Tiga Jenis
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.
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.
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 O H 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.
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-and-stick 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).
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.
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 4
10
4times more soluble than
AgCl (an “insoluble” compound):
Solubility of NaCl in H
2O at 20
C = 365 g/L
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 CaBr2
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)
7.42
10
22formula units of copper(II)
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
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 50-mL beaker of the solution.
Molarity (M) is the most common unit of concentration.
It expresses the concentration in units of moles of solute
Sample Problems
Glycine (C
2H
5NO
2) 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
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?
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)
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
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
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
Solving Dilution
Problems
To solve dilution problems, we use the fact that the 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?
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 (Ag
2CrO
4) forms
Figure 4.7 depicts the reaction at the
macroscopic level (photos), the atomic level
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)
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 NO
3-(
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,
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
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
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)
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(NO3)2, the other
two possible cation-anion combinations are NaNO3 and PbI2.
According to Table 4.1, NaNO3 is soluble, but PbI2 is not.
The total ionic equation shows the reaction that occurs as Pb2+ 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:
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
)
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.
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:
Using the molar ratio to convert amount (mol) of MgCl2 to
amount (mol) of Mg(OH)2
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?
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?
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):
Hg(NO
3)
2(
aq
) + Na
2S(
aq
)
HgS(
s
) +
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
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?
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?
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
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 H3O+ [or (H2O)H+], associates
with other water molecules to give species such as H5O2+ [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 H3O+
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
Sample Problem
Nitric acid is a major chemical in the fertilizer and
explosives industries. How many H+(aq) ions are in 25.3
Proton Transfer in
Acid-Base 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
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 H3O+.
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,
H
3O
+ion acts as the acid and
donates a proton to OH
-ion, which acts as the base
Gas-Forming Reactions: Acids
with Carbonates (or Sulfites)
When an ionic carbonate, such as K2CO3, is treated with an acid, suchas HCl, one of the products is carbon dioxide, as the molecular equation shows:
Square brackets around a species, in this case H2CO3, mean it is very
unstable: H2CO3 decomposes immediately into water and carbon
dioxide. Combining these two equations cancels [H2CO3] and gives the
overall equation:
Writing the total ionic equation with H3O+ ions from the HCl shows the
actual species in solution and the key event in a gas-forming reaction,
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
