Acids and Bases Notes Part 2

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Acids and Bases Notes Part 2
2
Acid Rain

• Many industrial processes produce gases such as
  NO, NO2, CO2, SO2, and SO3. These compounds can
  dissolve in atmospheric water to produce acidic
  solutions that fall to the ground in the form of
  rain or snow. Marble found in many buildings and
  statues is composed of calcium carbonate, when
  acid snow or rain falls on these structures a
  great deal of damage is caused. (page 475)

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Neutralization Reactions

• Neutralization the reaction of an acid with a
  base to produce water and a salt. Occurs when
  H3O and OH- ions are supplied in equal numbers
  by the reactants.
• Salt- an ionic compound composed of a cation from
  a base and an anion from an acid.

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Examples

• H2SO4 2KOH ? K2SO4
  2H2O
• HCl NaOH ? NaCl
  H2O
•  
• H3PO4 3NH4OH ? (NH4)3PO4
  3H2O
• salts water

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Titrations

• Titration A neutralization reaction of an acid
  by a base or vice versa it is usually used to
  find the concentration, molarity, of an unknown
  acid or base. The concentration of the other is
  known.
• Titrant solution of known concentration (in
  buret). The titrant is added to a solution of
  unknown concentration until the substance being
  analyzed is just consumed (stoichiometric point
  or equivalence point).

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Titrations

• Indicators-Indicators change color at the end
  point of a titration. Examples phenolphthalein
  and methyl orange
• Equivalence point-the point at which the two
  solutions used in a titration are present in
  chemically equivalent amounts.
• Strong acid Strong base at equivalence point
  pH7 http//wwwchem.uwimona.edu.jm1104/software
  /titr.html

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• Strong Acid-Strong Base Titrations Simple
  reaction H OH- ? H20
• The pH is easy to calculate because all reactions
  go to completion. At the equivalence point, the
  solution is neutral (pH7.00).

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Practice Problems 1. Suppose you know 20. mL
of a 0.0050M NaOH solution is required to reach
the equivalence point in the titration of 10. mL
of an unknown molarity of HCl. What is the
molarity of the

• Start with the balanced equation for the
  reaction.
• HCl NaOH ? NaCl H2O

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• Write out all givens and unknowns before
  beginning problem.
• ?M of known NaOH? 0.0050M
• ?moles of known NaOH?
• ?liters of known NaOH? 20.mL
• ?M of unknown HCl?
• ?moles of unknown HCl?
• ?liters of unknown HCl? 10.mL

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• Determine the moles of known solution used.
• ?M of known NaOH? 0.0050M
• ?moles of known NaOH?
• ?liters of known NaOH? 20.mL
• (Remember that Molarity Moles )
• Liters
• 0.005M NaOH ?moles
• .020 L
• ?moles (.020 L) (.0050 M)
• ?moles 0.00010 moles NaOH

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So now we know

• ?M of known NaOH? 0.0050M
• ?moles of known NaOH? 0.00010 moles
• ?liters of known NaOH? 20.mL
• ?M of unknown HCl?
• ?moles of unknown HCl?
• ?liters of unknown HCl? 10.mL

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• Determine the moles of unknown solution used.
  (Using the moles of known solution used and the
  equation for a mole ratio.)
• ?moles of unknown HCl?
• HCl NaOH ? NaCl H2O
• ?mol HCl?0.00010 mol NaOH X 1 mol HCl
  0.00010mol HCl
• 1 mol NaOH

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So now we know

• ?M of known NaOH? 0.0050M
• ?moles of known NaOH? 0.00010 moles
• ?liters of known NaOH? 20.mL
• ?M of unknown HCl?
• ?moles of unknown HCl? 0.00010 moles
• ?liters of unknown HCl? 10.mL

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• Determine the molarity of the unknown.
• ?M of unknown HCl?
• ?moles of unknown HCl? 0.00010 moles
• ?liters of unknown HCl? 10.mL
• ?M 0.00010 moles HCl
• 0.010 L
• 0.010M HCl

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So now we know it all!!!

• ?M of known NaOH? 0.0050M
• ?moles of known NaOH? 0.00010 moles
• ?liters of known NaOH? 20.mL
• ?M of unknown HCl? 0.010M
• ?moles of unknown HCl? 0.00010 moles
• ?liters of unknown HCl? 10.mL

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More Examples

• Complete examples 8 and 9 from your notes and
  then complete your homework problems.