Unit 10 Solutions

Solution: a homogenoueous mixture of 2 or more substances

2 parts to a solution

  • The solvent does the dissolving; normally the component present in largest amount
  • The solute is dissolved; normaly the component present in smallest amount

Examples of all types of Solutions: Solids dissolve into solids, liquids can dissolve into liquids, gases can dsissolve into liquids, etc

Salt in WaterAlcohol in Water Oxygen & Carbon Dioxide in Nitrogen

Silver in GoldOxygen in WaterHydrogen gas in Platinum

Mercury in Silver

Descriptive terms

Miscible: when 2 or more liquids can dissolve into one another (water and alcohol)

Immiscible: when 2 or more cannot dissolve into one another (water & oil)

Soluble: when a solid or gas can dissolve into one another

Insoluble: when a solid or gas cannot dissolve into one another

We will focus on aqueous solutions-solutions where water is the solvent

Ways of Measuring the Concentration of Solutions

Qualitative terms:

  • Dilute : small amount of solute compared to the amount of solvent
  • Concentrated: large amount of solute compared to the amount of solvent

Quantitative terms:

  • Molarity = moles of soluteabbreviated M; molarity changes with

Liters of solutiontemperature due to expansion/contraction

of solvent changing volume; most molar solutions

are made at 25 C; 1.0 M means 1 molar

  • Mass Percent = Mass of solute x 100 often expressed as parts per million(ppm)

Total Mass of solution or (ppb) 1 mg/1L = 1ppm

Can transfer easily to mole fraction from mass percent

  • Mole fraction – moles of solute over total moles of solution

Mole fraction of component χA = NA______

NA + NB….

sum of all mole fractions equals 1

  • Molality = moles of solute used in measuring colligative properties

Kilograms of solvent does not change with

temperature / mass does not change

with temp; abbreviated m; 1 m

means 1 molal

  • Density (g/ml) can be used to convert between the different methods of calculating concentration, especially between molarity and molality
  • Density of aqueous solution is usually identical to that of pure water (1g/1mL) at normal temperatures

Example 1: A solution is prepared by mixing 1.00 g ethanol( C2H5OH) with 100.0 g water to give a final volume of 101 ml. Calculate the molarity,

mass percent, mole fraction, and molality of ethanol in this solution.

Example 2:The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/ml.

Calculate the mass percent, and molality of the sulfuric acid.

Energy of Making Solutions

  • Solutions form when attractive forces between solute and solvent particles are comparable with those that exist between

the solute particles themselves or the solvent particles themselves.

  • Like Dissolves Like polar solvents dissolve ionic/polar solutes OR Nonpolar solvents dissolve nonpolar solutes

Example : NaCl is soluble in water. the strong ion-dipole forces between the ions and water are asstrong as ionic bonds and Hydrogen Bonding

Example : Oil is immiscible in water. The weak dipole-induced dipole forces between water & oil are not as strong as the weak LDF

and Hydrogen bonding of water

  • Interactions where the solvent completely envelops the solute is called solvation or when water is the solvent, hydration.

Enthalpy(heat) of solution (ΔHsoln ) is the energy change for making a solution. ΔHsoln = ΔH1 + ΔH2 + ΔH3

  • Most easily understood if broken into “3” steps. All 3 steps are added together to calculate the heat of solution.
  • It can be positive(endothermic) or negative (exothermic)

1.Break apart solvent. ΔH1

requires energy to overcome intermolecular forces.Expanding the solvent! ΔH1 >0  Endothermic

the values are large in polar solvents and low in nonpolar solvents

2. Break apart Solute. ΔH2

Requires energyto overcome attractive forces of particles. Expanding the solute!ΔH2 >0  Endothermic

the values are large in ionic & polar solutes and low in nonpolar solutes

3. Mixing the solute and SolventΔH3 ΔH3 < 0  Exothermic

requires the release of energy when solute & solvent interact to fomr a solution

the values are large in polar solutes-polar solvent and low in nonpolar solutes-nonpolar solvent

  • Molecules can attract each other ; ΔH3 is large and negative.
  • Molecules can’t attract; outcome is ΔH3 is small and negative.

 All 3 terms can add together to get a positive or negative sum = ΔHsoln

ΔHsoln can be exothermic or slightly endothermic

Types of Solvent and Solutes

Case 1

  • Oil and water do not mix
  • Oil(large nonpolar with LDF)
  • ΔH1 for solute is usually small & positive but large & positive due to size of oil molecules
  • ΔH2 for water is large & positive due to H- bonds
  • ΔH3 is small and negative due to little to no interactions between polar and nonpolar molecules
  • So ΔHsoln is large and positive- Does not usually happen  too much energy expended

Case 2

  • Salt in water – mix
  • ΔH1 for solute is usually large & positive due to electrostatic forces
  • ΔH2 for water is large & positive due to H- bonds
  • ΔH3 is large and negative due to ion-dipole forces
  • ΔHsoln is small and positive in this case approximately 3 kJ/mol
  • Remember, When heat of reaction is negative, reaction is spontaneous
  • When ΔHsoln small and positive, what makes salt soluble?
  • Entropy??? The increased disorder overrides the small cost of heat of solution.
  • Of course, solution formation takes place based on 2 factors: tendency towards a lower enthalpy(exothermic) and a higher entropy(favored)

Example 3: Calculate the enthalpy of solution for NaOH. Hf (s) = -425.6 kJ/mol and Hf (aq) 1m = -470.1 kJ/mol

Structure and Solubility

  • To be soluble in polar solvents, the molecules must be polar or ionic
  • To be soluble in non-polar solvents the molecules must be non polar.

ΔH1solute / ΔH2solvent / ΔH3interactions / ΔHsoln / Result
Polar solvent, polar solute / Large / Large / Large, negative / Small / Solution forms*
Polar solvent, nonpolar solute / Small / Large / Small / Large, positive / No solution forms
Nonpolar solvent, nonpolar solute / Small / Small / Small / Small / Solution forms*
Nonpolar solvent, polar solute / Large / Small / Small / Large, positive / No solution forms

Example 4: Decide whether liquid hexane or liquid methanol is the more appropriate solvent for the substance grease (C20H42) and potassium iodide.

Solubility

Solution formation is a dynamic equilibrium process

Dissolution

Solute + Solvent ↔ Solution

Crystallization

  • Defined as the maximum amount of solute that will dissolve in a sprcific amount of solvent at a specific temperature

This max. amount forms a saturated solution.

Example: 35.7 g of NaCl dissolves per 100 ml at 0ºC

  • Saturated solution: a solution that is in a dynamic equilibrium with an undissolved solute; think of it as being completely “full”,

no additional solute can fit into the solvent, it contains the maximum anout that can dissolve

  • Unsaturated solution: is not “full”, additional solute can be added to solvent and still be dissolved; it contains less than the maximum amount
  • Supersaturated solution: solutions that contain a greater amount of solute than needed to form a saturated solution;

unstable; seed crystal disturbs system and crystallization takes place  needs to be heated at high temp and cooled slowly

Pressure Effects

Changing the pressure doesn’t affect the amount of solid or liquid that dissolves (solubility)

  • They are incompressible.

Pressure DOES EFFECT the amount of gas that can dissolve in a liquid.

  • The dissolved gas in solution is at equilibrium with the gas above the liquid.
  • The equilibrium is dynamic- rate at which the gas molecules enter the solution equals the rate at which they escape from solution and enter gas phase
  • If you increase the pressure of the gas molecules, more will dissolve. The equilibrium is disturbed.
  • The system reaches a new equilibrium with more gas dissolved.

The solubility of a gas is directly proportional to its partial pressure above the solution (assuming there is no reaction between the gas and the solvent)

Henry’s Law C = kP

C = solubility/concentration of gas

K = proportionality constant dependent on the gas-liquid mixture, varies with temperature

P = partial pressure of gas

Example 5: A certain soft drink is bottled so that a bottle at 25 °C contains CO2 gas at a pressure of 5.0 atm over the liquid. Assuming that the

partial pressure of CO2 in the atmosphere is 4.0 x 10-4 atm, calculate the equilibrium concentrations of CO2 in the bottle both before and

after the bottle is opened. The Henry’s law constant for CO2 in aqueous solution is 3.1 x 10-2 mol/atm•L at 25°C.

Temperature Effects

  • Increased temperatureusuallyincreases the rate at which a solid dissolves.Most solution foramtion is endothermic & decreases when the solution process is exothermic
  • It is difficult to predict. A graph of experimental data (Solubility Curves)will show the relationship between temperature and the solubility of a solid in the solvent.
  • Gases are predictable: As temperature increases, the solubility of gases decreases
  • Environmental concern: Thermal pollution

Colligative Properties

Physical properties that depend on the quantity of the particles in the solution, not the kind of particles

  • Properties of solution differ from properties of pure solvents
  • These properties are: Vapor Pressure, Boiling Point, Freezing Point and Osmotic Pressure

Example: Pure water freezes at 0ºC but a Salt Water solution would freeze at a lower temperature

Vapor Pressure LOWERING of Solutions

  • A volatile substance has a measurable vapor pressure, a nonvolatile substance has no vapor pressure
  • When comparing the vapor pressure of a pure solvent with those of their solutions, addition of a nonvolatile solute to solvent will lowers the vapor pressure

WHY? The molecules of the solvent must overcome the forces of both the solvent
molecules and the solute molecules. The more solute particles present the less the solvent can evaporate AND fewer solvent molecules are at surface

Raoult’s Law (use with nonvolatile solutes)

  • Solutions with a nonvolatile solute -the solute doesn’t contribute to the vapor pressure.
  • States that the vapor pressure of the solution is directly proportional to the mole fraction of the solvent

Psoln = χsolvent x Pºsolvent

  • Psoln = Vapor pressure of the solution
  • χsolvent = mole fraction of solvent
  • Psolvent = vapor pressure of the pure solvent
  • i= van’t Hoff factor

  • Water has a higher vapor pressure than a solution

In its linear form,Psoln = χsolvent x Pºsolvent

Y = m x + b(0)

Example 6:Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1.26 g/ml at 25 °C.

Calculate the vapor pressure at 25C of a solution made by adding 50.0 ml of glycerin to 500.0 ml of water.

The vapor pressure of pure water at 25 °C is 23.8 torr.

Can use this information to experimentally determine molar mass of a substance:

  • If mass of substance is given and Raoult’s law determines moles of solute present ,one can calculate molar mass M= mass/mole

Electrolytes in Solutions (NEED TO BE TAKEN INTO CONSIDERATION WHEN CALCULATING CHANGES IN ANY COLLIGATIVE PROPERTY)

  • Since colligative properties only depend on the number of molecules.
  • Ionic compounds should have a bigger effect.; when they dissolve they dissociate.
  • Individual Na and Cl ions fall apart.

1 mole of NaCl makes 2 moles of ions.

1mole Al(NO3)3 makes 4 moles ions.

  • Electrolytes have a bigger impact on vapor pressure lowering because they make more pieces.
  • Relationship is expressed using the van’t Hoff factor ( i ) : factor equal to the moles of ions present

i = Moles of particles in solution

Moles of solute dissolved

  • The expected value can be determined from the formula.
  • The actual value is usually less because at any given instant some of the ions in solution will be paired.

Ion pairing increases with concentration. ( the joining of oppositely charged ions due to electrostatic attraction. The greater the charge on an ion the greater its tendency to pair in solution)

  • The more particles dissolve,d the more the property is affected. i=1 for a a nonelectrolyte (molecular compound) i= number of particles fromed when one formula unit of an ionic compound dissolves in the solvent
  • NaCl (i= 2)
  • Psoln = i (χsolvent) x (Pºsolvent)

Example: Glucose is only1 molecule, NaCl has 2 ions that split apart, FeCl3 has 3 ions that split apart vapor pressure will be lowered , lowered 2 times as expected, lowered 3 times as expected

What if the solute is volatile (NONIDEAL)?

  • Must add together each substances vapor pressure

Modified Raoult’s Law

Ptotal soln = PA + PB = χ AP0 A + χ BP0B

Ptotal = vapor pressure of mixture

χ A = mole fraction of A χ B = mole fraction of B

P0A= vapor pressure of A P0B= vapor pressure of B

Ideal solution- a liquid-liquid solution that obeys Raoult’s law

  • Near ideal behavior is when 2 volatile liquids dissolve in one another and the solute-solute, solvent-solvent, and solute-solvent interactions are very similar
  • Can use Raoult’s law to see if the solution is ideal
  • If it is ideal, (solute and solvent are alike) and predicted vapor pressure will be correct
  • Hexane and heptane or benzene and methylbenzene

Deviations: If it is not, the observed vapor pressure will be lower or higher than what was predicted

Negative deviation from Raoult’s law.

  • ΔHsoln is large & negative (exothermic).
  • Vapor pressure of solution is lower than expected(calculated) the real measured vapor pressure was lower than expected
  • Interactions between solute-solvent bonds are greater than those in the solvent- solvent bonds or solute- solute bonds
  • Acetone and water

Positive Deviation

  • ΔHsoln is large &positive(endothermic)
  • Vapor pressure of solution is greater than expected(calculated) the real measured vapor pressure is larger than what was calculated
  • Interactions between the solute - solvent bonds are weaker than those of the solvent-solvent bonds and solute- solute bonds
  • Ethanol (polar) and hexane(nonpolar)

Example 7:At 40 C, the vapor pressure of pure heptane is 92.0 torr and the vapor pressure of pure octane is 31.0 torr.

Consider a solution that contains 1.00 mol of heptane and 4.00 mol of octane. Calculate the total vapor pressure of

each component and the total vapor pressure above the solution. Is this considered an ideal solution?

Think of the type of substances in the solution.

Other Colligative Properties

  • Dissolved particles affect vapor pressure so they affect phase changes as well.
  • Electrolytes have a bigger impact on melting and freezing points per mole because they make more pieces.

Boiling Point Elevation

  • Because a non-volatile solute lowers the vapor pressure it raises the boiling point proportionally to the amount of solute added.
  • The equation is: ΔTb = iKbmsolute

ΔTb is the change in the boiling point

Kb is a molal boiling point constant specific to the solvent.

msolute is the molality of the solute

i = van’t Hoff Factor

One can calculate molar mass of an unknown compound if compound is soluble in a solvent of a known Kb or Kf

Example 8:Antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile nonelectrolyte.

Calculate the boiling point of a 25.0 mass percent solution of ethylene glycol in water. Kb = .52 °C/m

Freezing Point Depression

  • Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point.
  • The equation is: ΔTf = Kfmsolute

ΔTf is the change in the freezing point

Kf is a molal freezing point constant specific to the solvent

msolute is the molality of the solute

i = van’t Hoff Factor

Example 9:When 15.0 g of ethyl alcohol, C2H5OH is dissolved in 750. g of formic acid, the freezing point

of the solution is 7.20 °C. The freezing point of pure formic acid is 8.40 °C.

Evaluate the Kf for formic acid.

Example 10: A 1.20 g sample of an unknown compound is dissolved in 50.0 g of benzene.

The solution freezes at 4.92 °C. Calculate the molecular mass of the compound.

The freezing point of pure benzene is 5.48 °C and the Kf is 5.12 °C/m.

Example 11:Estimate the freezing point of .20m solution of Cr(NO3) 3 in water. Kf = 1.86 °C/m

Example 12: List the following solutions in order of their freezing points (from lowest to highest)

.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, 0.050 m HC2H3O2, 0.10 m C12H22O1

Osmotic Pressure

  • Osmosis- selective passage of solvent molecules through a porous semi permeable membrane from a dilute to more concentrated solution
  • Osmotic pressure(π) –the pressure created by the movement of the solvent through the membrane

Equal to the pressure applied in order to prevent osmosis

  • π = iMRT

M= molarity

R= universal gas constant(.08206 L·atm/mol ·K)

T= Kelvin temperature

i = van’t Hoff Factor

Colloids

Heterogenous mixture

Particles do not settle over time

Particle size is intermediate to solutions and suspensions

Exhibit the tyndall effect

To destroy colloids, heat or add additional electrolytes

Examples are: fog, aerosol sprays, smoke, whipped cream, soap, suds, milk, mayo, paint, butter, and cheese