Chapter 18 Solutions. I. Solutions A. Characteristics of solutions 1. Homogeneous mixture 2. Contains a solute and solvent 3. Can be a gas, liquid or.

Chapter 18

Solutions

I. Solutions

A. Characteristics of solutions

1. Homogeneous mixture

2. Contains a solute and solvent

3. Can be a gas, liquid or solid

4. Soluble - able to dissolve

5. Insoluble - not able to dissolve

6. Miscible - 2 liquids able to mix

7. Immiscible - 2 liquids not able to mix

B. Dissolving mechanisms

1. Solute-solute attraction is broken up, requiring energy

2. Solvent-solvent attraction is broken up, requiring energy

3. Solute-solvent attraction is formed, releasing energy

C. Solvation

1. When solvent particles surround solute particles to form a

solution

2. Depends on polarity and bonding

D. Factors that affect rate of dissolving

1. Agitating the mixture

2. Increase surface area

3. Increase temperature

4. Heat of solution (exothermic or endothermic reactions)

E. Solubility

1. Ability to dissolve

2. Saturated solutions - maximum amount of solute that can

dissolve

3. Unsaturated solutions - less solute than a saturated

solution

4. Supersaturated solution - has more solute than a

saturated solution at the same temperature

a. Formed at higher tempb. Cooled slowly

5. As temperature is increased solids dissolve faster

6. Lower temp. gases dissolve better

7. Pressure affects gases not solids

F. Henry’s Law

1. At a given temp. the solubility (s) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid

2. S1 = S2

P1 P2

units = g/L

3. Ex. If 0.24 g of a gas dissolves in 1.0 L

of water at 1.5 atm pressure, how much gas will dissolve if the pressure is raised to 6.0 atm? T remains constant.

II. Concentration of Solutions A. Molarity

1. Concentration: amount of solute that is dissolved in a given quantity of solvent

a. Dilute solution: contains a small amount of solute, large amounts of solvent

b. concentrated solution:

Contains a small amount of solvent and a large

amount of solute

2. Molarity (M)

a. Way to express concentration

b. Number of moles of solute in liters of solution

B. Making dilutions 1. Moles of solute before dilution = moles of solute

after dilution

2. M1V1 = M2V2

3. Example:

How many milliters of a stock solution of 2.00 M MgSO4 would you need to prepare 100.0 mL of 0.400 M MgSO4?

a. Ex.

What is the percent by mass of sodium carbonate in a water solution containing 0.497 g

NaCO3 in 58.3 g of solution?

C. Percent Solutions

b. The % mass of MgCl2 in a water solution is 1.47%. How many g of solute are dissolved in each 500.00 g of solution?

a. Ex.

What is the %v/v of ethanol (C2H6O) in the final

solution when 85 mL of ethanol is diluted to a volume of 250 mL with water?

III. Colligative Properites

A. Definition: Properties that depend only on the number of particels

dissolved in a given mass of solvent

B. Vapor pressure

1. A vapor that is dynamic equilibrium with its liquid

in a closed system

2. Dynamic equilibrium: when the forward and the reverse

reactions are equal

3. If it contians nonvolatile solutes (does not dissociate)

4. If it contains ionic compounds it dissociates completely.

C. Boiling point – the temperature difference between a solutions

boiling point and a pure solvent’s boiling point

D. ΔTb (boiling point elevation) – directly proportional to molality

E. molality

1. moles of solute per kg of solution

2. m = moles of solute

kg of solution

F . For nonelectrolyte(does’t dissociate)

1. ΔTb = Kbm Kb – molal boiling point elevation

constant units oC/m

m- molality2. 1 m or .512 oC

.512 oC 1 m for when water is the solvent

3. What is the boiling point elevation when 31.5 g of methanol

(C10H20O) is dissolved in 258 g of acetic acid? Kb for acetic acid is 2.93 oC/m.

4. How many grams of styrene glycol (C8H10O2) must be dissolved in 98.7g

of bezene to raise the boiling point by 8.57 oC?Kb for benzene is 2.67oC/m.

G. Freezing point depression (ΔTf)

1. The temperature difference between the freezing point

of a solution and the freezing point of its pure solvent.

2. For nonelectrolytes

ΔTf = Kfm

3. When water is the solvent Kf = 1.86 oC/m

4. Calculate the freezing point of a solution containing 5.70 g of sugar(C12H22O11) in

50.0 g of water.

5. If 13.4g of the medication scopolamine, C17H21NO4, is

dissolved in 50.3 g of water, how much will the freezing point be lowered?

IV. Molar fraction and Electrolytes

A. Mole fraction

1. The ratio of the moles of solute in solution to the total number of moles of both solvent and solute

2. Mole expressions

a. XA = nA

nA + nB

(for solute)

b. XB = nB

nA + nB

(for solvent)

3. Ex. Compute the mole fraction of each component in a solution of 1.25 mole of ethylene glycol

(EG) and 4.00 mole water.

B. Boiling point elevations

∆Tb = (# particles)(Kb)(m)

C. Freezing point depression

∆Tf = (# particles)(Kf)(m)

D. Molar mass

1. boiling point elevation

m = ∆Tb

Kb

2. freezing point depression

m = ∆Tf

Kf