introductory chemistry: concepts & connections 4th edition by … · 2011. 7. 24. · introductory...

INTRODUCTORY CHEMISTRYConcepts and Critical Thinking

Sixth Edition by Charles H. Corwin

1Chapter 14© 2011 Pearson Education, Inc.

Chapter 14Solutionsby Christopher Hamaker

1Chapter 14


Solutions• A solution is a homogeneous mixture.• A solution is composed of a solute dissolved in a

solvent.• Solutions exist in all three physical states:


Gases in Solution• Temperature affects the solubility of gases.• The higher the temperature is, the lower the

solubility of a gas is in solution.• An example is carbon dioxide in soda:

– Less CO2 escapes when you open a cold soda than when you open a warm soda.


Pressure and Gas Solubility• Pressure also influences the solubility of gases.• According to Henry’s law, the solubility of a gas

is directly proportional to the partial pressure of the gas above the liquid.

• If we double the partial pressure of a gas, we double the solubility.


Henry’s Law• We can calculate the solubility of a gas at a new

pressure using Henry’s law.

= new solubilitysolubility x new pressureold pressure• What is the solubility of oxygen gas at 25 °C and a

partial pressure of 1150 torr if the solubility of oxygen is 0.00414 g/100 mL at 25 °C and 760 torr?

1150 torr760 torr

= 0.00626 g/100 mL 0.00414 g/100 mL x


Polar Molecules• When two liquids make a solution, the solute is

the lesser quantity, and the solvent is the greater quantity.

• Recall that a net dipole is present in a polar molecule.

• Water is a polar molecule.


Polar and Nonpolar Solvents• A liquid composed of polar molecules is a polar

solvent. Water and ethanol are polar solvents.• A liquid composed of nonpolar molecules is a

nonpolar solvent. Hexane is a nonpolar solvent.


Like Dissolves Like• Polar solvents dissolve in one another.• Nonpolar solvents dissolve in one another.• This is the like dissolves like rule.• Methanol dissolves in water, but hexane does not

dissolve in water.• Hexane dissolves in toluene, but water does not

dissolve in toluene.


Miscible and Immiscible• Two liquids that completely

dissolve in each other are miscible liquids.

• Two liquids that are not miscible in each other are immiscible liquids.

• Polar water and nonpolar oil are immiscible liquids and do not mix to form a solution.


Solids in Solution• When a solid substance dissolves in a liquid, the

solute particles are attracted to the solvent particles.

• When a solution forms, the solute particles are more strongly attracted to the solvent particles than other solute particles.

• We can also predict whether a solid will dissolve in a liquid by applying the like dissolves like rule.


Like Dissolves Like for Solids• Ionic compounds, like sodium chloride, are

soluble in polar solvents and insoluble in nonpolar solvents.

• Polar compounds, like table sugar (C12H22O11), are soluble in polar solvents and insoluble in nonpolar solvents.

• Nonpolar compounds, like naphthalene (C10H8), are soluble in nonpolar solvents and insoluble in polar solvents.


Chemistry Connection: Colloids• Why is the flashlight beam visible in only one

container?• The solution, at the right of this slide, is a colloid.• A colloid is a solution with large solute particles

(ranging from 1 to 100 nm).• The solute particles in a

colloid are large enough to scatter light via a phenomenon known as the Tyndall effect.


The Dissolving Process• When a soluble crystal is placed into a solvent, it

begins to dissolve.

• When a sugar crystal is placed in water, the water molecules attack the crystal and begin pulling part of it away and into solution.

• The sugar molecules are held within a cluster of water molecules called a solvent cage.


Dissolving of Ionic Compounds• When a sodium chloride crystal is placed in water,

the water molecules attack the edge of the crystal.• In an ionic compound, the water

molecules pull individual ions off of the crystal.

• The anions are surrounded by the positively charged hydrogens on water.

• The cations are surrounded by the negatively charged oxygen on water.


Rate of Dissolving• There are three ways we can speed up the rate of

dissolving for a solid compound:1. Heating the solution

• This increases the kinetic energy of the solvent, and the solute is attacked faster by the solvent molecules.

2. Stirring the solution• This increases the interaction between solvent and solute

molecules.

3. Grinding the solid solute• There is more surface area for the solvent to attack.


Solubility of Solids and Temperature• The solubility of a compound is the maximum

amount of solute that can dissolve in 100 g of water at a given temperature.

• In general, a compound becomes more soluble as the temperature increases.


Saturated Solutions• A solution containing exactly the maximum

amount of solute at a given temperature is a saturated solution.

• A solution that contains less than the maximum amount of solute is an unsaturated solution.

• Under certain conditions, it is possible to exceed the maximum solubility of a compound. A solution with greater than the maximum amount of solute is a supersaturated solution.


Supersaturated Solutions• At 55 °C, the solubility of NaC2H3O2 is 100 g per

100 g water.• If a saturated solution at 55 °C is cooled to 20 °C,

the solution is supersaturated.

• Supersaturated solutions are unstable. The excess solute can readily be precipitated.


Supersaturation• A single crystal of sodium acetate added to a

supersaturated solution of sodium acetate in water causes the excess solute to rapidly crystallize from the solution.


Concentration of Solutions• The concentration of a solution tells us how

much solute is dissolved in a given quantity of solution.

• We often hear imprecise terms such as a “dilute solution” or a “concentrated solution.”

• There are two precise ways to express the concentration of a solution:

1. Mass/mass percent2. Molarity


Mass Percent Concentration• Mass percent concentration compares the mass of

solute to the mass of solvent.• The mass/mass percent (m/m %) concentration is

the mass of solute dissolved in 100 g of solution.

mass of solutemass of solution x 100% = m/m %

g soluteg solute + g solvent x 100% = m/m %


Calculating Mass/Mass Percent• A student prepares a solution from 5.00 g NaCl

dissolved in 97.0 g of water. What is the concentration in m/m %?

5.50 g NaCl5.00 g NaCl + 97.0 g H2O

x 100% = m/m %

5.00 g NaCl102 g solution x 100% = 4.90 %


Mass Percent Unit Factors• We can write several unit factors based on the

concentration 4.90 m/m % NaCl:

4.90 g NaCl100 g solution 4.90 g NaCl

100 g solution

4.90 g NaCl95.1 g water 4.90 g NaCl

95.1 g water

95.1 g water100 g solution 95.1 g water

100 g solution


Mass Percent Calculation• What mass of a 5.00 m/m % solution of dextrose

contains 25.0 grams of dextrose?• We want grams solution; we have grams dextrose.

= 500 g solution

100 g solution5.00 g dextrose25.0 g dextrose x


Molar Concentration• The molar concentration, or molarity (M), is the

number of moles of solute per liter of solution, and is expressed as moles/liter.

• Molarity is the most commonly used unit of concentration.

moles of soluteliters of solution = M


Calculating Molarity• What is the molarity of a solution containing 24.0

g of NaOH in 0.100 L of solution?• We also need to convert grams NaOH to moles

NaOH (M = 40.00 g/mol).

= 6.00 M NaOHx24.0 g NaOH0.100 L solution1 mol NaOH

40.00 g NaOH


Molarity Unit Factors• We can write several unit factors based on the

concentration 6.00 M NaOH:

6.00 mol NaOH1 L solution 6.00 mol NaOH

1 L solution

6.00 mol NaOH1000 mL solution 6.00 mol NaOH

1000 mL solution


Molar Concentration Problem• How many grams of K2Cr2O7 are in 250.0 mL of

0.100 M K2Cr2O7?• We want mass K2Cr2O7; we have mL solution.

= 7.36 g K2Cr2O7

0.100 mol K2Cr2O71000 mL solution250.0 mL solution x

x294.2 g K2Cr2O71 mol K2Cr2O7


Molar Concentration Problem, Continued

• What volume of 12.0 M HCl contains 9.15 g of HCl solute (M = 36.46 g/mol)?

• We want volume; we have grams HCl.

= 20.9 mL solution

1 mol HCl 36.46 g HCl9.15 g HCl x

x 1000 mL solution12.0 mol HCl


Critical Thinking: Water Fluoridation• Cities often add fluoride to drinking water.• Tooth enamel is made mostly of the mineral

hydroxyapatite, Ca10(PO4)6(OH)2.• Fluoride prevents tooth decay by converting some

of the hydroxyapatite to Ca10(PO4)6F2, which is more resistant to acid.

• Typically, fluoridation levels are less than 1 mg/L.


Dilution of a Solution• Rather than prepare a solution by dissolving a

solid in water, we can prepare a solution by diluting a more concentrated solution.

• When performing a dilution, the amount of solute does not change, only the amount of solvent.

• The equation we use is: M1 x V1 = M2 x V2.– M1 and V1 are the initial molarity and volume, and M2

and V2 are the new molarity and volume.


Dilution Problem• What volume of 6.0 M NaOH needs to be diluted

to prepare 5.00 L if 0.10 M NaOH?• We want final volume and we have our final

volume and concentration.

M1 x V1 = M2 x V2(6.0 M) x V1 = (0.10 M) x (5.00 L)

V1 = = 0.083 L(0.10 M) x (5.00 L)

6.0 M


Solution Stoichiometry• In Chapter 10, we performed mole calculations

involving chemical equations: stoichiometryproblems.

• We can also apply stoichiometry calculations to solutions.

molarity known ⇒ moles known ⇒moles unknown ⇒ mass unknown

solutionconcentration

balancedequation

molar mass


Solution Stoichiometry Problem• What mass of silver bromide is produced from the

reaction of 37.5 mL of 0.100 M aluminum bromide with excess silver nitrate solution?

AlBr3(aq) + 3 AgNO3(aq) → 3 AgBr(s) + Al(NO3)3(aq)

• We want g AgBr; we have volume of AlBr3.

= 2.11 g AgBr

37.5 mL soln x3 mol AgBr1 mol AlBr3

0.100 mol AlBr31000 mL soln

x1 mol AgBr

187.77 g AgBrx


Chapter Summary• Gas solubility decreases as the temperature

increases.

• Gas solubility increases as the pressure increases.

• When determining whether a substance will be soluble in a given solvent, apply the like dissolves like rule.– Polar molecules dissolve in polar solvents.

– Nonpolar molecules dissolve in nonpolar solvents.


Chapter Summary, Continued• Three factors can increase the rate of dissolving

for a solute:1. Heating the solution2. Stirring the solution3. Grinding the solid solute

• In general, the solubility of a solid solute increases as the temperature increases.

• A saturated solution contains the maximum amount of solute at a given temperature.


Chapter Summary, Continued• The mass/mass percent concentration is the mass

of solute per 100 grams of solution.

• The molarity of a solution is the moles of solute per liter of solution.

moles of soluteliters of solution = M

mass of solutemass of solution x 100% = m/m %


Chapter Summary, Continued• You can make a solution by diluting a more

concentrated solution.

M1 x V1 = M2 x V2• We can apply stoichiometry to reactions involving

solutions using the molarity as a unit factor to convert between moles and volume.

Slide Number 1SolutionsGases in SolutionPressure and Gas SolubilityHenry’s LawPolar MoleculesPolar and Nonpolar SolventsLike Dissolves LikeMiscible and ImmiscibleSolids in SolutionLike Dissolves Like for SolidsChemistry Connection: ColloidsThe Dissolving ProcessDissolving of Ionic CompoundsRate of DissolvingSolubility of Solids and TemperatureSaturated SolutionsSupersaturated SolutionsSupersaturationConcentration of SolutionsMass Percent ConcentrationCalculating Mass/Mass PercentMass Percent Unit FactorsMass Percent CalculationMolar ConcentrationCalculating MolarityMolarity Unit FactorsMolar Concentration ProblemMolar Concentration Problem, ContinuedCritical Thinking: Water FluoridationDilution of a SolutionDilution ProblemSolution StoichiometrySolution Stoichiometry ProblemChapter SummaryChapter Summary, ContinuedChapter Summary, ContinuedChapter Summary, Continued

introductory chemistry: concepts & connections 4th edition by … · 2011. 7. 24. · introductory...

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