Solutions (ch.16) Solution – a homogeneous mixture

of pure substances

The SOLVENT is the medium in which the SOLUTES are dissolved. (The solvent is usually the most abundant substance.)

– Example: • Solution: Salt Water• Solute: Salt• Solvent: Water

READ ONLY SECTION:

HOW DO SUBSTANCES DISSOLVE?– “According to the kinetic theory, the water

molecules in each glass of tea are always moving. Some moving water molecules collide with sugar crystals. When this happens, energy is transferred to the sugar molecules at the surface of the crystal.” (Holt, p. 192)

The process of dissolution is favored by:

1) A decrease in the energy of the system (exothermic)

2) An increase in the disorder of the system (entropy)

Liquids Dissolving in Liquids

Liquids that are soluble in one another (“mix”) are MISCIBLE.– “LIKE dissolves LIKE”

POLAR liquids are generally soluble in other POLAR liquids.

NONPOLAR liquids are generally soluble in other NONPOLAR liquids.

LIKE DISSOLVES LIKE : demo

Factors affecting rate of dissolution: think iced tea vs. hot tea &

the type of sugar you use: cubes or granulated

1) Surface area / particle size– Greater surface area, faster it dissolves

2) Temperature– Most solids dissolve faster @ higher temps

3) Agitation– Stirring/shaking will speed up dissolution

Saturation: a solid solute dissolves in a solvent until the soln is SATURATED

Unsaturated solution – is able to dissolve more solute

Saturated solution – has dissolved the maximum amount of solute

Supersaturated solution – has dissolved excess solute (at a higher temperature). Solid crystals generally form when this solution is cooled.

ROCK CANDY, YUM!!

Applying Concepts QUESTION

When the crystallization has stopped, will the solution be saturated or unsaturated?

answer

ANSWER: SATURATED

Solution has the maximum amount of solute for a given quantity of solvent at a constant temperature and pressure.

SOLUBILITY

Solubility = the amount of solute that will dissolve in a given amount of solvent

Factors Affecting Solubility The nature of the solute and solvent:

different substances have different solubilities Temperature: many solids substances

become more soluble as the temp of a solvent increases; however, gases are less soluble in liquids at higher temps.

Pressure: Only affects the solubility of gases. As pressure increases, the solubility of gases increases.

Concentration of Solution

Concentration refers to the amount of solute dissolved in a solution.

*MOLARITY

nsol' L

solute mol(M)Molarity

Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:a) solid Na2SO4

b) 5.00 M Na2SO4

L 2.50

xM 665.0

nsol' L

solute mol(M)Molarity

mol 1.6625x

g236mol 1

g 1.142SONa mol 6625.1 42

Dissolve 236 g of Na2SO4 in enough water

to create 2.50 Lof solution.

MOLARITY BY DILUTION

When you dilute a solution, you can use this equation:

2211 VMVM

Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:a) 5.00 M Na2SO4

L) M)(2.50 665.0()M)(V (5.00 1

mL 333 L 333.0V1

2211 VMVM

Add 0.333 L of Na2SO4 to 2.17 L of water.

Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:b) solid Na2SO4

L 2.50

xM 665.0

nsol' L

solute mol(M)Molarity

mol 1.6625x

g236mol 1

g 04.142SONa mol 6625.1 42

Dissolve 236 g of Na2SO4 in enough water

to create 2.50 Lof solution.

MASS PERCENT

100nsol' of mass total

solute mass % mass

MASS PERCENT

Example: What is the percent of NaCl in a solution made by dissolving 24 g of NaCl in 152 g of water?

100g 176

g 24%14%6.13

*MOLALITY

solvent kg

solute mol (m)molality

MOLALITY Example: What is the molality of a

solution that contains 12.8 g of C6H12O6 in 187.5 g water?

solvent kg

solute molm

g 180.18

mol 1OHC g8.12 6126 mol 07104.0

kg 1875.0

mol 0.07104m m 379.0

MOLALITY Example: How many grams of H2O must be

used to dissolve 50.0 g of sucrose to prepare a 1.25 m solution of sucrose, C12H22O11?

solvent kg

solute molm

g 342.34

mol 1OHC g 50.0 112212 mol 0.1461

x

mol 0.1461m 1.25

OH 117gkg 1168.0 2x

Colligative Properties of Solutions (chapter 16)

Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.

Colligative Properties

Lowering vapor pressure Raising boiling point Lowering freezing point Generating an osmotic pressure

2 to focus on…

Lowering vapor pressure Raising boiling point Lowering freezing point Generating an osmotic pressure

Boiling Point Elevation

a solution that contains a nonvolatile solute has a higher boiling pt than the pure solvent; the boiling pt elevation is proportional to the # of moles of solute dissolved in a given mass of solvent.

Like when adding salt to a pot of boiling water to make pasta

Boiling Point Elevation

Tb = kbmwhere: Tb = elevation of boiling pt

m = molality of solute (mol solute/kg solvent)

kb = the molal boiling pt elevation constant

kb values are constants; see table 16.3 pg. 495

kb for water = 0.52 °C/m

Ex: What is the normal boiling pt of a 2.50 m glucose, C6H12O6, solution?

“normal” implies 1 atm of pressure Tb = kbm

Tb = (0.52 C/m)(2.50 m)

Tb = 1.3 C

Tb = 100.0 C + 1.3 C = 101.3 C

Freezing/Melting Point Depression

The freezing point of a solution is always lower than that of the pure solvent.

Like when salting roads in snowy places so the roads don’t ice over or when making ice cream

Freezing/Melting Point Depression

Tf = kfmwhere: Tf = lowering of freezing point

m = molality of solute

kf = the freezing pt depression constant

kf for water = 1.86 °C/m kf values are constants;

see table 16.2 pg. 494

Ex: Calculate the freezing pt of a 2.50 m glucose solution.

Tf = kfm

Tf = (1.86 C/m)(2.50 m)

Tf = 4.65 C

Tf = 0.00C - 4.65 C = -4.65C

Textbook pg. 495 #34

Calculate the freezing-point depression (ΔTf) of a benzene solution containing 400. g of benzene and 200. g of acetone, C3H6O (solute).

Kf for benzene is 5.12 °C/m

Answer

ΔTf = Kf x m

ΔTf = (5.12 °C/m) x (m)

m =1

200. 58.09

8.61 0.400

molg x

mol solute gm

kg solvent kg

So, ΔTf = (5.12 °C/m) x (8.61 m) = 44.1 °C

Do the following on your paper

1. What is the boiling point of each solution?a) 0.50 mol glucose in 1000. g water

b) 1.50 mol NaCl in 1000. g water 2. What is the freezing point of each solution?

a) 1.40 mol Na2SO4 in 1750 g water

b) 0.060 mol MgSO4 in 100. g water

answers

ANSWERS

1a) 100.26 °C 1b) 100.78 °C 2a) -1.49 °C 2b) -1.1 °C

More calcs.

Ex: When 15.0 g of ethyl alcohol, C2H5OH, is dissolved in 750 grams of formic acid, the freezing pt of the solution is 7.20°C. The freezing pt of pure formic acid is 8.40°C. Determine Kf for formic acid.

Tf = kfm

1.20 C= (kf)( 0.4340 m)

kf = 2.76 C/m

mol 3255.0g 46.08

mol 1OHHC g 0.15 52

m 4340.0kg 75.0

mol 0.3255

EXTRA NOTES

Important info for lab, etc.

Electrolytes and Colligative Properties

• Colligative properties depend on the # of particles present in solution.• Because ionic solutes dissociate into ions, they have a greater effect on freezing pt and boiling pt than molecular solids of the same molal conc.

Electrolytes and Colligative Properties For example, the freezing pt of water is

lowered by 1.86°C with the addition of any molecular solute at a concentration of 1 m.– Such as C6H12O6, or any other covalent

compound

However, a 1 m NaCl solution contains 2 molal conc. of IONS. Thus, the freezing pt depression for NaCl is 3.72°C…double that of a molecular solute.– NaCl Na+ + Cl- (2 particles)

Electrolytes - Boiling Point Elevation and Freezing Point Depression

The relationships are given by the following equations:

Tf = kf ·m·n or Tb = kb·m·n

Tf/b = f.p. depression/elevation of b.p.m = molality of solute

kf/b = b.p. elevation/f.p depression constantn = # particles formed from the dissociation of

each formula unit of the solute

Ex: What is the freezing pt of a 1.15 m sodium chloride solution?

NaCl Na+ + Cl- n=2

Tf = kf·m·n

Tf = (1.86 C/m)(1.15 m)(2)

Tf = 4.28 C

Tf = 0.00C - 4.28 C = -4.28C