Understanding Reactions in Solution
There are literally millions of known chemical reactions; the rich variety of reactions derives
from the fact that there are many elements, each with different properties, and their compounds
can combine or exchange atoms in different ways. Fortunately, the properties of the elements
themselves change in a fairly regular and periodic fashion, and this helps in systematizing many
of these reactions.
A very important category of chemical reactions are those which occur in solution. These are
reactions between molecules and/or ions dissolved in liquids. Such reactions are responsible for
most life processes, for example, and are also very common in industrial chemistry. The liquid
(for example, water) is called the solvent, and the things dissolved in it are called solutes.
For reactions in aqueous solutions, there are two kinds of solutes -- those that dissociate into
ions, and those that dissolve as intact molecules. Dissociation occurs when an ionic or molecular
compound (e.g., NaCl or HCl) breaks up into cations and anions (Na+ + Cl-, or H+ + Cl-) upon
dissolving. Such compounds are known as strong electrolytes, since they ionize completely and
make the solution a good ionic conductor (This is why you don't swim in the ocean during an
electrical storm). Soluble salts, as well as strong acids and bases, are strong electrolytes. Weak
electrolytes dissociate very little in water. Weak acids and weak bases are weak electrolytes.
For example, acetic acid, CH3COOH, a weak acid and the "active ingredient" in vinegar,
dissolves in water as approximately 99.5% neutral CH3COOH and 0.5% ions (CH 3COO- + H+).
The neutral molecules do not contribute to conductivity, so these solutions are very weakly
conductive. There are also non-electrolytes (sugar and ethanol are good examples) which
dissolve but do not dissociate to any significant extent in water.
We will do a demonstration in class involving a pickle and a 110 volt AC power source. The
pickle sparks and gives off steam as its dissolved ions (Na+ and Cl -) conduct alternating current.
The same experiment repeated with a cucumber shows that non-electrolytes such as sugar
molecules do not contribute to the conductivity of solutions.
1
Properties of Acids and Bases. Acids and bases react with each other to give soluble and
insoluble salts (plus water). Acids and bases can be classified either as "strong", i.e., those
which ionize fully in water, or "weak", those which ionize only a little in water.
Common acids are compounds which generate H+. Strong acids ionize completely in water to
make H+ and the corresponding anion (e.g., HCl → H + + Cl -), whereas weak acids only ionize
a little bit, and are mostly present in solution as neutral molecules. For acids derived from
oxides, the strong ones are generally those which have at least two more oxygen atoms than
hydrogen atoms, e.g.
Strong acids
HCl
HNO3
H2SO4
HClO4
Weak acids
H2CO3 carbonic acid
H2SO3 sulfurous acid
H3PO4 phosphoric acid
HCN hydrocyanic acid
hydrochloric acid
nitric acid
sulfuric acid
perchloric acid
(hydrogen cyanide)
All the hydrogen halides (HCl, HBr, HI) are strong acids, except HF, which is a weak acid.
Strong bases are generally soluble hydroxides of metals from the left side of the periodic table,
such as NaOH and Ba(OH) 2. These compounds ionize completely in water to give metal cations
(e.g., Na+, Ba2+) and hydroxide ions (OH -). There are also weak bases, like NH3 (ammonia) that
dissociate very little in water:
NaOH → Na+(aq) + OH-(aq)
complete dissociation - strong electrolyte
Ba(OH)2 → Ba2+(aq) + 2 OH- (aq)
complete dissociation - strong electrolyte
NH3(aq) + H2O → NH4+(aq) + OH-(aq) very little dissociation - weak electrolyte
2
Reactions in Solution. In the simplest of these reactions, two compounds react to give new
compounds, with no change in oxidation numbers. Examples of these are acid-base reactions,
e.g.,
HF(aq) + NaOH(aq) → H2O + NaF(aq)
and precipitation reactions, for example,
AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)
AgCl is an insoluble salt that precipitates during the reaction.
Note that in both cases, we could have written net ionic equations, taking the spectator ions
(those ions that are present both on the reactant and product side as dissociated ions) out of the
molecular equations. In the first reaction, the full ionic equation was:
HF(aq) + Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) + F-(aq)
Na+ was a spectator ion, since it appears as Na+(aq) on both sides, so the net ionic equation is:
HF(aq) + OH-(aq) → H2O(l) + F-(aq)
In the second equation the spectator ions were Na+ and NO3-, both of which remain in solution.
The net ionic equation is therefore:
Ag+(aq) + Cl-(aq) → AgCl(s)
Solubility rules. How do we know which compounds will be insoluble in precipitation
reactions? There are some simple rules in your book. The essence of them is that singly charged
ions (e.g., Na+, NH4+, Cl-, NO3-, etc.) generally form soluble salts (exceptions are halides of Ag,
Hg, and Pb). For example, NaCl, AgNO3, (NH4)2SO4, and Ba(OH)2 are all soluble. Salts in
which both the cation and anion are multiply charged tend to be less soluble. We define an
insoluble (or sparingly soluble) salt as one which dissolves to an extent of 0.02 mole/liter or less.
Examples of insoluble salts are CaCO3, BaSO 4, Mg3(PO4)2, and Fe(OH) 3. The reason for these
trends is that there is a trade-off between lattice energy and solvation energy which determines
whether the dissolution process is exothermic or not. Breaking up the lattice into ions is
energetically unfavorable, but solvating the ions with polar water molecules is favorable. The
sum of these two energies (shown as ∆H = ? in the picture below) determines the solubility. If
the lattice contains multiply charged ions, their electrostatic attraction for each other is so strong
that it is difficult to break up the solid. Likewise, if the cations and anions are well-matched in
size (as in LiF and Mg(OH)2), the lattice energy tends to be high. These salts are therefore
insoluble. Salts with large, singly charged anions such as ClO4-, NO3-, CH3COO-, etc., have
small lattice energies and therefore are usually quite soluble.
3
+
-
+
O
H H
-
+
-
ionic solid
O
H H
HO H
H
∆H=?
OH
+ H
O
H
solvated ions
H
H
endothermic (lattice
energy)
∆H>0
O
+
H2 O
exothermic
(solvation
energy)
+
-
∆H<0
gaseous ions
The table below gives the aqueous solubilities of some salts, their lattice energies, and the
solvation energies of the ions. Can you explain the solubilities of these salts in terms of the
trends you see in lattice and solvation energies?
Compound
NaCl
CaF2
NaF
CaCl2
CaCO3
Solubility (M)
10
2.1 x 10-4
1.0
6.7
6.7 x 10-5
Lattice Energy (kJ/mol)
786
2,630
923
2,258
2,810
Solvation Energy (kJ/mol)
Cation
Anion
-406
-363
-1577
-505
-406
-505
-1577
-363
-1577
-1246
Note that cations tend to be smaller than anions. Smaller ions have stronger attraction to the
dipolar water molecules (because the distance is shorter), so the cation term often dominates the
solvation energy.
Mg2+
Ba2+
Cl-
ClO4-
Solvation energy (kJ): -1921
-1305
-363
-238
4
Acid-base reactions. There are three ways to write the equations:
1. Balanced molecular equation, for example
Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2 H2O(l)
(insoluble)
2. Total ionic equation: write down all the ions (for strong acids and bases, which ionize
completely).
Ba2+(aq) + 2 OH-(aq) + 2 H+(aq) + SO42-(aq) → BaSO4(s) + 2 H2O(l)
(note: H2SO4 →2 H+ + SO42-)
3. Net ionic equation: remove spectator ions from the total ionic equation (in this example,
there are no spectator ions, so it is the same thing as the total ionic equation).
Another example: nitric acid + ammonia → ammonium nitrate
(strong acid + weak base → soluble salt)
1. Balanced molecular equation:
HNO3(aq) + NH3(aq) → NH4NO3(aq)
2. Total ionic equation:
H+(aq) + NO3-(aq) + NH3(aq) → NH4+(aq) + NO3-(aq)
3. Net ionic equation: (note that NO3- appears on both sides of the reaction -- spectator ion)
H+(aq) + NH3(aq) → NH4+(aq)
(strong acid + ammonia → ammonium cation)
Displacement reactions. One element "displaces" another from a compound. These are really a
special case of oxidation-reduction, which we will discuss in more detail later. Examples of such
reactions are:
active metal + acid → hydrogen + salt of active metal
e.g.
Mg + 2 HCl → MgCl2 + H2(g)
Zn + H2SO4 → ZnSO4 + H2(g)
and
e.g.
active metal + salt of less active metal → less active metal + salt of active metal
Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)
This last reaction we have written as occurring in aqueous solution (this is what the "aq" means).
Some of the ions, such as SO 42-, in this reaction remain unchanged -- that is, they are "spectator"
ions. For this kind of reaction, normally we leave out the spectator ions in describing it.
5
Cu2+ (aq) + SO42-(aq) + Zn (s) → Cu (s) + SO42- (aq) + Zn2+ (aq) (is better written as:)
Cu2+ (aq) + Zn (s) → Cu (s) + Zn2+ (aq)
Net ionic equation or net reaction.
In this reaction, the more active metal (Zn) displaces the less active metal (Cu) from the solution.
We can put the metals in a series, in which any metal will displace those above it. This is called
the "activity series," or more generally, the electrochemical series.
Li
Na
Mg
Al
Zn
H2
Cu
Ag
Au
most active (best reducing agents)
least active
Metals above H2 in the series are sufficiently active to liberate hydrogen from water (or acid)
solutions. We saw a couple of extreme examples of this in the reactions of alkali metals with
water. Metals below H2 are not reactive with water or acids. The coinage metals (Cu, Ag, Au)
are all unreactive in this way, as are other relatively electronegative metals such as Pt. Cu is
more active than Ag, so it will displace Ag+ from solutions giving silver metal:
Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)
This can be demonstrated by using a solution of silver nitrate and potassium cyanide, which
contains the Ag(CN)2- ion. The copper on the surface of pennies displaces silver ions to make a
silver coating.
The less active metals (Cu, Ag, Au, Sn, Pb, Hg, Fe, etc.) were all known to the ancients, because
they can be made as pure elements simply by heating their ores, or by reducing them with
elements like carbon. For example, Hg occurs as the mineral cinnabar (HgS), which can be
decomposed to mercury just by heating it up. Iron, which was probably first made from ores like
hematite (Fe2O3) about 3000 B.C., can be made in reasonably pure form by heating the oxide
with carbon (coal or "coke", as it is called in the steelmaking industry). On the other hand, the
more active metals, like aluminum and sodium, cannot be made in this way, and were isolated as
elements fairly late in history. Elements ending with the suffix "ium" are relatively recent
discoveries (made in pure form after 1850 or so), and these are typically the ones which are more
active than hydrogen.
6
Oxidation-Reduction Reactions. This is an extremely important class of reactions which
involves the transfer of electrons between atoms, and a concomitant change in oxidation
numbers. Displacement reactions are a special case of this reaction type. Oxidation involves an
increase in the formal oxidation number (loss of electrons). Reduction involves a decrease in
oxidation number, and a gain of electrons.
We define the oxidation number (or oxidation state) as the number of electrons gained or lost
by an atom when it forms an ionic compound (for covalent compounds, the O.N. indicates a
partial shift of electrons, but not a complete transfer from one atom to another). For most
elements, there is more than one stable oxidation state. For example, the element chlorine can
have an oxidation number of 0 (as elemental Cl2), an oxidation number of -1 (in NaCl, where we
have Cl - ions), and an oxidation number of +1 in the hypochlorite ion (OCl-). It can also have
oxidation numbers of +3, +5, and +7 in other compounds. For many elements there is only one
common oxidation number (apart from 0 in the elemental state). For example, the alkali metals
have an oxidation number of +1 in most of their compounds. The alkali earths generally have an
oxidation number of +2.
In a pure element, the oxidation number is always zero. In an ion, the sum of the oxidation states
of the elements which comprise it must equal the overall charge on the ion. For a molecule of a
neutral compound, the sum of all the oxidation states must equal zero.
You should not take oxidation numbers too literally. They are an accurate representation of the
true electron density only for very ionic compounds (like NaF). However, they are a useful
construct because they help us count electrons in oxidation-reduction reactions. For example, in
the reaction of sodium + chlorine the oxidation numbers of both Na and Cl change when they
react, as chlorine takes electrons from sodium.
Calculating oxidation numbers: Normally we want to calculate the oxidation state of the central
atom in an ion or molecule, e.g., the I atom in HIO4. There are certain guidelines to help us do
this:
1. Oxygen almost always has an oxidation number of -2 in compounds (exceptions are
peroxides, where it is -1, and hypofluorite, where it is +1, but these are rare).
2. Hydrogen is usually +1 (H is -1 when bonded to an electropositive metal, as in the
compound CaH2; such compounds are called hydrides.).
3. Alkali metals are always +1 and alkali earths are always +2.
4. Halogens are -1 except when bonded to more electronegative elements (O and F).
7
Examples:
CO2: Since O is -2, C must be +4.
NO3-: Since O is -2, N must be +5
H2SO4: Since O is -2 and H is +1, S must be +6.
PbCl2: Since Cl is -1, Pb must be +2.
HClO4: Since O is -2 and H is +1, Cl must be +7.
NH4+: Since H is +1, N must be +3.
Note: Oxidation state as calculated in these examples is different from formal charge, which
we will learn how to calculate a little later.
In oxidation-reduction (or redox) reactions, the oxidizing agent starts out oxidized and gets
reduced, and the reducing agent starts out reduced and gets oxidized. Example:
Fe(s) + 3 Cl2(g) → 2 FeCl3(s)
oxid. nos. 0
0
+3 -1
The oxidation number of Fe goes from 0 to +3 (it is the reducing agent, and therefore gets
oxidized). For Cl, the oxidation number goes from 0 to -1. Therefore, it is the oxidizing agent
(which gets reduced). A good way to remember which one is the oxidizing agent is to think
about oxygen itself (which is an oxidizing agent). In most reactions, oxygen starts out as O2
(oxid. no. 0) and gets reduced to O2- (oxid. no. -2).
Note that many redox reactions occur in water, and under acidic or basic conditions (so H+ and
OH- may be involved). An example of a simple aqueous redox reaction not involving these is:
2 FeBr3(aq) + 3 Cl2(g) → 2 FeCl3(aq) + 2 Br2(aq)
Let's look at the oxidation numbers associated with this reaction. Fe is +3 on both sides of the
reaction. Therefore, it is not involved in oxidation-reduction. Cl goes from 0 to -1, therefore it is
the oxidizing agent. Br goes from -1 to 0, and it is therefore the reducing agent. We could also
express this reaction as a total ionic equation or a net ionic equation, which would show more
clearly that Fe3+ is really not involved:
Total ionic equation:
2 Fe3+(aq) + 6 Br-(aq) + 3 Cl2(g) → 2 Fe3+(aq) + 6 Cl-(aq) + 3 Br2(l)
Net ionic equation:
6 Br-(aq) + 3 Cl2(g) → 6 Cl-(aq) + 3 Br2(l)
(Fe3+ is a spectator ion)
We can consider this net ionic equation as being composed of two parts, one involving only Br
and the other involving only Cl:
8
2 Br-(aq) →
oxid. state
2 (-1)
Cl2(g) + 2e- →
Br2(l) + 2e2e-
0
2 Cl-
These two reactions are called "half-reactions". They are added to produce the total reaction.
Half-reactions have the general formulas:
X → Y + e- (oxidation half reaction: X gets oxidized)
X + e-→ Y
(reduction half reaction: X gets reduced)
Typical examples of these are:
Mg → Mg2+ + 2e- (oxidation of magnesium)
2 H+ + 2e- → H2(g) (reduction of aqueous acid to hydrogen)
Adding these, the electrons cancel, and we get the overall reaction:
Mg + 2H+ → Mg2+ + H2 (magnesium displaces hydrogen from acid)
Another useful application of half-reactions is in balancing redox equations. Usually, it is much
simpler to write out the two half reactions than it is to write down the full balanced equation.
However, we can take advantage of the fact that the electrons must cancel to get the full
equation. That is, the number of electrons lost by the reducing agent must be the same as the
number gained by the oxidizing agent. We multiply the two half reactions be appropriate factors
to get the electrons to cancel.
Examples of half reactions:
MnO4- + 8 H+ + 5e- → Mn2+ + 4 H2O
+7
+2
(purple)
(colorless)
Cr2O72- + 14 H+ + 6 e- → 2 Cr3+ + 7 H2O
2 H+ + NO3- + e- → NO2 + H2O
(Cr goes from O.N. +6 to +3)
(N goes from O.N. +5 to +4)
All of the above are examples of reduction half reactions. Examples of oxidations are:
Cu → Cu2+ + 2eZn → Zn2+ + 2e2 I- → I2 + 2eMn2+ + 4 H2O → MnO4- + 8H+ + 5eNote that this last example is one of the reductions above written in reverse. All reductions can
be written in reverse as oxidations, and vice-versa.
9
Balancing redox equations . We are usually given incomplete information, for example that
MnO4- is an oxidizing agent, which is reduced to Mn2+, and I- is a reducing agent, which is
oxidized to I2. Therefore:
MnO4- + I- → Mn2+ + I2
(not balanced)
Our task is to balance this reaction, using H+ or OH- as necessary, depending on whether the
reaction happens in acid or in base.
Step 1 (of six steps) is to determine the oxidation numbers of atoms which change oxidation
states.
MnO4- + I- → Mn2+ + I2
oxid. nos.
+7
-1
+2
0
We can then say how many electrons each atom gains or loses.
Step 2 is to write out the oxidation and reduction half-reactions:
MnO4- → Mn2+
+7
+2
(Mn must gain 5 electrons) therefore:
MnO4- + 5 e- → Mn2+
2 I- → I2 + 2 e2(-1)
0
(therefore 2 I- must lose 2 electrons)
Step 3: Balance oxidation and reduction half reactions so that electrons lost = electrons gained.
In this example, to get the electrons to cancel, we multiply the Mn reaction by 2 and the I
reaction by 5:
2 MnO4- + 10 e- → 2 Mn2+
10 I→ 5 I2 + 10 eNote that the reactions do not need to be mass balanced, and we can add them at this point:
2 MnO4- + 10I- → 2 Mn2+ + 5 I2
Step 4: Compare charge on either side, and use H+ (if in acid) or OH- (if in base) to balance
charge:
left side: -12
right side: +4
In acid, we would need to add 16 H+ to the left side to get the charge to balance:
16 H+ + 2 MnO4- + 10I- → 2 Mn2+ + 5 I2
Step 5: Add water to balance the hydrogen and oxygen simultaneously. By inspection we see
that we would need 8 H2O on the right.
16 H+ + 2 MnO4- + 10I- → 2 Mn2+ + 5 I2 + 8 H2O
Step 6: Check and make sure everything (charge and all the elements) is balanced. Both sides
have a total charge of +4, have 16 H, 2 Mn, 8 O, and 10 I.
10
From this overall equation, we can now extract the balanced half reactions. Note that we could
have gotten these by applying all the steps without adding half reactions in step 3. The Mn half
reaction is:
8 H+ + MnO4- + 5 e- → Mn2+ + 4 H2O
and the I-/I2 half-reaction is:
2 I- → I2 + 2 eOnce we have balanced half reactions, we can use them in other reactions without worrying
about mass and charge balance. For example, we could use the Mn half reaction with any other
one, such as oxidation of Mg or reduction of H+ (above). We know that MnO4- will always
require 5 electrons and 8 H+ when it does an oxidation in acidic solution.
Another example: Cu is oxidized by nitrate ion (NO3-) to Cu2+. NO3- is reduced to NO2. Try
to balance this one in acidic solution.
Cu + NO3- → Cu2+ + NO2
Step 1: oxidation nos. are
Step 2: half reactions are
Step 3: balance electrons
0
+5
+2
+4
2+
Cu → Cu + 2 e
2(NO3- + e- → NO2)
Cu + 2 NO3- + 2e- → Cu2+ + 2NO2 + 2e-
Step 4: charge balance using H+. Left side has charge -2, right side has charge +2 (add 4 H+ to
left)
4 H+ + Cu + 2 NO3- → Cu2+ + 2NO2
Step 5: add water to balance H and O (we need 2 H2O to balance 4 H+)
4 H+ + Cu + 2 NO3- → Cu2+ + 2NO2 + 2 H2O
Step 6: double check!
11
Practice Problems: Precipitation-Dissolution, Acid-Base, Redox, and Complexation Reactions
Activity Series: Li>K>Ba>Sr>Ca>Na>Mg>Al>Mn>Zn>Cr>Fe>Ni>Sn>Pb>H2>
Sb>Cu>Hg>Ag>Pd>Pt>Au
Write the NET IONIC REACTION for each, or “NR” if there is no reaction
1.)
Ba(OH)2(aq) and HCl (aq)
2.)
Na2CO3 (aq) and BaCl2 (aq)
3.)
AgNO3 (aq) and HCl (aq)
4.)
NaCl (aq) and NH4C2H3O2 (aq)
5.)
Na2CO3 (aq) and HClO4 (aq)
6.)
Ca(s) and liquid water
7.)
Cu(s) and Cr(NO3)3 (aq)
8.)
Al(s) and AgNO3 (aq)
9.)
AgNO3 (aq) and excess KCN (aq)
10.)
Au(NO3)3 (aq) and excess KCl (aq)
12
11.)
NaNO2 (aq) and H2SO4 (aq)
12.)
CaCl2 (aq) and Na3PO4(aq)
13.)
Pb(NO3)2 (aq) and Cu (s)
14.)
Zn (s) and HClO2 (aq)
15.)
FeCl3 (aq) and NaOH (aq)
16.)
Ca(OH)2 (s) and HF (aq)
17.)
SbCl3 (aq) and H2S (g)
18.)
CaSO4 (s) and liquid water
19.)
NaF (aq) and H2SO4 (aq)
20.)
HNO3 (aq) and CaCO3 (s)
21.)
FeSO4 (aq) and H2 (g)
22.)
Hg (l) and HNO3 (aq)
13