chapter 4

Chapter 4

AP Chemistry 2012

Dissolving

Molecular Compounds• Compound remains together

and is surrounded by water– Except for strong acids!

Ionic Compounds and Strong Acids

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

Electrolytes • A substance that COMPLETELY dissociated

when dissolved in water.– This includes all water soluble ionic compounds

and strong acids.Dissolving

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• 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

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

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• 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

• 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

• 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• For each of the following reactions, write a

balanced chemical equation and identify each as a precipitation or neutralization reaction.

1. Aqueous sodium sulfate is combined with aqueous barium nitrate.

2. Aqueous sulfuric acid is combined with aqueous potassium hydroxide.

3. Aqueous calcium hydroxide is combined with aqueous sodium phosphate.

4. Aqueous calcium hydroxide is combined with aqueous nitric acid.

Answers1. Na2SO4 (aq) + Ba(NO3)2(aq) BaSO4(s) + 2NaNO3(aq)

Precipitation

2. H2SO4(aq) + 2KOH(aq) 2H2O(l) + K2SO4(aq)Neutralization

3. 3Ca(OH)2(aq) + 2Na3PO4(aq) Ca3(PO4)2(s) + 6NaOH(aq)Precipitation

4. Ca(OH)2(aq) + 2HNO3(aq) 2H2O(l) + Ca(NO3)2(aq)Neutralization

• 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

• 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.

Answers

1. SO42- (aq) + Ba2+ (aq) BaSO4 (s)

2. H+(aq) + OH-(aq) H2O (l)

3. 3Ca2+(aq) + 2PO43-(aq) Ca3(PO4)2 (s)

4. OH- (aq) + H+ (aq) H2O (l)

Oxidation-Reduction (Redox) Reactions

• Oxidation – loss of electrons • Reduction – gain of electrons• Always occur together.• OIL RIG LEO says GER

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

• Consider the reactions:Cu(s) + Mg2+(aq)

No ReactionMg(s) + Cu2+(aq)

Mg2+

(aq) + Cu(s)

• The activity series lists metals from most likely to be oxidized to least likely.

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

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

€

Molarity = amount of solute (mol)volume of solution (L)

€

M =molL

Calculating Molarity

1. Determine the number of moles of solute (this may be given to you or you may need to use molar mass to convert)

2. Determine the volume in liters (sometimes it will be given in milliliters)

3. Divide the number of moles by liters.

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:

€

5 mol1 L

or 1L

5 mol

Example• A student needs 0.4 moles of NaOH for a

reaction. If the student has a solution of 1.5M NaOH, how many milliliters will the student need?

• Solution:– Start with your given (0.4 moles)

€

0.4mol 1L1.5M

1000mL1L

= 267mL

Molarity as conversion factor