SolutionsSolutions

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 effects the solubility of gases.

• The higher the temperature, the lower the solubility of a gas in solution.

• An example is carbon dioxide in soda:– Less CO2 escapes when you open a cold soda than

when you open the soda warm

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 & 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 it 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 & 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.

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 place 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.

• Heating the solution:– This increases the kinetic energy of the solvent and the

solute is attacked faster by the solvent molecules.

• Stirring the solution:– This increases the interaction between solvent and

solute molecules.

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

Solubility 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 55C, the solubility of NaC2H3O2 is 100 g per

100 g water.

• If a saturated solution at 55C is cooled to 20C, 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:– mass/mass percent

– 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

× 100% = m/m %

g soluteg solute + g solvent

× 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

× 100% = m/m %

5.00 g NaCl102 g solution

× 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 sucrose

contains 25.0 grams of sucrose?

• We want grams solution, we have grams sucrose.

100 g solution5.00 g sucrose

= 500 g solution25.0 g sucrose ×

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

number of moles of solute per liter of solution, 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

18.0 g of NaOH in 0.100 L of solution?

• We also need to convert grams NaOH to moles NaOH (MM = 40.00 g/mol).

= 4.50 M NaOH×18.0 g NaOH

0.100 L solution1 mol NaOH

40.00 g NaOH

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

concentration 4.50 M NaOH:

4.50 mol NaOH1 L solution 4.50 mol NaOH

1 L solution

4.50 mol NaOH1000 mL solution 4.50 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 K2Cr2O7 1000 mL solution250.0 mL solution × ×

294.2 g K2Cr2O7

1 mol K2Cr2O7

Molar Concentration Problem• What volume of 12.0 M HCl contains 7.30 g of

HCl solute (MM = 36.46 g/mol)?

• We want volume, we have grams HCl.

= 16.7 mL solution

1 mol HCl 36.46 g HCl

7.30 g HCl × ×1000 mL solution

12.0 mol HCl

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 × V1 = M2 × 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 × V1 = M2 × V2

(6.0 M) × V1 = (0.10 M) × (5.00 L)

V1 = = 0.083 L(0.10 M) × (5.00 L)

6.0 M

Solution Stoichiometry• In Chapter 10, we performed mole calculations

involving chemical equations, stoichiometry problems.

• 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 ×3 mol AgBr

1 mol AlBr3

0.100 mol AlBr3

1000 mL soln×

1 mol AgBr

187.77 g AgBr×

Conclusions

• 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 dissolved in nonpolar solvents.

Conclusions Continued• Three factors can increase the rate of dissolving

for a solute:– Heating the solution

– Stirring the solution

– 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.

Conclusions 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

× 100% = m/m %

Conclusions Continued

• You can make a solution by diluting a more concentrated solution:

M1 × V1 = M2 × V2

• We can apply stoichiometry to reactions involving solutions using the molarity as a unit factor to convert between moles and volume.