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Chapter 4

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Chapter 4

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  1. Chapter 4 AP Chemistry 2012

  2. 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!

  3. Electrolytes • A substance that COMPLETELY dissociated when dissolved in water. • This includes all water soluble ionic compounds and strong acids. Dissolving

  4. 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.

  5. 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

  6. How do the ions in a dissolved compound compare to those in an insoluble one?

  7. 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!

  8. 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)

  9. 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.

  10. 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.

  11. 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.

  12. 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

  13. 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”

  14. 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.

  15. 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)

  16. Oxidation-Reduction (Redox) Reactions • Oxidation – loss of electrons • Reduction – gain of electrons • Always occur together. • OIL RIG LEO says GER

  17. 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)

  18. 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.

  19. 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.

  20. Molarity • Units for concentration. • Moles of solute per liters of solution. • Symbolized M or mol/L

  21. 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.

  22. 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:

  23. 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