gaseous SO2;
the net ionic equation, in which SO32- acts as the base and
forms the unstable H2SO3, is
Reactions Between Weak
Acids and Strong Bases
Utama Reaksi
Kimia
Introduction
Artists did not have an affordable, stable blue pigment until
1704, when a paint maker accidentally produced Prussian blue
in his laboratory.
The pigment quickly became extremely popular due to its deep
color and because it faded in light much less than other blue
pigments; it has since been used by countless artists, including
van Gogh and Picasso.
This first modern synthetic pigment is produced by a simple
reaction occurring in water and is just one of a myriad of
products formed by aqueous reactions.
In nature, aqueous reactions occur unceasingly in the gigantic
containers we know as oceans. And, in every cell of your body,
thousands of reactions taking place right now enable you to
function. With millions of reactions occurring in and around you,
it would be impossible to describe them all.
Fortunately, it isn’t necessary because when we survey even a
small percentage of reactions, especially those in aqueous
solution, a few major classes emerge.
SOLUTION CONCENTRATION
AND THE ROLE OF WATER AS
A SOLVENT
A solution consists of a smaller quantity of one substance,
the solute, dissolved in a larger quantity of another, the
solvent; in an aqueous solution, water serves as the
solvent.
For any reaction in solution, the solvent plays a key role
that depends on its chemical nature.
Some solvents passively disperse the substances into
individual molecules.
But water is much more active, interacting strongly with
the substances and even reacting with them in some
cases. Let’s focus on how the water molecule interacts
with both ionic and covalent solutes.
The Polar Nature of
Water
On the atomic scale, water’s great solvent power arises from the
uneven distribution of electron charge and its bent molecular shape,
which create a polar molecule:
Uneven charge distribution. The electrons in a covalent bond are
shared between the atoms. In a bond between identical atoms—as in
H2, Cl2, O2—the sharing is equal and electron charge is distributed
evenly between the two nuclei
In covalent bonds between different atoms, the sharing is uneven
because one atom attracts the electron pair more strongly than the
other atom does.
For example, in each OH bond of water, the shared electrons are
closer to the O atom because an O atom attracts electrons more
strongly than an H atom does.
This uneven charge distribution creates a polar bond, one with partially
charged “poles.”
In Figure 4.1B, the asymmetrical shading
shows this distribution, and the symbol
indicates a partial charge.
The O end is partially negative,
represented by red shading and -, and
the H end is partially positive, represented
by blue shading and +. In the ball-andstick model of Figure 4.1C, the polar arrow
points to the negative pole, and the tail,
shaped like a plus sign, marks the positive
pole.
2. Bent molecular shape. The sequence of
the HOH atoms in water is not linear: the
water molecule is bent with a bond angle
of 104.5 (Figure 4.1C).
3. Molecular polarity. The combination of
polar bonds and bent shape makes water
a polar molecule: the region near the O
atom is partially negative, and the region
between the H atoms is partially positive
(Figure 4.1D)
Ionic Compounds in
Water
In an ionic solid, oppositely charged ions are held together by
electrostatic attractions.
Water separates the ions by replacing these attractions with others
between several water molecules and each ion.
Picture a granule of a soluble ionic compound in water: the
negative ends of some water molecules are attracted to the
cations, and the positive ends of other water molecules are
attracted to the anions (Figure 4.2).
Dissolution occurs because the attractions between each type of
ion and several water molecules outweigh the attractions between
the ions.
Gradually, all the ions separate (dissociate), become solvated—
surrounded closely by solvent molecules—and then move randomly
in the solution.
For an ionic compound that doesn’t dissolve in
water, the attraction between ions is greater than
the attraction between the ions and water.
Actually, these so-called insoluble substances do
dissolve to a very small extent, usually several
orders of magnitude less than so-called soluble
substances. For example, NaCl (a “soluble”
compound) is over 4104 times more soluble than
AgCl (an “insoluble” compound):
Solubility of NaCl in H2O at 20 C = 365 g/L
Solubility of AgCl in H2O at 20 C = 0.009 g/L
How Ionic Solutions Behave: Electrolytes and
Electrical Conductivity
Calculating the Number of
Moles of Ions in Solution
From
the formula of the soluble ionic compound, we know
the number of moles of each ion in solution.
For example, the equations for dissolving KBr and CaBr 2
in water to form solvated ions are
KBr(s) K+(aq) + Br-(aq)
CaBr2(s) Ca2+(aq) + 2Br-(aq)
(“H2O” above the arrows means that water is the solvent,
not a reactant.) Note that 1 mol of KBr dissociates into 2
mol of ions—1 mol of K+ and 1 mol of Br-—and 1 mol of
CaBr2 dissociates into 3 mol of ions—1 mol of Ca2+ and 2
mol of Br-.
Sample Problems
What amount (mol) of each ion is in each
solution?
a) 5.0 mol of ammonium sulfate dissolved
in water
b) 78.5 g of cesium bromide dissolved in
water
c) 7.421022 formula units of copper(II)
nitrate dissolved in water
Covalent Compounds in
Water
Water
dissolves many covalent (molecular) compounds also. Table
sugar (sucrose, C12H22O11), beverage (grain) alcohol (ethanol,
CH3CH2OH), and automobile antifreeze (ethylene glycol,
HOCH2CH2OH) are some familiar examples. All contain their own
polar bonds, which interact with the bonds of water.
However, most soluble covalent substances do not separate into
ions, but remain intact molecules. For example,
HOCH2CH2OH(l) HOCH2CH2OH(aq)
As a result, their aqueous solutions do not conduct an electric
current, and these substances are nonelectrolytes.
Many other covalent substances, such as benzene (C6H6) and
octane (C8H18), do not contain polar bonds, and these substances
do not dissolve appreciably in water
Expressing Concentration
in Terms of Molarity
When working quantitatively with any solution, it is
essential to know the concentration— the quantity of
solute dissolved in a given quantity of solution (or of
solvent).
Concentration is an intensive property (like density or
temperature) and thus is independent of the solution
volume: a 50-L tank of a solution has the same
concentration (solute quantity/solution quantity) as a 50mL beaker of the solution.
Molarity (M) is the most common unit of concentration.
It expresses the concentration in units of moles of solute
per liter of solution:
Sample Problems
Glycine (C2H5NO2) has the simplest structure of
the 20 amino acids that make up proteins. What
is the molarity of a solution that contains 53.7 g
of glycine dissolved in 495 mL of solution?
Calculate the molarity of the solution made
when 6.97 g of KI is dissolved in enough water
to give a total volume of 100. mL
Calculate the molarity of a solution that
contains 175 mg of sodium nitrate in a total
volume of 15.0 mL.
Amount-Mass-Number Conversions
Involving Solutions
Sample Problems
1. Biochemists often study reactions in solutions
containing phosphate ion, which is commonly found
in cells. How many grams of solute are in 1.75 L of
0.460 M sodium hydrogen phosphate?
2. In biochemistry laboratories, solutions of sucrose
(table sugar, C12H22O11) are used in high-speed
centrifuges to separate the parts of a biological cell.
How many liters of 3.30 M sucrose contain 135 g of
solute?
3. What amount (mol) of each ion is in 35 mL of 0.84
M zinc chloride?
Preparing and Diluting
Molar Solutions
Notice that the volume term in the denominator
of the molarity expression in Equation 4.1 is the
solution volume, not the solvent volume.
This means that you cannot dissolve 1 mol of
solute in 1 L of solvent to make a 1 M solution.
Because the solute volume adds to the solvent
volume, the total volume (solute 1 solvent)
would be more than 1 L, so the concentration
would be less than 1 M.
Preparing a Solutions
Correctly preparing a solution of a solid solute requires four steps.
Let’s prepare 0.500 L of 0.350 M nickel(II) nitrate hexahydrate
[Ni(NO3)26H2O]:
Weigh the solid. Calculate the mass of solid needed by
converting from volume (L) to amount (mol) and then to mass (g):
Transfer the solid. We need 0.500 L of solution, so we choose a
500-mL volumetric flask (a flask with a fixed volume indicated by
a mark on the neck), add enough distilled water to fully dissolve
the solute (usually about half the final volume, or 250 mL of
distilled water in this case), and transfer the solute. Wash down
any solid clinging to the neck with some solvent
Dissolve the solid. Swirl
the flask until all the solute
is dissolved. If necessary,
wait until the solution is at
room temperature. (As we
discuss in Chapter 13, the
solution process may be
accompanied by heating or
cooling.)
Add solvent to the final
volume. Add distilled water
to bring the solution
volume to the line on the
flask neck; cover and mix
thoroughly again.
Diluting a Solutions
A concentrated solution (higher molarity) is converted to
a dilute solution (lower molarity) by adding solvent, which
means the solution volume increases but the amount
(mol) of solute stays the same.
As a result, the dilute solution contains fewer solute
particles per unit volume and, thus, has a lower
concentration than the concentrated solution (Figure 4.6).
Chemists often prepare a concentrated solution (stock
solution) and store it and dilute it as needed
Solving Dilution
Problems
solve dilution problems, we use the fact that the
To
amount (mol) of solute does not change during the
dilution process.
Therefore, once we calculate the amount of solute
needed to make a particular volume of a dilute solution,
we can then calculate the volume of concentrated
solution that contains that amount of solute.
This two-part calculation can be combined into one step
in the following relationship:
where M and V are the molarity and volume of the dilute
(subscript “dil”) and concentrated (subscript “conc”)
solutions
Sample Problems
Isotonic saline is 0.15 M aqueous NaCl. It
simulates the total con centration of ions in
many cellular fluids, and its uses range from
cleaning contact lenses to washing red blood
cells. How would you prepare 0.80 L of isotonic
saline from a 6.0 M stock solution?
A chemist dilutes 60.0 mL of 4.50 M potassium
permanganate to make a 1.25 M solution. What
is the final volume of the diluted solution?
WRITING EQUATIONS FOR
AQUEOUS
IONIC REACTIONS
Chemists use three types of equations to
represent aqueous ionic reactions.
Let’s examine a reaction to see what each type
shows. When solutions of silver nitrate and
sodium chromate are mixed, brick-red, solid
silver chromate (Ag2CrO4) forms
Figure 4.7 depicts the reaction at the
macroscopic level (photos), the atomic level
(blow-up circles), and the symbolic level with the
three types of equations (reacting ions are in red
type):
The molecular equation (top) reveals the least about the species
that are actually in solution because it shows all the reactants and
products as if they were intact, undissociated compounds. Only
the designation for solid, (s), tells us that a change has occurred:
2AgNO3(aq) + Na2CrO4(aq)
Ag2CrO4(s) + 2NaNO3(aq)
• The total ionic equation (middle) is much more accurate
because it shows all the soluble ionic substances as they actually
exist in solution, where they are dissociated into ions. The
Ag2CrO4(s) stands out as the only undissociated substance:
2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + CrO42-(aq)
Ag2CrO4(s) + 2Na+(aq) + 2NO3-(aq)
The charges also balance: four positive and four
negative for a net zero charge on the left side, and
two positive and two negative for a net zero charge
on the right.
Notice that Na+(aq) and NO3-(aq) appear unchanged
on both sides of the equation. These are called
spectator ions (shown with pale colors in the atomic
level scenes).
They are not involved in the actual chemical change
but are present only as part of the reactants; that is,
we can’t add an Ag+ ion without also adding an anion,
in this case, the NO3- ion.
The net ionic equation (bottom) is very useful because it
eliminates the spectator ions and shows only the actual
chemical change:
2Ag+(aq) + CrO42-(aq) Ag2CrO4(s)
The formation of solid silver chromate from silver ions and
chromate ions is the only change.
To make that point clearly, suppose we had mixed solutions of
potassium chromate, K2CrO4(aq), and silver acetate,
AgC2H3O2(aq), instead of sodium chromate and silver nitrate.
The three ionic equations would be
Thus, the same change would have occurred, and only the
spectator ions would differ—K+(aq) and C2H3O2-(aq) instead
of Na+(aq) and NO3-(aq).
PRECIPITATION
REACTIONS
In a precipitation reaction, two soluble ionic compounds react
to form an insoluble product, a precipitate. The reaction you
saw between silver nitrate and sodium chromate is one example.
Precipitates form for the same reason that some ionic
compounds don’t dissolve: the electrostatic attraction between
the ions outweighs the tendency of the ions to remain solvated
and move throughout the solution.
When the two solutions are mixed, the ions collide and stay
together, and a solid product “comes out of solution.” Thus, the
key event in a precipitation reaction is the formation of an
insoluble product through the net removal of ions from solution.
Figure 4.8 (on the next page) shows the process for calcium
fluoride.
Predicting Whether a
Precipitate Will Form
To
predict whether a precipitate will form when
we mix two aqueous ionic solutions, we refer to
the short list of solubility rules in Table 4.1.
Let’s see how to apply these rules. When
sodium iodide and potassium nitrate are each
dissolved in water, their solutions consist of
solvated, dispersed ions:
Three steps help us predict if a precipitate forms:
Note the ions in the reactants. The reactant ions are
Na+(aq) + I-(aq) + K+(aq) + NO3-(aq) ?
Consider all possible cation-anion combinations. In
addition to NaI and KNO3, which we know are soluble, the
other cation-anion combinations are NaNO3 and KI.
Decide whether any combination is insoluble. According to
Table 4.1, all compounds of Group lA(l) ions and all nitrate
compounds are soluble. No reaction occurs because all the
cation-anion combinations—NaI, KNO3, NaNO3, and KI—are
soluble:
Na+(aq) + I-(aq) + K+(aq) + NO3-(aq)
Na+(aq) + NO3-(aq) + K+(aq) + I-(aq)
All the ions are spectator ions, so when we eliminate
them, we have no net ionic equation.
Stability Rules
Now, what happens if we substitute a solution of lead(II) nitrate,
Pb(NO3)2, for the KNO3 solution? The reactant ions are Na +, I-, Pb2+,
and NO3-.
In addition to the two soluble reactants, NaI and Pb(NO 3)2, the other
two possible cation-anion combinations are NaNO 3 and PbI2.
According to Table 4.1, NaNO3 is soluble, but PbI2 is not.
The total ionic equation shows the reaction that occurs as Pb 2+ and
I- ions collide and form a precipitate:
2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq)
2Na+(aq) + 2NO3-(aq) + PbI2(s)
And the net ionic equation confirms it:
Pb2+(aq) + 2I-(aq) PbI2(s)
Sample Problem
Does a reaction occur when each of these pairs
of solutions is mixed? If so, write balanced
molecular, total ionic, and net ionic equations,
and identify the spectator ions.
1. Potassium fluoride(aq) + strontium
nitrate(aq)
2. Ammonium perchlorate(aq) + sodium
bromide(aq)
Stoichiometry of
Precipitation Reactions
Solving stoichiometry problems for any reaction that
takes place in solution, such as a precipitation reaction,
requires the additional step of converting the volume of
reactant or product in solution to amount (mol):
1. Balance the equation.
2. Find the amount (mol) of one substance using its molar
mass (for a pure substance) or the volume and
molarity (for a substance in solution).
3. Relate that amount to the stoichiometrically equivalent
amount of another substance.
4. Convert to the desired units.
Summary
Sample Problem
Magnesium is the second most abundant metal in
seawater, after sodium. The first step in its industrial
extraction involves the reaction of the magnesium ion
with calcium hydroxide to precipitate magnesium
hydroxide. What mass of magnesium hydroxide is formed
when 0.180 L of 0.0155 M magnesium chloride reacts
with excess calcium hydroxide?
Solution
Write the balanced equation:
MgCl2(aq) + Ca(OH)2(aq) Mg(OH)2(s) + CaCl2(aq)
Using the molar ratio to convert amount (mol) of MgCl 2 to
amount (mol) of Mg(OH)2
Converting from amount (mol) of Mg(OH)2 to mass (g):
Tugas Mandiri - 4
1. It is desirable to remove calcium ion from hard water to
prevent the formation of precipitates known as boiler scale
that reduce heating efficiency. The calcium ion is reacted
with sodium phosphate to form solid calcium phosphate,
which is easier to remove than boiler scale. What volume of
0.260 M sodium phosphate is needed to react completely
with 0.300 L of 0.175 M calcium chloride?
2. To lift fingerprints from a crime scene, a solution of silver
nitrate is sprayed on a surface to react with the sodium
chloride left behind by perspiration. What is the molarity of
a silver nitrate solution if 45.0 mL of it reacts with excess
sodium chloride to produce 0.148 g of precipitate?
Limiting Reactant Problem
for a Precipitation Reaction
Mercury and its compounds have uses from fillings for
teeth (as a mixture with silver, copper, and tin) to the
production of chlorine. Because of their toxicity,
however, soluble mercury compounds, such as
mercury(II) nitrate, must be removed from industrial
wastewater. One removal method reacts the
wastewater with sodium sulfide solution to produce
solid mercury(II) sulfide and sodium nitrate solution. In
a laboratory simulation, 0.050 L of 0.010 M mercury(II)
nitrate reacts with 0.020 L of 0.10 M sodium sulfide.
a) How many grams of mercury(II) sulfide form?
b) Write a reaction table for this process.
Hg(NO3)2(aq) + Na2S(aq) HgS(s) +
2NaNO3(aq)
Finding the amount (mol) of HgS formed from Hg(NO3)2:
Finding the amount (mol) of HgS from Na2S:
Hg(NO3)2 is the limiting reactant because it forms fewer
moles of HgS.
Converting the amount (mol) of HgS formed from
Hg(NO3)2 to mass (g):
With Hg(NO3)2 as the limiting reactant, the reaction table
is:
A large excess of Na2S remains after the reaction. Note
that the amount of NaNO3 formed is twice the amount of
Hg(NO3)2 consumed, as the balanced equation shows.
Sample Problems
Despite the toxicity of lead, many of its
compounds are still used to make pigments.
a) When 268 mL of 1.50 M lead(II) acetate
reacts with 130. mL of 3.40 M sodium
chloride, how many grams of solid lead(II)
chloride can form?
b) Using the abbreviation “Ac” for the acetate
ion, write a reaction table for the process.
Tugas Mandiri - 5
Iron(III) hydroxide has been used to adsorb
arsenic and heavy metals from contaminated
soil and water. Solid iron(III) hydroxide is
produced by reacting aqueous solutions of
iron(III) chloride and sodium hydroxide.
a) What mass of iron(III) hydroxide is formed
when 155 mL of 0.250 M iron(III) chloride
reacts with 215 mL of 0.300 M sodium
hydroxide?
b) Write a reaction table for this process.
ACID-BASE
REACTIONS
Aqueous acid-base reactions occur in processes as
diverse as the metabolic action of proteins and
carbohydrates, the industrial production of fertilizer, and
the revitalization of lakes damaged by acid rain
These reactions involve water as reactant or product, in
addition to its common role as solvent. Of course, an
acid-base reaction (also called a neutralization
reaction) occurs when an acid reacts with a base, but
the definitions of these terms and the scope of this
reaction class have changed over the years. For our
purposes at this point, we’ll use definitions that apply to
substances found commonly in the lab:
Acids and the Solvated
Proton
Acidic
solutions arise when certain covalent H-containing
molecules dissociate into ions in water. In every case,
these molecules contain a polar bond to H in which the
other atom pulls much more strongly on the electron pair.
A good example is HBr. The Br end of the HBr bond is
partially negative, and the H end is partially positive.
When hydrogen bromide gas dissolves in water, the poles
of H2O molecules are attracted to the oppositely charged
poles of the HBr. The bond breaks, with H becoming the
solvated cation H+(aq) and Br becoming the solvated
anion Br-(aq):
HBr(g) H+(aq) + Br-(aq)
The solvated H+ ion is a very unusual species. The H atom is a proton
surrounded by an electron, so H+ is just a proton. With a full positive
charge concentrated in such a tiny volume, H + attracts the negative
pole of water molecules so strongly that it forms a covalent bond to
one of them. We can show this interaction by writing the solvated H +
ion as an H3O+ ion (hydronium ion) that also is solvated:
The hydronium ion, which we write as H 3O+ [or (H2O)H+], associates
with other water molecules to give species such as H 5O2+ [or (H2O)2H+],
H7O3+ [or (H2O)3H+], H9O4+ [or (H2O)4H+], and so forth. Figure 4.11
shows H7O3+, an H3O+ ion associated (dotted lines) with two H2O
molecules. As a general notation for these various species, we write
H+(aq), but later in this chapter and in much of the text, we’ll use H 3O+
(aq).
Acids and Bases as
Electrolytes
Strong acids and strong bases
dissociate completely into ions.
Therefore, like soluble ionic compounds,
they are strong electrolytes and
conduct a large current, as shown by
the brightly lit bulb (Figure).
Weak acids and weak bases dissociate
very little into ions. Most of their
molecules remain intact. Therefore,
they are weak electrolytes, which
means they conduct a small current.
Because a strong acid (or strong base)
dissociates completely, we can find the
molarity of H+ (or OH-) and the amount
(mol) or number of each ion in solution.
Sample Problem
Nitric acid is a major chemical in the fertilizer and
explosives industries. How many H+(aq) ions are in 25.3
mL of 1.4 M nitric acid?
Proton Transfer in AcidBase Reactions
When we take a closer look (with color) at the reaction
between a strong acid and strong base, as well as several
related reactions, a unifying pattern appears. Let’s examine
three types of reaction to gain insight about this pattern.
Reaction Between a Strong Acid and a Strong Base
When HCl gas dissolves in water, the H + ion ends up bonded
to a water molecule. Thus, hydrochloric acid actually consists
of solvated H3O+ and Cl- ions:
If we add NaOH solution, the total ionic equation shows that
H3O+ transfers a proton to OH- (leaving a water molecule
written as H2O, and forming a water molecule written as
HOH):
Without the spectator ions, the net ionic equation shows
more clearly the transfer of a proton from H3O+ to OH-:
This equation is identical to the one we saw earlier
with the additional H2O molecule left over from the H 3O+.
Figure 4.13 shows this process on the atomic level and
also shows that, if the water is evaporated, the spectator
ions, Cl- and Na+, crystallize as the salt NaCl.
Thus, an acid-base reaction is a proton-transfer
process. In the early 20th century, the chemists
Johannes Brønsted and Thomas Lowry stated:
• An acid is a molecule (or ion) that donates a
proton.
• A base is a molecule (or ion) that accepts a
proton.
Therefore, in an aqueous reaction between strong
acid and strong base, H3O+ ion acts as the acid and
donates a proton to OH- ion, which acts as the base
and accepts it.
Gas-Forming Reactions: Acids
with Carbonates (or Sulfites)
When an ionic carbonate, such as K2CO3, is treated with an acid, such
as HCl, one of the products is carbon dioxide, as the molecular equation
shows:
Square brackets around a species, in this case H 2CO3, mean it is very
unstable: H2CO3 decomposes immediately into water and carbon
dioxide. Combining these two equations cancels [H 2CO3] and gives the
overall equation:
Writing the total ionic equation with H 3O+ ions from the HCl shows the
actual species in solution and the key event in a gas-forming reaction,
removal of ions from solution through formation of both a gas and
water:
Writing the net ionic equation, with the intermediate
formation of H2CO3, eliminates the spectator ions, Cl- and K+,
and makes it easier to see that proton transfer takes place
as each of the two H3O+ ions transfers one proton to the
carbonate ion:
In essence, this is an acid-base reaction with carbonate ion
accepting the protons and, thus, acting as the base.
Ionic sulfites react similarly to form water and gaseous SO2;
the net ionic equation, in which SO32- acts as the base and
forms the unstable H2SO3, is
Reactions Between Weak
Acids and Strong Bases