Chapter 13. A solution forms when one substance disperses uniformly throughout another. The reason...

26
Properties of solutions Chapter 13

Transcript of Chapter 13. A solution forms when one substance disperses uniformly throughout another. The reason...

Properties of solutions

Chapter 13

A solution forms when one substance disperses uniformly throughout another.

The reason substances dissolve is due to intermolecular forces.

Ion-Dipole forces between water and ionic solids.

Dispersion forces between C6H14 and CCl4 The interactions between solute and solvent

molecules are know as solvation. When the solvent is water they are know as

hydration.

The Solution Process

NaCl dissolves in water because the water molecules have strong enough attractive forces for the Na+ and Cl- ions.

The water molecules must also move away from each other to make space for the Na+ and Cl- ions.

Energy Changes in Solvation

Heat of Solvation

As a solid dissolves the concentration of solute particles in the solution increases.

This leads to a certain amount of crystallization (the opposite of solvation)

When the solute concentration reaches a certain point the rate of solvation and the rate of crystallization will be equal

At this point the solution is said to be saturated.

Saturated Solutions and Solubility

The amount of solute needed to form a saturated solution in a given quantity of solvent is known as the solubility of that solute.

The solubility of a compound is the maximum amount of that compound that will dissolve in a given amount of solvent at a specific temperature.

The solubility of NaCl at 0 oC is 35.7 g per 100 mL of water.

If we dissolve less than this amount the solution is said to be unsaturated.

Under certain conditions it is possible for a solution to contain more than the amount of solute needed to form a saturated solution.

Along with the chemical nature of the solvent and solute the solubility of a substance depends on…

Temperature Pressure (For gases only)

Factors Affecting Solubility

One factor that affects solubility is the chemical nature of both the solute and the solvent.

The stronger the attractions between solute and solvent molecules the higher the solubility.

If we try to dissolve a polar solute into a polar solvent it would work very well.

This is due to favorable dipole-dipole interactions between solute and solvent particles.

If we tried to dissolve a polar solute into a nonpolar solvent it would not work.

This is because there would be little to know attractive forces between solute and solvent particles.

Solute-Solvent Interactions

The solubility of solids and liquids is affected so little by ambient pressure that we do not need to account for it.

However the solubility of gases is highly affected by pressure.

The solubility of a gas increases as pressure increases.

Soda Henry’s Law: Sg = kPg

Pressure Effects

Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4 atm over the liquid at 25 oC. The Henry’s Law constant for CO2 in water at this temperature is 3.1 x 10-2 mol/L-atm.

Using Henry’s Law

The solubility of most solid solutes in water increases as the temperature of the solution increase.

The solubility of gases in water decreases with increasing temperature.

Poor fishies…

Temperature Effects

Mass Percentage : % Mass

Parts Per Million: ppm

Mole Fraction

Ways to express solution concentration

Molarity: M

Molality: m

A solution is made by dissolving 13.5 g of glucose (C6H6O6) in 0.100 kg of water. What is the mass percentage of solute in this solution?

11.9 %

A 2.5 g sample of ground water was found to contain 5.4 μg of Zn2+. What is the concentration of Zn2+ in parts per million?

2.2 ppm

Practice Problems

A solution is made by dissolving 4.35 g of glucose in 25.0 mL of water at 25 oC. Calculate the molality of the glucose in solution.

0.964 m

An aqueous solution of hydrochloric acid contains 36% HCl by mass. Calculate the mole fraction of HCl in solution and the molality of HCl in solution.

Mole fraction = 0.22

Molality = 15 m

A solution contains 5.0 g of toluene (C7H8) dissolved in 225 g of benzene (C6H6) and has a density of 0.876 g/mL. Calculate the molarity of toluene in solution.

0.2 M

The physical properties of many solutions differ from those of the pure solvent.

Example: Pure water freezes at 0 oC but aqueous

solutions freeze at lower temperatures. Pure water boils at 100 oC but aqueous

solutions boil at higher temperatures.

Colligative Properties

We saw in a previous chapter that all liquids have a vapor pressure associated with them.

We find that the vapor pressure of a pure solvent is higher than the vapor pressure of a solution made using that solvent.

Raoult’s Law ◦ PA = XAPo

A

Vapor Pressure Lowering

Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 oC. Calculate the vapor pressure at 25

oC of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of water at 25 oC is 23.8 torr.

The boiling point of a solution is higher than that of the pure solvent.

The change in boiling point is denoted as ΔTb

We can solve for the change in boiling point using the equation…

ΔTb = Kbm Where Kb is the molal boiling point elevation

constant and m is the molal concentration of the solution.

Boiling Point Elevation and Freezing Point Depresson

Automotive antifreeze consists of ethylene glycol (C2H6O2), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0% by mass solution of ethylene glycol in water.

Kb = 0.51oC/m Kf = 1.86 oC/m

Example

List the following aqueous solutions in order of their expected freezing point: 0.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, 0.050 m CH3COOH, 0.10 m C12H22O11.

Osmosis is the movement of materials across a permeable membrane.

The pressure required to prevent osmosis of a pure solvent is called the osmotic pressure.

Isotonic – Two solutions with identical osmotic pressure.

Hypotonic vs Hypertonic

π = MRT

Osmosis

The average osmotic pressure of blood is 7.7 atm at 25oC. What molarity of glucose would be isotonic with blood?

A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.25 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357 oC higher than that of the pure solvent. Calculate the molar mass of the solvent.

Determining Molar Mass by Freezing Point Depresson