Unit 10: Part 1: Polarity and Intermolecular Forces

Name: ______

Block: ______Unit 10: Part 1: Polarity and Intermolecular Forces

Intermolecular Forces of Attraction and Phase Changes  Intramolecular Bonding: attractive forces that occur between atoms WITHIN a molecule; these are true chemical bonds, of two types: 1. ionic bond = electrostatic attraction between metal cation and non-metal anion 2. covalent bond = electron pair sharing between non-metal and non-metal

 Intermolecular Forces of Attraction: attractive forces that occur BETWEEN molecules

4 types of Intermolecular Forces of attraction: o Dipole - Dipole: strongest type of intermolecular forces of attraction between 2 polar molecules with dipole moments o Hydrogen Bonding: strong intermolecular forces of attraction between hydrogen and highly electronegative oxygen, nitrogen or fluorine of 2 different polar molecules *both molecules must have dipole-dipole forces* o London Dispersion Forces: weak intermolecular forces of attraction between NON- polar molecules; larger mass molecules have higher London Dispersion forces. o Van der Waal's Forces: weak, temporary intermolecular forces of attraction between molecules, assume present in all molecules - polar/non-polar

What is an Intermolecular Force?  Force between molecules (weak force)  Differs from an intramolecular force (strong force) which forms Covalent Bonds

Intramolecular Intermolecular Forces Forces Covalent Bonds H-Bonds Dipole-dipole London Dispersion 400 kcal 12-16 kcal 2-0.5 kcal Less than 1 kcal

Notice: covalent bonds are almost 40 times the strength

What creates an Intermolecular force? • Unequal distribution of electrons • Created as a result of differences in: • Electronegativity

Phase change = when energy enters or leaves a compound to cause changes from solid, liquid, or gas phases; substance overcomes weak intermolecular forces of attraction

6 phase changes: 1. melting = solid - liquid 4. condensation = gas - liquid 2. freezing = liquid - solid 5. deposition = gas - solid 3. evaporation = liquid - gas 6. sublimation = solid - gas

 Normal melting point = temperature where substance is in equilibrium between solid and liquid phases at standard pressure of 1 atm.  Normal boiling point = temperature where substance is in equilibrium between liquid and gas phases at standard pressure of 1 atm.

In order for a substance to move between the states of matter; for example, to turn from a solid into a liquid, which is called fusion, or from a liquid to a gas (vaporization), energy must be gained or lost. (As we move from solid to gas it is gained and from gas to a solid it is lost. Why? Molar Heat of Vaporization requires more energy to change phase from liquid to gas phase. Gas molecules have high kinetic energy and distance between gas molecules is very high, requiring more energy to overcome the intermolecular forces of attraction.)

Changes in the states of matter are often shown on phase diagrams, and you will probably see at least one of two different types of phase diagrams. Let’s start with the phase diagram for water. The phase diagram for water is a graph of pressure versus temperature. Each of the lines on the graph represents an equilibrium position, at which the substance is present in two states at once. For example, anywhere along the line that separates ice and water, melting and freezing are occurring simultaneously.

The intersection of all three lines is known as the triple point (represented by a dot and a T on the figure). At this point, all three phases of matter are in equilibrium with each other. Point X represents the critical point, and at the critical point and beyond, the substance is forever in the vapor phase. This diagram allows us to explain strange phenomena, such as why water boils at a lower temperature at higher altitudes, for example. At higher altitudes, the air pressure is lower, and this means that water can

2 reach the boiling point at a lower temperature. Interestingly enough, water would boil at room temperature if the pressure was low enough! Example What happens to water when the pressure remains constant at 1 atm but the temperature changes from - 10ºC to 75ºC?

Explanation Looking at the phase change diagram for water and following the dashed line at 1 atm, you can see that water would begin as a solid (ice) at 0ºC and begin melting. So from 10º C to 75º C water would be in a liquid phase until it reaches 100ºC.

The second type of phase change graph you might see on the SAT II Chemistry exam is called a heating curve. This is a graph of the change in temperature of a substance as energy is added in the form of heat. The pressure of the system is assumed to be held constant, at normal pressure (1 atm). As you can see from the graph below, at normal pressure water freezes at 0ºC and boils at 100ºC. The plateaus on this diagram represent the points where water is being converted from one phase to another; at these stages the temperature remains constant since all the heat energy added is being used to break the attractions between the water molecules.

Part 2: Solutions Vocabulary Solution - a homogeneous mixture of two or more substances in a single physical state  Solvent‐ substance that does the dissolving  Solute‐ substance that is dissolved  Soluble‐ when a substance is able to dissolve in another substance  Insoluble‐ when a substance cannot dissolve in another substance  Alloy‐ solution containing two or more metals  Miscible‐ when two liquids can dissolve in one another in any amount  Immiscible‐ when two liquids do not mix together  Aqueous solution‐ a solution where the solvent is water  Concentration‐ a solution that contains a large amount of solute  Dilute‐ a solution that contains a little solute  Saturated‐ a solution that contains a maximum amount of solute  Unsaturated‐ a solution that contains below the maximum amount of solute  Supersaturated‐ a solution that contains above the maximum amount of solvent  Solubility‐ describes the maximum amount of solute that can be dissolved in a solvent at a given temperature

The rate at which a solution is formed is affected by:

 Surface area (particle size)‐

More surface area, smaller particles dissolves FASTER

 Temperature‐ Increase temperature will increase the dissolving rate  Agitation

Example stirring, shaking, mixing will increase the dissolving rate

Measuring Solution Concentrations Molarity  Molarity (M)- the number of moles of solute dissolved in each liter of solution M = mol L  Calculate the molarity of a solution formed by mixing 10.0 g of sulfuric acid

(H2SO4) with enough water to make 100.0 mL of solution. 1. Calculate the moles of sulfuric acid.

10.0 g H2SO4 1 mol H2SO4 = 0.102 mol H2SO4 98.08 g H2SO4

2. Calculate liters of solution.

100.0 mL 1 L = 0.1000L 1000 mL

3. Calculate molarity.

M = mol 0.102 mol H2SO4= 1.02 M L 0.1000L

 What mass of sodium nitrate is needed to produce 500.0 mL of a 0.50 M solution?

M = mol 0.50 M = . X = 0.25 mol NaNO3 L 0.5000L

0.25 mol NaNO3 85.00 g = 21.25 g NaNO3 1 mol

 Molality (m)- moles of solute per kilogram of solution (mol/kg)  What is the molality of saltwater that contains 684 g of NaCl in 20.0 mL of water? Step 1 convert to moles 684 g NaCl 1 mol NaCl = 11.70 mol NaCl 58.45 g NaCl Step 2 convert to kilograms (1 ml of water is equal to 1 g of water) 20.0 g 1 Kg = 0.0200 kg 1000 g Step 3 convert to molality

m = mol 11.70 mol NaCl = 585 m kg 0.0200kg

Dilutions  Dilutions- many solutions come as concentrated stock solutions and must be diluted before use.

M1V1 = M2V2  What volume of a 12.0 M stock solution of hydrochloric acid is required to make 250.0 mL of a 0.10 M solution?

M1 = 12.0 M M2 = 0.10 M

V1 = ? V2 = 250.0 mL

12.0 M X = (0.10 M) (250.0 mL)

12.0 M X = (0.10 M) (250.0 mL) 12.0 M 12.0 M

X = 2.08 mL

You would add 2.08 mL to a volumetric flask. You would than add 247.02 mL to the flask for a final volume of 250.0 mL. You now have 250.0 mLof 0.10 M solution.

Solubility - We can predict miscibility using the rule “LIKE DISSOLVES LIKE”

 Polar + Polar = miscible  Polar + ionic = miscible  Non-polar + Non-polar = miscible  Non-polar + ionic = immiscible  Polar + non-polar = immiscible

To determine polarity

Polar Rules Non-polar Rules  Hydroxyl group – OH

 Organic changes CxHy

 Polar solutes dissolves in polar  Dissolves in non-polar substance solvent

 Asymmetric molecule w/ polar bonds  Symmetric molecule w/polar bonds

 Lone pairs on central atom  No polar bonds (look at  ionic bonds electronegativity difference)

Polar Examples Non-polar Examples

 Water, Salts, Sugar, Acids, Bases,  Butter, oil, lard, Some Paints Ammonia

Soap is an emulsifier:  Has a polar and non-polar end.  Non-polar end dissolves in oil, Polar end dissolves Oil in water: water and oil appear to mix

Substances that DO NOT dissolve in water are hydrophobic Substances that DO dissolve in water are hydrophilic

Solubility Charts

 The __Solubility_ of a solute dissolved in 100g of water is tested at different temperatures.

 The amount in grams is plotted on a graph based on the __Saturation_.

 Then the data points are Connected by a line or curve.  The curve represents the maximum amount of a solute dissolved in 100g of water for ALL temperatures between 0°C and 100°C.

Q: Why is the scale only between 0°C and 100°C? Water boils and turns to a gas at 100° C At 40°C, 100g of water can dissolve how much solute? Between 44 to 45 g Will 50 grams dissolve in 100g of water at 75°C? Yes

 For any point _below__ the solubility curve the solution is _Unsaturated_.  For any point _on__ the solubility curve the solution is _Saturated_.

Is a solution with 70g of solute dissolved at 40°C saturated, unsaturated, or supersaturated?

 For any point _above__ the solubility curve the solution is _Supersaturated

Solubility Questions: 1. At 60°C, 25g of solute is dissolved in 100g of water. What is the name of the solute? KClO3 .

2. If 50g of KCl are dissolved in 100g of water, at 80°C, is the solution saturated, unsaturated, supersaturated? Usaturated

3. What is the solubility of KNO3 at 45°C in 200g of water?

70 g = x g so 140 g of KNO3 100g 200 g 4. What is the solubility of NaCl at 99°C in 50g of water?

40 g = x g so 20 g of NaCl 100g 50 g

Dilution Problems

1) How much concentrated 18 M sulfuric acid is needed to prepare 250 mL of a 6.0 M solution? M1V1 = M2V2

M1 = 18 M M2 = 6.0 M V1 = ? V2 = 250.0 mL

18 M X = (6.0 M) (250.0 mL)

18 M X = (6.0 M) (250.0 mL) 18 M 18 M

X = 83 mL

2) How much concentrated 12 M hydrochloric acid is needed to prepare 100. mL of a 2.0 M solution? M1V1 = M2V2

M1 = 12 M M2 = 2.0 M V1 = ? V2 = 100 mL

12 M X = (2.0 M) (100 mL)

12 M X = (2.0 M) (100 mL) 12 M 12 M

X = 17 mL

Molarity Problems

1) What is the molarity of a solution in which 58 g of NaCl are dissolved in 1.0 L of solution? Step 1 convert to moles 58 g NaCl 1 mol NaCl = 1 mol NaCl 58.45 g NaCl Step 2 convert to Liters (it is in Liters already)

Step 3 convert to molality m = mol 1 mol NaCl = 1 M L 1 L

2) What is the molarity of a solution in which 10.0 g of AgNO3 is dissolved in 500. mL of solution? Step 1 convert to moles

10.0 g AgNO3 1 mol AgNO3 = 0.0589 mol AgNO3 169.88 g AgNO3

Step 2 convert to Liters 500.0 mL 1 L = 0.5000 L 1000 mL

Step 3 convert to molality m = mol 0.0589 mol AgNO = 0.118 M L 0.5 L

3) How many grams of KNO3 should be used to prepare 2.00 L of a 0.500 M solution? M = mol 0.500 M = . X = 1.00 mol KNO3 L 2.00 L

1.00 mol KNO3 101.11 g = 101.11 g KNO3 1 mol

4) To what volume should 5.0 g of KCl be diluted in order to prepare a 0.25 M solution?

5.0 g KCl 1 mol = 0.067 mol KCl 74.55 g

M = mol 0.25 M = 0.067 mol = 0.268 L KCL L X