Ch 10 Solutions

(Colligative Properties)

Masterson & Hurley

• Water is known as “the universal solvent”

• This is due to water’s unusual properties, it is:

1. Bent in shape

2. A Highly polar molecule

3. And it causes ionic and polar substances to be soluble, (remember “like dissolves like”)

4. Most ions are soluble in water: cations separate from anions

• Covalent compounds (like alcohols) do not separate into ions, they stay together because they are covalently bonded.

• Alcohols become less soluble because of the parent chain, as more carbons are added it becomes less polar (less like water).

• This causes less hydrogen bonding between the alcohol and the water.

Electrolytes are ionic, nonelectrolytes are covalent

• The van’t Hoff factor = i

• Ionic compounds have a factor greater than one.

• NaCl = 2 = 1 Na+ and 1 Cl-

• H2SO4 = 3 = 2 H+ and 1 SO42-

• CH3OH = 1 (covalent, does NOT dissociate)

Ch 10 Solutions

Masterson

10.1 Concentration Units

10.2 Principles of Solubility

Memorize!

• Solution: is a homogeneous mixture, the components are uniformly intermingled.

• Solvent: the substance present in the largest amount (usually the component doing the dissolving).

• Solute: the other substance(s) (usually the component being dissolved).

• Aqueous solutions: are solutions with water as the solvent.

Common Metric Conversions

WATCH UNITS!

1000 mL = 1 L

Molarity (M)

• The concentration of a solute is a measure of the amount of solute in a given volume of solution.

• Concentration can have various units, Molarity (M) is commonly used by chemists:

Molarity (M) = moles of solute = mol

Liters of solution L

• Calculate the standard solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.

• 11.5 g NaOH x 1 mol NaOH = 0.288 mol

40.0 g NaOH

• M = mol = 0.288 mol = 0.192 M NaOH

L 1.50 L

Concentration of Ions

• If a solution is 1.0 M NaCl it contains 1.0 mol Na+ ions and 1.0 mole of Cl- for example…

• Give the concentration of each type of ion in the following solution:

a. 0.50 M Co(NO3)2

Answer: Co(NO3)2 Co2+ + 2 NO3-

1 mol Co2+ ions = 0.50 M Co2+

2 mol NO3- ions = 2 (0.50 M) NO3- = 1.0 M NO3-

Calculate the number of moles of Cl- ions in 1.75 L of 1.0 x 10-3 M ZnCl2

• M * L = mol

• ZnCl2 1 mol of Zn2+ + 2 mols of Cl-

• (1.0 x 10-3 M)(1.75 L) = 0.00175 moles

• (0.00175 moles)(2 mol Cl-) = 0.00350 mol Cl-

• The dilution process is used to save time and space in the lab, often concentrated solutions (called stock solutions) are diluted by adding water to achieve the desired molarity.

• M1 * V1 = M2 * V2

• What volume of 16M H2SO4 must be used to prepare 1.5 L of a 0.10 M solution?

• (0.10 M)(1.5 L) = (16M)(V2)

= 0.0094 L of concentrated H2SO4 or 9.4 mL

• Therefore we must take 9.4 mL of concentrated H2SO4 and dilute to the 1.5 L mark with water…

• (Remember to do what you oughta add acid to water…)

• In performing calculations associated with dilutions remember that the moles of a solute are not changed by the dilution:

• M1 * V1 = M2 * V2

• M1 * V1 = mol solute before dilution

= mol solute before dilution M2 * V2

• Heat can either help or hurt a solute to be dissolved to form a solution.

• Heat can act as a reactant or product.

• Heat is given the symbol ∆H

• If ∆H = - it is exothermic, it gives off heat

• If ∆H = + it is endothermic, it absorbs heat (feels cool)

Na+ + NO3- NaNO3 ∆H = +22.8 kJ

22.8 kJ + Na+ + NO3- NaNO3

In this case heat is a reactant & is absorbed.

So to increase the reaction from left to right, we can add heat, known as Le Chatelier’s Principle

• Now if…

• 2 H+ + SO4 - H2SO4 ∆H = -22.8 kJ

• 2 H+ + SO4 - H2SO4 + 22.8 kJ

In this case heat is a product.

So to increase the reaction from left to right, we can decrease heat, known as Le Chatelier’s Principle

When The Original Concentration is: Start With:

Mass % 100 g solution

Molarity (M) 1.00 L solution

Molality (m) 1000 g solvent

Mole fraction (X) 1 mol (solute + solvent)

Compound Molality Weight % Mole Fraction

NaI 0.15

C2H5OH 5.0

C12H22O11 0.15

10.2 Principles of Solubility

• Henry’s Law: Cg = k Pg

• Cg = concentration in the solution

• K = constant

• Pg = partial pressure of the gas over the solution.

• Pressure has an effect on solubility for a gas-liquid systems.

• Raising the pressure increases the solubility of a gas.

• Carbonated drinks & the “bends.”

• The solubility of most solids increases with temperature.

– Sugar and hot vs. iced tea.

• Gases on the other hand, always decrease in solubility with increasing temperature.

– Carbonated drinks on the counter vs. in the fridge.

10.3 Colligative Properties of NONelectrolytes

Vapor Pressure

• Vapor Pressure: Demo of water & isopropyl alcohol.

• The vapor pressure is greater for the alcohol.

• The water molecules have a strong attraction to each other due to hydrogen bonding.

• Alcohol has weaker hydrogen bonding, so they move from a liquid to a gas phase more easily, compared to water.

• If one has pure water and salt water, the water will move from a liquid to a gas phase much easier than the solution.

• This is because of the vapor pressure of water is more than that of the solution.

• The more Van’t Hoff factors the less the vapor pressure:

• NaCl = 2 = 1 Na+ and 1 Cl-

• H2SO4 = 3 = 2 H+ and 1 SO42-

• CH3OH = 1 (covalent, does NOT dissociate)

• The sulfuric acid would have the least vapor pressure (most difficult to evaporate)

Raoult’s Law

• Raoult’s Law:

• The vapor pressure of a solution is directly proportional to the mole fraction of the solvent present.

• Psoln = XsolventP°solvent

• Psoln = is the observed vapor pressure of solution

• Xsolvent =is the mole fraction of the solvent

• P°solvent = is the vapor pressure of pure solvent

• An ideal solution, a plot of P vs X at constant temperature gives a straight line with a slope of Pº.

• Ideal solutions obey Raoult’s law. Real solutions best approximate the behavior of ideal solutions.

• The vapor pressure of a solution containing 53.6 g of glycerin in 133.7 g of ethanol is 113 torr at 40ºC.

• Calculate the vapor pressure of ethanol.

Boiling Point Elevation and

Freezing Point Lowering

• Boiling point elevation: ∆Tb = kb m

• Freezing point elevation: ∆Tf = kf m

• Antifreeze is 50.0 mL of C2H6O2 (d = 1.12 g/mL) in 50.0 g of water. Calculate the freezing point of mixture. (kf = 1.86ºC/m)

Osmotic Pressure

• Some units to consider:

• Pressure: 1 atm = 760 mmHg

= 760 torr

• Temperature: K = 273 + ºC

• Memorize: PV = nRT

• (use this in non-STP)

• R = 0.0821 L atm

mol K

• STP = 0ºC (273 K) & 1 atm

Summary of Colligative Properties

Lab Question…

Ch 10.4 Colligative Properties of Electrolytes