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Chapter 4 Types of Chemical Reactions and Solution Stoichiometry

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Chapter 4 Types of Chemical Reactions and Solution Stoichiometry Water is the dissolving medium of the common solvent: Some properties Water is bent or v-shaped – PowerPoint PPT presentation

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Title: Chapter 4 Types of Chemical Reactions and Solution Stoichiometry


1
Chapter 4Types of Chemical Reactions and
Solution Stoichiometry
  • Water is the dissolving medium of the common
    solvent Some properties
  • Water is bent or v-shaped
  • The OH bonds are covalent
  • Water is a polar molecule
  • Hydration occurs when salts dissolve in water

2
The Water Molecule is Polar
3
Polar Water Molecules Interact with the Positive
and Negative Ions of a Salt
4
  • A solute (substance being dissolved)
  • Dissolves in water or other solvent
  • Changes phase if different from the solvent
  • Present in lesser amount (if the same phase as
    the solvent)
  • A solvent
  • Retains its phase (if different from the solute)
  • Present in greater amount (if the same phase as
    the solute)
  • Rule for predicting solubility
  • Like dissolves Like

5
The Nature of Aqueous Solutions Strong and Weak
Electrolytes
  • Strong electrolytes Conduct current very
    efficiently. Strong electrolytes are completely
    ionized when they are dissolved in water (soluble
    salts, strong acids, strong bases).
  • Weak electrolytes Conduct only a small current.
    Weak electrolytes are substances that exhibit a
    small degree of ionization in water (weak acids,
    weak bases).
  • Non electrolytes Permit no current to flow. Non
    electrolytes are substances that dissolve in
    water but do not produce any ions. (pure water,
    sugar solution).

6
BaCI2 Dissolving
7
HCI (aq) is Completely Ionized
8
Acetic Acid in Water
9
The Composition of Solutions
  • To perform stoichiometric calculation in solution
  • we need-
  • The nature of the reaction
  • The amounts of chemicals present in the solutions
    (concentration)
  • The concentration of a solution can be expressed
    by Molarity (M).
  • Molarity (M) moles of solute per volume of
    solution in liters
  • M molarity (moles of solute)/(liters of
    solution)

10
  • Example If you weigh out 5.84 g of NaCl and
    dissolved it in 500. mL of water, what is the
    molarity of the solution?
  • 5.84 g NaCl x (1 mol NaCl)/(58.44 g NaCl)
    0.100 mol
  • 500. mL x 1 L/1000 mL 0.500 L
  • Molarity (0.100 moles)/(0.500 L) 0.200 M

11
Concentration of Ions
  • Example Give the concentration of each type of
    ion in the
  • following solution
  • a. 0.50 M Co(NO3)2 b. 1 M Fe(ClO4)3
    c. 1.75 M Al2(SO4)3
  • a. Solution reaction Co(NO3)2(s) H2O Co2(aq)
    2NO3-(aq)
  • Each mole of Co(NO3)2 that is dissolved, the
    solution contains 1
  • mol Co2 ions and 2 mol NO3- ions. Thus a
    solution of 0.50 M
  • Co(NO3)2 contains 0.50 M Co2 and (2 x 0.50) M or
    1.0 M NO3-.
  • b. Fe(ClO4)3(s) H2O Fe3(aq) 3ClO4-(aq)
  • 1 M Fe(ClO4)3 contains 1 M Fe3 ions and 3 M
    ClO4- ions.
  • c. Al2(SO4)3(s) H2O 2Al3(aq) 3SO42-(aq)
  • 1.75 M Al2(SO4)3 contains (2 x 1.75) M or 3.50 M
    Al3 ions and (3
  • x 1.75) M or 5.25 M SO42- ions.

12
Standard Solution
  • A standard solution is a solution whose
    concentration is accurately known.
  • Finding volume or weight
  • How many grams of K2Cr2O7 are required to prepare
    1.00 L of a 0.20 M solution?
  • M moles/L
  • Moles M x L 0.200 M x 1.00 L 0.200 mol
    K2Cr2O7.
  • 0.200 mol K2Cr2O7 x (294.20 g)/(1 mol) 58.8 g

13
  • What volume is needed to prepare a 1.20 M
    solution from 10.0 g CaCl2 ? (mm 111.0 g)
  • M moles/L gt L moles/M
  • 10.0 g CaCl2 x (1 mole CaCl2 )/111.0 g CaCl2)
    0.090 moles
  • L (0.090 moles)/(1.20 M) 0.0750 L
  • 75.0 mL

14
Preparation of a Standard Solution
15
Dilution
  • The desired molarity solutions are often prepared
    from concentrated stock solutions (routinely used
    solutions prepared in concentrated form) by
    adding water. This process is called dilution.
  • Dilution with water does not alter the numbers of
    moles of solute present.
  • Moles of solute before dilution
  • moles of solute after dilution
  • M1V1 M2V2

16
(a) A Measuring Pipet (b) A
Volumetric (transfer) Pipet
17
  • Example What volume of 16 M sulfuric acid
  • must be used to prepare 1.5 L of 0.10 M H2SO4
  • solution?
  • V1 Volume before dilution ?
  • M1 Concentration before dilution 16 M
  • V2 Volume after dilution 1.5 L
  • M2 Concentration after dilution 0.10 M
  • M1V1 M2V2
  • V1 M2V2 / M1 (0.10 M x 1.5 L)/ 16 M
  • 0.0094 L 9.4 mL

18
Types of Chemical Reactions
  • Solution Reactions are classified as follows
  • Precipitation reactions
  • Acid base reactions
  • Oxidation reduction reactions
  • Other classification
  • Formation reaction
  • Decomposition reaction
  • Single replacement
  • Double replacement

19
Precipitation Reactions
  • When two solutions are mixed, an insoluble
    substance (solid) sometimes forms and separates
    from the solution. Such a reaction is called a
    precipitation reaction, and the solid that forms
    is called a precipitate.
  • Example K2CrO4(aq) Ba(NO3)2(aq) ? Products
    (Soluble) (Soluble)
  • Contains the ions K, CrO42-, Ba2, NO3-
  • Product must contain both anions and cations
  • Most ionic products contain one type of cation
    and one type of anion

20
  • Precipitation Reactions continued
  • The possible combinations are-
  • K2CrO4, KNO3, BaCrO4, Ba(NO3)2
  • (reactant) (reactant)
  • The only real possibility for the solid yellow
  • products are- KNO3 or BaCrO4
  • Both K and NO3- ions are colorless but CrO42-
  • ion is yellow, so the yellow solid is BaCrO4.
  • K2CrO4(aq) Ba(NO3)2(aq) ? BaCrO4(s) 2KNO3(aq)

21
The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)
22
Describing Reactions in Solutions
  • Regular balanced equation or molecular equation
    (gives the overall reaction stoichiometry but not
    the actual forms of reactants and products)
  • K2CrO4(aq) Ba(NO3)2(aq) ? BaCrO4(s)
    2KNO3(aq)
  • Complete ionic equation (represents as ions all
    reactants and products that are strong
    electrolyte)
  • 2K(aq) CrO42-(aq) Ba2(aq) 2NO3-(aq)
  • ? BaCrO4(s) 2K(aq) 2NO3-(aq)
  • Net ionic equation (Ions participate in
    reaction)
  • Ba2(aq) CrO42-(aq) ? BaCrO4(s)

23
  • Spectator ions The ions that do not participate
    directly in the reaction, present in solution
    both before and after the reaction. K and NO3-
    ions.
  • 2K(aq) CrO42-(aq) Ba2(aq) 2NO3-(aq)
  • ? BaCrO4(s) 2K(aq) 2NO3-(aq)

24
Simple Rules for Solubility
  • 1. Most nitrate (NO3?) salts are soluble.
  • 2. Most alkali (group 1A) salts and NH4 are
    soluble.
  • 3. Most Cl?, Br?, and I? salts are soluble (NOT
    Ag, Pb2, Hg22)
  • 4. 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.

25
Stoichiometry of Precipitation Reactions
  • Step 1 Identify the species present in the
    combined solution, and determine what reaction
    occurs.
  • Step 2 Write the balanced net ionic equation
    for the reaction.
  • Step 3 Calculate the moles of the reactants
    (use volume and molarity).
  • Step 4 Determine which reactant is limiting.
  • Step 5 Calculate the moles of product or
    products, as required.
  • Step 6 Convert to grams or other units, as
    required.

26
  • Example 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.
  • Ag NO-3 Na Cl-
  • Ag(aq) Cl-(aq) ? AgCl(s)
  • 1.50 L x (0.100 mol Ag) 0.150 mol Ag
  • Ag and Cl- react in a 11 ratio, 0.150 mol Cl-
    ions and 0.150 mol NaCl x (58.45 g NaCl)/(1 mol
    NaCl) 8.77 g NaCl

27
Acid-Base Reactions
  • Arrhenius
  • Acid is a substance that produces H ions when
    dissolved in water
  • Base is substances that produces OH- ions
  • Bronsted-Lowry
  • An acid is a proton donor
  • A base is a proton acceptor
  • Balanced equation HCl NaOH ? NaCl H2O
  • Complete ionic equation H Cl- Na OH- ?
  • Na Cl- H2O
  • Net ionic equation H OH- ?H2O

28
Neutralization Reaction
  • Example What volume of a 0.100 M HCl solution
    is needed to neutralize 25.0 mL of 0.350 M NaOH?
  • The species available for reaction are H,
    Cl-, Na, OH-
  • NaCl is soluble, so Na, Cl- are spectator ions,
    net ionic equation, H(aq) OH-(aq) ? H2O(l)
  • Moles of OH- ions 25.0 mL x (1L)/(1000 mL) x
    (0.350 mol OH- )/(L NaOH) 8.75 x 10-3 mol OH-
  • Hand OH- ions react in a 11 ratio, 8.75 X
    10-3 mol H ions is required to neutralize OH-
    ions present.
  • V x (0.100 mol H)/L 8.75 x 10-3 mol H
  • V (8.75 x 10-3 mol H)/(0.100 mol H)/L
  • 8.75 x 10-2 L x (1000 mL)/1L 87.5 mL

29
Acid-Base Titration
  • Acid-base reaction is called a neutralization
    reaction (the acid and base are destroyed to
    leave a neutral solution).
  • Volumetric analysis is a technique for
    determining the amount of a certain substance by
    doing a titration.
  • A titration involves delivery of a measured
    volume of a solution of known concentration (the
    titrant) into a solution containing the substance
    being analyzed (the analyte).
  • Equivalence point or stoichiometric point
    enough titrant added to react exactly with the
    analyte.
  • Indicator A substance added at the beginning of
    the titration that changes color at or near the
    equivalence point.
  • End point The point where the indicator actually
    changes color.

30
Oxidation-Reduction Reactions
  • In oxidation-reduction (redox) reaction, one
    species loses electrons while another species
    gains electrons.
  • Both processed must occur simultaneously. We
    cant have a net ionic reaction that contain only
    oxidation or only reduction.
  • The number of electron lost by the oxidation
    process must equal the number gained by the
    reduction process. We cannot create or destroy
    electrons.
  • Cu ? Cu2 2e- (oxidation)
  • 2Ag 2e- ? 2Ag (reduction)
  • Cu 2Ag ? Cu2 2Ag (net ionic reaction)

31
Oxidation States
  • Oxidation states or oxidation numbers provides a
    way to keep track of electrons (accounting for
    electrons) in redox reactions.
  • Rules for Assigning Oxidation States
  • The oxidation state of an atom in an element is O
    (element in its elemental form is O).
  • The oxidation state of a monatomic ion is the
    same as its charge.
  • In its compound, fluorine is always assigned an
    oxidation state of 1.

32
  • Oxidation States continued
  • 4. Oxygen is assigned an oxidation state 2 in
    covalent compound (except in peroxides where it
    is 1)
  • 5. Hydrogen is assigned an oxidation state 1 in
    its covalent compounds with non metals.
  • 6. Sum of the oxidation states must be zero for
    an electrically neutral compound, for an ion
    equals the charge of the ion.

33
Assigning Oxidation States
  • Example Assign oxidation states to all atoms in
    the
  • following.
  • a. CO2 ? O -2 (for each oxygen) C 4
  • SF6 ? F -1 (for each fluorine) S 6
  • NO3- ? O -2 (each oxygen) N 5
  • MnO4- ? O -2 (each oxygen) Mn 7
  • HSO3- ? O -2 (each oxygen) H 1, S 4
  • H2O ? H 1 (each) O -2
  • Li3N ? Li 1 (each) N -3

34
Oxidation-Reduction
  • Oxidation is an increase in oxidation state or a
    loss of electron.
  • Reduction is a decrease in oxidation state or a
    gain of electron.
  • Oxidizing Agent is the electron acceptor
    (reduced).
  • Reducing Agent is the electron donor (oxidized)
  • PbO(s) CO(g) ? Pb(s) CO2(g)
  • 2 -2 2 -2 0 4 -2 (each)
  • Pb ? reduced, C ? oxidized, PbO ? oxidizing
    agent,
  • CO ? reducing agent

35
A Summary of an Oxidation-Reduction Process
36
Balancing Oxidation-Reduction Equations
  • Redox reactions in acidic solution
  • Step 1 Write separate equation for the oxidation
    and
  • reduction half reactions.
  • Step 2 For each half reaction-
  • Balance all the elements except hydrogen and
    oxygen
  • Balance oxygen using H2O
  • Balance hydrogen using H
  • Balance the charge using electrons
  • Step 3 If necessary, multiply one or both
    balanced half- reactions by an integer to
    equalize the number of electrons transferred
  • Step 4 Add the half reactions and cancel
    identical species
  • Step 5 Check that the elements and charges are
    balanced

37
Balancing Oxidation-Reduction Reactions
  • Example Balance the equation for the reaction
  • between permanganate and iron (II) ion in acidic
  • solution
  • MnO4-(aq) Fe2(aq) Acid Fe3(aq) Mn2(aq)
  • Half reactions MnO4- ? Mn2 (reduction)
  • Fe2 ? Fe3 (oxidation)
  • Balance half reactions MnO4- ? Mn2
  • Manganese is balanced
  • Balance oxygen by adding 4H2O to the right side
    of the equation

38
  • ..continued..
  • c. Balance hydrogen by adding 8H to the left
    side
  • 8H(aq) MnO4-(aq) ? Mn2(aq) 4H2O
  • Equalize the charge by adding electrons
  • 5e- 8H(aq) MnO4-(aq) ? Mn2(aq) 4H2O(l)
  • Fe2(aq) ? Fe3(aq) e- x 5
  • 5 Fe2(aq) ? 5 Fe3(aq) 5e-
  • Add half reactions
  • 5Fe2(aq) MnO-4(aq) 8H(aq)
  • ? 5Fe3(aq) Mn2(aq) 4H2O(l)
  • Check that elements and charges are balanced.

39
Balancing Oxidation-Reduction Reactions in Basic
Solution
  • Step 1 Use the half reaction method as
  • specified for acidic solution.
  • Step 2 Add a number of OH- ions that is equal
  • to the number of H ions (to both
    sides).
  • Step 3 Form H2O by combining H and OH-.
  • Step 4 Check that elements and charges are
    balanced.

40
  • Example
  • ClO- CrO2- ? Cl - CrO42- (basic solution)
  • 1 3 -1 6
  • First half reaction
  • ClO- 2e- ? Cl-
  • ClO- 2e- ? Cl- H2O
  • ClO- 2e- 2H ? Cl- H2O
  • Second half reaction
  • CrO2- ? CrO42-
  • CrO2- ? CrO42- 3e-
  • CrO2- 2H2O ? CrO42- 3e-
  • CrO2- 2H2O ? CrO42- 3e- 4H

41
  • Continued
  • Multiply first half reaction by 3 and second half
    reaction
  • by 2 then add two half reactions.
  • 3ClO- 6e- 6H ? 3Cl- 3H2O
  • 2CrO2- 4H2O ? 2CrO42- 6e- 8H
  • 3ClO- 2CrO2- H2O ? 3Cl- 2CrO42- 2H
  • Add OH- to both sides to remove H
  • 3ClO- 2CrO2- H2O 2OH- ? 3Cl- 2CrO42-
    2H 2OH-
  • 3ClO- 2CrO2- H2O 2OH- ? 3Cl- 2CrO42-
    2H2O
  • 3ClO- 2CrO2- 2OH- ? 3Cl- 2CrO42- H2O

42
Summary
  • Like dissolves Like
  • Strong, weak and non electrolytes
  • Molarity moles of solute/Liters of solution
  • Standard Solution concentration accurately known
  • Dilution M1V1 M2V2
  • Precipitation, acid-base and oxidation reduction
    reactions
  • Spectator ions
  • Acids produces H or proton donor
  • Bases produces OH- and proton acceptor
  • Acid-Base titration Neutralization reaction
  • Oxidation-Reduction reactions
  • Oxidation States
  • Balancing redox equations
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