Chapter 15

Solutions

Mixtures

• Hetereogeneous Mixture – is one in which the substances making up the mixture are not spread uniformly throughout the mixture.

• Homogeneous Mixture – is one in which the substances making up the mixture are spread uniformly throughout the mixture.

Solutions

• Solutions are homogeneous mixtures made up of very small particles that are, in fact, individual molecules, atoms, or ions.

• Typical properties of solutions:1. The solution is a homogenous mixture if it

has been well stirred during its formation.2. The dissolved particles will not come out

of the solution no matter how long the solution is allowed to stand. (Assuming the solution is covered so evaporation does not occur.)

Solutions

3. The solution is clear and transparent.4. Because of the extremely small size of

the dissolved particles, the solution will pass through the finest filters. Therefore, filtration cannot be used to separate one substance making up the solution from another.5. A solution is a homogeneous mixture that is considered to be a single phase even though the components may have been in different phases before the solution was formed.

Solute and Solvent

• The substance that is considered to be the dissolved substances in a solution is called the solute.

• The substance in a solution in which the solute is dissolved is called the solvent.

• Solutions in which water is the solvent are called aqueous solutions.

• Solutions in which alcohol is the solvent are called tinctures.

Types of Solutions

1. Gaseous Solutions – consist of gases or vapors dissolved in one another. Ex: air, iodine vapor

2. Liquid Solutions – consist of a liquid solvent in which a gas, liquid, or solid is dissolved. Ex: sugar water, carbonated beverage, antifreeze

3. Solid Solutions – are mixtures of solids uniformly spread throughout one another at the atomic or molecular level. Ex: brass, mercury.

• Alloy – solid solution of two or more metals (brass)• Amalgam – an alloy in which one metal is mercury.

Degree of Solubility

• The solubility of a solute is a measure of how much solute can be dissolved in a given amount of solvent.

• Factors affecting solubility:1. The nature of the solute and solvent

2. Temperature

3. Pressure

Factors Affecting the Rate of Solution

• The rate of solution is a measure of how fast a substance dissolves.

• It is the quantity of solute that will dissolve during one unit of time.

• Factors that determine the rate of solution:1. Size of particles

2. Stirring

3. Amount of solute already dissolved

4. Temperature

Solubility and the Nature of the Solvent and Solute

• Polar molecules are best dissolved by polar solvents.

• Nonpolar molecules are best dissolved by nonpolar solvents.

• “Like dissolves like.”

Energy Changes During Solution Formation

• Generally in all cases where a solid dissolves in a liquid, the change is endothermic.

• That is, heat is absorbed.

• The beaker often feels cold.

Saturated and Unsaturated Solutions

• A saturated solution is a solution that has dissolved in it all the solute that it can normally hold at given conditions. Abbreviated sat’d.

• A solution that contains less solute than it can hold at a certain temperature and pressure is said to be unsaturated. Abbreviated unsat’d.

Supersaturated Solutions

• Under special conditions, there are some solutions that can actually hold more solute than is present in their saturated solutions.

• These solutions are said to be supersaturated.

Dilute and Concentrated Solutions

• A dilute solution is one in which the amount of solute dissolved is small compared to the amount of solvent,

• A concentrated solution is one in which a relatively large amount of solute is dissolved.

Percentage by Mass

• Percentage by Mass is a unit of concentration that states the number of parts of mass of solute per hundred parts by mass of solution.

• A 15% aqueous solution of sodium chloride must contain 15 g of NaCl and 85 g of H2O.

% Mass = [mass solute / total mass] x 100

% Mass = [15g NaCl / (15g + 85g)] x 100

Percentage by Mass

• A mass of 25 g of NaCl is dissolved in 200 g H2O. What is the concentration of this solution expressed as percentage by mass?

25 g NaCl / (200 g H2O + 25 g NaCl) =

0.1111 x 100% =

11% NaCl

Percentage by Mass

• A solution of sodium chloride dissolved in water is 12.0% sodium chloride by mass. How many grams of sodium chloride are in a 120 g sample of this solution?

12.0% NaCl = [g NaCl / (120 g NaCl and H2O)] x 100

14.4 g NaCl

Calculating the Mass of a Solution

• Suppose that a solution of hydrogen chloride gas dissolved in water has a concentration that is 30% hydrogen chloride by mass. How much solute is there in 100 mL of this solution if the density of the solution is 1.15 g/mL?

100 mL x 1.15 g/mL = 115 g solution

30% HCl = (g HCl / 115 g Solution) x 100

= 34.5 g HCl

Molarity

• The molarity of a solution is the number of moles of solute in one liter of solution.

solution

solute

liter

molesMolarity

Molarity

• Some sucrose is dissolved in water. How many grams of sugar are dissolved in 200. mL of solution if its concentration is 0.150 M?

M = Moles / L

0.150 mol / L x 0.2 L = 0.03 mol sucrose

0.03 mol Sucrose x ( 342.34 g C12H22O11/ 1 mol) =

10.3 g Sucrose

Molality

• Molality is the number of moles of solute dissolved in one kilogram of solvent

solvent

solute

ramki

molesMolality

log

Molality

• Suppose that 0.25 moles of sugar are dissolved in 1000 grams of water. What is the molal concentration of this solution?

m = moles / kg

m = 0.25 moles / 1 kg

m = 0.25 moles / kg

Molality

• A mass of 23 grams of ethyl alcohol, C2H5OH, is dissolved in 500 mL of water. What is the molal concentration of this solution?**Hint: Need to know the density of water (1 g/mL)

m = mol / kg so…

Need to calculate mol:23 g C2H5OH x (1 mol / 46.08 g) = 0.50 mol

Need to calculate kg:500 mL x 1 g / mL = 500 g water = 0.5 kg

m = 0.50 mol / 0.5 kg = 1 mol / kg

Mole Fraction

• Just like mass %• Use moles instead of grams

• Ҳ = nindividual / ntotal

– Unit-less

• Determine the mole fraction of a solution containing 15.0 g NaCl in 50.0 g H2O

Dilutions

• Used when making solutions

• M1V1 = M2V2

• Example: To what volume will you have to dilute a 75 mL of a 12 M HCl solution to make a 1.0 M HCl solution?

Boiling-Point Elevation and Freezing-Point Depression

• Properties that depend upon only the number, and not on the identity, of the solute particles in an ideal solution are called colligative properties.

• Because of their direct relationship to the number of solute particles, the colligative properties are very useful for characterizing the nature of the solute after it is dissolved in a solvent and for determining molar masses of substances.

• Examples: freezing-point depression, boiling-point elevation, osmotic pressure

Boiling-Point Elevation and Freezing-Point Depression

• The normal boiling point of a liquid occurs at the temperature where the vapor pressure is equal to 1 atmosphere.

• We have seen that a nonvolatile solute lowers the vapor pressure of the solvent.

• Therefore, such a solution must be heated to a higher temperature than the boiling point of the pure solvent to reach a vapor pressure of 1 atmosphere.

• This means that a nonvolatile (no tendency to escape the vapor phase) solute elevates the boiling point of the solvent.

Boiling-Point Elevation and Freezing-Point Depression

• The magnitude of the boiling-point elevation depends on the concentration of the solute.

• The change in the boiling point can be represented by the equation:

ΔT = ikbmsolute

• Where ΔT is the boiling point elevation, or the difference between the boiling point of the solution and that of the pure solvent, kb is a constant that is characteristic of the solvent and is called the molal boiling-point elevation constant, msolute is the molality of the solute in the solution, and “i” is the number of ions the solute breaks into

Boiling-Point Elevation and Freezing-Point Depression

• When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent.

• The equation for freezing point depression is analogous to that for boiling point elevation:

ΔT=ikfmsolute

• Where ΔT is the freezing-point depression, or the difference between the freezing point of the pure solvent and that of the solution, and kf is a constant that is characteristic of a particular solvent and is called the molal freezing-point depression constant.

• Like the boiling point elevation, the observed freezing point depression can be used to determine molar masses and to characterize solutions.

• “i” is the number of particles the solute breaks into

Practice with “i”

• Determine “i” for the following substances:– NaCl

– Li2O

– CaBr2

BP/FP Example:

• Determine the boiling point and freezing point of a solution containing 50.0g KCl in 600.mL water. (**Hint: 1g = 1mL H2O). Kb = 0.52⁰/m Kf = 1.86⁰C/m

• Step 1: calculate molality (m) using m = mol solute/kg solvent• Step 2: determine “i”• Step 3: choose Kb or Kf depending on whether you are solving for the melting or boiling point• Plug into the equation

BP/FP Practice

• Which solution would have the higher boiling point:– A 500g water solution containing 40.0g NaCl

– A 500g water solution containing 40.0g K2O

Forces

• Intramolecular forces are forces within a molecule that hold the atoms of the molecule together.

• Intermolecular forces are forces between molecules that cause them to aggregate and form a solid or a liquid.

Energy

• Remember that temperature is a measure of the random motions (average kinetic energy) of the particles of a substance.

• The energy required to melt 1 mole of a substance is called the molar heat of fusion.– For ice, the molar heat of fusion is 6.02 kJ/mol.

• The energy required to change 1 mole of liquid to its vapor is called the molar heat of vaporization.– For water, the molar heat of vaporization is 40.6

kJ/mol

Sample Problems

1. Calculate the energy required to melt 8.5 grams of ice at 0 degrees Celsius. The molar heat of fusion for ice is 6.02 kJ/mol.

8.5 g water x (1 mol/18.02 g) = 0.471 mol

0.471 mol x (6.02 kJ/ 1 mol) = 2.8 kJ

Sample ProblemsCalculate the energy (in kJ) required to heat 25 g of liquid water from 25 degrees Celsius to 100 degrees Celsius and change it to steam at 100 degrees Celsius. The specific heat capacity of liquid water is 4.184 J/g°C and the molar heat of vaporization of water is 40.6 kJ/mol

• q = mc∆T (Heating Liquid)• q = (25g)(4.18 J/g x ºC)(100 ºC – 25 ºC)• 7,800 J = 7.8 kJ

• 25 g water x (1 mol/18.02 g) = 1.4 mol water (Boiling Liquid to vapor)

• 1.4 mol x (40.6 kJ/ 1 mol) = 57 kJ – **40.6 is the molar heat of vaporization of water

• 7.8 + 57 = 65 kJ

Sample Problems3. Calculate the total energy required to melt 15 g of ice at 0 degrees

Celsius, heat the water to 100 degrees Celsius, and vaporize it to

steam at 100 degrees Celsius.

The specific heat capacity of liquid water is 4.184 J/g°C, the molar heat of fusion is 6.02 kJ/mol and the molar heat of vaporization of water is 40.6 kJ/mol

INTERMOLECULAR FORCES

14.3

Intermolecular Forces

Dipole-Dipole AttractionMolecules with dipole moments can attract each

other by lining up so that the positive and negative ends are close to each other.The polarity of a molecule was discussed in Section 12.3

Hydrogen BondingOccur between molecules in which hydrogen is

bonded to nitrogen, oxygen, or fluorine.Great polarity of the bondClose approach of the dipoles because of the

small size of the hydrogen atom.See Section 12.2 for a discussion of electronegativity.

Intermolecular Forces

• London Dispersion Forces– Forces that exist between noble gas atoms

and nonpolar molecules.– Present in all molecules.

Practice Problem

Predict which substance in each of the following pairs will have the largest boiling point.

a. H2O or CH3OH

a. Water because water contains 2 polar bonds while methanol has only one. Therefore, the bonding among the water molecules is much stronger

b. CH3OH or CH3CH2CH2CH2OH

a. Each of these has 1 polar bond however butanol is much larger so will have a larger BP

Changes of State

• Vaporization or Evaporation is the process by which molecules of a liquid overcome strong intermolecular forces and form a gas.

• Melting is the process by which the molecules of a solid loosen to become a liquid

THE SOLID STATE: TYPES OF SOLIDS

14.5

Solids

• Crystalline Solid – those with a regular arrangement of their components.– Many varieties– Highly ordered– Regularly shaped crystals

Crystalline Solids

BONDING IN SOLIDS

14.6

Bonding in solids

Ionic SolidsStable substances with high melting pointsHeld together by strong forces that exist between

oppositely charged particles (ions). Molecular Solids

Low melting pointsWeak intermolecular forcesPolar: dipole-dipole forcesNonpolar: London forces

Atomic SolidsLow melting points – weak forcesHigh melting points – strong forces

Bonding in metals

• Electron Sea Model – a regular array (pattern) of metal atoms in a “sea” of valence electrons that are shared among the atoms in a nondirectional way and that are quite mobile in the metal crystal

• The electrons are not attached to any one atom

• Allows metals to conduct electricity

Metals - Alloys

An alloy is a substance that contains a mixture of elements and has metallic properties.

Substitutional Alloy some of the host metals are replace by other metal atoms of similar sizes.Ex: Brass, Sterling Silver, and Pewter

Interstitial Alloy is formed when some of the interstices (holes) among the closely packed metal atoms are occupied by atoms much smaller than the host atoms.Ex: Steel (C + Fe)