SOLUTIONS

SOLUTION – A homogeneous mixtureSOLVENT – The major component of a solutionSOLUTE – The minor component(s) of a solution

3G-1 (of 15)

LIKES DISSOLVE LIKES

Polar Component with a Nonpolar Component

Polar molecules attract with LDF, DDANonpolar molecules attract with LDF

To make a solution, polar and nonpolar molecules attract with LDFThese are too weak to overcome the polar molecules’ LDF, DDA

Polar

Nonpolar

3G-2 (of 15)

LIKES DISSOLVE LIKES

Two Nonpolar Components

Nonpolar molecules attract with LDF

To make a solution, different nonpolar molecules attract with LDFAll molecules equally attract each other a solution forms

3G-3 (of 15)

LIKES DISSOLVE LIKES

Two Polar Components

Polar molecules attract with LDF, DDA

To make a solution, different polar molecules attract with LDF, DDAAll molecules equally attract each other a solution formsUnless…

3G-4 (of 15)

LIKES DISSOLVE LIKES

Hydrogen-Bonding Solvent

The solute must be able to H-Bond with the solvent to dissolve

H O H O H O H H H

S Cl Cl

No H-Bonding

H-Bonding

3G-5 (of 15)

LIKES DISSOLVE LIKES

Hydrogen-Bonding Solvent

The solute must be able to H-Bond with the solvent to dissolve

H O H O H O H H H

H N H H

H-Bonding

H-Bonding

H H N H

H-Bonding

3G-6 (of 15)

LIKES DISSOLVE LIKES

Hydrogen-Bonding Solvent

The solute must be able to H-Bond with the solvent to dissolve

H O H O H O H H H

H-Bonding

H C H O

H-Bonding

3G-7 (of 15)

Solvent Solute Solubility

C6H6 C8H18

C6H6 CH2Br2

CH2Cl2 C8H18

CH2Cl2 CH2Br2

H2O C8H18

H2O CH2Br2

H2O CH3OH

H2O CH3CCH3

║

O

HighLowLowHighLowLow (a little)HighHigh

3G-8 (of 15)

When water-soluble polar molecules dissolve in water, the MOLECULES separate from each other, and exist as intact, neutral molecules

These solutions do not conduct electricity because no ions are formed

NONELECTROLYTE – A water-soluble compound whose solution does not conduct electricity

3G-9 (of 15)

Solute molecules surrounded by water molecules are said to be HYDRATED

C2H5OH (l) → C2H5OH (aq)H2O

C6H12O6 (s) → C6H12O6 (aq)H2O

Alcohols (CxHyOH) and sugars (Cx(H2O)y) are nonelectrolytes

3G-10 (of 15)

When acid molecules dissolve in water, the waters rip the acid molecules into IONS

These solutions conduct electricity because ions are formed

ELECTROLYTE – A water-soluble compound whose solution conducts electricity

Acids are electrolytes

3G-11 (of 15)

IONIZATION – The formation of ions during the dissolving process

HCl (g) → H+ (aq) + Cl- (aq)H2O

Strong acids produce many ions in solution, and their solutions are good conductors

HCl, HBr, HI, and acids with at least 2 more O’s than H’s are strong acids

Strong acids are STRONG ELECTROLYTES

3G-12 (of 15)

Weak acids produce few ions in solution, and their solutions are poor conductors

All other acids are weak acids

Weak acids are WEAK ELECTROLYTES

3G-13 (of 15)

HF (g) → HF(aq)H2O

When water-soluble ionic compounds dissolve in water, the IONS separate from each other

These solutions conduct electricity because ions are formed

Soluble ionic compounds are strong electrolytes

ION-DIPOLE ATTRACTIONS between the ions and the water molecules pull the ions into solution

3G-14 (of 15)

DISSOCIATION – The separation of ions from an ionic crystal during the dissolving process

NaCl (s) → Na+ (aq) + Cl- (aq)H2O

3G-15 (of 15)

SATURATED – A solution that contains as much dissolved solute as possible

SOLUTION EQUILIBRIUM – When the rates of dissolving and crystallizing are equal

Dissolving Rate Crystallizing Rate

A saturated solution is in solution equilibrium

SOLUBILITY – The maximum amount of solute that will dissolve in a specific amount of solvent

3H-1 (of 8)

(1) TemperatureIncreasing the temperature generally increases a solid’s solubilityIncreasing the temperature decreases a gas’s solubility

FACTORS AFFECTING SOLUBILITY

(2) PressureNo effect on a solid’s solubilityIncreasing the pressure increases a gas’s solubility

HENRY’S LAW – The amount of gas dissolved in a liquid is directly proportional to the pressure of the gas in contact with the liquidpgas = kCgas (Cgas is concentration of dissolved gas)

3H-2 (of 8)

CONCENTRATION UNITS

(1) MASS PERCENT – The mass of solute per mass of solution, times 100

Mass Percent = Mass Solute x 100 ___________________

Mass Solution

Find the mass percent of KCl in a solution made with 20.0 g KCl dissolved in 80.0 g H2O.

= 20.0 % 20.0 g KCl x 100 _____________________

100.0 g Solution

3H-3 (of 8)

(2) MOLE FRACTION (X) – The moles of solute per moles of solution

Find the mole fraction of KCl in a solution made with 20.0 g KCl dissolved in 80.0 g H2O.

x mol KCl_______________

74.55 g KCl

= 0.2683 mol KCl

20.0 g KCl

x mol H2O_______________

18.02 g H2O

= 4.440 mol H2O

80.0 g H2O

= 0.0570 0.2683 mol KCl ____________________________________

0.2683 + 4.440 mol Solution

3H-4 (of 8)

(3) MOLARITY (M) – The moles of solute per liter of solution

M = n ____

V

Find the molarity of a solution with 6.24 grams of magnesium chloride dissolved in enough water to make 500.0 mL of solution.

= 0.131 M MgCl2 0.06554 mol MgCl2 _________________________

0.5000 L Solution

n = quantity of matter (moles), V = volume (liters)

x mol MgCl2_________________

95.21 g MgCl2

= 0.06554 mol MgCl2

6.24 g MgCl2

3H-5 (of 8)

Used to easily calculate moles of solute by only measuring the volume of solution

Electrolyte solutions really consist of individual ions

Each MgCl2 formula unit has 1 Mg2+ ion and 2 Cl- ions

0.131 M x 1 = 0.131 M Mg2+

0.131 M x 2 = 0.262 M Cl-

Find the molarities of each ion in a 0.10 M Al2(SO4)3 solution

3H-6 (of 8)

Find the mass of potassium nitrate needed to prepare 250. mL of a 0.200 M potassium nitrate solution.

x 101.11 g KNO3 ___________________

mol KNO3

= 5.06 g KNO3

0.05000 mol KNO3

M = n ___

V

MV = n

x 0.250 L solution = 0.05000 mol KNO3

0.200 mol KNO3_____________________

L solution

3H-7 (of 8)

(4) MOLALITY (m) – The moles of solute per kilogram of solvent

m = mol solute _______________

kg solvent

Find the molality of a solution with 11.8 grams of glucose (C6H12O6, m = 180.16 g/mol) dissolved in 150.0 grams of water.

= 0.437 m C6H12O6 0.06550 mol C6H12O6 ___________________________

0.1500 kg H2O

x mol C6H12O6______________________

180.16 g C6H12O6

= 0.06550 mol C6H12O6

11.8 g C6H12O6

_

3H-8 (of 8)

Used because it does not change when the temperature changes

COLLIGATIVE PROPERTIES OF SOLUTIONS

COLLIGATIVE PROPERTIES – Properties that depend only on the number of dissolved solute particles, not what they are

VOLATILE – A substance (or solute) that evaporates easily NONVOLATILE – A substance (or solute) that does not evaporate

3I-1 (of 11)

(1) Vapor Pressure Lowering

A liquid’s equilibrium vapor pressure is lowered by having a nonvolatile solute dissolved in it

The pressure exerted by a vapor in equilibrium with its liquid is called the EQUILIBRIUM VAPOR PRESSUREWith solute particles dissolved in the liquid, there are less solvent molecules on the surface, therefore less solvent molecules can evaporate, so the rate of evaporation decreasesNow vapor molecules condense faster than liquid molecules evaporate, so the amount of vapor decreasesWhen the two rates are again equal, there are less vapor molecules than before, so a lower equilibrium vapor pressure

3I-2 (of 11)

10

2030

9.217.531.8

Temp (ºC) EVP of pure H2O (torr)

9.117.431.6

EVP of a dilute salt water solution (torr)

3I-3 (of 11)

(Temp °C)

EVPPure Water

Water Solution

RAOULT’S LAW – The equilibrium vapor pressure of a solution is proportional to the mole fraction of the solvent in the solution

psolution = Xsolventpºsolvent

psolution = equilibrium vapor pressure of the solution Xsolvent = mole fraction of the solvent in the solutionpºsolvent = equilibrium vapor pressure of the pure solvent

3I-4 (of 11)

1882 FRANÇOIS-MARIE RAOULT

A solution will obey Raoult’s Law if(1) the solution is dilute(2) the solvent and solute particles are about the same size(3) the solvent and solute particles have about the same attractive forces

Find the equilibrium vapor pressure of a solution at 20ºC that is prepared with 0.500 moles of sucrose dissolved in 35.00 moles of water, if the equilibrium vapor pressure of water at 20ºC is 17.5 torr.

= 0.9859 35.00 mol H2O ____________________________________

35.00 + 0.500 mol solution

psoln = Xsolvpºsolv = (0.9859)(17.5 torr)

= 17.3 torr

3I-5 (of 11)

Find the EVP of a solution at 20ºC that is prepared with 0.250 moles of Na2SO4 dissolved in 10.00 moles of water, if the EVP of water at 20ºC is 17.5 torr.

When a solute is an electrolyte (a salt or acid), the fact that it dissociates or ionizes must be considered

nsolute = 3 x 0.250 mol

3I-6 (of 11)

= 0.9302 10.00 mol H2O __________________________________

10.00 + 0.750 mol solution

psoln = Xsolvpºsolv = (0.9302)(17.5 torr)

= 16.3 torr

= 0.750 mol

Find the EVP of a solution at 23ºC that is prepared with 0.300 moles of KCl dissolved in 15.00 moles of water, if the EVP of water at 23ºC is 21.1 torr.

3I-7 (of 11)

nsolute = 2 x 0.300 mol

= 0.9615 15.00 mol H2O __________________________________

15.00 + 0.600 mol solution

psoln = Xsolvpºsolv = (0.9615)(21.1 torr)

= 20.3 torr

= 0.600 mol

In a solution, if both the solvent and solute are volatile, vapor pressure is produced from both components in the solution

psolution = psolvent + psolute

psolution = Xsolventpºsolvent + Xsolutepºsolute

3I-8 (of 11)

Find the EVP of a solution prepared with 0.300 moles of acetone and 0.100 moles of chloroform if the EVP of acetone is 293 torr and the EVP of chloroform is 345 torr.

psolution = Xsolventpºsolvent + Xsolutepºsolute

(0.7500)(293 torr) + (0.2500)(345 torr)

219.8 torr + 86.25 torr = 306.05 torr = 306 torr

3I-9 (of 11)

0.300 mol acetone________________________

0.400 mol solution

= 0.7500

0.100 mol chloroform___________________________

0.400 mol solution

= 0.2500

If the EVP of a solution equals the sum of the EVP’s of the 2 components, the solution obeys Raoult’s Law

IDEAL SOLUTION – A solution that obeys Raoult’s Law

3I-10 (of 11)

POSITIVE DEVIATION – Occurs when the attractions between solvent and solute molecules are WEAKER than the attractions in either pure component

NEGATIVE DEVIATION – Occurs when the attractions between solvent and solute molecules are STRONGER than the attractions in either pure component

Solution formation is endothermic Solution formation is exothermic

3I-11 (of 11)

(2) Boiling Point Elevation

A liquid’s boiling point is raised by having a nonvolatile solute dissolved in it

A solvent boils when its EVP equals the prevailing atmospheric pressure

With solute particles dissolved in the solvent, its EVP is lowered

Only when the temperature increases will the EVP again equal atmospheric pressure

the boiling point of the solution is now higher than the boiling point of the solvent

Atmospheric Pressure

100ºC101ºC102ºC

3J-1 (of 16)

(Temp °C)

EVP

100

Pure Water

Water Solution

3J-2 (of 16)

To determine the increase in boiling point from the solvent to the solution:

ΔTb = Kbmi

ΔTb = increase in the boiling point

Kb = molal boiling point constant of the solvent

m = molality of the solution

i = van’t Hoff factor of the solute = moles of particles in solution _____________________________________

moles of dissolved solute

1 3

C6H12O6 MgBr2 HCl Al2(SO4)3

2 5i =

3J-3 (of 16)

Find the boiling point of a solution that is prepared with 0.150 moles of sodium chloride dissolved in 90.0 grams of water.The boiling point of pure water is 100.00ºC and molal boiling point constant for water is 0.51 Cºkg H2O/mol solute.

0.150 mol NaCl____________________

0.0900 kg H2O

= 1.667 m NaCl

= 101.7ºC100.00ºC + 1.7 Cº

(0.51 Cºkg H2O/mol solute)(1.667 mol solute/kg H2O)(2) = 1.7 Cº

ΔTb = Kbmi

ΔTb =

3J-4 (of 16)

3J-5 (of 16)

All colligative properties can be used to calculate the molar mass of a nonvolatile solute

Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with 4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon disulfide has a boiling point of 47.5ºC.

molar mass X = grams X_____________

moles X

molar mass N.E. = 4.80 grams N.E. ____________________

? moles N.E.

3J-6 (of 16)

Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with 4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon disulfide has a boiling point of 47.5ºC. Pure carbon disulfide has a boiling point of 46.3ºC and molal boiling point constant of 2.37 Cºkg CS2/mol solute.

ΔTb = Kbmi ΔTb = 47.5ºC – 46.3ºC = 1.2 Cº

3J-7 (of 16)

ΔTb = Kb (mol solute) i _______________

(kg solvent)

(kg solvent) ΔTb = (mol solute)_____________________

Kb i

= (0.1500 kg CS2)(1.2 Cº) _____________________________________

(2.37 Cºkg CS2/mol solute)(1)

= 0.0759 mol solute

Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with 4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon disulfide has a boiling point of 47.5ºC. Pure carbon disulfide has a boiling point of 46.3ºC and molal boiling point constant of 2.37 Cºkg CS2/mol solute.

4.80 grams N.E._______________________

0.0759 moles N.E.

= 63 g/mol

3J-8 (of 16)

(3) Freezing Point Depression

A liquid’s freezing point is lowered by having a nonvolatile solute dissolved in it

A solvent freezes when the EVP of its liquid equals the EVP of its solidWith solute particles dissolved in the liquid solvent, its EVP is lowered Only when the temperature decreases will the EVP of the liquid solution equal the EVP of the solid solvent

0ºC-1ºC-2ºC

the freezing point of the solution is now lower than the freezing point of the solvent

3J-9 (of 16)

(Temp °C)

EVP

0

Pure Water

Water SolutionIce

3J-10 (of 16)

To determine the decrease in freezing point from the solvent to the solution:

ΔTf = Kfmi

ΔTf = decrease in the freezing point

Kf = molal freezing point constant of the solvent

m = molality of the solution

i = van’t Hoff factor of the solute

3J-11 (of 16)

Find the freezing point of a solution that is prepared with 0.150 moles of calcium chloride dissolved in 97.0 grams of water. The freezing point of pure water is 0.00ºC and molal freezing point constant for water is 1.86 Cºkg H2O/mol solute.

0.150 mol CaCl2_______________________

0.09700 kg H2O

= 1.546 m CaCl2

= -8.83ºC0.00ºC - 2.88 Cº

(1.86 Cºkg H2O/mol solute)(1.546 mol solute/kg H2O)(3) = 8.63 Cº

ΔTf = Kfmi

ΔTf =

3J-12 (of 16)

Find the molar mass of a nonelectrolyte if a solution prepared with 10.0 grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a freezing point of 3.51ºC.

3J-13 (of 16)

molar mass X = grams X_____________

moles X

molar mass N.E. = 10.00 grams N.E. ____________________

? moles N.E.

Find the molar mass of a nonelectrolyte if a solution prepared with 10.0 grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a freezing point of 3.51ºC. Pure benzene has a freezing point of 5.48ºC, a molal freezing point constant of 5.12 Cºkg benzene/mol solute, and a density of 0.780 g/mL.

ΔTf = Kfmi ΔTf = 5.48ºC – 3.51ºC = 1.97 Cº

3J-14 (of 16)

ΔTf = Kf (mol solute) i _______________

(kg solvent)

x 0.780 g benzene_____________________

mL benzene

= 78.00 g benzene

100.0 mL benzene

= 0.07800 kg benzene

Find the molar mass of a nonelectrolyte if a solution prepared with 10.0 grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a freezing point of 3.51ºC. Pure benzene has a freezing point of 5.48ºC, a molal freezing point constant of 5.12 Cºkg benzene/mol solute, and a density of 0.780 g/mL.

3J-15 (of 16)

= (0.07800 kg C6H6)(1.97 Cº) ______________________________________

(5.12 Cºkg C6H6/mol solute)(1)

= 0.03001 mol solute

ΔTf = Kfmi ΔTf = 5.48ºC – 3.51ºC = 1.97 Cº

(kg solvent) ΔTb = (mol solute)_____________________

Kb i

ΔTf = Kf (mol solute) i _______________

(kg solvent)

10.0 grams N.E._________________________

0.03001 moles N.E.

= 333 g/mol

Find the molar mass of a nonelectrolyte if a solution prepared with 10.0 grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a freezing point of 3.51ºC. Pure benzene has a freezing point of 5.48ºC, a molal freezing point constant of 5.12 Cºkg benzene/mol solute, and a density of 0.780 g/mL.

3J-16 (of 16)

Cells in an ISOTONIC SOLUTION have equal flow rates of water in and out

Cells in a HYPERTONIC SOLUTION have water flowing out faster than flowing in

Cells in a HYPOTONIC SOLUTION have water flowing in faster than flowing out

3K-1 (of 6)

(4) OSMOTIC PRESSURE

SEMIPERMEABLE MEMBRANE – A membrane that will allow small molecules to pass through but not big ones

water / not hydrated salt ionswater, salt ions / not proteins

Because the membrane blocks salt ions, water goes out of the tube slower than it goes in

The extra mass of salt water builds up pressure on the membrane, forcing water through faster

OSMOTIC PRESSURE (π) – The pressure on a semipermeable membrane needed to equalize the passage of water across the membrane between two solutions of different concentrations

3K-2 (of 6)

To determine the osmotic pressure in a solution:

πV = inRT

π = Osmotic Pressure (atm) V = Volume of Solution (L)

i = van’t Hoff factor

n = Difference in Quantity of Solute between the 2 solutions (mol)R = Universal Gas Constant (0.08206 Latm/molK)T = Temperature (K)

3K-3 (of 6)

To determine the osmotic pressure in a solution:

πV = inRT

π = inRT _______

V π = iMRT

n/V is molarity

3K-4 (of 6)

Find the osmotic pressure that would develop in a solution that is prepared at 22ºC with 0.0300 moles of glucose dissolved in enough water to make 100. mL of solution.

π = (1)(0.0300 mol)(0.08206 Latm/molK)(295.2 K) __________________________________________________________

0.100 L

= 7.27 atm

π = inRT _______

V

22ºC + 273.2 K = 295.2 K

3K-5 (of 6)

Find the molar mass of a protein if a solution is prepared by dissolving1.00 x 10-3 grams of the protein in enough water to make 1.00 milliliter of solution, and the osmotic pressure at 25ºC is 1.12 torr.

25ºC + 273.2 = 298.2 K

πV = n_____

iRT

= (0.001474 atm)(0.00100 L) _________________________________________

(1)(0.08206 Latm/molK)(298.2 K)

= 6.024 x 10-8 mol

1.00 x 10-3 g protein__________________________________

6. 024 x 10-8 moles protein

= 16,600 g/mol

π = inRT _______

V

x 1 atm_____________

760.0 torr

= 0.001474 atm

1.12 torr

3K-6 (of 6)