Electrons arent shared evenly (oxygen is more electronegative)
Electrons spend more time close to O than to H.
Slide 3
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.
Slide 4
When water surrounds an ionic crystal, the H end attracts the
anion and the O end attracts the cation. This process is called
hydration.
Slide 5
Hydration causes salts (ionic compounds) to dissolve. H 2 O
also dissolves polar covalent substances such as C 2 H 5 OH. H 2 O
doesnt dissolve nonpolar covalent substances because there is not
enough attraction between the water and the nonpolar molecule.
Slide 6
Hydration
Slide 7
Show the association of the ions with some water molecules when
1 formula unit of KCl dissolves in excess water. K+ Cl -
Slide 8
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.
Slide 9
Svante Arrhenius determined that the extent to which a solution
can conduct an electrical current depends directly on the number of
ions present.
Slide 10
Solubility- is usually shown as g/given volume solvent or
moles/given volume solution
Slide 11
Strong electrolytes 1.soluble salts 2.strong acids completely
ionize HCl(aq), HNO 3 (aq), H 2 SO 4 (aq)
Slide 12
Ex. Show how HCl dissociates when dissolved in water. HCl H + +
Cl -
Slide 13
Acid (Arrhenius) a substance that produces H + ions in water
solution
Weak electrolytes -only ionize slightly (weak acids and bases)
HC 2 H 3 O 2 H + + C 2 H 3 O 2 99% 1%
Slide 16
Slide 17
Ammonia (NH 3 ) -weak base NH 3 + H 2 O NH 4 + + OH
Slide 18
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
Slide 19
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 Na 2 SO 4 1 mol Na 2 SO 4 = 0.165 mol Na 2 SO 4 142.06g Na 2
SO 4 0.165 mol = 1.32 M 0.125 L
Slide 20
Ex. How many grams of Na 2 SO 4 are required to make 350 mL of
0.50 M Na 2 SO 4 ? 0.350L 0.50 mol Na 2 SO 4 142.06g Na 2 SO 4 =
24.9g 1 L 1 mol Na 2 SO 4
Slide 21
Ex. What volume of 1.000 M KNO 3 must be diluted with water to
prepare 500.0 mL of 0.250 M KNO 3 ? Dilution problem (M 1 V 1 = M 2
V 2 ) (1.000M)(V 1 ) = (0.250M)(500.0mL) V 1 = 125 mL Remember,
this formula can only be used for dilution! Never use it for a
chemical reaction (stoichiometry)!
Slide 22
Read procedure for using volumetric flasks and types of pipets.
We will be using both in several labs this year.
Slide 23
Slide 24
Lets Review Equation Writing from Chemistry I Some reactions
fit neatly into a certain category of reaction type, some do
not.
Slide 25
DECOMPOSITION REACTIONS
Slide 26
Reaction where a compound breaks down into two or more elements
or compounds. Heat, electrolysis, or a catalyst is usually
necessary.
Slide 27
A compound may break down to produce two elements. Ex. Molten
sodium chloride is electrolyzed. 2NaCl(l) 2Na + Cl 2
Slide 28
A compound may break down to produce an element and a compound.
Ex. A solution of hydrogen peroxide is decomposed catalytically. 2H
2 O 2 2H 2 O + O 2
Slide 29
A compound may break down to produce two compounds. Ex. Solid
magnesium carbonate is heated. MgCO 3 MgO + CO 2
Slide 30
Metallic carbonates break down to yield metallic oxides and
carbon dioxide.
Slide 31
Metallic chlorates break down to yield metallic chlorides and
oxygen.
Slide 32
Hydrogen peroxide decomposes into water and oxygen.
Slide 33
Sulfurous acid decomposes into water and sulfur dioxide.
Slide 34
Carbonic acid decomposes into water and carbon dioxide.
Slide 35
Hydrated salts decompose into the salt and water. Na 2 CO 3 H 2
O Na 2 CO 3 + H 2 O
Slide 36
ADDITION REACTIONS Also known as Synthesis, Combination, or
Composition
Slide 37
Two or more elements or compounds combine to form a single
product.
Slide 38
A Group IA or IIA metal may combine with a nonmetal to make a
salt.
Slide 39
A piece of lithium metal is dropped into a container of
nitrogen gas. 6Li + N 2 2Li 3 N
Slide 40
Two nonmetals may combine to form a molecular compound. C + O 2
CO 2
Slide 41
When an element combines with a compound, you can usually sum
up all of the elements on the product side. Ex. PCl 3 + Cl 2 PCl 5
This is a trick that works because the common positive oxidation
states of P are + 3 and +5.
Slide 42
Two compounds combine to form a single product. Ex. Sulfur
dioxide gas is passed over solid calcium oxide. SO 2 + CaO CaSO
3
Slide 43
A metallic oxide plus carbon dioxide yields a metallic
carbonate. (Carbon keeps the same oxidation state)
Slide 44
A metallic oxide plus sulfur dioxide yields a metallic sulfite.
(Sulfur keeps the same oxidation state)
Slide 45
A metallic oxide plus water yields a metallic hydroxide. A
nonmetallic oxide plus water yields an acid.
Slide 46
Double Replacement (metathesis)
Slide 47
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.
Slide 48
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!
Slide 49
Simple Rules for Solubility 1.Most nitrate (NO 3 ) salts are
soluble. 2.Most alkali (group 1A) salts and NH 4 + are soluble.
3.Most Cl , Br , and I salts are soluble (NOT Ag +, Pb 2+, Hg 2 2+
) 4.Most sulfate salts are soluble (NOT BaSO 4, PbSO 4, HgSO 4,
CaSO 4 ) 5.Most OH salts are only slightly soluble (NaOH, KOH are
soluble, Ba(OH) 2, Ca(OH) 2 are marginally soluble) 6.Most S 2 , CO
3 2 , CrO 4 2 , PO 4 3 salts are only slightly soluble.
Slide 50
SOLUBILITY SONG To the tune of My Favorite Things from The
Sound of Music My Favorite Things Nitrates and Group One and
Ammonium, These are all soluble, a rule of thumb. Then you have
chlorides, theyre 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!
Slide 51
Ex. Solutions of silver nitrate and lithium bromide are mixed.
AgNO 3 (aq) + LiBr(aq) AgBr(s) + LiNO 3 (aq)
Slide 52
Formation of a gas: Gases may form directly in a double
replacement reaction or can form from the decomposition of a
product such as H 2 CO 3 or H 2 SO 3. H 2 CO 3 H 2 O and CO 2 H 2
SO 3 H 2 O and SO 2 NH 4 OH NH 3 and H 2 O
Slide 53
Ex. Excess hydrochloric acid solution is added to a solution of
potassium sulfite. 2HCl(aq) + K 2 SO 3 (aq) H 2 SO 3 H 2 O(l) + SO
2 (g) + 2KCl(aq) Remember that sulfurous acid decomposes into water
and sulfur dioxide!
Slide 54
Ex. A solution of sodium hydroxide is added to a solution of
ammonium chloride. Remember that ammonium hydroxide does not exist.
NaOH(aq) + NH 4 Cl(aq) NaCl(aq) + NH 4 OH NH 3 (g) + H 2 O(l)
Slide 55
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".
Slide 56
Ex. Dilute solutions of lithium hydroxide and hydrobromic acid
are mixed. LiOH(aq) + HBr(aq) LiBr(aq) +H 2 O(l)
Slide 57
Ex. Gaseous hydrofluoric acid reacts with solid silicon
dioxide. 4HF(g) + SiO 2 (s) SiF 4 (g) + 2H 2 O(l) This reaction
occurs when glass is etched.
Slide 58
Single Replacement
Slide 59
Reaction where one element displaces another in a compound. One
element is oxidized and another is reduced. A + BC B + AC
Slide 60
Active nonmetals replace less active nonmetals from their
compounds in aqueous solution. Each halogen will displace less
electronegative (heavier) halogens from their binary salts.
Slide 61
Activity Series of Nonmetals Most Active F 2 Cl 2 Br 2 Least
Active I 2
Slide 62
Ex. Chlorine gas is bubbled into a solution of potassium
iodide. Cl 2 (g) + 2KI(aq) 2KCl(aq) + I 2 (s)
Slide 63
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.
Slide 64
Group I,II,+III Transition Metals Hydrogen Jewelry Metals
Slide 65
Ex. Magnesium turnings are added to a solution of iron(III)
chloride. 3Mg(s) + 2FeCl 3 (aq) 2Fe(s)+3MgCl 2 (aq)
Slide 66
Ex. Sodium is added to water. 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H
2 (g) Alkali metal demo
Slide 67
COMBUSTION REACTIONS
Slide 68
-Elements or compounds combine with oxygen to produce the
oxides of each element. The oxide of H is H 2 O, oxide of S is
usually SO 2, oxide of C is CO 2, etc.
Slide 69
Hydrocarbons or alcohols combine with oxygen to form carbon
dioxide and water.
Slide 70
Nonmetallic hydrides combine with oxygen to form oxides and
water.
Slide 71
Nonmetallic sulfides combine with oxygen to form oxides and
sulfur dioxide.
Slide 72
Ex. Carbon disulfide vapor is burned in excess oxygen. CS 2 +
3O 2 CO 2 + 2SO 2
Slide 73
Ex. Ethanol (C 2 H 5 OH) is burned completely in air. C 2 H 5
OH + 3O 2 2CO 2 + 3H 2 O
Slide 74
Diethyl ether (C 2 H 5 OC 2 H 5 ) is burned in air. C 2 H 5 OC
2 H 5 + 6O 2 4CO 2 + 5H 2 O
Slide 75
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
Slide 76
Ex. Separate Ag +, Ba 2+, Fe 3+ 1.Add Cl to remove Ag + as
AgCl. 2.Add SO 4 2- to remove Ba 2+ as BaSO 4. 3.Add OH or S 2- to
remove Fe 3+ as Fe(OH) 3 or Fe 2 S 3.
Slide 77
Ex. Separate Pb 2+, Ba 2+, Ni 2+ 1.Add Cl to remove Pb 2+ as
PbCl 2. 2.Add SO 4 2 to remove Ba 2+ as BaSO 4. 3.Add OH or S 2 to
remove Ni 2+ as Ni(OH) 2 or NiS.
Slide 78
Quantitative analysis- determines how much of a component is
present. Gravimetric analysis- quantitative procedure where a ppt
containing the substance is formed, filtered, dried &
weighed.
Slide 79
Ex. The zinc in a 1.2000g sample of foot powder was
precipitated as ZnNH 4 PO 4. Strong heating of the ppt yielded
0.4089 g of Zn 2 P 2 O 7. Calculate the mass percent of zinc in the
sample of the foot powder. 0.4089gZn 2 P 2 O 7 1 mol Zn 2 P 2 O 7 2
mol Zn 65.37g = 304.7 g 1 mol Zn 2 P 2 O 7 1 mol Zn 0.1754g Zn 100
= 14.62% Zn 1.200g sample
Slide 80
Ex. A mixture contains only NaCl and Fe(NO 3 ) 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(NO 3 ) 3 c. the mass percent of Fe(NO 3
) 3 in the sample 0.128g Fe(OH) 3 1 mol Fe(OH) 3 1 mol Fe 55.85g
Fe= 0.0669g Fe 106.9g Fe(OH) 3 1 mol Fe(OH) 3 1 mol Fe 0.0669g Fe 1
mol Fe 1 mol Fe(NO 3 ) 3 241.9g Fe(NO 3 ) 3 = 0.290g Fe(NO 3 ) 3
55.85g Fe 1 mol Fe 1 mol Fe(NO 3 ) 3 0.290g 100 = 63.6% Fe(NO 3 ) 3
0.456g
When a strong acid reacts with a strong base the net ionic rxn
is: H + (aq) + OH (aq) H 2 O(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.) HC 2 H 3 O 2 (aq)
+ OH (aq) H 2 O(l) + C 2 H 3 O 2 (aq)
Slide 84
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.
Slide 85
Volumetric Analysis titration- process in which a solution of
known concentration (standard solution) is added to analyze another
solution (analyte).
Slide 86
Titrations are most often used for acids and bases, but can be
used for other types of reactions, also.
Slide 87
titrant- solution of known concentration (usually in buret)
equivalence point or stoichiometric point- point where just enough
titrant has been added to react with the substance being
analyzed
Slide 88
Indicator - chemical which changes color at or near the
equivalence point End point- point at which the indicator changes
color
Slide 89
Ex. 54.6 mL of 0.100 M HClO 4 solution is required to
neutralize 25.0 mL of an NaOH solution of unknown molarity. What is
the concentration of the NaOH solution? HClO 4 + NaOH H 2 O + NaClO
4 0.0546 L HClO 4 0.100 mol HClO 4 1 mol NaOH = 1 L HClO 4 1 mol
HClO 4 0.00546 mol NaOH 0.00546 mol NaOH = 0.218 M NaOH 0.025L
Slide 90
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
Slide 91
1. NaCl(aq) + AgNO 3 (aq) NaNO 3 (aq) + AgCl(s) 2. Na + (aq) +
Cl (aq) + Ag + (aq) + NO 3 (aq) Na + (aq) + NO 3 (aq) + AgCl(s) 3.
Cl (aq) + Ag + (aq) AgCl(s)
Slide 92
Oxidation-Reduction Reactions Redox Rxns - reactions in which
one or more electrons are transferred.
Slide 93
Electronegativity - attraction for shared electrons most
electronegative F>O>N=Cl elements Phone Call These are most
likely to have negative oxidation numbers.
Slide 94
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
Slide 95
Review oxidation state rules on page 167. N 2 O PBr 3 HPO 3 2-
P 4 O 6 NH 2 - +1-2 +3-1 +1+3-2 +3 -2 -3 +1
Slide 96
Noninteger states are rare, but possible. Fe 3 O 4 8/3 -2 O =
4(-2) = -8 Fe = 8/3 = 2 2/3 or Fe 2+, Fe 3+, Fe 3+
Slide 97
Oxidation - loss of electrons - increase in oxidation number
Reduction - gain of electrons - decrease in oxidation number
Slide 98
OIL RIG Oxidation Is Loss (of e ), Reduction Is Gain (of e
)
Slide 99
LEO the lion goes GER Lose Electrons = Oxidation, Gain
Electrons = Reduction
Slide 100
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.
Slide 101
2KI + F 2 2KF + I 2 +1-1 0 +1-1 0 oxidized I reduced F OA F2F2
RA KI
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 H 2 O. 4.Balance H by adding H +.
Slide 104
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.
Slide 105
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.
Slide 106
Practice: Sn 2+ + Cr 2 O 7 2 Sn 4+ + Cr 3+ (acidic solution) Sn
2+ Sn 4+ Cr 2 O 7 2 Cr 3+ Sn 2+ Sn 4+ Cr 2 O 7 2 2Cr 3+ Sn 2+ Sn 4+
Cr 2 O 7 2 2Cr 3+ +7H 2 O Sn 2+ Sn 4+ Cr 2 O 7 2 + 14H + 2Cr 3+ +7H
2 O Sn 2+ Sn 4+ +2e Cr 2 O 7 2 + 14H + +6e 2Cr 3+ +7H 2 O 3(Sn 2+
Sn 4+ +2e ) Cr 2 O 7 2 + 14H + +6e 2Cr 3+ +7H 2 O 3Sn 2+ + Cr 2 O 7
2 + 14H + +6e - 3Sn 4+ +6e + 2Cr 3+ +7H 2 O 3Sn 2+ + Cr 2 O 7 2- +
14H + +6e - 3Sn 4+ +6e - + 2Cr 3+ +7H 2 O 3Sn 2+ + Cr 2 O 7 2 + 14H
+ 3Sn 4+ + 2Cr 3+ +7H 2 O
Slide 107
MnO 4 2- + I - MnO 2 + I 2 (basic solution) MnO 4 2- MnO 2 I -
I 2 MnO 4 2- MnO 2 2I - I 2 MnO 4 2- MnO 2 + 2H 2 O 2I - I 2 MnO 4
2- +4H + MnO 2 + 2H 2 O 2I - I 2 MnO 4 2- +4H + + 2e - MnO 2 + 2H 2
O 2I - I 2 +2e - MnO 4 2- +4H + + 2e - +2I - MnO 2 + 2H 2 O +I 2
+2e - MnO 4 2- +4H + +2I - MnO 2 + 2H 2 O +I 2 MnO 4 2- +4H + + 4OH
- +2I - MnO 2 + 2H 2 O +I 2 + 4OH - MnO 4 2- + 4H 2 O +2I - MnO 2 +
2H 2 O +I 2 + 4OH - MnO 4 2- + 2H 2 O +2I - MnO 2 + I 2 + 4OH
-
Slide 108
OXIDATION-REDUCTION TITRATIONS Most common oxidizing agents:
KMnO 4 & K 2 Cr 2 O 7
Slide 109
Potassium permanganate used to disinfect ponds and fish in
Egypt.
Slide 110
MnO 4 - in acidic solution: MnO 4 - + 8H + + 5e Mn 2+ + 4H 2 O
Purple colorless When you titrate with MnO 4 , the solution is
colorless until you use up all of the reducing agent (substance
being oxidized).
Slide 111
In calculations, work redox titrations like acid-base
titrations. You must have a balanced reaction to know the mole
ratio.