Physical Properties of Solutions

Chapter 12

SolutionStoichiometry

end of Chapter 4

Problems: 12.12, 12.15, 12.16, 12.17, 12.18, 12.21, 12.22, 12.28, 12.36, 12.38, 12.51, 12.54, 12.55, 12.57, 12.60, 12.65, 12.76, 12.77, 12.1124.12, 4.60, 4.70, 4.72, 4.74, 4.77, 4.88

12.1

A solution is a homogenous mixture of 2 or more substances

The solute is(are) the substance(s) present in the smaller amount(s)

The solvent is the substance present in the larger amount

An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.

A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.

nonelectrolyte weak electrolyte strong electrolyte4.1

A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.

An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.

A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.

Sodium acetate crystals rapidly form when a seed crystal isadded to a supersaturated solution of sodium acetate.

12.1

12.2

Three types of interactions in the solution process:• solvent-solvent interaction• solute-solute interaction• solvent-solute interaction

Hsoln = H1 + H2 + H3

“like dissolves like”

Two substances with similar intermolecular forces are likely to be soluble in each other.

• non-polar molecules are soluble in non-polar solvents

CCl4 in C6H6

• polar molecules are soluble in polar solvents

C2H5OH in H2O

• ionic compounds are more soluble in polar solvents

NaCl in H2O or NH3 (l)

12.2

Concentration UnitsThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

Percent by Mass

% by mass = x 100%mass of solutemass of solute + mass of solvent

= x 100%mass of solutemass of solution

12.3

Mole Fraction (X)

XA = moles of A

sum of moles of all components

Concentration Units Continued

M =moles of solute

liters of solution

Molarity (M)

Molality (m)

m =moles of solute

mass of solvent (kg)

12.3

What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?

m =moles of solute

mass of solvent (kg)M =

moles of solute

liters of solution

Assume 1 L of solution:5.86 moles ethanol = 270 g ethanol927 g of solution (1000 mL x 0.927 g/mL)

mass of solvent = mass of solution – mass of solute

= 927 g – 270 g = 657 g = 0.657 kg

m =moles of solute

mass of solvent (kg)=

5.86 moles C2H5OH

0.657 kg solvent= 8.92 m

12.3

Convert % mass to Molarity

• What is the Molarity of a 95% acetic acid solution? (density = 1.049 g/mL)

If you assume 1 L, that amount of solution = 1049 g

95% of the solution is acetic acid

1049 g solution x 0.95 = 997 g solute

997 g X 1 mol/60.05 g = 16.6 mol solute

Since we assumed 1 L, that’s 16.6 mol / 1 L or

16.6 M

Temperature and Solubility

Solid solubility and temperature

solubility increases with increasing temperature

solubility decreases with increasing temperature

12.4

Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities.

Suppose you have 90 g KNO3 contaminated with 10 g NaCl.

Fractional crystallization:

1. Dissolve sample in 100 mL of water at 600C

2. Cool solution to 00C

3. All NaCl will stay in solution (s = 34.2g/100g)

4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g

12.4

Temperature and Solubility

Gas solubility and temperature

solubility usually decreases with

increasing temperature

12.4

Pressure and Solubility of Gases

12.5

The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).

c = kP

c is the concentration (M) of the dissolved gas

P is the pressure of the gas over the solution

k is a constant (mol/L•atm) that depends onlyon temperature

low P

low c

high P

high c

Chemistry In Action: The Killer Lake

Lake Nyos, West Africa

8/21/86CO2 Cloud Released

1700 Casualties

Trigger?

• earthquake

• landslide

• strong Winds

Solution Stoichiometry (Chapter 4)

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

M = molarity =moles of solute

liters of solution

What mass of KI is required to make 500. mL ofa 2.80 M KI solution?

volume KI moles KI grams KIM KI M KI

500. mL = 232 g KI166 g KI

1 mol KIx

2.80 mol KI

1 L solnx

1 L

1000 mLx

4.5

4.5

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

Dilution

Add Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=4.5

How would you prepare 60.0 mL of 0.2 MHNO3 from a stock solution of 4.00 M HNO3?

MiVi = MfVf

Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L

4.5

Vi =MfVf

Mi

=0.200 x 0.06

4.00= 0.003 L = 3 mL

3 mL of acid + 57 mL of water = 60 mL of solution

Gravimetric Analysis

4.6

1. Dissolve unknown substance in water

2. React unknown with known substance to form a precipitate

3. Filter and dry precipitate

4. Weigh precipitate

5. Use chemical formula and mass of precipitate to determine amount of unknown ion

TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

the indicatorchanges color

4.7

What volume of a 1.420 M NaOH solution isRequired to titrate 25.00 mL of a 4.50 M H2SO4 solution?

4.7

WRITE THE CHEMICAL EQUATION!

volume acid moles acid moles base volume base

H2SO4 + 2NaOH 2H2O + Na2SO4

4.50 mol H2SO4

1000 mL solnx

2 mol NaOH

1 mol H2SO4

x1000 ml soln

1.420 mol NaOHx25.00 mL = 158 mL

M

acid

rx

coef.

M

base

Chemistry in Action: Metals from the Sea

CaCO3 (s) CaO (s) + CO2 (g)

Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)

CaO (s) + H2O (l) Ca2+ (aq) + 2OH (aq)-

Mg2+ (aq) + 2OH (aq) Mg(OH)2 (s)-

Mg2+ + 2e- Mg

2Cl- Cl2 + 2e-

MgCl2 (l) Mg (l) + Cl2 (g)

Now back to Chapter 12…Now back to Chapter 12…

Colligative Properties of Nonelectrolyte Solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Vapor-Pressure Lowering

Raoult’s law

If the solution contains only one solute:

X1 = 1 – X2

P 10 - P1 = P = X2 P 1

0

P 10 = vapor pressure of pure solvent

X1 = mole fraction of the solvent

X2 = mole fraction of the solute12.6

P1 = X1 P 10

Fractional Distillation Apparatus

12.6

Boiling-Point Elevation

Tb = Tb – T b0

Tb > T b0 Tb > 0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

Tb = Kb m

m is the molality of the solution

Kb is the molal boiling-point elevation constant (0C/m)

12.6

Freezing-Point Depression

Tf = T f – Tf0

T f > Tf0 Tf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

Tf = Kf m

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m)

12.6

12.6

What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

Tf = Kf m

m =moles of solute

mass of solvent (kg)= 2.41 m=

3.202 kg solvent

478 g x 1 mol62.01 g

Kf water = 1.86 0C/m

Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C

Tf = T f – Tf0

Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C

12.6

Colligative Properties of Nonelectrolyte Solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

12.6

Vapor-Pressure Lowering P1 = X1 P 10

Boiling-Point Elevation Tb = Kb m

Freezing-Point Depression Tf = Kf m

Osmotic Pressure () = MRT

Colligative Properties of Electrolyte Solutions

12.7

0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

0.1 m NaCl solution 0.2 m ions in solution

van’t Hoff factor (i) = actual number of particles in soln after dissociation

number of formula units initially dissolved in soln

nonelectrolytesNaCl

CaCl2

i should be

12

3

Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision?

a) sand, SiO2

b) Rock salt, NaCl

c) Ice Melt, CaCl2

Change in Freezing Change in Freezing Point Point

Boiling-Point Elevation Tb = i Kb m

Freezing-Point Depression Tf = i Kf m

Osmotic Pressure () = iMRT

Colligative Properties of Electrolyte Solutions

12.7

Change in Freezing Change in Freezing Point Point

Common Applications of Common Applications of Freezing Point Freezing Point DepressionDepression

Propylene glycol

Ethylene glycol – deadly to small animals

Change in Boiling Change in Boiling Point Point

Common Applications of Common Applications of Boiling Point ElevationBoiling Point Elevation

At what temperature will a 5.4 molal At what temperature will a 5.4 molal solution of NaCl freeze?solution of NaCl freeze?

SolutionSolution

∆∆TTFPFP = K = Kff • m • i • m • i

∆ ∆TTFPFP = (1.86 = (1.86 ooC/molal) • 5.4 m • 2C/molal) • 5.4 m • 2

∆ ∆TTFP FP = 20.1= 20.1 ooCC

FP = 0 – 20.1 = -20.1FP = 0 – 20.1 = -20.1 ooCC

Freezing Point Freezing Point DepressionDepression

The Cleansing Action of Soap

12.8

Osmotic Pressure ()

12.6

Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.

A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.

Osmotic pressure () is the pressure required to stop osmosis.

dilutemore

concentrated

HighP

LowP

Osmotic Pressure ()

= MRT

M is the molarity of the solution

R is the gas constant

T is the temperature (in K) 12.6

A cell in an:

isotonicsolution

hypotonicsolution

hypertonicsolution

12.6

Chemistry In Action: Desalination

A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.

Colloid versus solution

• collodial particles are much larger than solute molecules

• collodial suspension is not as homogeneous as a solution

12.8

Colloids

• Brownian motion • Tyndall Effect

Suspensions

• These are mixed, but not dissolved in each other

• Will settle over time

• Particles are bigger than 1 micrometer (larger than colloid)

• Examples: dust in air, muddy water