Chapter 4
description
Transcript of Chapter 4
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Chapter 4
Types of Chemical Reactions and Solution
Stoichiometry
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Parts of solutions
Solution – homogeneous mixtureSolute – part that dissolvesSolvent – causes the dissolvingSoluble – can be dissolvedMiscible – liquids dissolve in each
other
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Saturation of SolutionsA solution that contains the maximum
amount of solute dissolved under existing conditions is saturated.
A solution that contains less solute than a saturated solution under existing conditions is unsaturated.
A solution that holds more solute than a saturated solution under the same conditions is supersaturated.
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Aqueous SolutionsDissolved in waterWater is a polar moleculeThe oxygen atoms have a partial
negative chargeThe hydrogen atoms have a partial
positive chargeThe angle is 105o
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HydrationThe process of breaking apart ions
of a salt.The “+” end of water attracts the
anionThe “-” end of water attracts the
cation
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SolubilityThe ability to dissolve in a given
amount of waterUsually g/100mLVaries greatlyDepends upon ion attractionWill dissolve nonionic substances if
they have polar bonds
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ElectrolytesElectrical current through a
substanceIons that are dissolved can moveSolutions are classified three ways
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Types of SolutionsStrong electrolytes –Completely ionized
when dissolved in water many ions – conduct wellWeak electrolytes – partially fall apart
into ions few ions- conduct electricity slightlyNon-electrolytes – don’t fall apart no ions – don’t conduct electricity
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Types of solutions continuedAcids – form H+ ion when dissolvedStrong acids fall apart completely H2SO4 HNO3 HCl HBr HI HClO4
Weak acids – do not dissociate completely
Bases – form OH- when dissolvedStrong bases – KOH NaOH
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DissociationAcids
HCl H+ (aq) + Cl-(aq) HNO3 H+(aq) + NO3
- (aq) H2SO4 H+ (aq) + HSO4
- (aq)
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Strong bases NaOH Na+ (aq) + OH- (aq)
KOH K+ (aq) + OH-(aq)
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NonelectrolytesDissolve in water but do not
produce any ionsExample is ethanol (C2H5OH) molecules disperse in the water
but doesn’t conduct electricity
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Composition of SolutionsConcentration
1. Molarity (M) – moles of solute per volume of solution in liters
2. M = moles of solute liters of solution
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Preparation of Molar SolutionsCalculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.
1.56 g HCl
1
1 mole HCl 36.5 g HCl = 0.0427 mole HCl
26.8 mL 1
1L 1000 mL
= .0268 L
M = 0.0427 mole HCl .0268 L
= 1.60 M HCl
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Concentrations of Ions Give the concentration of each type of ion in 0.50 M
Co(NO3)2Co(NO3)2 (s)
Co+2 (aq) + 2NO3- (aq)
Co+2 1 x 0.50 M = 0.50 M Co+2
NO3-
2 x 0.50 M = 1.0 M NO3
-
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Calculate the number of moles of Cl- ions in 1.75 L of 1.0 x 10-3 M ZnCl2.
ZnCl2 Zn+2 (aq) + 2Cl- (aq)
2 x 1.0 x 10-3 = 2.0 x 10-3 M Cl-
1.75 L 2.0 x 10-3 mole Cl- L
= 3.5 x 10-3 mole Cl-
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Standard solution – a solution where the concentration is accurately known
How much solid K2Cr2O7must by weighted out to make a solution? A chemist needs 1.00 L of an aqueous 0.200 M K2Cr2O7 solution.
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Dilution1. Water is added to achieve a particular M2. Moles of solute after = moles of solute
before3. M1V1=M2V2
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution?
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Types of Solution ReactionsPrecipitation Reactions
1. A solid forms from two solutions
2. Precipitate – the insoluble solid
KNO3 (aq) + BaCl2 (aq)
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Three Types of Equations Used to Describe Reactions in a Solution
The formula equation gives the overall reaction.
The complete ionic equation represents as ions all the reactants and products that are strong electrolytes.
The net ionic equation includes only those ions that undergo a change.
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Write the formula equation, complete ionic equation, and the net ionic equation for
KCl (aq) + AgNO3 (aq)
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Stoichiometry of Precipitation Reactions
Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.
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When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed.
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Acid-Base ReactionsAcid is a proton donor (H+)Base is a proton acceptor usually OH— accepts a H+
H+(aq) + OH− (aq)
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Cations are surrounded and bound by water molecules protons are also solvated by water molecules
Two ways to show this
1. H+ (aq)2. H3O+ (aq) – hydronium ion
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Types of acid donorsMonoprotic HCl, HNO3 donates ____ H+
Diprotic H2SO4 donates ____ H+
Triprotic H3PO4 donates ____ H+
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Neutralization ReactionsWhat volume of a 0.100 M HCl solution is
needed to neutralize 25.0 mL of 0.350 M NaOH? HCl (aq) + NaOH (aq) NaCl(aq) + H2O (l)
H+(aq) + OH−(aq) H2O (l)
.0250 L 0.350 mole OH− =
L NaOH
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Acid-Base TitrationsVolumetric analysis
Process of determining the amount of a substance by titration
Titration Process of delivering solutions to one anotherEquivalence point (stoichiometric point) Point at which the titration has occurred Endpoint
Point at which the indicator changes color
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What volume of 0.812 M HCl, in milliliters, is required to titrate 1.33 g of NaOH to the
equivalence point?
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Oxidation-Reduction ReactionsElectrons are transferredSpontaneous redox rxns can transfer
energy Electrons (electricity) Heat
LEO the lion says GER
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Lose Electrons = OxidationGain Electrons = Reduction
Redox Reactions Examples:0 0 +1 --1
2Na + Cl2 2NaClEach sodium atom loses one electron oxidation
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Each chlorine atom gains one electron: 0 -1 Cl + e__ Cl
reduction
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Rules for Assigning Oxidation Numbers Rules 1 & 2
1. The oxidation number of any uncombined element is zero
2. The oxidation number of a monatomic ion equals its charge
3. The oxidation number of oxygen in compounds is -2
4. The oxidation number of hydrogen in compounds is +1
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5. The sum of the oxidation numbers in the formula of a compound is 0.
6. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its charge
ex. NO3—
SO42—
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Reducing Agents and Oxidizing Agents
The substance reduced is the oxidizing agentThe substance oxidized is the reducing agent
0 +1 Na Na + e --
Sodium is oxidized – it is the reducing agent0 -- 1
Cl + e-- Cl
Chlorine is reduced – it is the oxidizing agent
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Oxidation-Number Changes in reactions
Can you identify what is being oxidized and what is being reduced?
2AgNO3 (aq) + Cu (s) Cu(NO3)2(aq) + 2Ag (s)
The oxidation number of Ag decreases from +1 to 0 (reduction), copper’s oxidation number increase from 0 to +2 (oxidation)
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Balancing Redox EquationsLook at the reaction between solid copper and
silver ions in aqueous solution:Cu (s) + Ag+ (aq) Ag (s) + Cu+2 (aq)
Cu + Ag+ Ag + Cu+2
1 e--
gained
0 +1 0 +2
2 e- lost
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Ultimately must have equal number of electrons gained and lost
Cu (s) + 2 Ag+ (aq) 2 Ag (s) + Cu+2 (aq)
Try balancing:
H+ (aq) + Cl_ (aq) + Sn (s) + NO3- SnCl6 (aq) + NO2 (g)+ H2O
(l)