Introductory Chemistry: Concepts & Connections

4 th Edition by Charles H. Corwin

Chapter 14

Solutions

Christopher G. Hamaker, Illinois State University, Normal IL © 2005, Prentice Hall

Solutions

• A

solution

is a homogeneous mixture.

• A solution is composed of a

solute solvent

.

dissolved in a • Solutions exist in all three physical states: Chapter 14 2

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 CO 2 escapes when you open a cold soda than when you open the soda warm Chapter 14 3

Pressure & 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.

Chapter 14 4

Henry’s Law

• We can calculate the solubility of a gas at an new pressure using Henry’s Law.

new pressure solubility × = new solubility old 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?

0.00414 g/100 mL × 1150 torr 760 torr Chapter 14 = 0.00626 g/100 mL 5

Polar Molecules

• When two liquids make a solution, the

solute

the lesser quantity, and the

solvent

is is the greater quantity.

• Recall, that a

net dipole

is present in a polar molecule.

• Water is a polar molecule.

Chapter 14 6

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.

Chapter 14 7

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.

Chapter 14 8

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.

Chapter 14 9

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

.

Chapter 14 10

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 (C 12 H 22 O 11 ), are soluble in polar solvents and insoluble in nonpolar solvents.

• Nonpolar compounds, like naphthalene (C 10 H 8 ), are soluble in nonpolar solvents and insoluble in polar solvents.

Chapter 14 11

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

.

Chapter 14 12

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.

Chapter 14 13

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.

Chapter 14 14

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.

Chapter 14 15

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

.

Chapter 14 16

Supersaturated Solutions

• At 55  C, the solubility of NaC 2 H 3 O 2 100 g water.

is 100 g per • 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.

Chapter 14 17

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.

Chapter 14 18

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 Chapter 14 19

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 solute mass of solution × 100% = m/m % g solute g solute + g solvent × 100% = m/m % Chapter 14 20

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 NaCl 5.00 g NaCl + 97.0 g H 2 O × 100% = m/m % 5.00 g NaCl 102 g solution × 100% = 4.90 % Chapter 14 21

Mass Percent Unit Factors

• We can write several unit factors based on the concentration 4.90 m/m% NaCl: 4.90 g NaCl 100 g solution 100 g solution 4.90 g NaCl 4.90 g NaCl 95.1 g water 95.1 g water 4.90 g NaCl 95.1 g water 100 g solution Chapter 14 100 g solution 95.1 g water 22

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.

25.0 g sucrose × 100 g solution 5.00 g sucrose = 500 g solution Chapter 14 23

Molar Concentration

• The molar concentration, or

molarity

(

M

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

moles of solute liters of solution =

M

• Molarity is the most commonly used unit of concentration.

Chapter 14 24

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).

18.0 g NaOH 0.100 L solution × 1 mol NaOH 40.00 g NaOH = 4.50

M

NaOH Chapter 14 25

Molarity Unit Factors

• We can write several unit factors based on the concentration 4.50

M

NaOH: 4.50 mol NaOH 1 L solution 1 L solution 4.50 mol NaOH 4.50 mol NaOH 1000 mL solution 1000 mL solution 4.50 mol NaOH Chapter 14 26

Molar Concentration Problem

• How many grams of K 2 Cr 2 O 7 0.100

M

K 2 Cr 2 O 7 ?

are in 250.0 mL of • We want mass K 2 Cr 2 O 7 , we have mL solution.

250.0 mL solution × 0.100 mol K 2 Cr 2 O 7 × 1000 mL solution 294.2 g K 2 Cr 2 O 7 1 mol K 2 Cr 2 O 7 = 7.36 g K 2 Cr 2 O 7 Chapter 14 27

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.

7.30 g HCl × 1 mol HCl 36.46 g HCl × 1000 mL solution 12.0 mol HCl = 16.7 mL solution Chapter 14 28

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:

M

1 × V 1 – =

M

2 × V 2

M

1

M

2 and V 1 and V 2 are the initial molarity and volume and are the new molarity and volume Chapter 14 29

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.

M

1 × V 1 =

M

2 × V 2 (6.0

M

) × V 1 = (0.10

M

) × (5.00 L) V 1 = (0.10

M

) × (5.00 L) 6.0

M

= 0.083 L Chapter 14 30

Solution Stoichiometry

• In Chapter 10, we performed mole calculations involving chemical equations,

stoichiometry

problems.

• We can also apply stoichiometry calculations to solutions.

solution concentration balanced equation molarity known  moles unknown  moles known  mass unknown Chapter 14 molar mass 31

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?

AlBr 3 (aq) + 3 AgNO 3 (aq) → 3 AgBr(s) + Al(NO 3 ) 3 (aq) • We want g AgBr, we have volume of AlBr 3 37.5 mL soln × 0.100 mol AlBr 3 1000 mL soln × 3 mol AgBr 1 mol AlBr 3 × 187.77 g AgBr 1 mol AgBr = 2.11 g AgBr Chapter 14 32

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.

Chapter 14 33

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.

Chapter 14 34

Conclusions Continued

• The

mass/mass percent

concentration is the mass of solute per 100 grams of solution: mass of solute mass of solution × 100% = m/m % • The

molarity

of a solution is the moles of solute per liter of solution.

moles of solute liters of solution =

M

Chapter 14 35

Conclusions Continued

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

M

1 × V 1 =

M

2 × V 2 • We can apply stoichiometry to reactions involving solutions using the molarity as a unit factor to convert between moles and volume.

Chapter 14 36