Types of Chemical Reactions and Solution Stoichiometry

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Types of Chemical Reactions and Solution
Stoichiometry

• Chapter 4

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Aqueous Solutions

• An aqueous solution is one in which water is the
  dissolving medium or solvent.
• Water consists of bent H2O molecules that are
  polar. A polar molecule has positive and negative
  ends.
• A process called hydration happens to substances
  dissolved in water where negative ions are
  attracted to the positive end of water molecules
  and positive ions are surrounded by the negative
  ends.

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The Nature of Aqueous Solutions

• A solution is a homogeneous mixture composed of a
  solvent and a solute.
• The solute is the substance that is in smaller
  amount that will match the phase of the solvent.
• The solvent is usually water.

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Electrical Conductivity

• A useful property for characterizing solutions is
  electrical conductivity.
• A solution that conducts electricity is an
  electrolyte.
• A strong electrolyte ionizes completely. Examples
  are soluble salts, strong acids, and strong
  bases. A soluble salt separates into ions. A
  strong acid almost completely separates into H
  and anions. Strong bases almost completely
  separate into OH- and cations.

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Electrolytes continued

• A weak electrolyte only slightly separates into
  ions. Examples are acetic acid ( a weak acid) and
  ammonia ( a weak base).
• A nonelectrolyte does not produce ions in
  solution and will not conduct electricity.

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The Composition of Solutions

• Chemicals in solution are measured in units of
  molar concentration or molarity (M), which is
  defined as moles of solute per liter of solution.
• M molarity moles of solute
• liters of
  solution

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Stoichiometry with Molarity

• In order for a chemist to convert to moles and
  therefore be able to do stoichiometric
  calculations the following equation is often
  used
• Liters of solution x molarity moles of solute

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Molality

• Another method for describing concentration of
  solutions is molality or molal concentration.
• m moles solute
• kg solvent

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Dilution

• When water is added to decrease the molarity of a
  solution the process is called dilution.
• To calculate the volume or molarity of a new
  solution use
• M1V1 M2V2

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Reaction Types I

• Direct Combination or Synthesis-2 substances
  combine to form 1 product.
• A B ? AB
• element element ? compound
• compound element ? compound
• compound compound ? compound
• 2Na Cl2 ? 2NaCl

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Reaction Types II

• Decompostion- 1 substance decomposes into 2 or
  more substances.
• AB ? A B
• Compound ? element or compound element or
  compound
• 2H2O ? 2H2 O2
• CaCO3? CaO CO2

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Reaction Types III

• Single Replacement Reactions- an uncombined
  element replaces an element in a compound.
• A BX ? AX B
• Element compound ? element compound
• 2NaCl Br2 ? 2NaBr Cl2

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Reaction Types IV

• Double Replacement reactions the positive ions
  in 2 compounds switch places to form 2 different
  compounds.
• AX BY ? BX AY
• compound compound ? comp. comp.
• CaCl2 2NaOH ? Ca(OH)2 2NaCl

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Reaction Types V

• Combustion-a hydrocarbon and oxygen combine to
  form CO2 and H2O.
• CH4 2O2 ? CO2 2H2O

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Types of Chemical Reactions

• Precipitate Reactions- a solid is formed from the
  reaction between 2 liquids.
• Acid-Base Reactions- a reaction which results in
  the formation of a salt and water (this is
  sometimes called a neutralization reaction.
• Oxidation-reduction a reaction in which 1 or
  more electrons are transferred.

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Expression of precipitate reactions

• A formula or molecular equation
• 3KOH(aq) Fe(NO3)3(aq) ? 3KNO3(aq)
  Fe(OH)3(s)
• An ionic equation
• 3K(aq) 3OH-(aq) Fe3(aq) 3NO3-(aq) ?
• 3K(aq) 3NO3-(aq) Fe(OH)3(s)
• A net ionic equation
• 3OH- Fe3 ? Fe(OH)3

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Oxidation-Reduction Reactions

• Oxidation-Reduction reactions are reactions in
  which 1 or more electrons are transferred.
• The reactions are also called redox reactions.
• Oxidation is the loss of electrons. Reduction is
  the gain of electrons.LEO says GER.
• An oxidizing agent is an electron acceptor.
• A reducing agent is an electron donor.

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Single Replacement Reactions

• Single replacement reactions are one form of
  oxidation-reduction reaction.
• Use the activity series to determine if the
  reaction will occur.

• Cu 2AgNO3 ? 2Ag Cu(NO3)2
• Cu ? Cu2 2e-
• 2Ag 2e- ? 2Ag
• Cu 2Ag ? Cu2 2Ag

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Assigning Oxidation States

• The oxidation state of an element is 0.
• The oxidation state of a monotomic ion is the
  same as its charge.
• The oxidation state of oxygen is 2, except in
  compounds containing the O2-2 like peroxide H2O2.
• In covalent compounds hydrogen is 1.
• Fluorine is always 1.
• The sum of the oxidation states of a compound
  must equal 0. An ion equals the charge of the ion.

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Examples of Oxidation States

• The 2 oxygen have a total oxidation state of 2,
  therefore the carbon must be 4.
• The fluorine has a total oxidation state of 6,
  therefore the sulfur must be 6.

• CO2
• SF6

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Examples of Oxidation States

• Oxygen has a total oxidation state of 6.
• Since the total charge is 1, the charge on
  the nitrogen must be 5.

• NO3-

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Oxidation States in Equations

• CH4 2O2 ? CO2
  2H2O
• -4 1ea. 0 4 -2ea. 1
  ea. -2
• The carbon loses 8e- and the oxygen gains 8e-.
  The carbon is oxidized, and the oxygen is reduced.

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Acids and Bases

• Arrheneus Definition
• An acid produces H when dissolved in water.
• A base produces OH- when dissolved in water.
• Bronsted-Lowry Definition
• An acid is a proton donor.
• A base is a proton acceptor.

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Examples

• Arrhenius Acids Bases
• HCl(aq) ? H(aq) Cl-(aq)
• NaOH(aq)? Na(aq) OH-(aq)
• Bronsted-Lowry Acid-Base
• NH3 H2O ? NH4 OH-
• (base) (acid) (c. Acid)
  (c. Base)

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Acid-Base Reactions

• Acids are recognized by the H in the beginning of
  their chemical formula.
• Bases frequently have a hydroxide ion.
• The neutralization reaction is
• Acid Base ? Salt Water
• HNO3 KOH ? KNO3 H2O
• In a neutralization reaction the acid and base
  react completely.

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Acid-Base Reactions

• Formula Equation
• HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
• Ionic Equation
• H Cl- Na OH- ? Na Cl- H2O
• Net Ionic Equation
• H OH- ? H2O

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Calculating an Acid-Base Reaction

• Determine the balanced net ionic equation for the
  reaction.
• Calculate moles of reactant
• Determine the limiting reactant where
  appropriate.
• Calculate the quantity of the required reactant
  or product.

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Acid-Base Titrations

• A titration involves delivery(from a buret) of a
  solution of known concentration (the titrant)
  into a solution containing the substance to be
  analyzed (the analyte).
• For example you would use a known concentration
  of base to titrate an unknown concentration of
  acid until it is neutralized.

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Titration Terms

• The point where the analyte is neutralized is the
  end point or equivalence point.
• An indicator will change colors at this point.
• A common acid-base indicator is phenolphthalein
  which is pink in base and colorless in acid.
• A standard solution is one that is of a known
  concentration.

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More titration terms

• A monoprotic acid is one that will release one
  proton (H) per molecule(eg HCl).
• A diprotic acid releases two protons per molecule
  (egH2SO4).

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Calculating the Molarity of the standard solution

• 1.3009 grams of KHC8H4O4 (potassium hydrogen
  phthalate or KHP) completely neutralizes 41.20 mL
  of a sodium hydroxide solution. Calculate the
  molarity of the NaOH solution. (The acid is
  monoprotic)
• The net ionic equation is
• HC8H4O4-(aq) OH-(aq) ? H2O (l) C8H4O42-(aq)

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Calculating the Molarity of the standard solution

• Calculate moles of reactants
• 1.3009 g KHP x 1 mole KHP 6.3071 x 10-3 mol
  KHP
• 204.22
  g KHP
• 6.3071 x 10-3 mol KHP x 1 mol OH- 6.3071x 10-3
  mol OH-
• 1 mol
  KHP

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Calculating the Molarity of the standard solution

• Calculate the molarity of the standard solution
• 6.3071 x 10-3 mol OH- x 103 mL 0.1546 M
• 41.20 mL 1 L

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Equivalents and Normality

• Equivalents are the number of moles of H or OH-
  ions contributed by an acid or base or in broader
  terms it is a mole of charges.
• Normality is the number of moles of equivalents
  per liter of solution.
• In a titration the
• Normality(known) volume(known)
  Normality(unknown)
• 
  volume(unknown)

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Equivalents and Normality

• To calculate equivalents of H in a volume of
  22.50 mL and molarity of 0.200 HCl, change
  molarity to equivalents/L
• 22.50 mL x 1 L x .200 eq.HCl x1 eq. H
• 103mL 1L 1 eq.HCl
• 
  4.50 x 10-3 eq.H

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Equivalents and Normality

• To calculate normality of the base divide the
  equivalents of the acid by the volume of the base
  used to neutralize it. If the 4.50 x 10-3 eq. HCl
  requires 17.50 mL of NaOH to neutralize it what
  is the normality of the NaOH.
• .00450 eq. H x1 eq OH- x 103mL .257 N base
• 17.50 mL 1 eq. H 1 L
• N eq./L