Reactions in Aqueous Solutions
There are three major types of reactions which
take place in aqueous (water) solutions:
Terminology:
•Solute – A substance that is dissolved
•Solvent – The substance doing the dissolving
•Precipitation Reactions (double displacement or
metathesis)
•Solution – The combination of the solute and
solvent (homogeneous mixture).
•Acid-Base Neutralization Reactions
•Aqueous solution – A solution in which water is
the solvent. Designated by the label (aq). (Note that
•Oxidation Reduction Reactions (combustion (not
aqueous), single replacement, etc.)
in many cases water does not directly take part in the
reaction but is the medium in which the reaction occurs.)
•Electrolyte – A substance, that when dissolved in
water, is capable of conducting electricity
(compare to weak electrolyte and nonelectrolyte)
•Soluble – Capable of being
dissolved in water
•Insoluble – Incapable of being
dissolved in water
•Dissociate – To split apart into
component ions when dissolved
in water (such as a soluble salt or
strong acid or base)
•Hydration – To have the
components of a substance
surrounded by water molecules
To work with reactions in aqueous solutions you must
be able to convert back and forth between solution
volume and the number of moles of the substance
(solute) that it contains.
The most common way of converting between
volume and moles is through the chemical unit of
concentration known as molarity.
Grams
moles
divide by molar mass
Volume
moles
multiply by molarity
Solution concentration: Tells how much
solute is dissolved per unit volume of solution.
Molarity = moles solute / liters of solution
The unit for molarity is mol/L and is designated by the
symbol, M.
Remember: solution = solute + solvent
Solutions are said to have a certain “molar”
concentration
Note: Soluble ionic compounds exist as ions in
solution
For instance: NaCl(aq) is really Na+(aq) and Cl-(aq)
Steps for creation of a solution of a particular molarity:
1. Mass out the amount of solute needed.
2. Place into a volumetric flask or graduated cylinder.
3. Fill with purified water to the desired volume.
1
Example. Determine the number of grams of
ammonium nitrate required to produce 750.mL of a
.12M solution. How many moles of ammonium nitrate
are in 125mL of this prepared solution?
Example. Determine the final molarity of a solution
formed from combining 120.mL of a .30M HCl solution
with 160.mL of a .15M HCl solution. Assume the
volumes are additive.
Answer:
NH4NO3 = 80.0434g/mol
n = MV = (.12M)(.750L) = .090mol
Answer:
m = 80.0434g/mol(.090mol) = 7.2g
Total moles of HCl = (.120L)(.30M) + (.160L)(.15M) =
.060moles
.12 mol/L (.125L) = .015mol
Total volume = .120L + .160L = .280L
New molarity = .060mol/.280L = .21M
Dilutions:
Used to produce a specific volume of a
desired concentration from a more
concentrated “stock” solution.
Produced by adding additional water to a
portion of an existing solution.
This method can only make the solution more
dilute, not more concentrated.
Sample Problems:
Calculate the volume of .20M stock solution needed
to produce 600.mL of a desired .05M solution.
Describe the steps for how you would produce this
solution.
Useful Equation:
Since MV = moles
M1V1 = M2V2
where the subscripts 1 and 2 represent the before and
after conditions respectively
Total moles before dilution = Total moles after dilution
Double Displacement
Reactions
Aka Metathesis reaction or precipitation
reaction
Answer:
M1V1 = M2V2
(.20M)(V1) = (.05M)(.600L)
V1 = .15L = 150mL
Place 150mL of the stock solution in a graduated
cylinder and then fill with water to the 600.mL mark.
Reactions of the form:
AB(aq) + CD(aq) → AD(s) + CB(aq)
Where AB and CD are ionic compounds, with A and C the cations
and B and D the anions.
2
Consider the initial conditions
AB(aq) + CD(aq) → AD(s) + CB(aq)
AB(aq) + CD(aq) → AD(s) + CB(aq)
Bneg
Apos
What happens in this type of
reaction?
Bneg
Apos
Cpos
Dneg
Apos
Dneg
Cpos
Bneg
Dneg
Remember that
aqueous (soluble)
ionic compounds exist
as ions in solution.
Cpos
The cations exchange places = Double displacement
Pos = cation
When these solutions are brought
together what are the potential
combinations of ions?
Dneg
Apos
Still Soluble
Bneg
Apos
Bneg
Cpos
neg = anion
Solubility Rules
Refer to the solubility chart on page 80 in your
book or page 20 in your lab manual.
Remember that the chart is hierarchical.
?
?
Cpos
Still Soluble
Dneg
The solubility chart lists the common behavior of certain ions to predict
the solubility of compounds. This chart must be memorized.
How do you know which combinations are soluble?
Example:
Write a balanced chemical equation for the
double displacement reaction between
sodium carbonate and calcium nitrate.
Determine the identity of the insoluble
precipitate.
Na2CO3(aq) + Ca(NO3)2(aq) → CaCO3(s) + 2NaNO3(aq)
Ins
o
ipit
rec
=p
uct
d
o
r
le P
lu b
ate
3
There are 3 different ways we can
write this equation:
“Molecular” Equation:
Na2CO3(aq) + Ca(NO3)2(aq) → CaCO3(s) + 2NaNO3(aq)
Complete (Total) Ionic:
2Na+(aq) + CO32-(aq) + Ca2+(aq) + 2NO3-(aq) → CaCO3(s) + 2Na+(aq) + 2NO3-(aq)
On Your Own:
Write the molecular, complete ionic and net
ionic equations for the reaction between silver
nitrate and potassium chromate
Note: chromate is CrO42- and silver’s charge is
Ag1+
Net Ionic:
Ca2+(aq) + CO32-(aq) → CaCO3(s)
(With the spectator ions removed)
K2CrO4(aq) + 2AgNO3(aq) → 2KNO3(aq) + Ag2CrO4(s)
2K+(aq) + CrO42-(aq) + 2Ag+(aq) + 2NO3-(aq) → 2K+(aq) +
2NO3-(aq) + Ag2CrO4(s)
AcidAcid-Base (neutralization)
Reactions
(pp82-87)
Reactions of the Form
2Ag+
(aq)
+
CrO42-(aq)
→ Ag2CrO4(s)
Acid + Base → Salt + Water
Note that if no precipitate forms, then no reaction
occurs.
Arrhenius Definition of acids and bases:
Acids:
Acids Dissociate to produce H+ ions (protons) in
solutions (HCl, HNO3, HClO4, etc.)
Hydrochloric acid
Bases:
Bases Produce OH- (hydroxide) in solution (NaOH,
Ba(OH)2, RbOH, etc.)
Sodium Hydroxide
NaOH → Na+(aq) + OH-(aq)
HCl → H+(aq) + Cl-(aq)
Sulfuric acid
H2SO4 → H+(aq) + HSO4-(aq) (complete)
HSO4-(aq) ↔ H+(aq) + SO42-(aq) (partial)
Note the equilibrium arrow
sulfuric acid is said to be a diprotic acid (capable of producing
two H+ ions)
Ammonia
NH3 + H2O ↔ NH4+(aq) + OH-(aq)
Note that ammonia produces hydroxide ions
indirectly through its interaction with water.
Many weak bases contain the NH2 group known as an
amine.
The prefixes mono, tri and poly are also used.
4
Differentiating Between Strong/Weak vs.
Concentrated/Dilute
strong/weak is based on the ability to dissociate (i.e.
produce H+ or OH- ions) in solution.
1.5M
HC2H3O2
1.5M HCl
concentrated/dilute is based on the concentration of
the solution (i.e. molarity, mass percentage, etc.)
H+ ions = 1.5mol/L
Although somewhat arbitrary, we will assume anything 1M and over is a
concentrated solution.
Although these solution may have the same molarity
(concentration), the amount of H+ in solution is very different
due to the relative strengths of the acids.
Example Reaction
Table 4.1 Common Strong Acids and Bases
Acid
Name of Acid
Base
Name of Base
HCl
Hydrochloric acid
LiOH
Lithium hydroxide
HBr
Hydrobromic acid
NaOH
Sodium hydroxide
HI
Hydroiodic acid
KOH
Potassium hydroxide
HNO3
Nitric acid
Ca(OH)2
Calcium hydroxide
HClO4
Perchloric acid
Sr(OH)2
Strontium hydroxide
H2SO4
Sulfuric acid
Ba(OH)2
Barium hydroxide
H+ ions << 1.5mol/L
In the reaction between rubidium hydroxide and nitric
acid, rubidium nitrate is formed. Write the
neutralization reaction for this combination.
HNO3(aq) + RbOH(aq) → RbNO3(aq) + H2O(l)
Note that the salt (rubidium nitrate) is formed from the cation
of the base and the anion of the acid.
Page 83 in your textbook
The strong acids and bases must be committed to memory.
Strong Acid – Strong Base Reactions
H+(aq) + OH-(aq) → H2O
Table 4.2: Types of Acid-Base Reactions
Reactants
Reacting
Species
Net Ionic Equation
Strong acid – strong base
H+
Weak acid – strong base
HB – OH-
HB(aq) + OH-(aq) → H2O + B-(aq)
Strong acid – weak base
H+ - B
H+(aq) + B(aq) → BH+(aq)
-
OH-
H+(aq)
+
OH-
(aq)
Example:
Write the molecular, complete ionic and net ionic equation for
the reaction between perchloric acid and potassium hydroxide.
→ H2O
Answer:
HClO4(aq) + KOH(aq) → KClO4(aq) + H2O(l)
H+(aq) + ClO4-(aq) + K+(aq) + OH-(aq) → K+(aq) + ClO4-(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
Page 85 in your textbook
5
Weak Acid – Strong Base Reactions
Strong Acid – Weak Base Reactions
HB(aq) + OH-(aq) → H2O + B-(aq)
H+(aq) + B(aq) → BH+(aq)
Example:
Example:
Write the molecular, complete ionic and net ionic equation for
the reaction between hydrofluoric acid and potassium
hydroxide.
Write the molecular, complete ionic and net ionic equation for
the reaction between hydrochloric acid and methylamine,
CH3NH2.
Answer:
Answer:
HF(aq) + KOH(aq) → KF(aq) + H2O(l)
HCl(aq) + CH3NH2(aq) → CH3NH3Cl(aq)
HF(aq) + K+(aq) + OH-(aq) → K+(aq) + H2O(l) + F-(aq)
H+(aq) + Cl-(aq) + CH3NH2(aq) → CH3NH3+(aq) + Cl-(aq)
HF(aq) + OH-(aq) → H2O(l) + F-(aq)
H+(aq) + CH3NH2(aq) → CH3NH3+(aq)
Example:
Problem:
Write the net ionic equation for the reaction between
hypochlorous acid and calcium hydroxide.
Answer:
This reaction is between a weak acid and a strong base,
therefore the reaction is of the form
HB(aq) + OH-(aq) → H2O + B-(aq)
If lithium sulfate salt is formed in a neutralization reaction, what
is the most likely acid and base that would have formed this
product?
Answer:
Given that the anion of the salt comes from the acid and the
cation from the base, back solving for the parent acid and base
would give
sulfuric acid, H2SO4 and lithium hydroxide, LiOH
giving,
2HClO(aq) + Ca(OH)2(aq) → Ca(ClO)2(aq) + 2H2O(l)
Which as a net ionic gives
HClO(aq)
(aq) + OH (aq)
aq) → ClO (aq)
(aq) + H2O(l)
Equivalence Point: The point in a reaction when
stoichiometrically equivalent amounts of reactant have been
combined. (i.e. There is no excess or limiting reagent)
Acid-Base Titrations
Titration: Using a solution of precisely known concentration
(standardized solution) to determine the concentration of
another solution. Commonly used in acid-base reactions.
Endpoint: The point in a titration in which an indicator that has
been placed in solution changes color.
Normally you want the endpoint and the equivalence point to
coincide with one another.
Image from SparkNotes
Types of acid-base indicators.
6
Example:
Example:
A 125.0mL sample of a solution of nitric acid is titrated with
.02500M potassium hydroxide (the titrant). If 75.00mL of the
hydroxide is used, what is the molarity of the nitric acid?
In a titration, 30.0mL of .0400M calcium hydroxide are required to
completely titrate a 100.mL sample of hydroiodic acid. Determine
the molarity of the acid solution.
Answer:
Answer:
Balanced Equation:
Balanced Equation:
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
Ca(OH)2(aq) + 2HI(aq) → CaI2(aq) + 2H2O(l)
The mole ratio is 1:1
Moles of Ca(OH)2 = (.0300L)(.0400M) = .00120moles
Moles of KOH used = (.02500M)(.07500L) = .001875mol
.00120mol Ca(OH)2 (2mol HI / 1mol Ca(OH)2 ) = .00240mol HI
Molarity of HNO3 = (.001875mol) / (.1250L) = .01500M
M = .00240mol / .100L = .0240M
Oxidation Reduction (Redox
(Redox)) Reactions
General Summary of Steps for Solving Titration
Problems
1. Write a balanced equation for the neutralization reaction.
Characteristics:
2. Convert to moles using the molarity and volume via mol = MV
-Electrons are lost/gained by particles in the reaction
3. Set up a mole ratio to determine the number of moles of the
reactant in question (Remember: It’s the coefficient of
What you want / What you know
-There must be a change in “oxidation number”.
-You must learn the rules for assigning oxidation numbers (page
20 lab manual; p 88-89 textbook)
-Oxidation numbers are for “book keeping purposes” and may not
represent actual charges
4. Use the moles determined in step 3 to calculate molarity from
the sample volume given via M = n/V (where n = moles)
Common Redox Reactions
Single Replacement (A + BC → B + AC):
Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)
Combustion (hydrocarbon + O2 → CO2 + H2O):
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Redox Reactions in acid/base aqueous medium:
14H+(aq) + Cr2O72-(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
-Oxidations and reductions must occur in pairs (i.e. electron loss
must be accompanied by electron gain)
Oxidation:
Formation of calcium chloride from its elements
-Lose electrons
-Gain oxygen (special case)
-Lose hydrogen (special case)
Calcium chloride is commonly
found in ice melting products
Example:
0
+2
Ca → Ca2+ + 2e-
Note electrons on
right side of arrow
Reduction:
Synthesis (A + B → AB):
Mg(s) + Cl2(g) → MgCl2(s)
-Gain electrons
-Lose oxygen (special case)
-Gain hydrogen (special case)
Decomposition (AB → A + B):
2H2O(l) → 2H2(g) + O2(g)
Example:
Not all of these take place in aqueous solutions but in each
case electrons are lost and gained somewhere in the reaction
equation.
0
-1
Cl2 + 2e- → 2Cl-
Note electrons on
left side of arrow
7
If the two half-reactions are combined:
0
0
+2 -1
You can remember what is oxidized and what is reduced by
these phrases:
Ca + Cl2 → CaCl2
LEO goes GER
Since the oxidation numbers of both Ca and Cl change, this is
a redox reaction.
Gains Electrons Reduced
Calcium is oxidized (and is the reducing agent because it
reduces the chlorine)
or
Chlorine is reduced (and is the oxidizing agent because it
oxidized the calcium)
OIL RIG
Loses Electrons Oxidized
GER!
Oxidation Is Loss
Reduction Is Gain
You must use the rules for assigning oxidation numbers to
determine the value to assign to each element in a reaction.
Assigning Oxidation Numbers: (Page 20 in your lab manual; pages
88-89 in your textbook)
1. In free elements (that is, in the uncombined state), each atom has an oxidation number of zero. Thus
each atom in H2, Br2, Na, Be, K, O2, and P4 has the same oxidation number: zero.
2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Thus Li+
has an oxidation number of +1; Ba2+ ion, +2; Fe3+, +3; I- ion, -1; O2- ion, -2; and so on. All alkali metals
have an oxidation number of +1, and all alkaline earth metals have an oxidation number of +2 in their
compounds. Aluminum has an oxidation number of +3 in all its compounds.
3. The oxidation number of oxygen in most compounds (for example, MgO and H2O) is -2, but in
hydrogen peroxide (H2O2) and peroxide ion (O22-), its oxidation number is -1.
4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds. In
these cases (for example, LiH, NaH, and CaH2), its oxidation number is -1.
5. Fluorine has an oxidation number of -1 in all its compounds. Other halogens (Cl, Br, and I) have
negative oxidation numbers when they occur as halide ions in their compounds. When combined with
oxygen-for example in oxoacids and oxoanions, they have positive oxidation numbers.
Example Problem:
Assign an oxidation number to each element in the following
species:
SF6
PCl3
CO32H2 O
XeCl4
NO2ClO4RbH
6. In a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero. In a polyatomic
ion, the sum of oxidation numbers of all the elements in the ion must be equal to the net charge of the
ion. For example, in the ammonium ion, NH4+, the oxidation number of N is 3- and that of H is +1. Thus
the sum of the oxidation numbers is -3 + 4(+1) = +1, which is equal to the net charge of the ion.
Answers:
+6 -1
SF6
+4 -1
XeCl4
+3 -1
PCl3
+3 -2
NO2-
+4 -2
CO32-
+7 -2
ClO4-
+1 -2
H2 O
+1 -1
RbH
Example:
Example:
Using oxidation numbers, determine which species is
oxidized, which is reduced, and the total number of
electrons that are transferred. Write the net ionic equation
for the reaction.
Write the oxidation and reduction half reactions for cupric
chloride decomposing to form copper metal and chlorine gas.
CuCl2(aq) → Cu(s) + Cl2(g)
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Answer:
Answer:
Reduction:
Magnesium is oxidized from a 0 to a +2
Hydrogen is reduced from a +1 to a 0
All total there are 2 electrons transferred.
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
The chloride ions are spectator ions
Cu2+(aq) + 2e- → Cu(s)
Cu goes from +2 to 0
Oxidation:
2Cl-(aq) → Cl2(g) + 2e-
Cl goes from (-1 to 0)x2
H+
*Note: Although it is
that is reduced, the whole molecule,
HCl, is said to be the oxidizing agent.
8
Balancing Redox Reactions in Acidic and Basic Medium
Example:
Balance the following redox reaction that takes place in acidic
solution. (Note that spectator ions are not included)
Many redox reactions take place in either acidic or basic
solutions. Acidic solutions are a source of H+ ions and basic
solutions provide OH- when balancing equations.
You must balance both elements AND charges in redox
reactions.
Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
Notice that the reaction has oxygen on the left (reactant) side but
not on the right (product) side. As we will see, the solution
provides the oxygen atoms necessary to conserve these atoms.
Use the balancing redox rules on page 21 of your lab manual or
in your textbook on pages 90-92.
Step 1:
1 Separate the reaction into oxidation/reduction half
reactions.
Fe2+(aq) → Fe3+(aq)
-
MnO4 (aq) →
Step 3:
3 For every oxygen, add an equivalent number of water
molecules on the opposite side to balance them out.
Fe2+(aq) → Fe3+(aq)
Mn2+(aq)
MnO4-(aq) → Mn2+(aq) + 4H2O(l)
Step 2:
2 Balance atoms other than oxygen and hydrogen.
Step 4:
4 Add H+ ions to balance the hydrogen added from
water.
Fe2+(aq) → Fe3+(aq)
Fe2+(aq) → Fe3+(aq)
MnO4-(aq) → Mn2+(aq)
8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l)
In this case the half-reactions are already balanced.
Step 5:
5 Balance charges between reactants and products by
adding electrons.
Fe2+(aq) → Fe3+(aq) + 1e-
Now we can add the two reaction equations together. Note that
the electrons cancel out so are not included in the final net
reaction.
5Fe2+(aq) + 8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
8H+(aq) + MnO4-(aq) + 5e- → Mn2+(aq) + 4H2O(l)
Step 6:
6 Ensure that the number of electrons lost and gained is
the same by multiplying half-reactions through where
necessary.
5Fe2+(aq)
→
5Fe3+(aq)
+
5e-
8H+(aq) + MnO4-(aq) + 5e- → Mn2+(aq) + 4H2O(l)
Notice that the Mn is reduced from +7 to +2 and each iron is
oxidized from +2 to +3.
The permanganate ion, MnO4-(aq), is said to be the oxidizing agent
(because it oxidizes the iron), and the iron is the reducing agent.
In a disproportionation reaction, an element in one oxidation
state is simultaneously oxidized and reduced.
9
As a final step, if the reaction were to have taken place in basic solution,
the hydrogen ions must be removed and replaced by hydroxide ions. This
is done by adding OH- to each side of the equation.
On Your Own:
Balance the following reaction that takes place in basic solution:
Cl2(g) + Cr(OH)3(s) → Cl-(aq) + CrO42-(aq)
5Fe2+(aq) + 8H+(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
5Fe2+(aq) + 8H+(aq) + 8OH-(aq) + MnO4-(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) + 8OH(aq)
5Fe2+(aq) + 8H2O(l) + MnO4-(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) + 8OH-(aq)
Simplify the water molecules by removing 4 from both sides to end up with
5Fe2+(aq) + 4H2O(l) + MnO4-(aq) → Mn2+(aq) + 5Fe3+(aq) + 8OH-(aq)
Answer:
5. Cl2(g) + 2e- → 2Cl-(aq)
Cr(OH)3(s) + H2O(l) → CrO42-(aq) + 5H+(aq) + 3e-
1. Cl2(g) → Cl-(aq)
Cr(OH)3(s) → CrO42-(aq)
6. 3Cl2(g) + 6e- → 6Cl-(aq)
2Cr(OH)3(s) + 2H2O(l) → 2CrO42-(aq) + 10H+(aq) + 6e-
2. Cl2(g) → 2Cl-(aq)
Cr(OH)3(s) → CrO42-(aq)
3Cl2(g) + 2Cr(OH)3(s) + 2H2O(l) → 6Cl-(aq)+ 2CrO42-(aq) + 10H+(aq)
7. Since it is in basic solution:
3. Cl2(g) → 2Cl-(aq)
3Cl2(g) + 2Cr(OH)3(s) + 2H2O(l) + 10OH-(aq) → 6Cl-(aq) + 2CrO42-(aq) + 10H+(aq) + 10OH-(aq)
Cr(OH)3(s) + H2O(l) → CrO42-(aq)
3Cl2(g) + 2Cr(OH)3(s) + 2H2O(l) + 10OH-(aq) → 6Cl-(aq) + 2CrO42-(aq) + 10H2O(l)
4. Cl2(g) → 2Cl-(aq)
simplifying the water:
Cr(OH)3(s) + H2O(l) → CrO42-(aq) + 5H+(aq)
3Cl2(g) + 2Cr(OH)3(s) + 10OH-(aq) → 6Cl-(aq) + 2CrO42-(aq) + 8H2O(l)
Continued…
Follow up question: What element(s) is/are oxidized and
reduced? What is the oxidizing/reducing agents?
Special Note about Single Replacement Reactions
A + BC → B + AC
Whether a reaction occurs or not depends on the relative “activities”
activities” of
manual)
ual)
the elements involved (See activity series p. 20 in your lab man
Answer:
In the above general equation, A will only replace B if A is a more “active”
element than B. If not, the reaction does not occur.
Chlorine is reduced (0 to -1)(the oxidizing agent)
Example:
Chromium is oxidized (+3 to +6)(chromium(III)hydroxide is the
reducing agent)
Mg(s) + CuCl2(aq) → Cu(s) + MgCl2(aq)
Since Mg is more active than Cu, a reaction occurs. (Note that Mg is
oxidized and Cu is reduced)
3Ag(s) + Fe(NO3)3(aq) → No Reaction
Since Fe is more active than Ag, no reaction occurs. (Note that the more
active metal remains “oxidized” in its preferred charge state)
10