Transcript Chapter 4

Types of Chemical Reactions
And Solution Stoichiometry
Chapter 4
Section 4.1: Water, The Common Solvent
Hydration of an ionic compound will occur
when the partial positive end of a water
becomes attracted to the anions in the
compound; likewise for the partial negative
center of the water and the cations.
Solubility depends on the strength of the
intermolecular attractions between the ions
and water, as well as the intramolecular
attractions of the cations and anions of the
compound.
NH4NO3(s)  NH4+(aq) + NO3-(aq)
What can dissolve in H2O?
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Soluble
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Alcohols
ex: C2H5OH
Sugars
ex: C6H12O6
Ionic compounds
ex: NaCl, KOH, LiBr
Insoluble
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Fats
ex: bacon grease
Oils
ex: cooking oil
Non-Polar Substances
ex: turpentine
Because of intermolecular forces: the OH group on
the sugars and alcohols is particularly attractive to a
water molecule.
Generally speaking: “Like Dissolves Like”
Section 4.2: Strong and Weak Electrolytes
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Solute + Solvent = Solution
Strong electrolytes conduct electricity
Weak electrolytes barely conduct electricity
Conductivity depends upon ionization
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Strong Electrolytes
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Soluble salts
Strong acids
Strong bases
All of these dissociate
completely in water.
HCl  H+ + Cl-
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Weak Electrolytes
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Weak acids
Weak bases
All of these partially
dissociate in water
HC2H3O2
H+ + C2H3O2
NaOH  Na+ + OHNon-Electrolytes are completely molecular
substances in water (not even a little dissociation);
Non polar substances.
Section 4.3: Composition of Solutions
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Concentration is measured in molarity,
molality, and many others.
Concentration DOES NOT directly express
the number of ions present in a solution.
MgCl2  Mg2+ + 2ClM= moles solute
liters solution
1.0 M
1.0 M
2.0 M
Sample Problems
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Calculate the number of moles of Cl- ions in 1.75 L of 1
x 10-3 M ZnCl2.
2.0 x103 mol
1.75 L x
 3.5x103 molCl 
1L
A chemist needs 1.0 L of 0.20 M K2Cr2O7 solution. How
much solid K2Cr2O7 must be weighed out to make this
solution?
1.0 L x 0.201 Lmol  0.20 m olK 2Cr2O7
0.20 m olK 2Cr2O7 x
294.20 gK 2Cr2O7
1mol
 58.8 g K 2Cr2O7
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Standard Solution: a solution whose
concentration is accurately known.
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Example: 0.1022 M HCl; 1.003 M NaOH
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ANSWER
NOW
Creating dilutions
Chemical analysis of a compound
Theoretical Calculations
What would you do to prepare a
standard solution? In your
answer, include specific pieces
of glassware, techniques, or
equipment you should use.
Dilutions
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Dilution is the process used to make the solution
less concentrated.
moles before dilution = moles after dilution
Because M =mol/L,
V1(M1) = V2(M2)
Lab Technique: Use a pipet to deliver the correct
amount of original solution to a volumetric flask.
Add some water, swirl. Fill to line, invert.
DO
NOW
You have a large quantity of 1.5 M
NaOH solution available. Dilute this
to 100.0mL of a 0.05 M solution.
Submit your calculations and store
your final product for use in our first
lab.
Section 4.4: Types of Chemical Reactions
There are more than just these few types,
but in this chapter we will cover…
 Precipitation
 Acid-base
 Oxidation-Reduction
Section 4.5: Precipitation Reactions
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Precipitation Reactions
(double displacement)
Forms a solid precipitate from aqueous reactants.
 Color of precipitate can help in identification
 Solubility rules help BUNCHES
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MORE…
Solubility RULES
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All compounds containing alkali metal cations and the ammonium
ion are soluble.
All compounds containing NO3-, ClO4-, ClO3-, and C2H3O2- anions are
soluble.
All chlorides, bromides, and iodides are soluble except those
containing Ag+, Pb2+, and Hg2+.
All sulfates are soluble except those containing Hg2+, Pb2+, Sr2+,
Ca2+, and Ba2+.
All hydroxides are only slightly soluble, except those containing an
alkali metal, Ca2+, Ba2+,and Sr2+. NaOH and KOH are the most
soluble hydroxides.
All compounds containing PO43-, S2-, CO32-, and SO32- are only
slightly soluble except for those containing alkali metals or the
ammonium ion.
Practice Predicting
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Potassium nitrate and barium chloride
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Sodium sulfate and lead (II) nitrate
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Potassium hydroxide and iron (III) nitrate
ALL REACTIONS
SHOULD BE
WRITTEN IN
NET IONIC
FORM
Section 4.7: Stoichiometry of Precipitation Reactions
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Stoichiometry in a precipitation reaction is
performed just like stoichiometry for a
molecular reaction.
You need to know which ion comes from
which molecular formula.
Sample problem
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Calculate the mass of solid NaCl needed to
add to 1.5 L of 0.1 M silver nitrate solution to
precipitate all Ag+ ions in the form of AgCl.
Net Ionic Eq:
1.5 L x
0.10 M Ag 
1L

Ag+ + Cl-  AgCl
 0.150m ol Ag 
1Cl 

0.150m ol Ag x 1 Ag   0.15 m olCl  0.15 m ol NaCl
0.150m ol NaCl x
58.45 g NaCl
1mol NaCl
 8.77 g NaCl
General Format
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Write the Net Ionic Equation
Calculate the moles present
Identify the Limiting Reactant*
Use Mole Ratio(s)
Fancy-fy your answer (put in correct units)
Try Me!
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What mass of precipitate will be produced
when 50.0 mL of 0.200M aluminum nitrate
is added to 200.0 mL of 0.100 M
potassium hydroxide?
Section 4.8: Acid-Base Reactions
Acids yield
H+
Bases yield
OH -
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Bases are
proton
acceptors
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Acids
are proton
donors
Definitions of acid
and base vary.
Arrhenius and
Bronsted/Lowry are
common theories.
Acid-Base rxns are
called
NEUTRALIZATIONS
Strong Acid-Strong Base
(HCl) (NaOH)
Both dissociate completely
H+ + OH-  H2O
Na+ and Cl- are spectators.
Weak Acid - Strong Base
(HC2H3O2)
(KOH)
Acetic acid will not dissociate
KOH will completely
HC2H3O2 + OH-  H2O + C2H3O2K+ is a spectator.
Stoichiometry sample
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What volume of 0.100 M HCl is needed to
neutralize 25 mL of 0.35 M NaOH?
H+ + OH-  H2O
0.025L NaOHx
0.35 molOH 
1LNaOH
2
x
1molH 
1molOH 
 8.75x10 L HCl
mLHCl
x 1.100
mol
Titrations…define me!
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Volumetric analysis
Titration
Titrant
Analyte
Equivalence point
Indicator
Endpoint
To complete a successful titration…
1.
2.
3.
The reaction between the titrant and the analyte
should be known (you should know WHAT
substances you have)
The equivalence point should be marked
accurately (you should use the right indicator)
Volume of the titrant needed to reach the
equivalence point should be recorded accurately
(you should use a buret!)
Effective Indicator Ranges
Titration Try Me Calc 1
A 50.0 mL sample of a sodium hydroxide
solution is to be standardized. 1.3009 M
of KHP (potassium hydrogen phthalate,
KHC8H4O4) is used as the titrant. KHP has
one acidic hydrogen. 41.20 mL of the
KHP solution is used to titrate the sodium
hydroxide solution to the endpoint. What
is the resulting concentration of the
analyte?
Titration Try Me Calc 2
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How many milliliters of a 0.610 M sodium
hydroxide solution are needed to
neutralize 20.0 mL of a 0.245 M sulfuric
acid solution?
Norton Tutorial
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Go to the website
http://www.wwnorton.com/college/chemistry
/chemistry3/ch/17/chemtours.aspx
Find the tutorial on Acid/Base ionization.
Complete the tutorial question form.
Section 4.9: Redox Reactions
What is it??
-A reaction that occurs in conjunction with a
transfer of electrons.
We assign oxidation states to individual atoms in a
reaction to observe the change in electrons.
Oxidation states
are written with
the +/- sign
before the quantity.
Ion charges are
written with the
+/- sign behind
the quantity.
The Oxidation State of…
An atom in element form
A monatomic ion
Fluorine in a compound
Oxygen in a compound
Quantity of Oxid. State Examples
Zero
Equal to the charge on
the ion
Na+, Cl-
-1 , always
HF, PF3
-2, except in peroxide
where it is -1
Hydrogen in a compound
Na(s), O2(g)
+1, always
H2O, CO2, H2O2
H2O, HCl, NH3
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Oxidation= an increase in the oxidation
state
Reduction = a decrease in the oxidation
state
oxidation
2Na(s) + Cl2(g)  2NaCl(s)
reduction
Metal Atom
Metal
Ion
Other Atom
Oxidized Substance:
•Loss of electrons
•Oxidation state increases
•Gets Smaller
•Called the Reducing Agent
Other Ion
Reduced Substance:
The metal is oxidized
and the other
substance is reduced.
•Gain of electrons
•Oxidation state decreases
•Gets Bigger
•Called the Oxidizing Agent
Section 4.10: Balancing Redox
1.
2.
3.
4.
5.
6.
7.
Write the ½ reactions
Balance the non-H and non-O atoms
Balance O by adding H2O where needed
Balance H by adding H+ where needed
Balance charge using eMultiply by coefficients until both e- are equal
for each ½ reaction
Add the ½ reactions together (cancel stuff)
Redox Sample Problem
Balance:
MnO4- + Fe2+  Fe3+ + Mn2+
Fe2  Fe3
MnO4  Mn2

4
2+ +2
2
MnO  Mn  4 H 2O
8H   MnO4  Mn2  4 H 2O
1- +7
8+


2+ +2

4
0
2
3+ +3
Check
Charges!
5e  8H  MnO  Mn  4H 2O
x5!
2
3
Fe  Fe  e
5Fe2  MnO4  8H   5Fe3  Mn2  4H 2O
 x5!
Redox Try Me Problem
As2O3(s) + NO3-  H3AsO4 + NO(g)
1.
2.
3.
4.
Repeat steps from old
procedure
Cancel out H+ by adding OHions
Re-write H+ and OH- as water
Add ½ reactions together
(and cancel stuff)
Redox Try Me Problem 2
Balance, in base:
Ag(s) + CN- + O2  Ag(CN)2-