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Chapter 4. AP Chemistry 2012. Dissolving. Molecular Compounds. Ionic Compounds and Strong Acids. Dissociate – break apart into ions. Each ion is surrounded by water. Compound remains together and is surrounded by water Except for strong acids!. Electrolytes .

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chapter 4

Chapter 4

AP Chemistry 2012

dissolving
Dissolving

Molecular Compounds

Ionic Compounds and Strong Acids

Dissociate – break apart into ions. Each ion is surrounded by water.

  • Compound remains together and is surrounded by water
    • Except for strong acids!
electrolytes
Electrolytes
  • A substance that COMPLETELY dissociated when dissolved in water.
    • This includes all water soluble ionic compounds and strong acids.

Dissolving

solubility
Solubility
  • How can we predict if an ionic compound will dissolve or not?
    • Memorize!
    • Compounds containing the following ions are ALWAYS soluble: NO3- (nitrate), CH3COO- (acetate), Group 1 cations, NH4+ (ammonium)
    • Page 125 of your textbook contains a complete list of rules.
precipitation reactions
Precipitation Reactions
  • Combination of 2 or more aqueous reactants to form an insoluble product (indicated by a ‘s’ in the equation and referred to as a PRECIPITATE)
  • Example:AgNO3(aq) + NaCl(aq)AgCl(s) + NaNO3(aq)

2 aqueous (aq) reactants

1 solid and 1 aqueous product

*You must learn solubility rules in order to predict the solid product formed in a precipitation reaction

example
Example
  • Predict the products formed by the following reaction (including states)
    • Aqueous ammonium carbonate is combined with calcium chloride.
  • Solution:
    • First write the formulas of reactants:(NH4)2CO3(aq) + CaCl2(aq)
    • The cations should switch the anions they are paired with in products:(NH4)2CO3(aq) + CaCl2(aq)  2NH4Cl + CaCO3
    • Use solubility rules to determine which product is INSOLUBLE (this will be the solid or precipitate)(NH4)2CO3(aq) + CaCl2(aq) 2NH4Cl (aq)+ CaCO3 (s)

DON’T FORGET TO BALANCE!

acids
Acids
  • Substances that produce H+ (hydrogen ion) when dissolved in water.
  • Formulas usually being with ‘H’
  • Memorize the names and formulas of the following strong acids (these seven acids dissociate completely into H+ and their anion in water):
    • Hydrochloric (HCl)
    • Hydrobromic (HBr)
    • Hydroiodic (HI)
    • Nitric (HNO3)
    • Sulfuric (H2SO4)
    • Chloric (HClO3)
    • Perchloric (HClO4)
bases
Bases
  • Substance that produce OH- (hydroxide ions) when dissolved in water.
  • Strong bases are soluble ionic compounds whose anion is OH-
    • Including all group 1 metal hydroxides
    • And heavy group 2 (Ca2+, Sr2+, and Ba2+) hydroxides.
neutralization reactions
Neutralization Reactions
  • Combining of acid and base to produce salt and water.
  • The pH of the product is around 7, and thus neutral.
  • Example:

HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)

* Notice how similar this reaction is to a precipitation reaction.

practice problems
Practice Problems
  • For each of the following reactions, write a balanced chemical equation and identify each as a precipitation or neutralization reaction.
  • Aqueous sodium sulfate is combined with aqueous barium nitrate.
  • Aqueous sulfuric acid is combined with aqueous potassium hydroxide.
  • Aqueous calcium hydroxide is combined with aqueous sodium phosphate.
  • Aqueous calcium hydroxide is combined with aqueous nitric acid.
answers
Answers
  • Na2SO4 (aq) + Ba(NO3)2(aq)  BaSO4(s) + 2NaNO3(aq)

Precipitation

  • H2SO4(aq) + 2KOH(aq) 2H2O(l) + K2SO4(aq)Neutralization
  • 3Ca(OH)2(aq) + 2Na3PO4(aq)  Ca3(PO4)2(s) + 6NaOH(aq)Precipitation
  • Ca(OH)2(aq) + 2HNO3(aq) 2H2O(l) + Ca(NO3)2(aq)Neutralization
slide13

Consider the reaction below:Pb(NO3)2(aq) + 2NaI(aq) PbI2(s) + 2NaNO3(aq)

  • Because aqueous ionic compounds dissociate we can more accurately represent the above reaction as:Pb2+(aq) + 2NO3+(aq) + 2Na+(aq) + 2I-(aq)  PbI2(s) + 2Na+(aq) + 2NO3-(aq)
  • Notice the solid product remains as a compound, only the aqueous species dissociate.
  • If you look closely at the reactants and products, you will notice that both the nitrate and sodium ions remain unchanged in the reaction.
    • We call ions that do not participate in a reaction “spectator ions”
net ionic equations
Net Ionic Equations
  • Contain only ions changed in the reaction.
    • Spectator ions are not included.
  • Example:

Pb2+(aq) + 2I-(aq)  PbI2(s)

  • Return to the practice problems from 2 slides ago and try to write the net ionic equation for each reaction.
answers1
Answers
  • SO42-(aq) + Ba2+ (aq)  BaSO4 (s)
  • H+(aq) + OH-(aq)  H2O (l)
  • 3Ca2+(aq) + 2PO43-(aq)  Ca3(PO4)2 (s)
  • OH-(aq) + H+ (aq)  H2O (l)
oxidation reduction redox reactions
Oxidation-Reduction (Redox) Reactions
  • Oxidation – loss of electrons
  • Reduction – gain of electrons
  • Always occur together.
  • OIL RIG LEO says GER
metal oxidation
Metal Oxidation
  • Metals are commonly oxidized to form metal ions.
    • Metals can react with:
      • Oxygen in air (rusting) 2Mg(s) + O2(g) 2MgO(s)
      • Acids (H+ will be reduced):Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)
      • Aqueous solutions of other metal ions:Mg(s) + Cu(NO3)2(aq) Mg(NO3)2(aq) + Cu(s)
activity series
Activity Series
  • Consider the reactions:

Cu(s) + Mg2+(aq) No Reaction

Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)

  • The activity series lists metals from most likely to be oxidized to least likely.
oxidation numbers
Oxidation Numbers
  • How can we tell if a substance is being oxidized or reduced?
    • Oxidation numbers help us keep track of electrons by assigning them to atoms. *They have NO real meaning!
    • See complete rules on page 137
    • Some important notes:
      • elements always have an oxidation state of zero
      • ions in ionic compounds have the same oxidation state as their charge.
      • In molecules, atoms at the start of the formula are usually positive while the ones at the end are negative.
molarity
Molarity
  • Units for concentration.
  • Moles of solute per liters of solution.
  • Symbolized M or mol/L
calculating molarity
Calculating Molarity
  • Determine the number of moles of solute (this may be given to you or you may need to use molar mass to convert)
  • Determine the volume in liters (sometimes it will be given in milliliters)
  • Divide the number of moles by liters.
molarity as a conversion factor
Molarity as a Conversion Factor
  • If you are asked to find the moles or liters of a solution with known molarity, use molarity as a conversion factor.
    • For example, a 5M solution can also be represented as:
example1
Example
  • A student needs 0.4 moles of NaOH for a reaction. If the student has a solution of 1.5MNaOH, how many milliliters will the student need?
  • Solution:
    • Start with your given (0.4 moles)

Molarity as conversion factor