1 Types of Chemical Reactions and Solution Stoichiometry Chapter 4.
1 Chapter 4 Types of Reactions & Solution Stoichiometry.
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Transcript of 1 Chapter 4 Types of Reactions & Solution Stoichiometry.
1
Chapter 4Types of Reactions &
Solution Stoichiometry
2
Unit essential Question:
How do chemicals react with one another in aqueous solutions?
3
Lesson essential questions (4.1-4.4):
1) How do water molecules interact with chemicals?
2) How is the concentration of a solution measured?
4
Water, the Common Solvent
Section 4.1
5
Aqueous solutions Dissolved in water. Good solvent- polar molecules. Hydration: ions in salts break
apart due to attraction to polar water molecules.
Example:
NH4NO3 (s) NH4+ (aq) + NO3
- (aq)
δ+
δ+
δ-
6
Hydration
H HOH
H OH
HO
H HO
HHO
HH
O
HH
OH
H
O
HH
O
7
Solubility Amount of substance that will dissolve in
a given amount of water. If they do dissolve, ions are separated,
and can move around. Water can also dissolve non-ionic
compounds if they have polar bonds.
8
“Like dissolves like” Polar substances generally dissolve
other polar and ionic substances
– Alcohol is slightly polar and dissolves (mixes) in water
Nonpolar substances dissolve other nonpolar substances
– Fat will not dissolve in water
9
The Nature of Aqueous Solutions: Strong & Weak
Electrolytes
Section 4.2
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Parts of Solutions Solution- homogeneous mixture. Solute- what gets dissolved. Solvent- what does the dissolving. Soluble- Able to be dissolved. Miscible- liquids dissolve in each other.
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Electrolytes Electrolytes- ionic compounds in solution that
conduct electricity. Strong electrolytes- completely dissociate (fall
apart into ions).– Many ions = conduct electricity well.
Weak electrolytes- partially dissociate into ions.– Few ions = conduct electricity slightly.
Non-electrolytes- don’t dissociate at all.– No ions = don’t conduct electricity.
12
Acid/Base Electrolytes Arrhenius acid- forms H+ ions when dissolved. Strong acids dissociate completely.
– Ex: H2SO4 HNO3 HCl HBr HI Weak acids do not dissociate completely.
– Ex: HC2H3O2
Arrhenius base - forms OH- ions when dissolved.
Strong bases also dissociate completely.– Ex: KOH NaOH (Groups 1 & 2
hydroxides)
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Sections 1&2 Homework
Pg. 170-171 #1,9,18,19
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Warm-Up
HNO3 is a strong acid. Write the chemical equation for a solution of HNO3. Will it conduct electricity?
15
Composition of SolutionsSection 4.3
16
Measuring Composition of Solutions To do stoichiometry:
– Need to know chemicals – Need to know amounts (concentrations)
Concentration- how much is dissolved. Molarity = Moles of solute
Liters of solution abbreviated M (molar) 1 M = 1mol solute / 1 liter solution
17
Molarity Calculations Can solve for:
– Amount or mass of solid to dissolve
– Moles of solute
– Volume of solution
– Standard solution• Solution whose concentration is accurately
known.
18
Examples Calculate the molarity of a solution
prepared by dissolving 11.5g of solid NaOH in water to make 1.50L of solution. (pg. 134)
Give the concentration of each ion in 0.50 M Co(NO3)2. (pg. 135)
#27 pg. 172
19
Dilutions Stock solution – a concentrated solution Dilution – number of moles of solute
stays the same, just adding more water
– M1V1 = M2V2
Example: #30 (a) pg. 172
mol1 x V1 = mol2 x V2L1 L2
20
Section 3 Homework
Pg. 171-172 #21-23,28,31
21
Lesson essential questions (4.5-4.7):
1) How do we identify and work with precipitation reactions?
22
Precipitation ReactionsSection 4.5
23
24
Precipitation Reactions Solid forms when two solutions of ionic
compounds are mixed. Precipitate (ppt) To help you remember: ‘If you’re not a
part of the solution, your part of the precipitate!’
25
Precipitation reactions NaOH(aq) + FeCl3(aq) NaCl(aq)
+ Fe(OH)3(s)
is really: Na+(aq)+OH-(aq) + Fe+3 + Cl-(aq)
Na+ (aq) + Cl- (aq) + Fe(OH)3(s)
So all that really happens is
OH- (aq) + Fe+3 (aq) Fe(OH)3 (s)
Also a double displacement reactionnet ionic equation!
26
Precipitation reaction Can predict products, but can only be certain
by experimenting. The anion and cation switch partners. Only occurs if a product is insoluble!
– Otherwise all the ions stay in solution- nothing has happened (spectators)
Memorize solubility rules! Pg. 144
27
Solubility Rules All nitrates, Na+, K+, NH4
+ are soluble.
You must know this for the AP exam!
Additional solubility rules on pg. 144.
28
Describing Reactions in Solutions
Section 4.6
29
Three Types of Equations 1. Formula Equation- write formulas, not
ions. K2CrO4(aq) + Ba(NO3)2(aq) 2. Complete Ionic equation- show dissolved
electrolytes as the ions. 2K+ + CrO4
-2 + Ba+2 + 2 NO3-
BaCrO4(s) + 2K+ + 2 NO3-
Spectator ions are those that don’t react- appear as ions on both sides.
30
Three Type of Equations 3. Net Ionic equation- show only ions
that react, not spectator ions
Ba+2 + CrO4-2 BaCrO4(s)
If all species in a reaction are aqueous (soluble), write NR!
31
Sections 5&6 Homework
Pg. 172-173 #36,42,44
32
AP Practice QuestionHow many moles of Na2SO4 must be added to 500 milliliters of water to create a solution that has a 2- molar concentration of the Na+ ion? (Assume the volume of the solution does not change.)
Think about what you need to answer this! Need to find moles Na+. Then find moles Na2SO4
•0.5 moles•1 mole•2 moles•5 moles
33
Stoichiometry of Precipitation Reactions
Section 4.7
34
Stoichiometry of Precipitation Steps for reference: pg.148
– Similar to other stoichiometry problems we’ve done!
Sample problem: What volume of 0.15M KCl is needed to precipitate out all of the lead ions from 100.mL of 0.20M Pb(NO3)2?
270mL KCl needed
35
Section 7 Homework
Pg. 173 #47,48,50,54
36
Acid-Base ReactionsSection 4.8
37
Lesson essential question (4.8):
1) How do we classify acids and bases?
2) What happens when acids and bases are mixed together?
38
Acid-Base Reactions For our purposes an acid is a proton donor, H+
(BrØnsted-Lowry theory). A base is a proton acceptor, usually OH-
acid + base salt + water H+ + OH- H2O Practice: Write the net ionic equation for the
acid/base rxn. below:
HNO3(aq) + NaOH(aq) ? Note: H2CO3 always breaks down into CO2 &
H2O
39
Acid-Base Reactions Follow same steps as precipitation reactions
for stoichiometry problems.– See p.149-150
Practice: What volume (in mL) of 0.100M HCl will react completely with 25.00mL of 0.200 M NaOH?– (1) Write net ionic equation– (2) Find moles you’re starting with– (3) Find moles needed– (4) Find volume needed
40
Acid-Base Reactions Also called neutralization reactions. Use titrations to determine concentrations. Titrant: solution of known concentration Analyte: solution of unknown concentration Equivalence Point: when enough titrant has been
added to exactly react with the analyte (neutralization is complete).– Stoichiometric amounts come from balanced equation!– Tells us how many moles of the titrant fully reacted
with the analyte- then can solve for moles of analyte!
41
Titration Solution of known concentration (titrant),
is added to the unknown (analyte), until the equivalence point is reached.
How do we know when the equivalence point has been reached?
– Add indicator to analyte at the beginning
42
Titration Where the indicator changes color is the
endpoint.– Ex: phenolphthalein used often
• Pink in base, colorless in acid As close as we can get to the equivalence
point; still assume complete neutralization.– The solution will not turn pink until one
drop after the equivalence point (when the solution is more basic).
Can also use titration for non acid/base substances to find amounts/concentrations.
43
AP Practice Question Which of the following best represents the
balanced net ionic equation for the reaction of lead(II) carbonate & concentrated hydrochloric acid? (All lead compounds are insoluble.)
a. Pb2CO3 + 2H+ + Cl- Pb2Cl + CO2 + H2O
b. PbCO3 + 2H+ + 2Cl- PbCl2 + CO2 + H2O
c. PbCO3 + 2H+ Pb+2 + CO2 + H2O
d. PbCO3 + 2Cl- PbCl2 + CO3-2
44
AP Practice Question
The conductivity of a solution of Ba(OH)2 is monitored as the solution is titrated with 0.10 M H2SO4. The original volume of the Ba(OH)2
solution is 25.0 mL. A precipitate of BaSO4 is formed during the titration. The data collected from the experiment is plotted in the graph above.
45
Question Continued
1) As the first 30.0 mL of 0.10 M H2SO4 are added to the Ba(OH)2 solution, two types of chemical reactions occur simultaneously. Write the balanced net-ionic equations for (i) the neutralization reaction and (ii) the precipitation reaction.
(i) Equation for neutralization reaction:
(ii) Equation for precipitation reaction:
OH- (aq) + 2H+ (aq) H2O (l)
Ba+2 (aq) + SO4-2 (aq) BaSO4
(s)
46
Question Continued2) The conductivity of the Ba(OH)2 solution decreases as the volume of added 0.10 M H2SO4 changes from 0.0 mL to 30.0 mL.
(i)Identify the chemical species that enable the solution to conduct electricity as the first 30.0 mL of 0.10 M H2SO4 are added.
(ii) On the basis of the equations you wrote in question 1, explain why the conductivity decreases.
OH- (aq) & Ba+2 (aq) (Can’t be anything from H2SO4 because the ions immediately react.)
[Ba+2] in sltn. decrease as they precipitate out, and [OH-] in sltn. decrease as they react to form H2O. Note: be specific in your answers!! Reference all species and reactions!
47
Question Continued
3) Using the information in the graph, calculate the molarity of the original Ba(OH)2 solution.
Think about what information can be determined from this point!
At equivalence point: complete neutralization
0.12M Ba(OH)2
48
Section 8 Homework
Homework: pg.173-174 #56, 58, 60, 64, 66
49
Oxidation – Reduction Reactions
Section 4.9
50
Lesson essential questions (4.9-4.10):
1) How can we identify redox reactions?
2) How do we assign oxidation states?
3) Why is balancing different for redox reactions?
51
52
Redox Reactions Ionic compounds are formed through
the transfer of electrons. An oxidation-reduction reaction involves
the transfer of electrons.
– One element gains, one loses Non-ionic compounds can also undergo
redox reactions.
53
Oxidation States = ‘charge’ A way of keeping track of the electrons.
Not necessarily true of what is in nature, but it works.
Need to memorize rules for assigning (pg.156):
The oxidation state of elements in their standard states is zero.
Oxidation state for monatomic ions are the same as their charge.
54
Oxidation states Oxygen is assigned an oxidation state of -2
in its covalent compounds except in peroxide (-1).
In compounds with nonmetals hydrogen is assigned the oxidation state +1.
In its compounds fluorine is always –1. The sum of the oxidation states must be
zero in compounds or equal the charge of the ion.
55
Practicing Oxidation StatesDetermine the oxidation states in the
following:
1)Cl22)SO4
-2
3)CaBr2
4)C6H12O6
Cl: 0
S: +6 O: -2Ca: +2 Br: -1C: 0 H: +1 O: -2
56
Section 9 Homework
Pg. 174 #67(c-e),68(a-c),72
57
Balancing Redox ReactionsSection 4.10
58
Oxidation-Reduction e- transferred, so the oxidation states
change. Oxidation is the loss of electrons.
– More positive oxidation state. Reduction is the gain of electrons.
– More negative oxidation state. OIL RIG LEO (the lion goes) GER
59
Agents Oxidizing agent- substance that gets reduced
(causes oxidation in another species).– Gains electrons.– More negative oxidation state.
Reducing agent- substance that gets oxidized (causes reduction in another species).– Loses electrons.– More positive oxidation state.
60
Identify the… Oxidizing agent Reducing agent Substance oxidized Substance reduced
#1: 2Na + Cl2 2NaCl
#2: CH4 + 2O2 CO2 + 2H2O
oxidizing agent, substance reduced
oxidizing agent, substance reduced
reducing agent, substance oxidized
reducing agent, substance oxidized
61
Half-Reactions All redox reactions can be thought of as
happening in two halves. One produces electrons - oxidation half. The other requires electrons - reduction
half. Ex: Fe (s) + CuSO4 (aq) Cu (s) + FeSO4 (aq) Net Ionic: Fe (s) + Cu+2 (aq) Cu (s) + Fe+2 (aq) Oxidation: Fe (s) Fe+2 (aq) + 2e- Reduction: Cu+2 (aq) + 2e- Cu (s)
62
Balancing Redox Equations Redox reactions may involve an acid or
base as a reactant. The number of electrons produced
must be the same as those required. 8 step procedure for acidic solution, 10
step procedure for basic solution. Called the half reaction method.
– Balance each half reaction, then combine for total balanced reaction
63
Balancing in Acidic Solution Write separate half reactions.
For each half reaction balance all species except H and O.
Balance O by adding H2O to one side.
Balance H by adding H+ to one side. Balance charge by adding e- to the
more positive side.
64
Balancing in Acidic Solution Multiply equations by a number to make
electrons equal. Add equations together and cancel
identical species. Reduce coefficients to smallest whole numbers.
Check that charges and elements are balanced.
65
Balancing in Acidic Solution Ex: Balance the following equation:
H+ (aq) + Cr2O7-2 (aq) + C2H5OH (l)
Cr+3 (aq) + CO2 (g) + H2O (l)
Reduction: 6e- + 14H+ + Cr2O7-2 2Cr+3 + 7H2O
Oxidation: C2H5OH + 3H2O 2CO2 + 12H+ + 12e-
Final: 16H+ + 2Cr2O7-2 + C2H5OH 4Cr+3 + 11H2O
+ 2CO2
*Note: there should NOT be any e- in the final balanced equation! If so, not balanced!
66
Basic Solution Do everything you would with acid, but
add one more step. Add enough OH- to both sides to
neutralize the H+. Any H+ and OH- on the same side form
water. Cancel out any H2O’s on both sides.
Simplify coefficients, if necessary.
67
Balancing in Basic Solution Assume previous example in acidic solution was
actually in a basic solution.
Had: 16H+ + 2Cr2O7-2 + C2H5OH 4Cr+3 + 11H2O +
2CO2
For any H+ ions, add same number of OH- ions to both sides. This forms water with H+. Cancel out waters on both sides.
Now: 16H2O + 2Cr2O7-2 + C2H5OH 4Cr+3 + 11H2O +
2CO2 + 16OH-
5 H2O
16H+, so add 16OH-
68
Practice Balancing Redox Rxns. Pg. 174 #74(b)
Pg. 174 #75(b)
Answer: 6Cl- + Cr2O7 + 14H+ 3Cl2 + 2Cr+3 + 7H2O
Answer: 2OH- + Cl2 OCl- + Cl- + H2O
69
Side Note: Redox Titrations Same as titrations discussed before, just
looking at redox reactions instead of acid/base reactions.
Permanganate ion is used often because it is its own indicator: MnO4
- is purple, Mn+2 is colorless. When reaction solution remains clear, MnO4
- is gone. Chromate ion is also useful, but color change,
orangish yellow to green, is harder to detect.
70
Section 10 Homework
Pg. 174-175 #73-76 ONLY letter a for each