Stoichiometry of Precipitation Reactions

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• Stoichiometry of Precipitation Reactions
• Procedure
• Example ??? g NaCl needed to precipitate all Ag
  from 1.50L of
• 0.100 M AgNO3?
• Net Ionic Equation Ag(aq) Cl-(aq)
  -------gt AgCl(s)

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• C. Example ??? g PbSO4 form when 1.25L 0.0500M
  Pb(NO3)2 and 2.00L 0.0250M Na2SO4 are mixed?
• Net ionic equation Pb2(aq) SO42-(aq)
  -------gt PbSO4(s)
• Acid-Base Reactions
• Acid Base models
• Arrhenius model correct, but too limiting
• Acid H producer in water
• Base OH- producer in water
• Bronsted-Lowry model includes Arrhenius model,
  but more general
• Acid proton (H) donor
• Base proton acceptor
• Example HC2H3O2 H2O -------gt C2H3O3-
  H3O

acid base
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• Neutralization Reactions
• Water is a nonelectrolyte (not highly ionized in
  solution)
• H(aq) OH-(aq) -------gt H2O(l)
• Reaction of an acid and a base (usually to form
  water) Neutralization
• H is a strong acid and will completely react
  with any weak base present
• H(aq) NH3(aq) -------gt NH4(aq)
• OH- is a strong base and will completely react
  with any weak acid present
• OH-(aq) HC2H3O2(aq) -------gt H2O(l)
  C2H3O2-(aq)
• Similar to a precipitation reaction, except the
  product is a liquid (H2O)
• Example ??? L of 0.100M HCl to neutralize 25.0
  ml 0.350M NaOH?
• Net Ionic Equation H(aq) OH-(aq)
  -------gt H2O(l)

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• D. Example ??? mol H2O is formed when 28.0ml
  of 0.250M HNO3 and 53.0ml of 0.320M KOH are
  reacted? What is the concentration of H/OH-
  left?
• Net Ionic Equation H(aq) OH-(aq)
  -------gt H2O(l)
• Calculate how much water is formed
• H was limiting, so calculate how much OH- is
  left over and concentration
• Acid-Base Titrations
• Definition volumetric analysis of the
  concentration of an unkown
• Titrant solution of known concentration whose
  volume is measured
• Analyte solution whose concentration is to be
  determined
• Equivalence Point amount of titrant just reacts
  with all analyte
• Indicator changes color at endpoint of the
  titration

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• Acid-Base Titration results in neutralization of
  all of the analyte
• a. H(aq) OH-(aq) -------gt H2O(l)
• Phenolphthalein indicator is colorless in acid
  and pink in base
• A buret accurately measures the amount of titrant
  added

The flask contains acid of unknown concentration
and phenolphthalein. The buret contains base of
a known concentration
Base is added dropwise. The faint pink color
goes away as you stir. The endpoint hasnt been
reached yet.
The endpoint has now been reached as the pink
color persists. Measuring the volume of base
dispensed aids calculation.
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• Standardizing the base solution Example
• 1.3009g KHP is dissolved in distilled water
  (doesnt matter how much) and titrated with
  41.20ml unknown NaOH solution.
• HP-(aq) OH-(aq) -------gt H2O(l)
  P2-(aq)
• Calculating the concentration of the analyte
  Example
• 0.3518g of sample was titrated with 10.59ml of
  0.1546M NaOH. Calculate the mass percent of
  HC7H5O2 (benzoic acid) in the sample
• HC7H5O2 (aq) OH-(aq) -------gt H2O(l)
  C7H5O2-(aq)

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• F. Reactions that give off gases
• Sometimes the product of a reaction is not a
  solid, but a gas
• We can still observe that something happened
  bubbles form

Compound that reacts with acid Equation for Formation of the gas Gas produced
Sulfides 2H S2- ? H2S H2S
Carbonates 2H CO32- ? H2CO3 ? H2O CO2 CO2
Bicarbonates H HCO3- ? H2CO3 ? H2O CO2 CO2
Sulfites 2H SO32- ? H2O SO2 SO2
Bisulfites H HSO32- ? H2O SO2 SO2
Cyanides H CN- ? HCN HCN
Compound that reacts with base Equation for Formation of the gas Gas produced
Ammonium salts NH4 OH- ? NH3 H2O NH3
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• III. Oxidation-Reduction Reactions
• Definition Reactions in which electrons are
  transferred (aka Redox Rxns)
• Example 2Na(s) Cl2(g) -------gt
  2NaCl(s)
• Each Na loses one electron to become Na
  (Oxidation)
• Each Cl gains one electron to become Cl-
  (Reduction)
• Oxidation States tool for keeping track of
  electrons in Redox reactions
• The total charge on all atoms must match the
  molecule or ion charge
• Oxidation states of a few elements help us
  calculate for all others

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• B. Example
• CO2
• O -2 so we have a total of -4 coming from the
  2 oxygens
• C must be 4 to balance the negative charge
• SF6
• F -1 so we have a total of -6 coming from the 6
  fluorines
• S must be 6 to balance the negative charge
• NO3-
• O -2 so we have -6 coming from the 3 oxygens
• N must be 5 in order to give us an overall 1-
  charge
• C. Recognizing what is happening in Redox
  Reactions

Radius decreases Radius increases
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• CH4(g) 2O2(g) -------gt CO2(g)
  2H2O(g)
• (-4)(1) (0)
  (4)(-2) (1)(-2)
• CH4 -----gt CO2 8e- Carbon oxidized, CH4
  reducing agent
• 1(-4) 1(4)
• 2O2 8e- -----gt CO2 2H2O Oxygen
  reduced oxidizing agent
• 2(0) 2(-2)
  2(-2)
• Example 2Al(s) 3I2(s) -----gt
  2AlI3(s)
• (0)
  (0) (3)(-1)
• a. Al is oxidized reducing agent
• b. I is reduced I2 is the oxidizing agent
• 3. Example Oxidized? Reduced? Oxidizing agent?
  Reducing agent?
• 2PbS(s) 3O2(g) -----gt 2PbO(s) 2SO2(g)
• 2PbO(s) CO(g) -----gt Pb(s) CO2(g)

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• Balancing Redox Equations Half-Reaction Method
  in Acidic Solution
• MnO4-(aq) Fe2(aq) -----gt Fe3(aq)
  Mn2(aq) (acidic solution)
• Identify and write equations for the two
  half-reactions
• MnO4- -----gt Mn2 (this is the reduction
  half-reaction)
• (7)(-2) (2)
• Fe2 -----gt Fe3 (this is the oxidation
  half-reaction)
• (2) (3)
• Balance each half-reaction
• Add water if you need oxygen
• Add H if you need hydrogen (since we are in
  acidic solution)
• Balance the charge by adding electrons
• MnO4- -----gt Mn2 4H2O
• 8H MnO4- -----gt Mn2 4H2O
• (7) (2)
• 5e- 8H MnO4- -----gt Mn2
  4H2O Balanced!
• Fe2 -----gt Fe3 1e- Balanced!

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• 3. Equalize the number of electrons in each
  half-reaction and add reactions
• 5(Fe2 -----gt Fe3 1e-) 5Fe2 -----gt
  5Fe3 5e-
• 5e- 8H MnO4- -----gt Mn2 4H2O
• 5Fe2 8H MnO4- ------gt 5Fe3 Mn2
  4H2O
• Species (including e-) on each side cancel out
  (algebra)
• Check that the charges and elements all balance
  DONE!
• Example H Cr2O72- C2H5OH -----gt
  Cr3 CO2 H2O
• Balancing Redox Equations Half-Reaction Method
  in Basic Solution
• Follow the Acidic Solution Method until you have
  the final balance eqn
• H cant exist in basic solution, so add enough
  OH- to both sides to turn all of the H in H2O
• Example Ag CN- O2 -----gt Ag(CN)2-
  (basic solution)
• Ag CN- -----gt Ag(CN)2- (oxidation
  half-reaction)
• Becomes Ag 2CN- -----gt Ag(CN)2- 1e-
  Balanced
• O2 -----gt (reduction
  half-reaction)
• Becomes 4e- 4H O2 -----gt 2H2O Balanced

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• Equalize the electrons in each half-reaction and
  add reactions
• 4(Ag 2CN- -----gt Ag(CN)2- 1e-)
• Becomes 4Ag 8CN- -----gt 4Ag(CN)2- 4e-
• Add to 4e- 4H O2 -----gt 2H2O
• Gives 4Ag 8CN- 4H O2 -----gt
  4Ag(CN)2- 2H2O DONE!
• Add OH- ions to both sides to remove H ions
• 4Ag 8CN- 4H O2 4OH- -----gt
  4Ag(CN)2- 2H2O 4OH-
• 4Ag 8CN- 4H2O O2 -----gt 4Ag(CN)2-
  2H2O 4OH-
• Cancel water molecules appearing on both sides of
  the equation
• 4Ag 8CN- 2H2O O2 -----gt 4Ag(CN)2-
  4OH-
• Check that everything balances REALLY DONE!