Solutions are homogeneous mixtures. The substance in the solution which is present in the largest quantity is called the solvent. The substance dissolved in the solvent is the solute.

· General Rule: "Like dissolves in like" (i.e., polar substances dissolve in each other)

· Some non-polar substances which appear to dissolve in water, are actually reacting with the water to produce polar substances

Ex. 2 Na + H₂O à 2 NaOH + H₂

CO₂ + H₂O ¾ H₂CO₃ ¾ H⁺ + HCO₃^-

· Dissolving Process: Separate and surround solute particles with solvent.

· Rate of dissolving varies based on:

1. Total change in energy

· Energy is required to overcome attractive forces (endothermic)

· Energy is given off when new attractive forces form (exothermic)

· This amount of energy may or may not be enough to compensate for the initial energy requirement to overcome solvent-solvent and solute-solute attractive forces.

2. Total change in entropy

· Entropy is a measure of the disorder or randomness of matter – an increase in entropy helps to make the process more energetically favorable (spontaneous)

Ex 1: Dissolving a solid in a liquid (NaCl & H₂O)

· For dissolving to occur sodium ions and chloride ions must overcome the ionic forces of attraction holding them in their crystal lattice

· The amount of energy required to do this is referred to as crystal lattice energy

· Solvent molecules must overcome their attractive forces (hydrogen bonding) to accommodate ions

· New attractive forces must form as sodium and chloride ions interact with the water

· The ions are hydrated (surrounded by water molecules) – the number of water molecules varies based on the size and charge of the ion (most cations have 4-9 waters of hydration in solution, with 6 being the most common)

These steps can be summarized as:

1. Solute expands (overcoming attractive forces)

2. Solvent expands (overcoming attractive forces)

3. Solute and solvent mix (new attractive forces forming)

· Note: for determining the value of total change in energy, these steps can be thought of as occurring separately. They actually occur simultaneously. (See Fig. 11.1)

· Even if the overall process is slightly endothermic, the increase in entropy may be sufficient to overcome this

Ex 2: Dissolving a liquid in a liquid (Ethanol in water)

· Miscibility is the term applied when one liquid completely dissolves in another

· The same three kinds of attractions must be considered, but solute-solute attractions are generally weaker in the liquid state.

· For dissolving to occur, there still must be an attractive force between solute and solvent (dipole-dipole, H-bonding, etc.)

· When the new attractive force is strong, large amounts of heat may be generated by the dissolving process – Ex. H₂SO₄ and H₂O

· Non-polar liquids can dissolve in other non-polar liquids – all types of interactions will involve London forces (solute-solute, solvent-solvent, and solute-solvent)

· London forces are relatively easy to overcome and to form, so non-polar molecules can easily slide past each other to mix.

· Note: Ionic compounds break down into ions in solution. Most covalent molecules remain intact.

Ex 3: Dissolving a gas in a liquid

· Polar gases will dissolve in water fairly easily as did polar liquids

· Non-polar gases which dissolve do so by reacting to form polar products (CO₂) or by dipole-induced dipole interactions (O₂)

· Most hydrogen halides react with water to produce hydronium ions and halide ions

Ex. HCl + H₂O à H₃O⁺ + Cl^-

· The covalent bond in hydrogen fluoride is too strong to be broken by water, so it hydrogen bonds to water when dissolving instead.

Solubility: Under set circumstances of T and P, each solute will have a definite amount which dissolves in a set amount of solvent

· Solubility is expressed as a concentration --common units include grams of solute/100 mL of solution, etc.

· When the maximum possible amount of solute has dissolved in a solution, the solution is said to be saturated.

Temperature can have an effect on the solubility of a solute in its solvent

· For exothermic dissolution processes, an increase in temperature decreases the solubility

· Many gases dissolve by exothermic processes – these gases are less soluble in warm water than in cold water (Ex. oxygen)

· For endothermic dissolution processes, an increase in temperature increases the solubility

Pressure can effect the solubility of a gas in its solvent

· An increase in pressure will increase the solubility, while a decrease in pressure will decrease the solubility (Ex. CO₂ in H₂O)

Mole fractions and molality are units of concentration commonly used for solutions

Molality (m) = moles of solute/kg of solvent

Mole fraction (for a two component solution)

X_A = Moles of component A

Moles of A + Moles of B

X_B = Moles of component B

Moles of A + Moles of B

Colligative Properties: Effects of the number of solute particles on the physical properties of the solvent

Vapor pressure: Caused by the pressure emitted from particles escaping (evaporating) from the surface of a liquid

· When solute is dissolved in a solvent, solute particles occupy a portion of the surface area of the liquid.

· This leaves less room for solvent molecules.

· The number of particles escaping decreases, so the vapor pressure decreases.

· The vapor pressure of a solvent is directly proportional to the mole fraction of the solvent in the solution.

· Raoult’s Law: P_solvent = X_solventP°_solvent

X_solvent = mole fraction of solvent

P°_solvent = vapor pressure of the pure solvent

Note: Solutions which contain volatile solute particles require use of Raoult’s law to calculate vapor pressure from both solute and the solvent.

P_solvent = X_solventP°_solvent

P_solute = X_soluteP°_solute

P_total = P_solvent + P_solute

Some solutions deviate from Raoult’s law due to attractive or repulsive forces between solute and solvent – attractions decrease vapor pressure while repulsions increase vapor pressure.

Boiling Point: the temperature where vapor pressure becomes equal to atmospheric pressure

· Dissolving solute particles decreases the vapor pressure of the solvent.

· Hence, more energy is required to raise the vapor pressure to the point where it is equal to atmospheric pressure (to boil).

· Boiling point elevation is proportional to the number of solute particles present (0.512 °C/m for aqueous solutions)

DT_b = K_bm

T_b = boiling point

K_b = molal boiling point elevation constant

m = molal concentration of solute

Freezing Point: the temperature at which molecular motion slows down enough for intermolecular attractive forces to lock molecules in place

· Solute particles block liquid solvent particles from coming close together to form a solid – (molecules must come close together for attractive forces to take effect)

· The freezing point of the solution is lower than that of the pure solvent – the temperature must get colder to slow molecules down further.

· Freezing point depression is proportional to the number of solute particles present (1.86 °C/m for aqueous solutions)

DT_f = K_fm

T_f = freezing point

K_f = molal freezing pt. depression constant

m = molal concentration of solute

Effects of electrolytes on colligative properties

When non-electrolytes dissolve, these molecules remain intact. Dissolving 1 mole of sugar molecules (C₁₂H₂₂O₁₁) produces one mole of solute particles.

When electrolytes dissolve, they dissociate into ions. Strong electrolytes are assumed to almost completely dissociate, while weak electrolytes only partially dissociate.

· The number of moles of solute particles present in an electrolytic solution must take into consideration the idea that more solute particles are produced by dissociation.

· Ideal behavior would predict that 1 mole of KCl would produce 2 moles of solute particles. (KCl à K⁺ + Cl^-)

· Some degree of ion-pairing occurs in solution, even for strong electrolytes. For brief moments in time, ions of opposite charge will associate. This results in non-ideal behavior.

Assuming ideal behavior of a strong electrolyte, calculate the freezing point of a 1.00 m solution of KBr in water.

1.00 m KBr à 2.00 m solute particles

DT_f = K_fm

= 1.86 °C/m(2.00 m)

= 3.72 °C

T_f = 0 °C - 3.72 °C = -3.72 °C

The actual observed freezing point of a 1.00 m solution of KBr in water is –3.29 °C, explained by the concept of ion pairing.

Assuming ideal behavior of a strong electrolyte, calculate the boiling point of a 1.50 m solution of K₂CO₃ in water.

Osmotic pressure

Osmosis – solvent moves through a semi-permeable membrane from low solute concentration to high solute concentration

· Solvent must collide with the membrane to move through it

· Movement of solvent is blocked by some solute particles; the side with more solute blocks solvent movement more creating a net movement toward this side

· Liquid level rises as the solvent passes into one compartment until back pressure stops the flow

· This pressure forces solvent molecules back through the membrane at the same rate as they are traveling into this compartment … equilibrium is reached with solvent molecules traveling in opposite directions at the same rate

· The pressure exerted by the flow of solvent through the membrane is referred to as osmotic pressure

· Osmotic pressure varies based on the number of solute particles present since the solvent is attempting to dilute solute

· Osmotic pressure can also be defined as the pressure required to prevent osmosis

· In a dilute solution solute particles are far apart and do not interact significantly – like gas particles in an ideal gas

Osmotic pressure can be calculated based on the ideal gas equation

P = nRT/V

P = osmotic pressure

n = moles of solute

V= volume (L) of solution

(n/V = molar concentration = M)

· So P = MRT

· Temperature influences osmosis because a rise in temperature increases the number of solvent-membrane collisions, increasing the likelihood of movement through the membrane

· Molarity of solute influences osmosis because a rise in molar concentration of solute on one side decreases the number of solvent-membrane collisions, decreasing the likelihood of backflow.

· For dilute aqueous solutions molarity is approximately equal to molality because density is approximately = 1 kg/L

Calculate the osmotic pressure of a 1.0 molal solution of a nonelectrolyte in water at 0°C.