Transcript aq - Humble ISD

AP CHEMISTRY
CHAPTER 4
SOLUTION STOICHIOMETRY
Electrons aren’t shared evenly (oxygen is
more electronegative)
Electrons spend more time close to O
than to H.
This uneven distribution of charge makes
water polar.
Because of this, water is a good solvent.
The positive end (H) attracts negative
ions or the negative end of another polar
molecule.
The negative end of water (O) attracts
the positive ions or the positive end of
another polar molecule.
When water surrounds an ionic
crystal, the H end attracts the anion
and the O end attracts the cation.
This process is called hydration.
Hydration causes salts (ionic
compounds) to dissolve. H2O also
dissolves polar covalent substances
such as C2H5OH. H2O doesn’t
dissolve nonpolar covalent substances
because there is not enough attraction
between the water and the nonpolar
molecule.
Hydration
Show the association of the ions
with some water molecules when
1 formula unit of KCl dissolves
in excess water.
ClK+
A solution is a homogeneous mixture.
In a solution, a solute dissolves in the
solvent.
If the solute ionizes in the solution,
electricity can be conducted and the
solute is said to be an electrolyte.
If the solute ionizes 100% or nearly
100%, it is called a strong electrolyte.
Lesser ionization occurs with weak
electrolytes.
Svante Arrhenius determined that
the extent to which a solution can
conduct an electrical current
depends directly on the number
of ions present.
Solubilityis usually shown as g/given
volume solvent or
moles/given volume solution
Strong electrolytes
1.soluble salts
2.strong acids –completely ionize
HCl(aq), HNO3(aq), H2SO4(aq)
Ex. Show how HCl dissociates
when dissolved in water.
HCl 
+
H
+
Cl
Acid (Arrhenius) – a substance
that produces H+ ions in water
solution
3.strong bases- completely ionize

-contain OH
-bitter taste and slippery feel
-NaOH, KOH
Weak electrolytes
-only ionize slightly (weak
acids and bases)
HC2H3O2  H+ + C2H3O2
99%
1%
Ammonia (NH3) -weak base
NH3 + H2O  NH4+ + OH
Molarity (M) = moles of solute
liters of solution
Molarity is the most common unit of
concentration used in Chemistry.
We may also see mM (millimolar) = moles of solute
mL of solution
Ex. Calculate the molarity of a solution
made by dissolving 23.4g of sodium
sulfate in enough water to form 125 mL
of solution.
23.4 g Na2SO4
1 mol Na2SO4 = 0.165 mol Na2SO4
142.06g Na2SO4
0.165 mol = 1.32 M
0.125 L
Ex. How many grams of Na2SO4 are
required to make 350 mL of 0.50 M
Na2SO4?
0.350L 0.50 mol Na2SO4 142.06g Na2SO4 = 24.9g
1L
1 mol Na2SO4
Ex. What volume of 1.000 M KNO3 must be
diluted with water to prepare 500.0 mL of
0.250 M KNO3?
Dilution problem (M1V1 = M2V2)
(1.000M)(V1) = (0.250M)(500.0mL)
V1 = 125 mL
Remember, this formula can only be
used for dilution! Never use it for a
chemical reaction (stoichiometry)!
Read procedure for using
volumetric flasks and types of
pipets. We will be using both in
several labs this year.
Let’s Review Equation Writing
from Chemistry I
Some reactions fit neatly into a
certain “category” of reaction type,
some do not.
DECOMPOSITION
REACTIONS
Reaction where a compound
breaks down into two or more
elements or compounds. Heat,
electrolysis, or a catalyst is
usually necessary.
A compound may break down
to produce two elements.
Ex. Molten sodium chloride is
electrolyzed.
2NaCl(l) 2Na + Cl2
A compound may break down
to produce an element and a
compound.
Ex. A solution of hydrogen
peroxide is decomposed
catalytically.
2H2O2  2H2O + O2
A compound may break down
to produce two compounds.
Ex. Solid magnesium carbonate
is heated.
MgCO3  MgO + CO2
Metallic carbonates break
down to yield metallic oxides
and carbon dioxide.
Metallic chlorates break down
to yield metallic chlorides and
oxygen.
Hydrogen peroxide decomposes
into water and oxygen.
Sulfurous acid decomposes into
water and sulfur dioxide.
Carbonic acid decomposes into
water and carbon dioxide.
Hydrated salts decompose
into the salt and water.
Na2CO3H2O  Na2CO3 + H2O
ADDITION REACTIONS
Also known as Synthesis, Combination, or Composition
Two or more elements or
compounds combine to form a
single product.
A Group IA or IIA metal may
combine with a nonmetal to
make a salt.
A piece of lithium metal is
dropped into a container of
nitrogen gas.
6Li + N2  2Li3N
Two nonmetals may combine to
form a molecular compound.
C + O2 →CO2
When an element combines
with a compound, you can
usually sum up all of the
elements on the product side.
Ex. PCl3 + Cl2  PCl5
This is a trick that works because the
common positive oxidation states of P
are + 3 and +5.
Two compounds combine to
form a single product.
Ex. Sulfur dioxide gas is passed
over solid calcium oxide.
SO2 + CaO  CaSO3
A metallic oxide plus carbon
dioxide yields a metallic
carbonate. (Carbon keeps the
same oxidation state)
A metallic oxide plus sulfur
dioxide yields a metallic sulfite.
(Sulfur keeps the same oxidation
state)
A metallic oxide plus water
yields a metallic hydroxide.
A nonmetallic oxide plus water
yields an acid.
Double Replacement
(metathesis)
Two compounds react to form two new
compounds. No changes in oxidation
numbers occur. All double replacement
reactions must have a "driving force"
that removes a pair of ions from solution.
Formation of a precipitate: A
precipitate is an insoluble substance
formed by the reaction of two aqueous
substances. Two ions bond together so
strongly that water can not pull them
apart. You must know your solubility
rules to write these net ionic equations!
Simple Rules for Solubility
Most nitrate (NO3) salts are soluble.
Most alkali (group 1A) salts and NH4+ are soluble.
Most Cl, Br, and I salts are soluble (NOT Ag+, Pb2+, Hg22+)
Most sulfate salts are soluble (NOT BaSO4, PbSO4, HgSO4,
CaSO4)
5. Most OH salts are only slightly soluble (NaOH, KOH are
soluble, Ba(OH)2, Ca(OH)2 are marginally soluble)
6. Most S2, CO32, CrO42, PO43 salts are only slightly soluble.
1.
2.
3.
4.
SOLUBILITY SONG
To the tune of “ My Favorite Things” from “The Sound of Music”
Nitrates and Group One and Ammonium,
These are all soluble, a rule of thumb.
Then you have chlorides, they’re soluble fun,
All except Silver, Lead, Mercury I.
Then you have sulfates, except for these three:
Barium, Calcium and Lead, you see.
Worry not only few left to go still.
We will do fine on this test. Yes, we will!
Then you have the--Insolubles
Hydroxide,
Sulfide and Carbonate and Phosphate,
And all of these can be dried!
Ex. Solutions of silver nitrate
and lithium bromide are mixed.
AgNO3(aq) + LiBr(aq)  AgBr(s) + LiNO3(aq)
Formation of a gas: Gases may form
directly in a double replacement reaction
or can form from the decomposition of a
product such as H2CO3 or H2SO3.
H2CO3  H2O and CO2
H2SO3  H2O and SO2
NH4OH  NH3 and H2O
Ex. Excess hydrochloric acid solution is
added to a solution of potassium sulfite.
2HCl(aq) + K2SO3(aq) 
H2SO3 H2O(l) + SO2(g) + 2KCl(aq)
Remember that sulfurous acid decomposes into water
and sulfur dioxide!
Ex. A solution of sodium hydroxide is
added to a solution of ammonium chloride.
Remember that ammonium hydroxide does not exist.
NaOH(aq) + NH4Cl(aq) 
NaCl(aq) + NH4OH NH3(g) + H2O(l)
Formation of a molecular substance:
When a molecular substance such as
water or acetic acid is formed, ions are
removed from solution and the reaction
"works".
Ex. Dilute solutions of lithium
hydroxide and hydrobromic acid are
mixed.
LiOH(aq) + HBr(aq)  LiBr(aq) +H2O(l)
Ex. Gaseous hydrofluoric acid
reacts with solid silicon dioxide.
4HF(g) + SiO2(s) 
SiF4(g) + 2H2O(l)
This reaction occurs when glass is etched.
Single Replacement
Reaction where one element
displaces another in a compound.
One element is oxidized and
another is reduced.
A + BC  B + AC
Active nonmetals replace less active
nonmetals from their compounds in
aqueous solution. Each halogen will
displace less electronegative (heavier)
halogens from their binary salts.
Ex. Chlorine gas is bubbled into a
solution of potassium iodide.
Cl2(g) + 2KI(aq) 2KCl(aq) + I2(s)
Activity Series of Nonmetals
Most Active
Least Active
F2
Cl2
Br2
I2
Active metals replace less active
metals or hydrogen from their
compounds in aqueous solution.
Use an activity series or a reduction
potential table to determine activity. The
more easily oxidized metal replaces the
less easily oxidized metal.
ACTIVITY SERIES OF METALS
Element
Lithium
Potassium
Group
Barium
Decreasing
Calcium
I,II,+III
Activity
Sodium
Magnesium
Aluminum
Zinc
Iron
Cadmium
Transition
Nickel
Metals
Tin
Lead
Hydrogen ( a nonmetal)
Hydrogen
Copper
Mercury
Jewelry Metals
Silver
Gold
Platinum
*Metals from Li to Na will replace H from water and
acids; metals from Mg to Pb will replace H from
acids only.
Ex. Magnesium turnings are added
to a solution of iron(III) chloride.
3Mg(s) + 2FeCl3(aq) 2Fe(s)+3MgCl2(aq)
Ex. Sodium is added to water.
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
Alkali metal demo
COMBUSTION REACTIONS
-Elements or compounds
combine with oxygen to produce
the oxides of each element.
The oxide of H is H2O, oxide of
S is usually SO2, oxide of C is
CO2, etc.
Hydrocarbons or alcohols
combine with oxygen to form
carbon dioxide and water.
Ethane(C2H6) is burned in air.
2C2H6 + 7O2  4CO2 + 6H2O
Ex. Ethanol (C2H5OH) is burned
completely in air.
C2H5OH + 3O2  2CO2 + 3H2O
Time to Practice!!!
Writing net-ionic equations
1. molecular equation
-overall reaction stoichiometry
2. complete ionic equation
-all strong electrolytes are represented as
ions
3. net ionic equation
-spectator ions are not included
1. NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl(s)
2. Na+ (aq) + Cl(aq) + Ag+(aq) + NO3(aq) 
Na+(aq) + NO3(aq) + AgCl(s)
3. Cl(aq) + Ag+(aq)  AgCl(s)
Ex. HCl + Ba(OH)2 
1. 2HCl + Ba(OH)2  2H2O + BaCl2
2. 2H+ + 2Cl− + Ba2+ + 2OH−
2H2O + Ba2+ + 2Cl−
3. 2H+ + 2OH−  2H2O
H+ + OH−  H2O
Time to Practice!!!
Selective precipitation- process
by which ions are caused to ppt
one by one in sequence to
separate mixtures of ions.
Qualitative analysis- process of
separating and identifying ions
Ex. Separate
+
Ag ,
2+
Ba ,
3+
Fe
(use your solubility chart)
1. Add Cl to remove Ag+ as AgCl.
2. Add SO42- to remove Ba2+ as BaSO4.
3. Add OH or S2- to remove Fe3+ as
Fe(OH)3 or Fe2S3.
Ex. Separate
2+
Pb ,
2+
Ba ,
2+
Ni
1. Add Cl to remove Pb2+ as PbCl2.
2. Add SO42 to remove Ba2+ as BaSO4.
3. Add OH or S2 to remove Ni2+ as
Ni(OH)2 or NiS.
Quantitative analysisdetermines how much of a
component is present.
Gravimetric analysisquantitative procedure where a
ppt containing the substance is
formed, filtered, dried &
weighed.
Ex. The zinc in a 1.2000g sample of foot
powder was precipitated as ZnNH4PO4.
Strong heating of the ppt yielded 0.4089 g of
Zn2P2O7. Calculate the mass percent of zinc
in the sample of the foot powder.
0.4089gZn2P2O7 1 mol Zn2P2O7 2 mol Zn
65.37g
=
304.7 g
1 mol Zn2P2O7 1 mol Zn
0.1754g Zn
1.200g sample
× 100 = 14.62% Zn
Ex. A mixture contains only NaCl and Fe(NO3)3. A 0.456g sample of
the mixture is dissolved in water, and an excess of NaOH is added,
producing a precipitate of Fe(OH)3. The ppt is filtered, dried, &
weighed. Its mass is 0.128g.
Calculate: a. the mass of the iron
b. the mass of Fe(NO3)3
c. the mass percent of Fe(NO3)3 in the sample
0.128g Fe(OH)3 1 mol Fe(OH)3 1 mol Fe
55.85g Fe=
106.9g Fe(OH)3 1 mol Fe(OH)3 1 mol Fe
0.0669g Fe
0.0669g Fe 1 mol Fe 1 mol Fe(NO3)3 241.9g Fe(NO3)3= 0.290g Fe(NO3)3
55.85g Fe 1 mol Fe
1 mol Fe(NO3)3
0.290g × 100 = 63.6% Fe(NO3)3
0.456g
Acid-Base Reactions
Bronsted-Lowry
acid-base definitions:
acid- proton donor
base- proton acceptor
When a strong acid reacts with a strong base the
net ionic rxn is:
H+(aq) + OH(aq)  H2O(l)
When a strong acid reacts with a weak base or a
weak acid reacts with a strong base, the reaction
is complete (the weak substance ionizes
completely.)
HC2H3O2(aq) + OH(aq)  H2O(l) + C2H3O2(aq)
neutralization reaction
- acid-base rxn
When just enough base is
added to react exactly with the
acid in a solution, the acid is said
to be neutralized.
Volumetric Analysis
titration- process in which a
solution of known concentration
(standard solution) is added to
analyze another solution
(analyte).
Titrations are most often used for acids and bases, but can be
used for other types of reactions, also.
titrant- solution of known
concentration (usually in buret)
equivalence point or
stoichiometric pointpoint where just enough titrant
has been added to react with the
substance being analyzed
Indicator - chemical which
changes color at or near the
equivalence point
End point- point at which the
indicator changes color
Ex. 54.6 mL of 0.100 M HClO4 solution is
required to neutralize 25.0 mL of an NaOH
solution of unknown molarity. What is the
concentration of the NaOH solution?
HClO4 + NaOH  H2O + NaClO4
0.0546 L HClO4 0.100 mol HClO4 1 mol NaOH =
1 L HClO4
1 mol HClO4
0.00546 mol NaOH
0.00546 mol NaOH = 0.218 M NaOH
0.025L
Oxidation-Reduction Reactions
Redox Rxns - reactions in which
one or more electrons are
transferred.
Electronegativity - attraction for
shared electrons
most
electronegative
elements
F>O>N=Cl
“Phone Call”
These are most likely to have
negative oxidation numbers.
Rules for Assigning Oxidation States
1. Oxidation state of an atom in an element = 0
2. Oxidation state of monatomic element = charge
3. Oxygen = 2 in covalent compounds (except in
peroxides where it = 1)
4. H = +1 in covalent compounds
5. Fluorine = 1 in compounds
6. Sum of oxidation states = 0 in compounds
Sum of oxidation states = charge of the ion
Review oxidation state rules on page
167.
+1-2
+3-1
N2O
PBr3
+3 -2
P4O6
+1+3-2
HPO32-3 +1
NH2-
Noninteger states are rare, but
possible.
8/3 -2
Fe3O4
O = 4(-2) = -8
Fe = 8/3 = 2 2/3 or Fe2+, Fe3+, Fe3+
Oxidation - loss of electrons
- increase in oxidation number
Reduction - gain of electrons
- decrease in oxidation number
OIL RIG
Oxidation Is Loss (of e),
Reduction Is Gain (of e)
LEO the lion goes GER
Lose Electrons = Oxidation,
Gain Electrons = Reduction
Oxidizing agent
- electron acceptor
- substance that is reduced
Reducing agent - electron donor
- substance that is oxidized
The terms oxidizing agent and reducing agent are not
tested on the AP test.
+1-1
0
+1-1
0
2KI + F2  2KF + I2
oxidized
I
reduced
F
OA
F2
RA
KI
+4 -2
+2 -2
0
2PbO2  2PbO + O2
oxidized:
O
reduced:
Pb
OA:
PbO2
RA:
PbO2
Balancing redox reactions by the
half-reaction method
1.Write skeleton half-reactions.
2.Balance all elements other than O
and H.
3.Balance O by adding H2O.
4.Balance H by adding H+.
5.Balance charge by adding e- to the
more positive side.
6.Make the # of e lost = # of e
gained by multiplying each half-rxn
by a factor.
7.Add half-reactions together.
8.Cancel out anything that is the
same on both sides.
9.If the reaction occurs in basic
solution, add an equal number of
hydroxide ions to both sides to cancel
out the hydrogen ions. Make water on
the side with the hydrogen ions. Cancel
water if necessary.
10.Check to see that charge and mass
are both balanced.
CH3OH + Cr2O72  HCO2H + Cr3+
(acidic solution)
CH3OH  HCO2H
CH3OH  HCO2H
Cr2O72  Cr3+
Cr2O72   2Cr3+
CH3OH + H2O HCO2H
CH3OH + H2O HCO2H + 4H+
Cr2O72   2Cr3+ +7H2O
Cr2O72  + 14H+  2Cr3+ +7H2O
CH3OH + H2O HCO2H + 4H+ + 4e Cr2O72  + 14H+ +6e  2Cr3+ +7H2O
3(CH3OH + H2O HCO2H + 4H+ + 4e ) 2(Cr2O72  + 14H+ +6e   2Cr3+ +7H2O)
3CH3OH + 3H2O + 2Cr2O72  + 28H+ +12e- 
3HCO2H + 12H+ + 12e  + 4Cr3+ +14H2O
3CH3OH + 3H2O + 2Cr2O72  + 16 28H+ +12e- 
3HCO2H + 12H+ + 12e  + 4Cr3+ + 11 14H2O
3CH OH + 2Cr O
2+
16H+  3HCO H + 4Cr3+ +11H O
MnO4 + I  MnO2 + I2 (basic solution)
MnO4  MnO2
MnO4  MnO2
MnO4  MnO2 + 2H2O
MnO4 +4H+  MnO2 + 2H2O
I  I2
2I  I2
2I  I2
2I  I2
MnO4  +4H+ + 3e  MnO2 + 2H2O
2I  I2+2e
2(MnO4  +4H+ + 3e  MnO2 + 2H2O)
3(2I  I2+2e)
2MnO4  +8H+ + 6e +6I  2MnO2 + 4H2O +3I2+6e
2MnO4  +8H+ + 6e +6I  2MnO2 + 4H2O +3I2+6e
2MnO4 +8H+ +6I   2MnO2 + 4H2O +3I2
2MnO4 +8H+ + 8OH +6I   2MnO2 + 4H2O +3I2 + 8OH
2MnO4 + 8H2O + 6I  2MnO2 + 4H2O + 3I2 + 8OH 
2MnO4 + 4 8H2O + 6I  2MnO2 + 4H2O + 3I2 + 8OH
2MnO4 + 4H2O + 6I  2MnO2 + 3I2 + 8OH
OXIDATION-REDUCTION
TITRATIONS
Most common oxidizing agents:
KMnO4 & K2Cr2O7
Potassium
permanganate used
to disinfect ponds
and fish in Egypt.
MnO4- in acidic solution:
MnO4- + 8H+ + 5e   Mn2+ + 4H2O
Purple
colorless
When you titrate with MnO4 , the
solution is colorless until you use up all
of the reducing agent (substance being
oxidized).
In calculations, work redox
titrations like acid-base titrations.
You must have a balanced
reaction to know the mole ratio.