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# Chapter 11c: Solutions and Their Properties - PowerPoint PPT Presentation

Chapter 11c: Solutions and Their Properties. Some Factors Affecting Solubility. Solubility The amount of solute per unit of solvent needed to form a saturated solution Miscible Mutually soluble in all proportions Effect of Temperature on Solubility

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### Chapter 11c: Solutions and Their Properties

Solubility

The amount of solute per unit of solvent needed to form a saturated solution

Miscible

Mutually soluble in all proportions

Effect of Temperature on Solubility

1. Most solid substances become more soluble as temperature rises

2. Most gases become less soluble as temperature rises

Effect of Pressure on Solubility

1. No effect on liquids or solids

2. The solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas over the solution, @ 25°C

Henry’s Law solubility = k x P

k = constant characteristic of specific gas, mol/Latm

P = partial pressure of the gas over the sol’n

a) Equal numbers of gas molecules escaping liquid and returning to liquid

b)   Increase pressure, increase # of gas molecules returning to liquid, solubility increases

c) A new equilibrium is reached, where the #’s of escaping = # of returning

Which of the following will become less soluble in water as the temperature is increased?

• NaOH(s)

• CO2(g)

The solubility of CO2 in water is 3.2 x 10-2 M @ 25°C and 1 atm pressure. What is the Henry’s-Law constant for CO2 in mol/L atm?

solubility = k x P

Physical Behavior of Solutions: Colligative Properties

o

• H2O b.p. 100.0o C f.p. 0.0o C

• 1.00 m NaCl b.p. 101.0o C f.p. -3.7o C

o

Colligative properties:

Properties that depend on the amount of a dissolved solute but not on its chemical identity

There are four main colligative properties:

1. Vapor pressure lowering

2. Freezing point depression

3. Boiling point elevation

4. Osmosis:

The vapor pressure of a solution is different from the vapor of the pure solvent.

Two different cases:

1. solute is non-volatile

solute has no vapor pressure of its own

example: dissolving a solid

vapor pressure of the solution is always lower than that of the pure solvent

2. solute is volatile

solute has its own vapor pressure

example: mixing 2 liquids

vapor pressure of the mixture is intermediate between the vapor pressures of the two pure liquids

Vapor-pressure Lowering of Solutions: Raoult’s Law of the pure solvent.

• Solutions with a Nonvolatile Solute

If the solute is nonvolatile and has no appreciable vapor pressure of its own (solid dissolved) the vapor pressure of the solution is always lowerthat that of the pure solvent.

Solutions with a Nonvolatile Solute!!! of the pure solvent.

Raoult’s Law: Psoln = Psolv· Xsolv

Psoln = vapor pressure of the solution

Psolv = vapor pressure of the pure solvent

Xsolv = mole fraction of the solvent in the solution

Vapor pressure lowering is a colligative property (only

dependent on amount of solute and not on its chemical identity!)

For ionic substances calculate the total moles of solute particles, 1 mol NaCl will result in 1 mol Na+ and 1 mol

Cl- = 2 moles of particles

1 mol Na2SO4 will give 3 moles of particles

Raoult’s Law applies to only Ideal solutions of the pure solvent.

1. Law works best when solute concentrations are low an d when solute and solvent particles have similar intermolecular forces.

2. Further complication is that at higher concentrations

ionic compounds are not 100% dissociated.

Example

1 mol NaCl is only 90% dissociated

10% is undissociated

resulting in less particles in solution than expected

Example 9 of the pure solvent.

What is the vapor pressure (in mm Hg) of a solution prepared by dissolving 5.00 g of benzoic acid (C7H6O2) in 100.00 g of ethyl alcohol (C2H6O) at 35°C? The vapor pressure of the pure ethyl alcohol at 35°C is 100.5 mm Hg

P of the pure solvent.soln = Psolv· Xsolv MM C7H6O2 = 122.12 g/mol

Psolv = 100.5 mm Hg MM C2H6O = 46.07 g/mol

1mol

a) 5 g C7H6O2 x ---------- = 0.0409 mol

122.12g

b) 100 g C2H6O x 1 mol/ 46.07 g = 2.17 mol

c) Xsolv = 2.17 mol / (0.0409 + 2.17 mol) = 0.982

Psoln = Psolv· Xsolv

= 100.5 mm Hg · 0.982 = 98.7 mm Hg

Solutions with a Nonvolatile Solute of the pure solvent.

Close-up view of part of the vapor pressure curve for a pure solvent and a solution of a nonvolatile solute. Which curve represents the pure solvent, and which the solution?

Why?

Reason for vapor pressure lowering of the pure solvent.

DG = DHvap - T DS DHvap = positive, disfavored

DS = positive, favored

DHvap is (nearly) the same for a pure solvent and a solvent in

a solution

DS is different

solvent in a solution has more disorder than pure solvent

entropy of a solution is higher than the pure solvent

entropy of the vapor in both cases the same

Entropy increase for vaporization from a solution is smaller than

vaporization from a pure solvent

Less entropy increase means less favored

Solutions with a Volatile Solute!! of the pure solvent.

For a mixture of 2 volatile liquids A and B the overall vapor pressure is the sum of the vapor pressure of the 2 components (Dalton’s law)

Ptotal = PA + PB

The vapor pressure for each component is calculated by

Raoult’s law: vapor pressure is equal to the mole fraction of A times the vapor pressure of pure A

Ptotal = PA + PB = (P0A · XA) + (P0B · XB)

P°A = vapor pressure of pure A XA = mole fraction of A

P°B = vapor pressure of pure B XB = mole fraction of B

Solutions with a Volatile Solute of the pure solvent.

Close-up view of part of the vapor pressure curves for two pure liquids and a mixture of the two. Which curves represent the pure liquids, and which the mixture?

P of the pure solvent.total should be intermediate to A & B

Raoult’s law applies only to ideal solutions

Most real solutions show deviations

Example 10 of the pure solvent.

What is the vapor pressure ( in mm Hg) of a sol’n prepared by dissolving 25.0 g of ethyl alcohol (C2H5OH) in 100.0 g of water at 25°C? The vapor pressure of pure water is 23.8 mm Hg and the vapor pressure of ethyl alcohol is 61.2 mm Hg at 25°C

Example 10 of the pure solvent.

P°H2O = 23.8 mm Hg

P°C2H5OH = 61.2 mm Hg

XH2O = mole fraction of A

25 g C2H5OH x 1 mol / 46.07 g = 0.543 mol C2H5OH

100.0 g H2O x 1 mol/ 18 g = 5.56 mol H2O

XH2O = 5.56 /(5.56 + 0.543) = 0.91

XC2H5OH = mole fraction of B

XC2H5OH = 0.543 / (0.543 + 5.56) = 0.09

Ptot = (23.8 x 0.91) + (61.2 x 0.09) = 27.2 mm Hg

A solution has a lower vapor pressure than the pure liquid.

To reach the atmospheric pressure (boiling point) the

temperature must be higher.

• Tb = Kb · m Boiling point elevation

• Tf = Kf · m Freezing point depression

Kb = molal boiling-point elevation constant

Kf = molal freezing-point depression constant

m = molality

Example 11 Solutions

What is the normal boiling point in °C of a solution prepared by dissolving 1.50 g of aspirin (C9H8O4) in 75.00 g of chloroform (CHCl3)? The normal boiling point of chloroform is 61.7 °C and Kb of chloroform is 3.63 °C kg/mol

Example 11 Solutions

Tb = Kb · m

m = mole solute / kg solvent

MM C9H8O4 = 180.16 g/mol

1.50 g C9H8O4 x 1 mol / 180.16 g = 0.00833 mol C9H8O4

75.00 g CHCl3 = 0.07500 kg CHCl3

m =0 .00833 mol C9H8O4 / 0.07500 kg CHCl3 = 0.111 m

Tb = 3.63 °C kg/mol · 0.111 mol/kg = 0.403 °C

Boiling point = 0.403 °C + 61.7 °C = 62.1 °C

Osmosis and Osmotic Pressure Solutions

Membranes are semipermeable materials

They allow water and other small molecules to

pass through, but they block the passage of

larger molecules or ions.

All living cells contain membranes Solutions

and osmosis is important in biological systems

Osmosis provides the primary means by which

water is transported into and out of cells

Osmosis is responsible for the ability of plant

roots to suck up water from the soil

Osmosis Solutions

Thermodynamic explanation Solutions

Every system wants to balance out the concentration

One side pure solvent

Other side solution ordered system

The system tries to get into a more disordered

more randomness state

The entropy will increase

Osmosis is similar to diffusion

Osmosis Pressure Solutions

1. The amount of pressure necessary to achieve equilibrium

2.  = MRT

 = osmotic pressure

M = molarity

R = gas constant, .08206 L atm/K mol

T = temperature in Kelvin

Isotonic sodium chloride solution Solutions

The total concentration of dissolved particles inside

red blood cells is 0.30 M.

What is the osmotic pressure at body temp (310 k) ?

Isotonic sodium chloride solution Solutions

The total concentration of dissolved particles inside

red blood cells is 0.30 M.

What is the osmotic pressure at body temp (310 k) ?

• = MRT

• = 0.30 mol/L x 0.08206 L atm/K mol x 310 K

• = 7.63 atm

Otherwise the blood cells would burst!

Therefore isotonic NaCl solutions are injected

Q: What is the mass% of an isotonic NaCl solution?

Since the molarity in blood cells is 0.3 M, we need 0.15 M

NaCl (0.15 M Na+ and 0.15 M Cl-)

Na = 23.0 amu

Cl = 35.5 amu NaCl = 58.5 amu

0.15 M NaCl = 58.5 x 0.15 = 9 g/L ; 0.9 mass%

Example 12 Solutions

What osmotic pressure in atm would you expect for a solution of 0.125 M C6H12O6 that is separated from pure water by a semipermeable membrane at 310 K?

 = MRT

= (0.125 mol/L)(.08206 L atm/K mol)(310 K) = 3.18 atm

Example 13 Solutions

A solution of unknown substance in water at 300 K gives rise to an osmotic pressure of 3.85 atm. What is the molarity of the solution?

 = MRT

M = /RT

M = 3.85 atm / [(.08206 L atm/K mol)(300 K)]

M = .156 mol/L

Some uses of colligative properties Solutions

1 Freezing-point depression

- sprinkling of salt to melt snow

- antifreeze in automobile cooling system

- de-icing of airplane wings

2 Osmosis

- desalination of seawater with reverse osmosis

3 Molar mass determination

can use any four colligative properties

most accurate is osmotic pressure, since the magnitude

of osmosis effect is so great

Example 14 Solutions

• What is the molar mass of sucrose if a solution prepared by dissolving 0.822 g of sucrose in water and diluting to a volume of 300.0 mL has an osmotic pressure of 149 mm Hg at 298 K?

Solutions = MRT

149 mm Hg x 1 atm / 760 mm Hg = 0 .196 atm

M =  /RT

= 0.196 atm / [(0.08206 L atm/K mol)(298 K)]

= 0.00802 mol/L

0.00802 mol/L x 1 L/1000 mL x 300 mL = 0.00241 mol

MM = mass of sucrose / moles of sucrose

= 0.822 g / 0.00241 mol = 341.08 g/mol

Summary Solutions