AP/IB ChemistryAP/IB ChemistryChapter 4:Chapter 4:

Aqueous SolutionsAqueous Solutions

andand

Solution StoichiometrySolution Stoichiometry

DefinitionsDefinitions

SolutionSolution: homogenous mixture of: homogenous mixture of

SolventSolvent::– Present in greater quantitiesPresent in greater quantities

SoluteSolute::

Aqueous SolutionAqueous Solution: solution in which: solution in which

SolubilitySolubility: amount of substance : amount of substance that dissolves in athat dissolves in a

Ionic Compounds in Ionic Compounds in WaterWater

SolvationSolvation: the clustering of : the clustering of solvent molecules around a solute solvent molecules around a solute particleparticle

Ions are solvated by waterIons are solvated by water– Why?Why?

Solvation of ions by water Solvation of ions by water prevents anions and cations fromprevents anions and cations from– Why?Why?

Solvation in WaterSolvation in Water

In-Class ProblemIn-Class Problem

Solvation of Ionic CompoundsSolvation of Ionic CompoundsWhat dissolved species are What dissolved species are present in a solution of:present in a solution of:– KCNKCN

– NaClONaClO44

– KK33POPO44

Electrolytes vs. Electrolytes vs. NonelectrolytesNonelectrolytes

ElectrolyteElectrolyte: substance that : substance that dissociates dissociates

NonelectrolyteNonelectrolyte: substance that : substance that

3 types of solutes in aqueous 3 types of solutes in aqueous solutions:solutions:

Electrolytes as Electrolytes as ConductorsConductors

Aqueous solutions canAqueous solutions can

Depends on number ofDepends on number of

Transport of ions through Transport of ions through solution causessolution causes

Molecular Compounds in Molecular Compounds in WaterWater

Aqueous solutions of molecular Aqueous solutions of molecular compounds do compounds do notnot– Ex:Ex:

If no ions in solution,If no ions in solution,

But… some molecular compounds But… some molecular compounds dissociate into ions in aqueous dissociate into ions in aqueous solutions:solutions:

Strong ElectrolytesStrong Electrolytes: completely : completely dissociate in solutiondissociate in solution

– STRONG acids and bases (considered molecular STRONG acids and bases (considered molecular compounds)compounds)

Weak ElectrolytesWeak Electrolytes: yield small : yield small concentration of ions when dissolved concentration of ions when dissolved and mostly remain in molecular formand mostly remain in molecular form– Ions exist inIons exist in– Weak acids and basesWeak acids and bases

Strong and Weak Strong and Weak ElectrolytesElectrolytes

Concentrations of Solutions

Concentration: amount of solute dissolved in a– Change amounts of to change concentration

Measures of solution concentration– Molarity: moles of solute per

• Units:

– Molality: moles of solute per• Units: mol/kg

In-Class Problem

Calculating MolarityCalculating Molarity

Calculate the molarity of a Calculate the molarity of a solution made by dissolving 5.00 solution made by dissolving 5.00 g of glucose in 100 mL of water.g of glucose in 100 mL of water.

In-Class Problems

Calculating Molar Concentration Calculating Molar Concentration of Ionsof Ions

What is the molar concentration of KWhat is the molar concentration of K++ ions in a 0.015 M solution of ions in a 0.015 M solution of potassium chloride?potassium chloride?

What is the molar concentration of KWhat is the molar concentration of K++ ions in a 0.015 M solution of ions in a 0.015 M solution of potassium carbonate?potassium carbonate?

Why?Why?

In-Class ProblemsIn-Class Problems

How many grams of sodium sulfate are How many grams of sodium sulfate are there in 15 mL of 0.50 M sodium there in 15 mL of 0.50 M sodium sulfate?sulfate?How many mL of 0.50 M sodium sulfate How many mL of 0.50 M sodium sulfate solution are needed to provide 0.038 solution are needed to provide 0.038 mol of this salt?mol of this salt?What volume of 2.50 M lead (II) nitrate What volume of 2.50 M lead (II) nitrate solution contains 0.050 mol of nitrate?solution contains 0.050 mol of nitrate?What volume of 2.50 M lead (II) nitrate What volume of 2.50 M lead (II) nitrate solution contains 1.30 g of Pbsolution contains 1.30 g of Pb2+2+??

DilutionsDilutions

A solution of lower concentration A solution of lower concentration is obtained byis obtained by

KeyKey: Number of moles : Number of moles in in both concentrated ( ) both concentrated ( ) solution and solution and ( ( ) ) solutionsolution

In-Class ProblemsIn-Class Problems

Diluting SolutionsDiluting SolutionsHow many mL of a 5.0 M KHow many mL of a 5.0 M K22CrCr22OO77 solution must be diluted to solution must be diluted to prepare 250 mL of a 0.10 M prepare 250 mL of a 0.10 M solution?solution?

If 10.0 mL of a 10.0 M stock If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to solution of NaOH is diluted to 250 mL, what is the concentration 250 mL, what is the concentration of the resulting solution?of the resulting solution?

Solution StoichiometrySolution Stoichiometry

In-Class ProblemsIn-Class Problems

Solution StoichiometrySolution StoichiometryHow many grams of sodium hydroxide How many grams of sodium hydroxide are needed to neutralize 20.0 mL of are needed to neutralize 20.0 mL of a 0.150 M sulfurous acid solution?a 0.150 M sulfurous acid solution?

How many liters of a 0.50 M HCl How many liters of a 0.50 M HCl solution are needed to react solution are needed to react completely with 0.100 mol of completely with 0.100 mol of lead(II) nitrate, forming a lead(II) nitrate, forming a precipitate of lead(II) chloride?precipitate of lead(II) chloride?

TitrationsTitrations

TitrationsTitrations

Suppose we know the concentration of a Suppose we know the concentration of a NaOH solutionNaOH solution

and we want to find the concentration of and we want to find the concentration of a HCl solutiona HCl solution

We know:We know:

What do we want?What do we want?

What do we do?What do we do?– Take a known volume of theTake a known volume of the

Titrations

What do we get?– Volume of– We know molarity of , so we can calculate moles of

Next step?– We also know– We can calculate moles of

Can we finish?– Knowing moles of and volume of we can calculate

In-Class ProblemsIn-Class Problems

Acid-Base TitrationsAcid-Base Titrations

What is the molarity of an sodium What is the molarity of an sodium hydroxide solution if 48.0 mL is hydroxide solution if 48.0 mL is needed to neutralize 35.0 mL of 0.144 needed to neutralize 35.0 mL of 0.144 M carbonic acid?M carbonic acid?

45.7 ml of 0.500 M sulfuric acid is 45.7 ml of 0.500 M sulfuric acid is required to neutralize a 20.0 mL required to neutralize a 20.0 mL sample of potassium hydroxide sample of potassium hydroxide solution. What is the concentration solution. What is the concentration of the potassium hydroxide solution?of the potassium hydroxide solution?

In-Class ProblemIn-Class Problem

Redox TitrationsRedox TitrationsAn iron ore is dissolved in acid and the iron An iron ore is dissolved in acid and the iron is converted to Feis converted to Fe2+2+. The sample is then . The sample is then titrated with 47.20 mL of 0.02240 M MnOtitrated with 47.20 mL of 0.02240 M MnO44

-- solution. The redox reaction that occurs solution. The redox reaction that occurs during titration is as follows:during titration is as follows:

MnOMnO44--(aq)(aq) + 5Fe + 5Fe2+2+

(aq)(aq) + 8H + 8H++(aq)(aq) Mn Mn2+2+

(aq)(aq) + 5Fe + 5Fe3+3+(aq)(aq) + 4H + 4H22OO(l)(l)

– How many moles of MnOHow many moles of MnO44-- were added to the solution? were added to the solution?

– How many moles of FeHow many moles of Fe2+2+ were in the sample? were in the sample?– How many grams of iron were in the sample?How many grams of iron were in the sample?– If the sample had a mass of 0.8890 g, what is the If the sample had a mass of 0.8890 g, what is the percentage of iron in the sample?percentage of iron in the sample?

Sample Test QuestionSample Test Question

70.5 mg of potassium phosphate is added to 70.5 mg of potassium phosphate is added to 15.015.0

mL of 0.050 M silver nitrate solution, mL of 0.050 M silver nitrate solution, resultingresulting

in the formation of a precipitate.in the formation of a precipitate.

Write the molecular equation for the Write the molecular equation for the reaction.reaction.

What is the limiting reagent in the What is the limiting reagent in the reaction?reaction?

Calculate the theoretical yield, in grams, Calculate the theoretical yield, in grams, of the precipitate that forms.of the precipitate that forms.