Download Ch 4 Types of Chemical Reactions and Solution Stoichiometry

CH 4 TYPES OF
CHEMICAL REACTIONS
AND SOLUTION
STOICHIOMETRY
AP Chemistry
2014-2015
Ch 3/4 Quiz
T 9/16
Ch 3/4 Exam
Th 9/18
4.1 WATER, THE COMMON SOLVENT
 Water has a high specific heat, high
heat of vaporization, and high
adhesive/cohesive forces
 The two O-H bonds in water are polar
covalent (oxygen is more
electronegative—partial negative
charge on oxygen, partial positive
charge on each hydrogen)
 Bond angle =~105° (the two unshared
electron pairs on oxygen are “space hogs”,
squishing the bond angle from a tetrahedral
109.5 despite four areas of electron density)
4.1 CONTINUED
 Partially positive hydrogens are attracted to negative
ions, partially negative oxygen is attracted to positive
ions  hydration
4.1 CONTINUED
 For ionic solids: when the hydration attraction are
greater than the crystal lattice attractions, the
compound is soluble
 Water can also dissolve nonionic substances (ex.
alcohols, sugars) if they exhibit polarity
 Fats (generally all nonpolar substances) do not
dissolve in water
 “Like dissolves like”
4.2 THE NATURE OF AQUEOUS SOLUTIONS:
STRONG AND WEAK ELECTROLY TES
A solution is homogeneous mixture where a
solute is dissolved in a solvent; if the solvent
is water the solution is considered “aqueous”
Properties of aqueous solutions
 Electrolytes are solutions that conduct an electric
current
 Strong electrolytes completely dissociate (ex. strong acids,
strong bases, soluble salts)
 Common strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4
 Common strong bases: oxides and hydroxides of group 1A and 2A
metals (2A metal salts tend to be less soluble than 1A metal salts)
 Weak electrolytes do not completely dissociate (only about
1% dissociation)—examples include acetic acid and ammonia
AQUEOUS SOLUTIONS CONTINUED
Nonelectrolytes are solutions in which the
solute dissolves but does not make ions. They
cannot conduct electricity. Ex. pure water,
sugar, alcohol, antifreeze, starch
Arrhenius—the extent to which a solution can
conduct an electric current depends on the
number of ions present
4.3 THE COMPOSITION OF SOLUTIONS
Molarity = moles of solute/liters of
solution (ex. 0.75 M NaCl means that for
every 1 L of solution there are 0.75
moles of NaCl present)
EXERCISE 4.1 MOLARIT Y I
Calculate the molarity of a solution prepared
by dissolving 11.5 g of solid NaOH in enough
water to make 1.50 L of solution. (answer =
0.192 M)
11.5 g / 40.00 g/mol = 0.288 mol NaOH
M = moles/volume = 0.283 moles / 1.50 L = 0.192 M
EXERCISE 4.2 MOLARIT Y II—IONS
 Calculate the concentration of the cobalt (II) ions
and the nitrate ions in a 0.50 M solution of cobalt (II)
nitrate. (answer = 0.50 M Co 2+ and 1.0 M NO 3 - )
0.50 M Co(NO3)2
[Co2+] = [Co(NO3)2] = 0.50 M
[NO3-] = 2[Co(NO3)2] = 1.00 M
EXERCISE 4.3 MOLES AND MOLARIT Y
Calculate the number of moles of Cl - ions in
1.75 L of 0.001 M ZnCl 2 . (answer = .0035
moles Cl - )
1.75 L 0.0001 moles 2 moles Cl1 mole ZnCl2
1L
= 0.0035 moles Cl-
EXERCISE 4.4 VOLUME AND MOLARIT Y
Typical blood serum is about 0.14 M NaCl.
What volume of blood contains 1.0 mg NaCl?
(answer = 0.12 mL)
1.0 mg NaCl 1 g NaCl
1000 mg
1 mole NaCl 1 L
0.14 moles
58.44 g
= 1.2 x 10-4 L
STANDARD SOLUTIONS
 A standard solution is a solution whose concentration is
accurately known. To prepare a solution of known
concentration, weigh out the solid as accurately as possible
and place it in a volumetric flask. Add enough just enough
distilled water to dissolve the solid and then fill to the mark
on the flask. Mix.
EXERCISE 4.5 STANDARD SOLUTIONS
To analyze the alcohol content of a certain wine, a
chemist needs 1.00 L of an aqueous 0.200 M
potassium dichromate solution. How many grams of
solid K 2 Cr 2 O 7 are needed to make this solution?
1.00 L 0.200 moles 294.20 g
1L
1 mole K2Cr2O7
= 58.8 g K2Cr2O7
DILUTION
M 1V 1 = M 2V 2
Generally, you measure out the quantity
of stock (concentration) solution that you
calculate from the formula above, and
place it in a volumetric flask (of the
correct volume for your dilution). Next
you fill the volumetric flask to the mark
on the neck with distilled water until the
meniscus is reached. Mix.
EXERCISE 4.6 DILUTION
What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of a 0.10 M sulfuric acid solution?
(answer = 9.4 mL)
(16 M)(x L) = (0.10 M)(1.5 L)
x = 0.0094 L or 9.4 mL
4.4 T YPES OF CHEMICAL REACTIONS
Review terms: chemical reaction,
balanced chemical equation, (s), (l),
(g), (aq)
Not all reactions fall neatly into one
category
4.5 PRECIPITATION REACTIONS
The formation of a
precipitate is the driving
force for some chemical
reactions. A precipitate is
an insoluble solid that is
formed when two aqueous
solutions are mixed. We
can separate the
precipitate from solution by
filtration in what is called
gravimetric analysis.
YOU HAVE TO KNOW THE SOLUBILIT Y
RULES
 Most alkali metal salts and ammonium salts are soluble
 Chloride, bromide, and iodide are soluble —except for A g + ,
Hg 2 2+ , and Pb 2+
 Nitrates, chlorates, perchlorates, and acetates are soluble
 Sulfates are soluble—except for Ca 2+ , Sr 2+ , Ba 2+ , A g + , Hg 2 2+ ,
and Pb 2+
 Carbonates, phosphates, chromates, dichromates, sulfides,
hydroxides, and oxides are insoluble —but the first rule (most
alkali metal salts and ammonium salts are soluble) takes
priority
 It can be assumed that ionic compounds that dissolve in
water are strong electrolytes.
EXERCISE 4.7 SOLUBILIT Y RULES
Predict what will happen when the following pairs of solutions
are mixed.
 Potassium nitrate and barium chloride
2 KNO3(aq) + BaCl2(aq)  2 KCl(aq) + Ba(NO3)2(aq)
No reaction
 Sodium sulfate and lead (II) nitrate
Na2SO4(aq) + Pb(NO3)2(aq)  PbSO4(s) + 2 NaNO3(aq)
Formation of a precipitate, lead (II) sulfate
 Potassium hydroxide and iron (II) nitrate
2 KOH(aq) + Fe(NO3)2 (aq)  2 KNO3(aq) + Fe(OH)2(s)
No reaction
4.6 DESCRIBING REACTIONS IN
SOLUTION
Complete balanced equation—gives the overall
reaction stoichiometry but does not specify
which reactants/products ionize
Complete ionic equation—same as above but
represents all strong electrolytes as ions
Net ionic equation—includes only the solution
components undergoing a change; spectator
ions are not included
 There is always a conservation of charge in net ionic
equations
EXERCISE 4.8 THREE T YPES OF
EQUATIONS
Aqueous potassium chloride is added to aqueous
silver nitrate to form a silver chloride precipitate plus
aqueous potassium nitrate.
 Write the complete balanced equation:
KCl(aq) + AgNO3(aq)  AgCl(s) + KNO3(aq)
 Write the complete ionic equation:
K+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq)  AgCl(s) + K+(aq) + NO3-(aq)
 Write the net ionic equation:
Cl-(aq) + Ag+(aq)  AgCl(s)
4.7 STOICHIOMETRY OF PRECIPITATION
REACTIONS
Precipitation reactions form an
insoluble salt; this salt is dried and
its mass is obtained
We can use stoichiometry to
determine the expected/needed
quantities of each compound
EXERCISE 4.9 PRECIPITATION REACTION
STOICHIOMETRY I
 Calculate the mass of solid sodium chloride that must be
added to 1 .50 L of a 0.100 M silver nitrate solution to
precipitate all of the silver ions in the form of AgCl. (answer
= 8.77 g NaCl)
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
1.50 L 0.100 moles
AgNO3 AgNO3
1L
AgNO3
1 mole
NaCl
1 mole
AgNO3
58.44 g
NaCl
1 mole
NaCl
= 8.77 g NaCl
EXERCISE 4.10 PRECIPITATION REACTION
STOICHIOMETRY II
 When aqueous solutions of sodium sulfate and lead (II) nitrate
are mixed, lead (II) sulfate precipitates. Calculate the mass
of lead (II) sulfate formed when 1 .25 L of 0.0500 M lead (II)
nitrate and 2.00 L of 0.0250 M sodium sulfate are mixed.
(answer = 15.2 g lead (II) sulfate)
Na2SO4(aq) + Pb(NO3)2(aq)  PbSO4(s) + 2 NaNO3(aq)
1.25 L
0.0500 moles 1 mole
PbSO4
Pb(NO3)2 Pb(NO3)2
1L
2.00 L
Na2SO4
1 mole
Pb(NO3)2
303.3 g
PbSO4
0.0250 moles 1 mole
PbSO4
Na2SO4
1 mole
PbSO4
303.3 g
PbSO4
1 mole
Na2SO4
1 mole
PbSO4
1L
= 19.0 g PbSO4
= 15.2 g PbSO4
4.8 ACID-BASE REACTIONS
 Acids are compounds that produce anions and hydronium
ions (H 3 O + ) when they react with water (you can also just
say hydrogen ions, H + , though hydronium is technically
more correct)—Arrhenius definition
 Bases are compounds that produce cations and hydroxide
ions when they react with water—Arrhenius definition
 NH 3 is therefore not an Arrhenius acid, but it IS a Bronsted-Lowry
acid because it is a proton acceptor
 In a neutralization reaction, an acid and a base in
equimolar (really equinormal) quantities react to produce
water and a salt. The salt may or may not be soluble.
EXERCISE 4.11 NEUTRALIZATION
 What volume of a 0.100 M HCl solution is needed to
neutralize 25.0 mL of 0.350 M NaOH? (answer =
0.0875 L = 87.5 mL HCl)
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
In other words, HCl is monoprotic and NaOH is monobasic,
so they react in a 1:1 ratio. M1V1 = M2V2 applies (not
always true for polyprotic acids/polybasic bases).
(0.100 M HCl)(V1 mL) = (0.350 M NaOH)(25.0 mL)
V1 = 87.5 mL
ACID-BASE TITRATIONS
Volumetric analysis is a
technique for
determining the amount
of a certain substance
by doing a titration
Review terms: titrant,
analyte, burette,
indicator
 The equivalence point
occurs when the
number of moles of
hydroxide ions is
equal to the number
of moles of hydrogen
(hydronium) ions—this
term is also used in
redox titrations
 The end point is
where you see a color
change when using
an indicator
Standardization is a procedure for establishing
the exact concentration of a reagent—ex. you
may prepare a solution to be 0.50 M but it
might be “off”; standardization helps you to
determine the true concentration
EXERCISE 4.12 STANDARDIZATION
 A student carries out an experiment to standardize a
sodium hydroxide solution. To do this the student
weighs out a 1.3009 g sample of potassium
hydrogen phthalate (KHC 8 H 4 O 4 abbreviated KHP,
molar mass 204.22 g/mol, has one acidic hydrogen).
The student dissolves the KHP in distilled water,
adds phenolphthalein as an indicator, and titrates
the resulting solution with the sodium hydroxide
solution to the phenolphthalein endpoint. The
difference between the final and initial burette
readings is 41.20 mL. Calculate the concentration of
the sodium hydroxide solution. (answer = 0.1546 M
NaOH)
EXERCISE 4.12 CONTINUED
KHP: mass = 1.3009 g
molar mass = 204.22 g/mole
NaOH: volume = 41.20 mL molarity = ?
KHP is monoprotic, so KHP and NaOH react in a 1:1 ratio.
Moles KHP = moles NaOH
Moles KHP = 1.3009 g / 204.22 g/mole = 0.00637 moles KHP
Moles NaOH = volume * molarity = 0.04120 L * V = .4120V
0.00637 = 0.04120V
V = 0.1546 M
4.9 OXIDATION-REDUCTION REACTIONS
Terms to know
 OIL RIG (oxidation is loss, reduction is gain, of
electrons)
 Oxidation is the loss of electrons, which increases
charge (more positive)
 Reduction is the gain of electrons, which decreases
charge (more negative)
 Oxidation number is the assigned charge on an atom
 Oxidizing agents get reduced, reducing agents get
oxidized—reduction and oxidation must be coupled
RULES FOR ASSIGNING OXIDATION
STATES
 Always zero for elements
 The oxidation state of a monatomic ions is the same as its
charge
 Fluorine is always -1 , oxygen is almost always -2 (exceptions—
peroxides where it is -1 , or OF 2 where it is +2)
 Hydrogen is almost always +1; metal hydrides are an
exception, where it is -1 (in these situations, hydrogen is
placed at the end of a chemical formula like LiH)
 The sum of the oxidation states must be zero for a neutral
compound; for polyatomic ions, the sum of the oxidation
states must equal the charge on the polyatomic ion
 It’s odd, but there can be non -integer oxidation states. Ex.
Fe 3 O 4 where the oxygens total -8, so iron’s charge must be
+8/3  Fe +8/3
EXERCISE 4.13 OXIDATION STATES
Assign oxidation states to each atom in the
following compounds/ions.
CO 2
SF 6
NO 3 -
C = +4
S = +6
N = +5
O= -2
F = -1
O = -2
EXERCISE 4.14 OXIDATION AND
REDUCTION I
For the following reaction, identify the atoms
that are oxidized and reduced.
2 Al(s) + 3 I 2 (s)  2 AlI 3 (s)
Aluminum: 0  +3, so aluminum was oxidized
Iodine: 0  -1, so iodine was reduced
Now go back and identify the oxidizing and
reducing agents .
Aluminum: reducing agent
Iodine: oxidizing agent
EXERCISE 4.15 OXIDATION AND
REDUCTION II
The following reactions are associated with
metallurgy. For each reaction, identify the atoms
that are oxidized and reduced. Then identify the
oxidizing and reducing agents.
2 PbS(s) +3 O 2 (g)  2 PbO(s) + 2 SO 2 (g)
Ox: S (-2  +4)
Red: O (0 -2)
Ox agent: oxygen
Red agent: sulfur
PbO(s) + CO(g)  Pb(s) + CO 2 (g)
Ox: C (+2  +4)
Red: Pb(+2  0)
Ox agent: lead
Red agent: carbon
4.10 BALANCING OXIDATION-REDUCTION
EQUATIONS
Divide the equation into oxidation and
reduction half reactions (OIL RIG)
Balance all elements besides hydrogen and
oxygen
Balance oxygen by adding water to the
appropriate side of the equation
Balance hydrogen by adding hydrogen ions to
the appropriate side of the equation
4.10 CONTINUED
 Balance the charge by adding electrons
 Multiply the half reactions to make the electrons
equal for oxidation/reduction reactions
 Cancel terms when you recombine the two half
reactions
 These rules are for acidic solutions; if this takes
place in a basic solution, you have one more step.
Neutralize any hydrogen ions by adding the same
number of hydroxide ions to each side; check your
water and cancel terms as necessary
 Check
SAMPLE PROBLEM 1
Balance the following equation using the half -reaction method
(acidic):
MnO 4 - (aq) + I - (aq)  Mn 2+ (aq) + I 2 (aq)
Reduction:
Mn: +7  +2; gained 5 electrons
4 H+ + 5 e- + MnO4-  Mn2+ + 2 H2O
Oxidation: I: -1  0; each iodine atom lost 1 electron (2 total)
2 I-  I2 + 2 eCombined: Need to multiply Mn equation by 2 and I equation
by 5 to get 10 electrons transferred; the number of
electrons lost in one equation must equal the
number of electrons gained in the other equation
SAMPLE PROBLEM 1 CONTINUED
Manganese: 8 H + + 10 e - + 2 MnO 4 -  2 Mn 2+ + 4 H 2 O
Iodine: 10 I -  5 I 2 + 10e Now, we add and cancel. In this equation, only the
electrons will cancel.
8 H + + 2 MnO 4 - + 10 I -  2 Mn 2+ + 4 H 2 O + 5 I 2
We would reduce if we could, but this equation is
finished.
SAMPLE PROBLEM 2
Balance the following equation using the half -reaction
method (basic):
Ag(s) + CN - (aq) + O 2 (g)  Ag(CN) 2 - (aq) + H 2 O
Reduction:
O: 0  -2; gained 2 electrons on two atoms
4 H+ + O2 + 4 e- 2 H2O
Oxidation: Ag: 0  +1; lost 1 electron
Ag + 2 CN-  Ag(CN)2- + eCombined:
4 H+ + O2 + 2 Ag + 4 CN-  2 Ag(CN)2- + e- + 2 H2O
This equation is only correct in acidic solution; let’s
make it basic now
4 H + + O 2 + 2 Ag + 4 CN -  2 Ag(CN) 2 - + e - + 2 H 2 O
To make this basic, we will add enough hydroxide ions to each
side to neutralize the four hydrogen ions present on the products
side. This will give us
4 OH - + 4 H + + O 2 + 2 Ag + 4 CN -  2 Ag(CN) 2 - + e - + 2 H 2 O + 4 OH Which simplifies to
2 H 2 O + O 2 + 2 Ag + 4 CN -  2 Ag(CN) 2 - + e - + 4 OH Because we made four molecules of water on the reactants’ side
and we had two molecules of water on the right, we have to
cancel two molecules of water from each side.
EXERCISE 4.15 BALANCING HALF
REACTIONS
Potassium dichromate reacts with ethanol (C 2 H 5 OH) as follows:
H + (aq) + Cr 2 O 7 2- (aq) + C 2 H 5 OH(l)  Cr 3+ (aq) + CO 2 (g) + H 2 O(l)
Balance this equation using the half -reaction method.
Cr: +6  +3 so 3 electrons are gained; 2 atoms present
14 H+ + 6e- + Cr2O72-  2 Cr3+ + 7 H2O
C: -2  +4 so 6 electrons are lost; 2 atoms present
3 H2O + C2H5OH  2 CO2 + 12 e- + 12 H+
We have 6 electrons gained in one reaction and 12 electrons
lost in the other. We can fix this by multiplying so that 12
electrons are transferred (x2 chromium equation)
14 H + + 6 e - + Cr 2 O 7 2-  2 Cr 3+ + 7 H 2 O
3 H 2 O + C 2 H 5 OH  2 CO 2 + 12 e - + 12 H +
Becomes
28 H + + 12 e - + 2 Cr 2 O 7 2-  4 Cr 3+ + 14 H 2 O
3 H 2 O + C 2 H 5 OH  2 CO 2 + 12 e - + 12 H +
Which we add to get
16 H + + 2 Cr 2 O 7 2- + C 2 H 5 OH  4 Cr 3+ + 11 H 2 O + 2 CO 2