Chapter 4 - Reactions in Aqueous Solutions
Aqueous Solutions

Most of the reactions we will encounter this year will be in aqueous media (water H2O).

Define solution - homogeneous mixture of two or more substances.

solute = substance present in the smaller amount

Solvent = substance present in the layer amount

examples of solutions air, tea, coffee, smoke etc.

e.g. let’s put a solid (e.g. NaCl) into water

Crystal structure => a 3-D array of atoms, ions, or molecules.

NaCl being ionic has cations and anions in the crystal structure. We’ll look at the dissolution process.

Water molecules “chew off” ions and surround them due to the electrostatic interactions that occur
between ions and (+), (-) ends (the dipoles) of water. The ions become hydrated (i.e., solvated by water;
general case = solvation).

Hydration helps prevent anions and cations from recombining. The ions are now free to move about in
the solution, whereas they were in a fixed arrangement in the solid.
Some properties of aqueous solutions:
1. Carry and electrical current (current carried by ions)
2. Alter the physical properties of the solvent (i.e. m.p., b.p.) We will study colligative properties later

Electrolyte 
conductor

Non-electrolyte  no electrical conduction when dissolved in water.
A substance that when dissolved in water, the resulting solution is a
Concentrations of Solutions

Many chemical reactions are carried out in solutions (usually water)

The quantities of reactants are given in terms of CONCENTRATION UNITS.
2

defn concentration  the amount of solute present in a given quantity of solution (or a fixed amount of
solvent).
3
Molarity  M = # moles of solute/L of solution
e.g.,
3.0 M HCl means 3.0 moles HCl(gas)/1L solution
0.742M NaNO3 means 0.742 moles of NaNO3/1L of solution
NOTE 0.742 M NaNO3 also means 0.0742 moles of NaNO3/0.100L sol’n or 0.371 moles NaNO3/ 500mL
solution.
Example #1

39.30g of KCl is dissolved in 600mL of solution. What is the molarity of the solution?
Number of moles of KCl = 39.30g x 1 mol/74.55g = 0.527 moles
M
= moles/volume (L)
= 0.5271 moles / 0.600 L (i.e. 600mL * 1L/1000 mL)
=0.879 moles / L
We can also write the concentration of KCl as [KCl] = [0.879] where the square brackets around the
substance and the number are understood to mean molarity (moles / L).
NOTE [K+] = [Cl -] = 0.879M since KCl is a strong electrolyte
Example #2

How many moles of solute are in 250 mL of 0.100M KNO3?
M = # moles /(L of sol’n) 
# of moles of solute = 0.100 moles KNO3/1 L x 250mL 1 L / 1000 mL
= 0.0250 moles of KNO3.

How do we make a solution of a known concentration?
Two ways
1.
By adding the appropriate mass of solute to a given volume of solution, e.g., we want 250. mL of a
0.01400M NaCl. What mass of NaCl do we need?
Moles of NaCl = 0.01400 moles NaCl/1L * 250. mL * 1L/1000mL
= 3.500 * 10-3 moles of NaCl
mass of NaCl = 3.500 * 10-3 moles of NaCl * 58.44g NaCl / 1mole NaCl
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= 0.2045g of NaCl
2.
By diluting a solution of a known concentration
What are we actually doing?
We are taking a quantity of moles from beaker 1 and making a new volume of solution in
beaker 2 from the same # of moles.
 beaker 1
# of moles = M1V1 (from def’n of molarity)
 beaker 2
# of moles (same because) = M2V2

Therefore, the number of moles of solute is unchanged. We are simply placing that same number of
moles of solute into a larger volume of solution.

Hence, to prepare a solution of lower concentration from a solution of higher concentration, we take an
accurately known volume of the concentrated solution and add solvent to make it up to a new volume.
M1 V1 = M2 V2
Example

Calculate the molarity (M) of an NaCl solution that is prepared by taking 50.00 mL of a 0.260 molar
solution and diluting to 100.0 ml.
M1 V1 = M2 V2
M1 = 0.260
M2 = ?
V1 = 50.00mL
V2 = 100.mL
0.260 moles / L x 50.00 mL = M2 x 100.0 mL
 M2 = 0.130 mol/L
Electrolytes

Two samples of white crystalline solids (e.g., NaCl and sucrose or table sugar). Both of the solutes have
the same appearance (i.e., they are both white crystalline solids). However, we observe a significant
difference in some of their properties when they are dissolved in water. The NaCl solution has the
ability to conduct electricity, whereas the sucrose solution does not transport charge. Why?

NaCl (aq) is an electrolyte solution (i.e., NaCl is an electrolyte). Sucrose, however, is a non-electrolyte. A
solution of sucrose in water doesn’t conduct electricity.
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NaCl (aq)  Na +(aq) + Cl - (aq)

NaCl (aq) ionizes in solution (recall our earlier discussion)

Electrolyte Types - ionic electrolytes are chiefly one of the following salts, acids, bases, surfactants.
They are sub-classed as weak vs. strong electrolytes.

Strong electrolytes are compounds that ionize or dissociate completely in aqueous solution
HCl (aq), NaCl (aq), NaI (aq), BaCl2 (aq), Na2SO4 (aq), HNO3 (aq)
HCl (aq)  H+ (aq)+ Cl- (aq)

100% ionized ( = 100 %).
Weak electrolytes don’t ionize completely e
CH3COOH (aq) ⇄ CH3COO - (aq) + H+(aq)

Note that the double arrow indicates that both the forward and the reverse reactions are occurring to
some degree, i.e., the reaction is in chemical equilibrium. The balance between the amount of
backward/forward reactions (the position of the equilibrium or whether or not the forward or the
reverse reactions occurs to a greater extent) determines the relative amounts of ionic species present.

Non-electrolytes are generally
1. molecular solids (e.g., urea, glucose, fructose, sucrose);
2. alcohols (CH3CH2OH (aq), iso-propyl alcohol)
Acids, Bases, Salts

Acids  substances that ionize and increase the concentration of H+ in aqueous solution

Examples
HCl (aq)  H+(aq) + Cl-(aq)
HCOOH (aq) ⇄ HCOO-(aq) + H+ (aq)

General acid
HA (aq)  H+(aq) + A-(aq)

These are monoprotic acids, they are capable of donating only one mole of protons per mole of acid.

Some acids are capable of donating more than 1 H+ /molecule of acid.
6
H2A (aq)  2 H+ (aq) + A2- (aq)

These acids are diprotic acids.  2H+ donated/molecule of acid.

Some common examples are H2SO4, H2SO3, and (COOH)2 (oxalic acid).

Triprotic acid 

note => in diprotic and triprotic acids, not all protons are donated with equal ease (we shall discuss later
in the acid-base chapters).
NOTE:
3H+ donated/molecule of acid, e.g., H3PO4
Some acids ionize completely (strong electrolytes)
HNO3 (aq)  H+ (aq) + NO3-(aq) (Strong acid)
while some only ionize partially
CH3CH2COOH (aq)  CH3CH2COO- (aq) + H+ (aq)
propanoic acid < 1% ionized in solution.
Bases  substances that ionize and increase the concentration of OH- in aqueous solution.
Examples
NaOH (aq)  Na+(aq) + OH-(aq)
NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)

These are monbasic substances, they are capable of producing only one mole of OH- per mole of base.

Dibasic  capable of producing more than 1 OH- /molecule of base.
e.g., Ca(OH)2 (aq)  Ca2+ (aq) + 2 OH- (aq)
NOTE:
Some bases acids ionize completely (strong electrolytes)
e.g. NaOH (aq)  Na+ (aq) + OH- (aq) (Strong base)
while some only partially dissociate in aqueous solution.
7
NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)
Salts

Salts are compounds that are formed by when an acid and a base react. In effect, th \e reaction replaces
one or more protons from the acid by the cations from the base. The products of the reaction between
an acid and a base are a salt and water (a neutralization reaction).
e.g.,
HCl (aq) + NaOH  NaCl (aq) + H2O (l)
HNO3 (aq) + KOH (aq)  KNO3 (aq) + H2O (l)
CH3COOH (aq) + LiOH (aq)  CH3COOLi (aq) + H2O (l)

In each of the above reactions, an acid reacts with a base, producing a salt and water (the salts are NaCl
(aq), KNO3 (aq), and CH3COOLi (aq)).
Identifying Strong and weak electrolytes

Most acids are weak electrolytes. There are only seven strong acids
1. HCl
2. HBr
3. HI
4. HNO3
5. H2SO4
6. HClO3
7. HClO4

Most salts are strong electrolytes.

The common strong bases are the alkali metal and the alkaline earth hydroxides Ca(OH)2, Sr(OH)2, and
Ba(OH)2.

Organic bases (e.g., amines) and ammonia are weak electrolytes.
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Metathesis Reactions

In the molecular equations for a number of ionic reactions, close inspection reveals that the different
cations appear to exchange nonmetal partners.
AX (aq) + BY (aq)  AY (aq) + BX (aq)

example
HNO3 (aq) + NaOH (aq)  NaNO3 (aq) + H2O (l)

In this reaction, the cations H+ and Na+ appear to change the respective anions, the NO3- (nitrate) and
the OH- (hydroxide). These reactions are termed metathesis reactions (Greek, to transpose). Metathesis
reactions function by removing ions from the solution; this constitutes the driving force for the
metathesis reaction.

There are three general driving forces for metathesis reactions.
1. The formation of a precipitate.
2. The formation of a soluble weak electrolyte or a soluble nonelectrolyte.
3. The formation of a gaseous product.
Examples
AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
(AgCl precipitates out of solution).
CH3COOLi (aq) + HCl (aq)  CH3COOH (aq) + LiCl (aq)
(CH3COOH is a weak electrolyte).
2 HBr (aq) + Na2S (aq)  H2S (g) + 2 NaBr (aq)
(gaseous H2S escapes from the solution).
Precipitation Reactions

In this class of metathesis reaction, the exchange of the anions between the metal cations results in the
formation of an insoluble solid (a precipitate).
9

Solubility rules for Ionic Compounds (Table 4.2 in textbook).

Some important points to note from the table,
1. All nitrates are soluble.
2. All chlorates are soluble.
3. All compounds of alkali metal cations are soluble.
4. All compounds of ammonium salts are soluble.
5. All perchlorates are soluble.
6. The solubility of sulfite compounds mostly follows that of the sulfates.

We can use the solubility table to predict whether or not a precipitate will form.
Ca(NO3)2 (aq) + Na2SO4 (aq)  CaSO4 (s) + 2 NaCl (aq)

We look at the anions in the metal compounds to see if there is the possibility that a precipitate will
form. Since the sulfate ion is in the system, there is the possibility of a precipitate forming.

From the solubility table, we see that the sulfate ion will precipitate with the calcium ion!

Another example.
AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq).

We again look at the solubility table. All nitrates are soluble (no possibility of precipitation here).
However, Cl- will precipitate with the Ag+ ion.
Net-Ionic Reactions

Let’s examine the precipitation and the neutralization reactions.
For the metathesis reaction,
10
AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq).

AgNO3 is a strong electrolyte
AgNO3 (aq)  Ag+ (aq) + NO3- (aq)

And NaCl is a strong electrolyte
NaCl (aq)  Na+ (aq) + Cl- (aq)

On the products side, NaNO3 is a strong electrolyte.
NaNO3 (aq)  Na+ (aq) + NO3- (aq)

Therefore, when we rewrite the equation taking into account of all the dissociated strong electrolytes,
AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq)  AgCl (s) + Na+ (aq) + NO3- (aq)

We immediately see that the nitrate ion and the sodium ion are the same on both sides of the reaction
arrow. Therefore, we can eliminate them from the reaction equation. We are left with the net ionic
equation!
Ag+ (aq) + Cl- (aq)  AgCl (s)

Another example
11

Let’s examine the following metathesis (neutralization) reaction.
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
HCl is a strong electrolyte.
HCl (aq)  H+ (aq) + Cl- (aq)

And NaOH is a strong electrolyte
NaOH (aq)  Na+ (aq) + OH- (aq)

On the products side, NaCl is a strong electrolyte.
NaCl (aq)  Na+ (aq) + Cl- (aq)
H+ (aq) + NO3- (aq) + Na+ (aq) + OH- (aq)  Na+ (aq) + NO3- (aq) + H2O (l)

We immediately see that, in this reaction, the nitrate ion and the sodium ion are again the same on both
sides of the reaction arrow. Therefore, we can eliminate them from the reaction equation.
H+ (aq) + NO3- (aq) + Na+ (aq) + OH- (aq)  Na+ (aq) + NO3- (aq) + H2O (l)

We are left with the net ionic equation!
H+ (aq) + OH- (aq)  H2O (l)
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Oxidation-Reduction Reactions

Corrosion reaction  the conversion of a metal into its metal compounds.
Example
4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s)

Note that the metal is undergoing corrosion, it is losing electrons, i.e., it is being oxidized.

Another example
2 Mg (s) + O2 (g)  2 MgO (s)
Mg (s) + 2 HCl (aq)  MgCl2 (aq) + H2 (g)
net ionic:
Mg (s) + 2 H+ (aq)  Mg2+ (aq) + H2 (g)

The magnesium goes from elemental Mg to Mg2+, whereas the hydrogen goes from H+ to H2 (g). How
did these changes occur? Note that in this second example, we are seeing an example of corrosion of
the metal by an acid solution (Mg is reacting with H+ (aq)).

Electrons are transferred from the Mg metal to the H+; Mg gets oxidized, while the H+ gets reduced.

LEO says GER (Loss Equals Oxidation; Gain Equals Reduction).

In order for us to discuss electron transfer reactions (i.e., oxidation reduction reactions), we must first
look at a systematic way of assigning the oxidation numbers to the atoms in elements and compounds.

Oxidation number  a fictitious charge assigned to atoms in elements and compounds as a
bookkeeping device for keeping track of the electrons in oxidation reduction equations.
ASSIGNMENT OF OXIDATION NUMBERS
#1 -
In any elemental form (atom or molecule), an atom is assigned an oxidation number of 0
e.g. He, Cu, N in N2, S in S8
13
#2 -
For a monatomic ion, the oxidation number equals the charge.
e.g. -1 for Cl in Cl-, +2 for Ca+2, -2 for S-2
#3 -
Fluorine’s oxidation number is -1 in any compound.
e.g. -1 for F in CF4, but 0 for F in F2
#4 -
Oxygen’s oxidation number is -2 except when combined with fluorine or in peroxides.
e.g. -2 for O in H2O and OH-, +2 for O in OF2, -1 for O in H2O2
#5 For elements in Groups IA, IIA & most of IIIA, oxidation numbers are positive and equal to the group
number.
e.g. +3 for Al in AlCl3, +1 for Na in NaCl, +2 for Mg in Mg SO4
#6 -
Hydrogen has a +1 oxidation number, except in metallic hydrides, in which it is -1.
e.g., +1 for H in H2O and CH3OH, -1 for H in NaH
#7 The sum of the oxidation numbers of the atoms in a neutral compound is zero; in a polyatomic ion,
the sum equals the charge.
e.g. see OH- and H2O in #4 & #6 above, +6 for S in SO4-2

Once we have assigned the oxidation numbers, we can follow the change in oxidation states for atoms
in elements and compounds and balance oxidation-reduction (REDOX) equations.
A General Method for the Balancing of REDOX Equations
#1 Assign oxidation numbers to all atoms in the equation. A polyatomic ion, which does not change
structure or charge in the reaction, can be treated as a single unit with an oxidation number equal to its
charge.
#2 Isolate, in partial equations, the ATOMS that have undergone a change of oxidation number. (A reduction
in number indicates a reduction; an increase in number, an oxidation.) Balance each partial equation for
charge, treating the oxidation numbers as charges and adding electrons where needed. NOTE: under no
14
circumstances should any atoms except those undergoing an oxidation number change appear in these
partial equations. Thus, there is no balancing for mass at this stage.
#3 Balance the partial equations so that:
electrons lost = electrons gained.
#4 Add the partial equations, eliminating the electrons and obtaining the net ionic REDOX equation.
#5 Translate the coefficients for the REDOX atoms to the full reaction equation.
#6 Balance, by inspection, all elements except H’s and O’s
#7 Balance the charge. If the reaction is in acid solution, add H+’s, if in base, add OH-‘s.
#8 Balance for H’s and O’s by adding H2O to the side deficient in H’s and O’s.
15
Example

Balance the following REDOX reaction equation. The reaction is carried out in acid solution.
Cr2O7-2 + I- + ClO4-  Cr(ClO4)3 + I2
#1 Assign oxidation numbers.
2x+6 7x-2
-1
-1
+3 3x-1
0
Cr2O7-2 + I- + ClO4-  Cr(ClO4)3 + I2
#2 Set up & balance for charge the partial equations.
Cr(+6) + 3e  Cr(+3)
= gain of e’s, reduction
I(-1)  I(0) + e
= loss of e’s, oxidation
#3 Balance equations so the e’s lost = e’s gained.
Cr(+6) + 3e

Cr(+3)
 3 I(0) + 3e
3 I(-1)
#4 Add equations, eliminating e’s.
Cr(+6) + 3e
3 I(-1)

Cr(+3)
 3 I(0) + 3e
Cr(+6) + 3 I(-1)  Cr(+3) + 3I(0)
#5 Translate coefficients to full equation.
In this case, multiply all of the coefficients in the net ionic equation by 2 to avoid fractional
coefficients in the full equation,
2 Cr(+6) + 6 I(-1)  2 Cr(+3) + 6I(0)
And thus
1 Cr2O7-2 + 6 I- + ? ClO4-  2 Cr(ClO4)3 + 3 I2
#6 Balance all elements except H’s & O’s - here Cl’s.
16
1 Cr2O7-2 + 6 I- + 6 ClO4-  2 Cr(ClO4)3 + 3 I2
#7 Balance for charge, adding H’s in this case, i.e. acid solution.
14 H+ + 1 Cr2O7-2 + 6 I- + 6 ClO4-  2 Cr(ClO4)3 + 3 I2
#8 Balance for H’s & O’s, here by adding H2O to the product side.
14 H+ + 1 Cr2O7-2 + 6 I- + 6 ClO4-  2 Cr(ClO4)3 + 3 I2 + 7 H2O
CHECK:
Charge:
Reagent
Product
14 H
14 H
2 Cr
2 Cr
31 O
31 O
6I
6I
6 Cl
6 Cl
0
0
Types of oxidation reduction reactions.
1. Displacement of gases by acids/salts.
Ca (s) + 2 HCl (aq)  CaCl2 (aq) + H2 (g)
2 Fe (s) + 6 HCl (aq)  2 FeCl3 (aq) + 3 H2 (g)
net ionic equations:
Ca (s) + 2 H+ (aq)  Ca2+ (aq) + H2 (g)
17
2 Fe (s) + 6 H+ (aq)  2 Fe3+ (aq) + 3 H2 (g)
2. Metal Displacement Reactions
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)
Fe (s) + Cu2+ (aq)  Fe2+ (aq) + Cu (s)
3. Halogen Displacement.
Cl2 (g) + 2 Br – (aq)  Br2 (g) + 2 Cl- (aq)
Cl2 (g) + 2 I– (aq)  I2 (s) + 2 Cl- (aq)
Solution Stoichiometry
Titrations

Volumetric analysis  technique based on volume measurements used to determine the quantity of a
substance in solution.

Titration  The process by which a solution of an accurately known concentration is added gradually to
a solution of an unknown concentration, until a reaction is complete.

Standard solution  solution of accurately known concentration.

Equivalence point  point at which unknown substance is completely consumed by reaction with the
added standard solution.

At the equivalence point reagents are present in stoichiometric amounts.

Note: volumetric analyses, acid/base & redox, require reactions with very high K values, i.e. Essentially
going to products completely. Also, we must have some way of determining whether or not the
reaction has been completed, i.e., We need some way of detecting the equivalence point.
We will look at two types of titrations
1. Acid base
2. Redox
18
Acid-Base titrations

How do we detect the equivalence point in an acid-base titration?

pH meters  voltmeters that measure pH directly.

Indicators  organic molecules that are dyes and that change color over a short pH range; extremely
weak acids/bases.
Example
H-Indicator + H2O
⇄ H3O+ + Indicator-
acid color
base color

End point  volume of titrant added at which indicator changes color.

Note that the indicators change colour at the first slight excess of the titrant!

Ideally, the end point of the titration is the same as the equivalence point. In most cases, the difference
between the end point volume and the equivalence point is so small, the we normally take the
equivalence point as being equal to the endpoint.
Example

In an acid base titration experiment, 20.00 mL of an aqueous solution of HCl requires 40.29 mL of a
0.1016 M NaOH solution. Calculate the original molarity of the HCl solution.
Solution.
Step # 1  Write down the complete titration reaction (not the net ionic reactions).
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
Step # 2  Calculate the number of moles of NaOH titrated into the HCl solution.
n NaOH = MNaOH x VNaOH
= 0.1016 M x 0.04029 L
19
= 4.093 x 10-3 mole
Step # 3  Obtain the stoichiometric relationship between the acid and the base at the equivalence point
(end-point) of the titration (use the complete metathesis reaction). Calculate the number of moles of acid
(or base, if the base is the unknown quantity).
nNaOH = nHCl
nHCl = 4.093 x 10-3 mole
Step # 4  Calculate the molarity of the solution being titrated.
MHCl = nHCl / VHCl
= 4.093 x 10-3 mole / 0.0200 L
= 0.2047 mole HCl / L sol’n.
Redox Titrations

To calculate the quantity of an unknown involved in a redox reaction, we can carry out the same
procedure. We have the reaction of a certain reactant (volume and concentration known) with a secon
reactant (volume known, concentration to be determined).
Example

In an redox titration experiment, 25.00 mL of an aqueous solution of FeCl2 requires 26.00 mL of a
0.02500 M solution of K2Cr2O7. Calculate the concentration of the Fe2+ in the original solution.
Solution.
Step # 1  Write down the complete redox reaction.
20
14 H+ (aq) + Cr2O72- (aq) + 6 Fe2+ (aq)  2 Cr3+ (aq) + 6 Fe3+ (aq) + 7 H2O (l)
Step # 2  Calculate the number of moles of the dichromate ion titrated into the FeCl2 solution.
n (Cr2O72-) = M(Cr2O72-) x V(Cr2O72-)
= 0.02500 M x 0.02600 L
= 6.500 x 10-4 mole Cr2O72-
Step # 3  Obtain the stoichiometric relationship between the substance being oxidized and the substance
being reduced at the equivalence point (end-point) of the titration (use the balanced redox reaction).
Calculate the number of moles of the substance being oxidized (or reduced) in the original solution.
n(Fe2+) = n (Cr2O72-) x 6 Fe2+/ 1 Cr2O72-
n(Fe2+) = 6.500 x 10-4 mole Cr2O72- x 6 Fe2+/ 1 Cr2O72-
= 0.003900 moles Fe2+
Step # 4  Calculate the molarity of the solution being titrated.
M (Fe2+) = M (FeCl2)
= 0.003900 moles Fe2+ / 0.02500 L
= 0.1560 mole HCl / L sol’n.