Chapter 4 Lecture Slides

Chapter 4:
Reaction in Aqueous
Solution
What is a Solution?
Solute
+
substance dissolved
typically smaller
quantity
Solvent
!
dissolving medium
typically larger
quantity
Solution
homogeneous
mixture
variable
composition
When water is used as
the solvent, the solution
is an aqueous solution
Speciation in Solutions: Ionic Compounds
Soluble Ionic Compounds - result in aqueous
solutions with individually dispersed ions
must consider stoichiometry as well:
AgNO3 (s) ! Ag+ (aq) + NO3– (aq)
vs.
Ba(NO3)2 (s) ! Ba2+ (aq) + 2 NO3– (aq)
Speciation in Solutions:
Molecular Compounds
Soluble Molecular Compounds
- result in aqueous solutions with individually
dispersed, neutral molecules.
Solution Composition:
How much solute is present?
◆
Solution Calculations
1. Calculate the molarity of a solution.
qualitatively – is a solution dilute or concentrated?
OR
Calculate the molarity of ions in a solution.
◆
quantitatively – molar concentration, or molarity (M)
mol solute
molarity = –––––––––––––––
volume solution
units of molarity are mol/L
2. Calculation to determine how to prepare a
solution of known concentration.
3. Solution reaction stoichiometry calculations;
including identifying limiting reactants.
4. Dilution calculations.
Solution Calculations
Solution Calculations
Calculate the molarity of a solution or of ions in a
solution.
Calculation to determine how to prepare a solution
of known concentration.
ex. 2.6009 g CoCl3•6H2O is dissolved in water
resulting in 250.0 mL of solution.
ex. Determine the mass (in g) of NaNO3 required
to prepare 100.0 mL of 0.125 M NaNO3 (aq).
Determine the molar concentration of this solution,
and the molar concentration of chloride ions in this
solution.
[CoCl3] = ????
[Cl–]
= ????
ex. Determine the mass (in g ) of solute in 75.0 mL
of 88.8 mM NiSO4 (aq).
Process of Preparing a Solution
of Specific Concentration
Solution Calculations
Solution reaction stoichiometry calculations:
ex. Determine the volume (in mL) of 0.758 M
Ba(OH)2 (aq) required to react completely with
10.00 mL of 1.07 M HCl (aq):
2 HCl (aq) + Ba(OH)2 (aq) ! BaCl2 (aq) + 2 H2O (l)
ex. 100.0 mL of 0.800 M AgNO3 (aq) and 100.0 mL
of 1.025 M BaCl2 (aq) are combined and allowed
to react accoding to the following equation:
2 AgNO3 (aq) + BaCl2 (aq) ! 2 AgCl (s) + Ba(NO3)2 (aq)
Determine the theoretical yield (in g) of AgCl (s).
Dilution Calculations
Dilution is the procedure of making a solution of
lower concentration (i.e. more dilute) from a
solution of higher concentration (i.e. more
concentrated).
mol
mol solute
M = ––––––––––––– = –––––
volume solution
V
so:
mol solute = M ! V
How do each of the following change from
Solution 1 (concentrated) to Solution 2 (dilute)?
!
Solution 1
concentrated
Dilution Calculations
Solution 2
dilute
mol solute
volume solution (V)
molarity (M)
Dilution Calculations
Aqueous Solutions and
Electrolyte Classifications
ex. 12.5 mL of 12.0 M HCl (aq) is transferred
to a flask and an additional 200.0 mL of
water is added. Determine the molar
concentration of the new, dilute HCl (aq).
Classification of solutes and resulting
solutions as:
strong electrolytes
weak electrolytes
nonelectrolytes
based on the ability of the solution to
conduct electricity.
ex. How would you prepare 750.0 mL of
0.250 M NH3 (aq) from a stock solution
with [NH3] = 4.85 M?
◆
◆
strong electrolytes are completely ionized in aqueous sol’n
solutions of strong electrolytes conduct electricity
The extent to which a solution can
conduct electricity is directly
dependent on the number of charged
particles (ions) present
◆
◆
weak electrolytes are only partially ionized in aqueous sol’n
solutions of weak electrolytes conduct electricity only a little
Identification of Electrolytes
◆
compounds that are strong electrolytes:
soluble ionic compounds
strong acids and strong bases
◆
compounds that are weak electrolytes:
weak acids and weak bases
◆
◆
◆
molecular compounds (even if soluble)
insoluble ionic compounds
nonelectrolytes are not ionized in aqueous sol’n
solutions of nonelectrolytes do not conduct electricity
Precipitation Reactions:
Predicting Products
Precipitation Reactions:
Formation of an Insoluble Solid
◆
Consider the reaction that occurs
between aqueous solutions of NaCl
& AgNO3:
◆
AgCl (s) + NaNO3 (aq)
If aqueous solutions of soluble ionic
compounds MX and NZ are mixed, what
new ionic compounds can form as products?
in general:
MX (aq) + NZ (aq) ! ??? + ???
NaCl (aq) + AgNO3 (aq)
↓
compounds that are nonelectrolytes:
◆
need to determine:
chemical formulas of products
physical states . . .
Are these ionic compounds soluble or
insoluble?
Solubility Rules
Predict the Products:
Are these precipitation reactions?
ex: Pb(NO3)2 (aq) + KI (aq) ! ???? + ????
ex: (NH4)2S (aq) + CuSO4 (aq) ! ???? + ????
ex: K2CO3 (aq) + Ca(OH)2 (aq) ! ???? + ????
a few notes and additions:
rule 1: applies to all Group IA cations & NH4+, so include Rb+ and Cs+
rule 2: applies to chlorates (ClO3–) and perchlorates (ClO4–) too
rule 3: F– salts tend to be less soluble than Cl–, Br– and I–
to rules 5, 6 & 7: applies to oxalates (C2O42–) and chromates (CrO42–) too
Writing Ionic Equations
◆
◆
◆
Balanced Chemical Equation
focus: the big picture
identity of reactants and products
physical states
stoichiometry
ex: NH4NO3 (aq) + MgCl2 (aq) ! ???? + ????
Writing Ionic Equations
an example: Pb(NO3)2 (aq) + KI (aq) ! ???? + ????
balanced chemical equation:
Pb(NO3)2 (aq) + 2 KI (aq) ! PbI2 (s) + 2 KNO3 (aq)
full ionic equation:
Pb2+(aq) + 2 NO3– (aq) + 2 K+(aq) + 2 I– (aq) ! PbI2 (s) + 2 K+(aq) + 2 NO3– (aq)
Full (or Complete, or Total) Ionic Equation
focus: complete speciation
identity of all species present in solution
net ionic equation:
Pb2+ (aq) + 2 I– (aq) ! PbI2 (s)
Net Ionic Equation
focus: actual chemical change
do not include spectator ions in net ionic
equation
notes:
all equations are balanced in terms of atoms & charge
ions shown with charges
polyatomic ions do not split up further
all physical states included
spectator ions: K+ and NO3–
Writing Ionic Equations
What do you “split up”
into ions when writing
ionic equations?
only strong electrolytes
in (aq)
soluble ionic cmpd’s
strong acid
strong bases
What do you not “split up”
when writing ionic
equations?
anything that is (s), (l), (g)
weak electrolytes and
nonelectrolytes - even
if in (aq)
Acids and Bases
Arrhenius Definitions:
◆ acids - compounds that produce an increase in
[H+] when dissolved in water
◆
Acids and Bases
bases - compounds that produce an increase in
[OH–] when dissolved in water
Brønsted-Lowry Definitions:
◆ acids - H+ donors
◆ bases - H+ acceptors
Lewis Definitions:
◆ acids - electron pair acceptors
◆ bases - electron pair donors
Acids and Bases
Brønsted-Lowry acid-base behavior:
acid ionization equation:
HNO2 (aq) + H2O (l) ! NO2– (aq) + H3O+ (aq)
notes: HNO2 - acid;
H2O - base
+
H3O (hydronium ion) and H+ used
interchangeably
Water is amphoteric; it can act
as an acid or a base.
base ionization equation:
NH3 (aq) + H2O (l) ! NH4+ (aq) + OH– (aq)
notes: H2O - acid;
NH3 - base
Acids and Bases
Identify the Brønsted-Lowry acid and base in the
following reaction:
H2SO4 (aq) + HPO42– (aq) ! HSO4– (aq) + H2PO4– (aq)
Strong vs. Weak Acids and Bases
Acid and base strength is based on the extent of
ionization that occurs when the substance is
dissolved in water.
Strong Acids:
◆
strong electrolytes - completely ionized in solution
◆
there are 6 strong acids - KNOW THEM!
HCl, HBr, HI, HNO3, HClO4, H2SO4 (diprotic)
Weak Acids:
◆
weak electrolytes - partially ionized (typically < 5%)
in aqueous solution
◆
any acid that is not a strong acid is a weak acid
some examples: HF, H2CO3, H3PO4, HNO2, HBrO4
Strong vs. Weak Acids and Bases
Strong Bases:
◆
strong electrolytes - completely ionized in solution
◆
the strong bases are the hydroxides of the alkali
metals & hydroxides of most alkaline earth metals:
LiOH, NaOH, KOH, RbOH, CsOH
Ca(OH)2, Sr(OH)2, Ba(OH)2
Weak Bases:
◆
weak electrolytes - partially ionized (typically < 5%) in
aqueous solution
◆
weak bases tend to be organic compounds that
contain nitrogen; ammonia and substituted amines
some examples: NH3, (CH3)NH2, (CH3)3N
C5H5N, N2H4, NH2OH
Acid - Base Neutralization Reactions:
general form:
Acid + Base ! Salt* + Water
* salt will have cation related to the base
+ anion related to the acid
example:
Acid - Base Neutralization Reactions:
Writing Ionic Equations
Strong Acid + Strong Base:
balanced chemical equation:
HNO3 (aq) + KOH (aq) ! KNO3 (aq) + H2O (l)
full ionic equation:
Write the balanced chemical equation for the
neutralization reaction that will occur between
aqueous solutions of nitrous acid and calcium
hydroxide.
Acid - Base Neutralization Reactions:
Writing Ionic Equations
H+ (aq) + NO3– (aq) + K+ (aq) + OH– (aq) ! K+ (aq) + NO3– (aq) + H2O (l)
net ionic equation:
H+ (aq) + OH– (aq) ! H2O (l)
spectator ions: K+ and NO3–
Acid - Base Neutralization Reactions:
Writing Ionic Equations
Weak Acid + Strong Base:
Strong Acid + Weak Base:
balanced chemical equation:
balanced chemical equation:
HCl (aq) + NH3 (aq) ! NH4Cl (aq)
HC2H3O2 (aq) + CsOH (aq) ! CsC2H3O2 (aq) + H2O (l)
full ionic equation:
HC2H3O2 (aq) +
Cs+
(aq) +
OH–
(aq) !
Cs+
(aq) +
C2H3O2–
(aq) + H2O (l)
net ionic equation:
HC2H3O2 (aq) + OH– (aq) ! C2H3O2– (aq) + H2O (l)
spectator ion: only Cs+
full ionic equation:
H+ (aq) + Cl– (aq) + NH3 (aq) ! NH4+ (aq) + Cl– (aq)
net ionic equation:
H+ (aq) + NH3 (aq) ! NH4+ (aq)
spectator ions: only Cl–
Other Examples of Acid-Base Chemistry:
Acidic and Basic Oxides
Other Examples of Acid-Base Chemistry:
Acidic and Basic Oxides
in general - oxides of metals are basic
basic oxides or base anhydrides
◆
metal oxides form basic solutions when dissolved
in water:
in general - oxides of nonmetals are acidic
acidic oxides or acid anhydrides
◆
SO2 (g) + H2O (l) ! H2SO3 (aq)
CaO (s) + H2O (l) ! Ca(OH)2 (aq)
◆
metal oxides react with acids:
CaO (s) + 2 HCl (aq) ! CaCl2 (aq) + H2O (l)
Other Examples of Acid-Base Chemistry:
Neutralization Reactions that Produce Gases
general form of reaction:
acid + base ! salt + water + gas
carbonates & hydrogen carbonates: gas formed is CO2
sulfates & hydrogen sulfates: gas formed is SO3
sulfites & hydrogen sulfites: gas formed is SO2
sulfides: gas formed is H2S
nonmetal oxides form acidic solutions when
dissolved in water:
◆
nonmetal oxides react with bases:
SO2 (g) + 2 NaOH (aq) ! Na2SO3 (aq) + H2O (l)
Other Examples of Acid-Base Chemistry:
Neutralization Reactions that Produce Gases
some examples:
Na2CO3 (s) + 2 HCl (aq) ! 2 NaCl (aq) + H2O (l) + CO2 (g)
full ionic equation:
Na2CO3 (s) + 2 H+ (aq) + 2 Cl– (aq) ! 2 Na+ (aq) + 2 Cl– (aq) + H2O (l) + CO2 (g)
net ionic equation:
Na2CO3 (s) + 2 H+ (aq) ! 2 Na+ (aq) + H2O (l) + CO2 (g)
spectator ion: Cl– (aq)
MgSO4 (s) + 2 HCl (aq) ! MgCl2 (aq) + H2O (l) + SO3 (g)
Ba(HSO3)2 (aq) + 2 HBr (aq) ! BaBr2 (aq) + 2 H2O (l) + 2 SO2 (g)
K2S (s) + 2 HNO3 (aq) ! 2 KNO3 (aq) + H2S (g)
The pH Scale
logarithmic scale of [H+] in solution
pH = "log[H+];
recall:
◆
◆
◆
◆
[H+] = 10–pH
pH Calculations:
Relative Acidity and Basicity of Solutions
◆
Relationship Between [H+] and pH
as [H+] changes by a factor of 10, the pH of the
solution changes by 1 unit
◆
higher [H+] corresponds to lower pH
◆
higher [H+] corresponds to more acidic solution
pH Calculations:
Relative Acidity and Basicity of Solutions
ex: Calculate the pH of 0.00283 M HNO3 (aq).
in any aqueous solution at 25°C:
[H+][OH–] = 1 x 10–14
pH = –log[H+];
[H+] = 10–pH
higher [H+] " more acidic solution "
lower pH
higher
[OH–]
[H+]
" lower
"
more basic solution " higher pH
ex: Will the pH of 0.00283 M HNO2 (aq) be less
than, greater than, or equal to the pH of
0.00283 M HNO3 (aq)? Why?
ex: Calculate the [H+] in a solution with pH = 3.61.
ex: Calculate the pH of 0.20 M Ba(OH)2 (aq) and
0.20 M NaOH (aq). Should they be the same?
Why or why not?
Oxidation-Reduction Reactions
aka Redox Reactions
oxidation: loss of electrons
reduction: gain of electrons
Consider the reaction of calcium
metal with oxygen:
2 Ca (s) + O2 (g) ! 2 CaO (s)
What happens to each element
during the course of this reaction?
Oxidation Numbers (Nox)
An oxidation number indicates the amount of electropositive
or electronegative character of an atom - particularly as part
of a polyatomic species.
Oxidation and Reduction Half Reactions
for the reaction of calcium with oxygen:
oxidation 1⁄2 reaction:
Ca ! Ca2+ + 2 e–
note: in oxidation 1⁄2 rxn e–’s are products
reduction 1⁄2 reaction: O2 + 4 e– ! 2 O2–
note: in reduciton 1⁄2 rxn e–’s are reactants
net redox reaction: 2 Ca + O2 ! 2 Ca2+ + 2 O2–
note: add together the half-reactions;
multiply as necessary to have same
number of electrons in each half
reaction before adding
Determining Oxidation Numbers:
ex. Determine Nox of sulfur in SO3.
Nox S + 3(Nox O) = 0
Nox S + 3(–2) = 0
∴ Nox S = +6
ex. Determine Nox of sulfur in SO32–.
Nox S + 3(Nox O) = –2
Nox S + 3(–2) = –2
∴ Nox S = +4
ex. Determine Nox N, H, P, and O in (NH4)3PO4.
consider NH4+ and PO43– separately:
Nox N + 4(Nox H) = +1
Nox P + 4(Nox O) = –3
Nox N + 4(+1) = +1
Nox P + 4(–2) = –3
∴ Nox N = –3
∴ Nox P = +5
Recognizing Oxidation and Reduction
Recognizing Oxidation and Reduction
Identify the oxidizing and reducing agents:
Compare oxidation numbers of elements in reactants
and products:
ex:
◆
if Nox increases - the element is oxidized
◆
if Nox decreases - the element is reduced
2 Al (s) + Cr2O3 (s) ! Al2O3 (s) + 2 Cr (s)
Recognizing Oxidation and Reduction
Consider the following reaction:
2 Ca3(PO4)2 + 6 SiO2 + 10 C ! P4 + 6 CaSiO3 + 10 CO
Identify the following:
element oxidized
element reduced
oxidizing agent
reducing agent
oxidizing agent
◆
reactant that facilitates
oxidation by taking on
(i.e. gaining) electrons
lost during oxidation
process
∴ oxidizing agent is the
reactant that contains
the element that is
reduced
reducing agent
◆
reactant that facilitates
reduction by providing
(i.e. losing) electrons
that are gained during
reduction process
∴ reducing agent is the
reactant that contains
the element that is
oxidized
Balancing Redox Equations: the Half-Reaction Method
in acidic solution
1. Identify the element oxidized and the element reduced.
2. Write the skeletal oxidation & reduction half-reactions.
3. Balance elements other than H and O.
4. Balance O’s by adding H2O.
5. Balance H’s by adding H+.
6. Balance charge by adding e–’s.
*at this point the individual half reactions are balanced*
7. Prepare to add the half-reactions together;
multiply as necessary so that the # of e–’s in
oxidation half reaction equals # of e–’s in the
reduction half reaction.
8. Add half reactions together; clean up.
Balance the following redox reaction that occurs in acidic
solution using the half-reaction method:
I2 + NO3– ! IO3– + NO2
step 2:
ox 1⁄2 rxn:
red 1⁄2 rxn:
step 3:
ox 1⁄2 rxn:
red 1⁄2 rxn:
step 4:
ox 1⁄2 rxn:
red 1⁄2 rxn:
step 5:
ox 1⁄2 rxn:
red 1⁄2 rxn:
I2 ! IO3–
NO3– ! NO2
Balance the following redox reaction that occurs in acidic
solution using the half-reaction method (continued):
I2 + NO3– ! IO3– + NO2
step 6:
ox 1⁄2 rxn:
red 1⁄2 rxn:
I2 ! 2 IO3
NO3– ! NO2
–
6 H2O + I2 ! 2 IO3–
NO3– ! NO2 + H2O
6 H2O + I2 ! 2 IO3– + 12 H+
2 H+ + NO3– ! NO2 + H2O
combination (or formation) reactions:
elements compounds
ex. 2 Al (s) + 3 Br2 (l) ! 2 AlBr3 (s)
◆
decomposition reactions:
compounds elements
ex. 2 NaH (s) ! 2 Na (s) + H2 (g)
6 H2O + I2 ! 2 IO3– + 12 H+ + 10 e–
+ 2 H+ + NO3– ! NO2 + H2O
step 7:
ox 1⁄2 rxn:
6 H2O + I2 ! 2 IO3– + 12 H+ + 10 e–
red 1⁄2 rxn: 10!(1 e– + 2 H+ + NO3– ! NO2 + H2O)
step 8:
6 H2O + I2 + 20 H+ + 10 NO3– ! 2 IO3– + 12 H+ + 10 NO2 + 10 H2O
net rxn: I2 + 8 H+ + 10 NO3– ! 2 IO3– + 10 NO2 + 4 H2O
Some Categories of Redox Reactions
Some Categories of Redox Reactions
◆
1
e–
◆
◆
combustion reactions:
ex. CH4 (g) + 2 O2 (g) ! CO2 (g) + 2 H2O (g)
halogen displacement reactions:
F2 > Cl2 > Br2 > I2
most
least
reactive
reactive
ex. F2 (g) + 2 NaCl (s) ! 2 NaF (s) + Cl2 (g)
Br2 (l) + 2 KI (s) ! 2 KBr (s) + I2 (s)
but Br2 (l) + 2 LiF (s) !
# 2 LiBr (s) + F2 (g)
Volumetric Analysis and Titrations
Some Categories of Redox Reactions
◆
displacement reactions with metals:
Titration - laboratory technique usually used to determine
the composition of a reactant or a solution.
reactions of metals with water:
2 M + 2 H2O ! 2 MOH + H2
M + 2 H2O ! M(OH)2 + H2
reactions of metals with acid:
unbalanced
M + H+ ! Mn+ + H2
metal displacement reactions:
M + M’X ! MX + M’
◆ more reactive metals are higher in
the activity series
◆
any metal can displace or reduce
any metal below it in the activity
series
◆
◆
add one reactant sol’n to the other in a slow, controlled way
stop when the stoichiometric point is reached
at the stoichiometric point - no limiting reactant and no
excess reactant;
reactants completely consumed and converted to products
Acid - Base Titration example:
Gravimetric Analysis
20.00 mL of HNO3 (aq) is titrated with Ba(OH)2 (aq).
18.62 mL of 0.254 M Ba(OH)2 is required to reach
the stoichiometric point. Determine [HNO3].
◆
Redox Titration example:
◆
24.0 mL of 0.0100 M KMnO4 (aq) are needed to
reach the stoichiometric point in the titration of a
1.00 g sample of impure H2C2O4. Determine the
mass % H2C2O4 in the sample.
2 MnO4– + 16 H+ + 5 C2O42– ! 2 Mn2+ + 10 CO2 + 8 H2O
lab technique used to analyze and determine
composition of a compound or mixture
involves specific reaction of one species to form
an insoluble solid that can be isolated, and its mass
determined
ex: An excess of potassium chromate solution is
added to 30.0 mL of a solution containing a
soluble barium salt. All the Ba2+ is removed from
solution as a yellow precipitate of BaCrO4 is
formed. Determine [Ba2+] in the original solution
if 2.58 g BaCrO4 is formed.
Ba2+ (aq) + CrO42– (aq) ! BaCrO4 (s)
Some Additional Problems:
ex. Saccharin, C7H5NO3S, is sometimes dispensed in tablet
form. Ten of these saccharin-containing tablets have a
total mass of 0.5894 g.
These ten tablets were dissolved in water, and then
oxidized so that all the sulfur present is converted to
sulfate ion.
Next, excess BaCl2 (aq) was added so that the sulfate
ion was removed from solution as BaSO4 (s). The mass
of BaSO4 collected was 0.5032 g.
Ba2+ (aq) + SO42– ! BaSO4 (s)
a. Determine the average mass of saccharin per tablet.
b. Determine the average mass % saccharin in these tablets.
Some Additional Problems:
ex. A mixture contains only NaCl and Fe(NO3)3. A
0.456 g sample of this mixture is dissolved in water,
and then an excess of NaOH is added, resulting in
the formation of Fe(OH)3 precipitate:
Fe3+ (aq) + 3 OH– (aq) ! Fe(OH)3 (s)
The precipitate is collected by filtration and dried;
the mass of Fe(OH)3 collected is 0.107 g.
a. Determine the mass of iron in the original sample.
b. Determine the mass of Fe(NO3)3 in the original sample.
c. Determine the mass % Fe(NO3)3 in the original sample.