Chapter 16Solutions

Killarney High School

Section 16.1Properties of Solutions

OBJECTIVES:–Identify the factors that determine the rate at which a solute dissolves.

Section 16.1Properties of Solutions

OBJECTIVES:–Calculate the solubility of a gas in a liquid under various pressure conditions.

Solution formationNature of the solute and the solvent

–Whether a substance will dissolve–How much will dissolve

Factors determining rate of solution...–stirred or shaken (agitation)–particles are made smaller– temperature is increased

Why?

Making solutionsIn order to dissolve, the solvent

molecules must come in contact with the solute.

Stirring moves fresh solvent next to the solute.

The solvent touches the surface of the solute.

Smaller pieces increase the amount of surface area of the solute.

Solution Formation Flash

Temperature and SolutionsHigher temperature makes the

molecules of the solvent move around faster and contact the solute harder and more often.–Speeds up dissolving.

Usually increases the amount that will dissolve (exception is gases)

How Much? Solubility- The maximum amount of

substance that will dissolve at a specific temperature (g solute/100 g solvent)

Saturated solution- Contains the maximum amount of solute dissolved

Unsaturated solution- Can still dissolve more solute

Supersaturated- solution that is holding more than it theoretically can; seed crystal will make it come out

LiquidsMiscible means that two liquids

can dissolve in each other–water and antifreeze, water and ethanol

Partially miscible- slightly–water and ether

Immiscible means they can’t–oil and vinegar

Solubility?For solids in liquids, as the

temperature goes up-the solubility usually goes up (Fig. 16.5, p.373)

For gases in a liquid, as the temperature goes up-the solubility goes down (Fig. 16.6, p.374)

For gases in a liquid, as the pressure goes up-the solubility goes up (Fig. 16.6, p.373)

Gases in liquids...Henry’s Law - says the solubility

of a gas in a liquid is directly proportional to the pressure of the gas above the liquid–think of a bottle of soda pop, removing the lid releases pres.

Equation: S1 S2

P1 P2

=

Cloud seedingEver heard of seeding the clouds

to make them produce rain?Clouds- mass of air

supersaturated with water vaporSilver Iodide (AgI) crystals are

dusted into the cloudThe AgI attracts the water,

forming droplets to attract others

Section 16.2Concentration of Solutions

OBJECTIVES:–Solve problems involving the molarity of a solution.

Section 16.2Concentration of Solutions

OBJECTIVES:–Describe how to prepare dilute solutions from more concentrated solutions of known molarity.

Section 16.2Concentration of Solutions

OBJECTIVES:–Explain what is meant by percent by volume [ % (v/v) ], and percent by mass [ % (m/v) ] solutions.

Concentration is...a measure of the amount of solute

dissolved in a given quantity of solventA concentrated solution has a large

amount of soluteA dilute solution has a small amount of

solute– thus, only qualitative descriptions

But, there are ways to express solution concentration quantitatively

Molarity - most importantThe number of moles of solute in 1

Liter of the solution.M = moles/Liter; such as 6.0 molarWhat is the molarity of a solution

with 2.0 moles of NaCl in 250 mL of solution?

Sample 16-2, page 378

Making solutionsPour in a small amount of solventThen add the solute (to dissolve it)Carefully fill to final volume.

–Fig. 18-10, page 509Also: M x L = molesHow many moles of NaCl are

needed to make 6.0 L of a 0.75 M NaCl solution?

Making solutions10.3 g of NaCl are dissolved in a

small amount of water, then diluted to 250 mL. What is the concentration?

How many grams of sugar are needed to make 125 mL of a 0.50 M C6H12O6 solution?

Dilution

Adding water to a solution

DilutionThe number of moles of solute doesn’t

change if you add more solvent!The # moles before = the # moles afterM1 x V1 = M2 x V2

M1 and V1 are the starting concentration and volume.

M2 and V2 are the final concentration and volume.

Stock solutions are pre-made to known Molarity

Practice2.0 L of a 0.88 M solution are diluted

to 3.8 L. What is the new molarity?You have 150 mL of 6.0 M HCl. What

volume of 1.3 M HCl can you make?Need 450 mL of 0.15 M NaOH. All

you have available is a 2.0 M stock solution of NaOH. How do you make the required solution?

Percent solutions...Percent means parts per 100, soPercent by volume:

= Volume of solute x 100% Volume of solution

indicated %(v/v)What is the percent solution if 25

mL of CH3OH is diluted to 150 mL with water?

Percent solutions Percent by mass:

= Mass of solute(g) x 100% Volume of solution(mL)

Indicated %(m/v) More commonly used 4.8 g of NaCl are dissolved in 82 mL of

solution. What is the percent of the solution?

How many grams of salt are there in 52 mL of a 6.3 % solution?

Section 16.3Colligative Properties of

SolutionsOBJECTIVES:

–Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution.

Section 16.3Colligative Properties of

SolutionsOBJECTIVES:

–Explain on a particle basis why a solution has an elevated boiling point, and a depressed freezing point compared with the pure solvent.

Colligative Properties

Depend only on the number of dissolved particles

Not on what kind of particle

Vapor Pressure decreasedThe bonds between molecules keep

molecules from escaping.In a solution, some of the solvent is

busy keeping the solute dissolved.Lowers the vapor pressureElectrolytes form ions when they are

dissolved = more pieces.NaCl Na+ + Cl- (= 2 pieces) More pieces = bigger effect

Boiling Point ElevationThe vapor pressure determines

the boiling point.Lower vapor pressure = higher

boiling point.Salt water boils above 100ºCThe number of dissolved

particles determines how much, as well as the solvent itself.

Freezing Point DepressionSolids form when molecules make

an orderly pattern.The solute molecules break up the

orderly pattern. Makes the freezing point lower.Salt water freezes below 0ºCHow much depends on the number

of solute particles dissolved.Simulation

Section 16.4Calculations Involving Colligative Properties

OBJECTIVES:–Calculate the molality and mole fraction of a solution.

Section 16.4Calculations Involving Colligative Properties

OBJECTIVES:–Calculate the molar mass of a molecular compound from the freezing point depression or boiling point elevation of a solution of the compound.

Molalitya new unit for concentrationm = Moles of solute

kilogram of solventm = Moles of solute

1000 g of solvent What is the molality of a solution

with 9.3 mole of NaCl in 450 g of water?

Why molality?The size of the change in boiling

point is determined by the molality.Tb = Kb x m x nTb is the change in the boiling pointKb is a constant determined by the

solvent (Table 16.2, page 387).m is the molality of the solution.n is the number of pieces it falls into

when it dissolves.

What about Freezing?The size of the change in freezing

point is also determined by molality.Tf = -Kf x m x nTf is the change in freezing pointKf is a constant determined by the

solvent (Table 16.3, page 388).m is the molality of the solution.n is the number of pieces it falls into

when it dissolves.

ProblemsWhat is the boiling point of a

solution made by dissolving 1.20 moles of NaCl in 750 g of water?

What is the freezing point?What is the boiling point of a

solution made by dissolving 1.20 moles of CaCl2 in 750 g of water?

What is the freezing point?

Mole fractionThis is another way to express

concentrationIt is the ratio of moles of solute to

total number of moles of solute + solvent (Fig. 16-19, p.386)

na

na + nb

X =Sample 18-8,

page 521

Molar MassWe can use changes in boiling

and freezing to calculate the molar mass of a substance

Find: 1) molality 2) moles, and then 3) molar mass

Sample 16-10, page 388

A solution of 7.50g of a nonvolatile compound in 22.60 g of water boils at 100.78 degrees celsius at 1 atm. What is the molar mass of the solute?

1. Use the BP elevation to calculate the molality of the solution.

Tb = Kb x m x n thus m = tb

Kb

Kb = 0.512 C x kg of waterMol of solute

0.78 C

0.512 C / mm = = 1.5 m

2. Now calculate the moles of solute I the solution

1.5 mol solute

Kg of water

0.0226 Kg 0.034 mol solute

3. Use the # of moles of solute and its mass to get the molar mass

Molar mass = mass of solute

Moles of solute

Molar mass = 7.50 g

0.0344 mol

Molar mass = 0.022 g/mol

Key EquationsNote the key equations to solve

problems in this chapter.