Transcript Chem 1201 - LSU Department of Chemistry

Watkins
Chapter 4
CHAPTER 4
Reactions in Aqueous Solution
•
•
•
•
•
Solutions - Composition & Properties
Acids, Bases and Salts
Ionic Equations
Redox
Solution Stoichiometry
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Chapter 4
Definitions
Solution homogeneous mixture of two
or more pure substances
(components); the solution
can be solid (alloys), gas (air)
or liquid.
Solvent component in XS (e.g.
water).
Solute(s) components dissolved in
solvent.
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Chapter 4
Composition of Solutions
Concentration tells how much solute is
in the solution.
Qualitative Terms
A concentrated solution has a large
quantity of solute relative to solvent.
A dilute solution has a small quantity of
solute relative to solvent.
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Chapter 4
Composition of Solutions
Quantitative Terms
amount of solute
Concentration =
amount of solution
amount of solute
Concentration =
amount of solvent
The most common for liquid solutions:
mMoles of solute
Moles
of
solute
Molarity =
= mLiters of solution
Liters of solution
n
M=
V
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Chapter 4
Solution Problem
Dissolve 25 grams of sucrose (C12H22O11) in enough
water to produce 300 mL of solution. What is the
molarity of the sugar solution?
M = n/V
F.W.(C12H22O11) = 342.3 g/mol
n (C12H22O11) = 25 g/342.3 g mol-1 = 0.073 mol
= 73 mmol
V(Soln) = 300 mL
M = n/V = 73 mmol/300 mL = 0.243 mmol/mL
= 0.243 M (molar)
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Chapter 4
Molarity Equations
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• M = n/V
Molarity of solution
• n = MV
Moles of solute
• V = n/M
Volume of solution
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Chapter 4
Dilution Problem
What volume of 10 M H2SO4(aq) is required to make
200 mL of 0.15M H2SO4(aq)?
Mi = ni/Vi
Add water
Initial Molarity = 10 M
Initial Volume = ? L = Vi
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Mf = nf/Vf
Final Molarity = 0.15 M
Final Volume = 0.200 L
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Chapter 4
Dilution Problem
What volume of 10 M H2SO4(aq) is required to make
200 mL of 0.15M H2SO4(aq)?
Mi = ni/Vi
Add Solvent
Mf = nf/Vf
not Solute
ni = nf
MiVi = MfVf
Vi = MfVf / Mi
Initial Molarity = 10 M
Initial Volume = ? L = Vi
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Final Molarity = 0.15 M
Final Volume = 0.200 L
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Chapter 4
Dilution Problem
What volume of 10 M H2SO4(aq) is required to make
200 mL of 0.15M H2SO4(aq)?
Mi = ni/Vi
Add Solvent
Mf = nf/Vf
not Solute
ni = nf
MiVi = MfVf
Vi = MfVf / Mi
Vi = (0.15×200)/10 = 3 mL
(add 197 mL of water)
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Chapter 4
Electrolyte Solutions
• Electrolyte = solution conducts electricity
• Electricity is the flow of charged particles
• Solution must contain ions: cations & anions
• What kinds of solutes produce ions in water???
• Acids, Bases, Salts
H3PO4(aq) = H+(aq) + H2PO4-(aq)
NH3(aq) = NH4+(aq) + OH-(aq)
NaCl(aq) = Na+(aq) + Cl-(aq)
(NH4)3PO4(aq) = 3NH4+(aq) + PO43-(aq)
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Chapter 4
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Chapter 4
Aqueous Acids & Bases
•Acid produces hydronium ions H+(aq)
– acid(aq) → H+(aq) + anions(aq)
– HCl(g) + H2O(l) → HCl(aq) = H+(aq) + Cl-(aq)
•Base produces hydroxyl ions OH-(aq)
– base(aq) → cations(aq) + OH-(aq)
– NaOH(s) + H2O(l) → NaOH(aq) = Na+(aq) + OH-(aq)
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Chapter 4
Classification of Acids
• Number of "ionizable protons":
– Monoprotic (HCl, C2H4O2 = HC2H3O2)
– Diprotic (H2S, H2SO4)
– Triprotic (H3PO4), etc.
• Acid Strength
– Strong Acid ionizes completely and produces a
strong electrolyte.
– Weak Acid ionizes incompletely and produces
a weak electrolyte.
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Chapter 4
Classification of Acids
Strong Acid ionizes completely
0%
100%
Before ionization: HCl(aq) → H+(aq) + Cl-(aq)
10-15 s
100%
0%
After ionization: HCl(aq) → H+(aq) + Cl-(aq)
Weak Acid ionizes incompletely
0%
100%
Before ionization: HF(aq) → H+(aq) + F-(aq)
3%
97%
After ionization: HF(aq) → H+(aq) + F-(aq)
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Chapter 4
Classification of Acids
Strong Acid ionizes completely
Every mole of acid molecules produces at least two
moles of ions.
The solution contains lots of ions, so the solution
conducts electricity very strongly.
The acid solution is a strong electrolyte.
Weak Acid ionizes incompletely
Every mole of acid molecules produces much less
than two moles of ions.
The solution contains few ions, so the solution
conducts electricity very weakly.
The acid solution is a weak electrolyte.
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Chapter 4
Classification of Acids
• Strong Acids (Memorize):
– HCl, HBr, HI,
HNO3, H2SO4, HClO3, HClO4
• Weak Acids: all others (millions of them)
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Chapter 4
Naming of Acids
(Chapter 2, pg 65 ff.)
• Hydronium Ion + Anion → Acid
–
–
–
–
–
–
–
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H+ + Cl- → HCl
H+ + C2H3O2- → HC2H3O2
H+ + SO42- → HSO4H+ + HSO4- → H2SO4
H+ + PO43- → HPO42H+ + HPO42- → H2PO4H+ + H2PO4- → H3PO4
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Naming of Acids
Chapter 4
(Chapter 2, pg 65 ff.)
Root names, prefixes and suffixes
Root: Element: chlor- (chlorine), sulf- (sulfur), etc.
Suffix: Anion –id or –ate → Acid –ic
Anion –ite
→ Acid –ous
chloride ion → hydrochloric acid
chlorate ion → chloric acid
acetate ion → acetic acid
sulfate ion → sulfuric acid
phosphate ion → phosphoric acid
sulfite ion → sulfurous acid
nitrite ion → nitrous acid
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Chapter 4
Naming of Acids
(Chapter 2, pg 65 ff.)
• -ate/ic, -ite/ous, oxyanions and oxyacids
Suffix refers to number of oxygen atoms in the formula:
suffix -ate/-ic = more O atoms
ClO4- (hy)perchlorate HClO4
ClO3- chlorate
HClO3
suffix -ite/-ous = fewer O atoms
ClO2- chlorite
HClO2
ClO- hypochlorite
HClO
perchloric
chloric
chlorous
hypochlorous
(Br and I also, but not F)
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Chapter 4
Naming of Acids
(Chapter 2, pg 65 ff.)
• -ate, -ite, oxyanions and oxyacids
Suffix refers to number of oxygen atoms in the formula:
SO42NO3CO32-
sulfate
nitrate
carbonate
H2SO4 sulfuric
HNO3 nitric
H2CO3 carbonic
SO32NO2-
sulfite
nitrite
H2SO3 sulfurous
HNO2 nitrous
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Chapter 4
Classification of Bases
• Strong - Solution is a strong electrolyte
• Weak - Solution is a weak electrolyte
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Chapter 4
Classification of Bases
• Strong - 1A & Heavy 2A Metal Hydroxides:
– MIOH (MI = Li, Na, K, Rb, Cs);
MII(OH)2 (MII = Ca, Sr, Ba but not Be or Mg)
• Weak - all others:
– Metal hydroxides; e.g., Mg(OH)2 , Fe(OH)3, Al(OH)3
– Amines (N-containing compounds, e.g., alkaloids):
• NH3(aq) + H2O(l) → NH4+(aq) + OH-(aq) [= NH4OH(aq)]
• C9H13N + H2O → C9H13NH+(aq) + OH-(aq)
(amphetamine)
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Chapter 4
Salts
Acid/Base Neutralization Reaction
acid(aq) → H+(aq) + anion(aq)
base(aq) → cation(aq) + OH-(aq)
}
}
acid + base → H+ + OH- + cation + anion
H2O
salt
acid + base → water + salt
HCl + NaOH → H2O + NaCl
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Chapter 4
Solubility of Salts
All salts are soluble to some extent in water
but each salt has a solubility limit.
A Saturated Solution contains the maximum
amount of dissolved salt.
357 g NaCl dissolves in one liter of water.
NaCl is very soluble (strong electrolyte).
0.1 g MgCO3 dissolves in one liter of water.
MgCO3 is only slightly soluble (weak electrolyte).
"insoluble"
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Soluble Ionic Compounds
Compounds with
Chapter 4
Important Exceptions
S1 cations:
NH4+, Li+, Na+,
K+, Rb+, Cs+
None
NO3-
None
ClO4-
None
C2H3O2-
None
Cl-
Compounds of Ag+, Hg22+, and Pb2+
Br-
Compounds of Ag+, Hg22+, and Pb2+
I-
Compounds of Ag+, Hg22+, and Pb2+
SO42Compounds of Sr2+, Ba2+, Hg22+ and Pb2+
Rule of Thumb: soluble salts contain anions of strong acids and/or cations of strong bases
Insoluble Ionic Compounds
Compounds with
Important Exceptions
S2-
S1 cations, Ca2+, Sr2+ and Ba2+
CO32-
S1 cations
PO43-
S1 cations
OH-
S1 cations, Ca2+, Sr2+ and Ba2+
Rule of Thumb: insoluble salts contain anions of weak acids and cations of weak bases
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Chapter 4
Solubility of Salts
General “Rules of Thumb”
• "Soluble" salts - have:
– cations of 1A & Heavy 2A metals, NH4+,
and/or anions of strong acids.
• "Insoluble" salts - have:
– cations of any other metal
and anions of weak acids.
• Complete Solubility Rules, pg. 127
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Chapter 4
Precipitation Reaction
A precipitate forms when solutions of two
"soluble" salts combine to produce a
supersaturated solution of an "insoluble" salt
AgNO3(aq) = Ag+(aq) + NO3-(aq)
NaCl(aq) = Na+(aq) + Cl-(aq)
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
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soln 1
soln 2
Add
soln 1+2
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Chapter 4
Precipitation Reaction
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
soln 1+2
The four ions can form four possible salts:
NaNO3 (aq)
Ag+
NO3-
AgNO3 (aq)
Na+
Cl-
NaCl (aq)
AgCl (s)
precipitation
reaction
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
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Chapter 4
Ionic Reactions in Solution - I
Net ionic Reaction
Write the Complete ionic reaction:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Show dissolved ions explicitely
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
→
AgCl(s) + Na+(aq) + NO3-(aq)
Identify spectator ions
Write the Net ionic reaction:
Ag+(aq) + Cl-(aq) → AgCl(s)
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Chapter 4
Ionic Reactions in Solution - II
Net ionic Reaction
• Write the Complete ionic reaction:
– H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
• Show dissolved ions explicitely
– 2H+(aq) + SO42-(aq) + 2Na+(aq) + 2OH-(aq)
→
2Na+(aq) + SO42-(aq) + 2H2O(l)
• Identify spectator ions (Na+ and SO42-)
• Write the Net ionic reaction:
2H+(aq) + 2OH-(aq) → 2H2O(l)
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Chapter 4
Ionic Reactions in Solution - III
Net ionic Reaction
• Write the Complete ionic reaction:
– 2H3PO4(aq) + 3Ca(OH)2(aq) → 2Ca3(PO4)2(s) + 6H2O(l)
• Show dissolved ions explicitely
– 6H+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6OH-(aq)
→
2Ca3(PO4)2(s) + 6H2O(l)
• There are no spectator ions! The complete ionic
reaction is the net ionic reaction.
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Chapter 4
REDOX Reactions
REDOX = Reduction-Oxidation Reaction
REDOX = Electron Transfer Reaction
Half
Rxs
Ca(s) → Ca2+(aq) + 2e-
{ 2e + 2H (aq) → H (g)
-
+
2
oxidation = loss
reduction = gain
Ca(s) + 2e- + 2H+(aq) → Ca2+(aq) + 2e- + H2(g)
Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g)
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REDOX Terminology
Electron Transfer Process
This reactant:
Molecule or
ion
This reactant:
e-
Molecule or
ion
• loses electrons
• is oxidized
• is the reducing agent
• gains electrons
• is reduced
• is the oxidizing agent
The reducing agent is always a reactant in the redox equation
The oxidizing agent is always a reactant in the redox equation
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Chapter 4
REDOX Balance
All Reactions must balance by Mass
1 Ca
1 Ca
2H
2H
Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g)
0 + 2(+1) = +2
1(+2) + 0 = +2
and by Charge
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Chapter 4
Recognizing REDOX
Every atom in every reactant and
product has an Oxidation Number.
A Reduction-Oxidation reaction is
one in which the Oxidation Numbers
of some atoms change from the
reactant side to the product side.
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Chapter 4
Oxidation Number Rules
1. Atoms in elements: ON = 0
H2(g), S8(s), Br2(l), C60(s), Fe(s), Hg(l)
2. Atom in monatomic ion: ON = charge
Cl- (-1), Fe3+ (+3), Hg2+ (+2)
3. Fluorine in all compounds: ON = -1
4. Oxygen in most compounds: ON = -2
5. Hydrogen: combined with ...
a) non-metals (molecules): ON = +1 [H2O]
b) metals (salts): ON = -1 [NaH, CaH2]
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Chapter 4
Oxidation Number Rules
6. The sum of the ON's of all atoms in a:
a) neutral molecule = 0
b) polyatomic ion = the charge
7. ON Limits: main Group Elements (1A - 8A):
a) Maximum ON = Group Number (GN)
b) Minimum ON = GN - 8
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Chapter 4
Oxidation Number Example 1
SO42ON(O) = y = -2
ON(S) = x = +6
Assume all four oxygen atoms have the same ON
Rule: ON(O) = -2 (probably)
Rule: Sum of ON's = -2
x + 4(-2) = -2
x = +6
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Check Limits
-2 ≤ ON(S) ≤ +6
-2 ≤ ON(O) ≤ +6
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Chapter 4
Oxidation Number Example 2
C3H6O
ON(C) = x = -4/3 ON(H) = y = +1 ON(O) = z = -2
Rule: ON(O) = -2
Rule: ON(H) = +1
Rule: Sum of ON's = 0
3(x) + 6(+1) + (-2) = 0
x = - 4/ 3
(no rule about integers)
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Check Limits
-4 ≤ ON(C) ≤ +4
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Chapter 4
Oxidation Number Example 3
NaO = (Na+)2(O22-)
ON(Na) = x = +1 ON(O) = y = -1
Rule: Sum of ON's = 0
If ON(O) = -2, then (x) + (-2) = 0
=> x = +2
But Max ON(Na) = +1
-2  ON(O)  +6
If x = +1, then (+1) + y = 0
y = -1
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ON(O) = -2
Oxide
ON(O) = -1
Peroxide
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Chapter 4
REDOX Activity
Activity Series (Table 4.5, pg. 143)
Metal
Oxidation Rx
Lithium
Li --> Li+ + e-
•••
active
Metals above will
reduce hydrogen ions
•••
Hydrogen H2 --> 2H+ + 2e•••
•••
Gold
Au --> Au3+ + 3e-
Metals below will not
reduce hydrogen ions
passive
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Chapter 4
REDOX Activity
2Li(s) + 2H+(aq) --> H2(g) + 2Li+ (aq)
Metal
Oxidation Rx
Lithium
Li --> Li+ + e-
•••
•••
Hydrogen H2 --> 2H+ + 2e•••
•••
Gold
Au --> Au3+ + 3e-
Au(s) + H+(aq) --> N.R.
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Chapter 4
REDOX Activity
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Metal
Oxidation Rx
Lithium
Li --> Li+ + e-
•••
•••
Hydrogen H2 --> 2H+ + 2e•••
•••
Gold
Au --> Au3+ + 3e-
43
oxidizing power
Higher
Metals
REDUCE
lower
Ions
reducing power
Activity Series (Table 4.5, pg. 143)
Lower
Ions
OXIDIZE
higher
Metals
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Chapter 4
REDOX Activity
2Li(s) + 2H+(aq) --> H2(g) + 2Li+ (aq)
Metal
Oxidation Rx
Lithium
Li --> Li+ + e-
•••
•••
Hydrogen H2 --> 2H+ + 2e•••
•••
Gold
Au --> Au3+ + 3e-
Metals
above ions
react
Metals
below ions
don't react
Au(s) + H+(aq) --> N.R.
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Chapter 4
Solution Stoichiometry
Titrations*
Acid/Base Reactions
Precipitation Reactions
REDOX Reactions
* When two solutions are added together quantitatively
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Chapter 4
All Reaction Tables
Most Important Rows:
The three "MOL" rows
aA + bB → cC + dD
I
C
-ax
-bx
+cx
+dx
F
The other rows differ for
weight-weight problems
and solution problems
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Chapter 4
Mass Reaction Tables
In mass reaction problems, the initial data and final
answers are generally masses of reagents.
aA +
im
mA
MM
MMA
I
InA=mA/MMA
C
-ax
F
InA-ax
fm fmA=FnA×MMA
Initial Masses
Formula Weights
Initial Moles
Change in Moles
Final Moles
Final Masses
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Chapter 4
Solution Reaction Tables
For solution reactions (titrations), the data are
Molarities and Volumes of reactants and products.
aA(aq) +
Initial Molarities
iM
iV
I
C
F
fV
fM
Initial Volumes
Initial Moles
Change in Moles
Final Moles
Final Volume
Final Molarities
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MA
VA
inA=MA×VA
-ax
inA-ax
fVA=S iV
fMA=fnA/fV
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Chapter 4
Titration Problem
Add 25 mL of 0.075 M H3PO4 solution to 15 mL of 0.105 M
NaOH solution. What are the final molarities of all solutes?
H3PO4(aq) + 3 NaOH(aq)
IM
IV
I
C
F
FV
FM
0.075
25
1.875
-x
1.875-x
40
fgA
0.105
15
1.575
-3x
1.575-3x
40
→
Na3PO4(aq)
0
0
0
+x
x
40
fgB
fgC
+ 3 H2O(l)
fgD
If LR = acid:
Fn(A) = 1.875-x = 0
x = 1.875
If LR = base
Fn(B) = 1.575-3x = 0
x = 1.575/3 = 0.525
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n=MV
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Chapter 4
Titration Problem
Add 25 mL of 0.075 M H3PO4 solution to 15 mL of 0.105 M
NaOH solution. What are the final molarities of all solutes?
H3PO4(aq) + 3 NaOH(aq)
IM
IV
I
C
F
FV
FM
0.075
25
1.875
-x
1.35
40
0.034
fgA
0.105
15
1.575
-3x
0
40
0
→
Na3PO4(aq)
0
0
0
+x
0.525
40
0.013
fgB
fgC
+ 3 H2O(l)
fgD
If LR = acid:
Fn(A) = 1.875-x = 0
x = 1.875
If LR = base
Fn(B) = 1.575-3x = 0
x = 1.575/3 = 0.525
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M=n/V
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Chapter 4
End Point Titration
End Point (Equivalence Point) Titration
aA(aq) + bB(aq) →
Fn(A) = Fn(B) = 0
products
MA
MB
VA
VB
MA×VA MB×VB
-ax
-bx
0
0
VA+VB VA+VB
0
0
fgC
fgD
Fn(A) = fgA
MA×VA - fgB
ax = 0 and Fn(B)
= MB×VB
- bx = 0
x = MA×VA/a
End Point
x=x
x = MB×VB/b
MA×VA = MB×VB
Both reactants are limiting! Both
a
b
reach stoichiometric amounts.
IM
IV
I
C
F
FV
FM
Equivalence Point
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Chapter 4
End Point Titration
End Point (Equivalence Point) Titration
aA(aq) + bB(aq) →
products
Another way to derive this equation is this:
at the end point, the actual moles of A (MA×VA) and moles of
B (MB×VB) are in the same ratio as the stoichiometric
coefficients:
MA×VA = a
b
MB×VB
End Point
MA×VA
a
= MB×VB
b
Equivalence Point
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Chapter 4
End Point Titration
What volume of 0.075 M H3PO4 solution will exactly
neutralize 15 mL of 0.105 M Ca(OH)2 solution?
For an acid-base reaction:
“neutralize” = “end point” = “equivalence point”
2H3PO4(aq) + 3Ca(OH)2(aq) → Ca3(PO4)2(s) +6H2O(l)
MA×VA
a
0.075×VA = 0.105×0.015
2
3
VA = 0.014 L = 14 mL
= MB×VB
b
The word “neutralize” is only used to describe an
acid/base end point titration, but any reaction can have
an end point (when both reactants react completely and
there is no excess of either).
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