Types of mixtures

 Heterogeneous mixture:  Heterogeneous mixtures do not blend smoothly. Individual substances can still be seen  Suspension:   A suspension is a heterogeneous mixture where particles will settle to the bottom if left undisturbed.

Example: sand in water.

 Colloids:   Colloids are heterogeneous mixtures where particles will not settle to the bottom.

Example: Milk

Homogeneous Mixtures

 Homogeneous mixtures are mixtures where a substance is dissolved in another substance. These are also called solutions.

 Parts of a solution:  Solute: The substance that is dissolved  Solvent: The substance that does the dissolving  When a solute readily dissolves in a solvent it is said to be soluble.

 Two liquids that can be combined to make a solution are said to be miscible.

 A solute that does not readily dissolve in a solvent is said to be insoluble.

 Two liquids that do no mix to form a new solution are said to be immiscible.

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Solution Concentration

Concentration is the measure of how much solute is dissolved in the solution.

There are many ways to measure concentration Percent by mass:  (Mass of solute/Mass of solution) x 100 Percent by volume:  (Volume of solute/Volume of solution) x 100 Molarity:  Moles of solute/Liter of solution Molality:  Moles of solute/Kilograms of solvent Mole Fraction:  Moles of solute/Moles of solute + Moles of solvent

Percent by mass

 In order to maintain a sodium chloride concentration similar to ocean water, an aquarium must contain 3.6g NaCl per 100.0 g of water. What is the percent by mass of NaCl in the solution?

 3.5%  What is the percent by mass of NaHCO 3 in a solution containing 20.0 g of NaHCO3 dissolved in 600.0 mL of H 2 O?

 3%

Percent by Volume

 What is the percent by volume of ethanol in a solution that contains 35mL of ethanol dissolved in 155mL of water?

 18%

Molarity

 A 100.5 mL intravenous solution contains 5.10 g of glucose. What is the molarity of the solution? The molar mass of glucoes is 180.6 g/mol  0.282M

 If you have 1500 g of a bleach solution with a percent by mass of NaOCl of 3.62 % how many grams of NaCOl are in the solution?

 54.3 g  What is the percent by volume of isopropyl alcohol in a solution that contains 24 mL of isopropyl alcohol in 1.1 L of water?

 2.1 %   Calculate the molarity of 1.6 L of a solution containing 1.55 g of KBr.

8.13 x 10 -3

Molality

 In the lab a student adds 4.5 g of NaCl to 100.0 g of water. Calculate the molaity of the solution.

 0.77 mol/kg

Preparing Solutions

 How would you prepare 1 L of a 1.5 molar CuSO 4 Solution?

 Diluting Solutions:  Dilution equation M 1 V 1 = M 2 V 2  Calculate the new volume if you wanted to dilute the above solution to 0.5 molar.

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Diluting Solutions

Remember Molarity = (Moles of solute/Liters of solvent) The Dilution Equation: M 1 V 1 = M 2 V 2 Example: What volume of a 2.00M CaCl 2 0.50 L of a 0.300M solution?

M 1 V 1 = 2.00M

= ?

M 2 V 2 = 0.300M

= 0.5L

(2.00M)(V 1 ) = (0.300M)(0.5L) V 1 = 0.075 L solution would you use to make

 What volume of a 3.00M KI solution would you use to make 0.300 L of a 1.25M KI solution?

 125mL   How many milliliters of a 5.0M H 2 SO 4 stock solution would you need to prepare 100.0mL of 0.25M H 2 SO 4 ?

5.0mL

Mole Fraction

 To express concentration as a mole fraction we need to know the number of moles of solute and the number of moles of solvent.

 The mole fraction is calculated by dividing the number of moles of either the solute or solvent by the total number of moles in the solution.

Example

 If we have a solution that contains 36 g of HCl and 64 g of H 2 O. What are the mole fractions of HCl and H 2 O?

 HCl:  H 2 O:

Practice

 What is the mole fraction of NaOH in an aqueous solution that contains 22.8% NaOH by mass?

 If the mole fraction of H 2 SO 4 is an aqueous solution is 0.325, what is the percent by mass of H 2 SO 4

Factors Affecting Solvation

 Many physical factors affect the solubility of a solute.

 Such as:  Temperature  Pressure  Polarity  Solvation is the process of surrounding solute particles with solvent particles.

Aqueous ionic solutions

 Many ionic compounds are soluble in water. Why?

 Water is a polar molecule.

 Why might an ionic compound not be water soluble?

Heat of solution

 The overall energy change that occurs during the formation of a solution is called the heat of solution.

 certain solvation processes are exothermic while others are endothermic.

Factors that affect solvation

 Agitation:  Stirring or shaking moves solvated particles away from the surface of the solid and allows more solute particles to move into solution.

 Surface Area:  Breaking the solute into small pieces increases the available surface area and increase the rate of solvation.

 Temperature:  As temperature increase so does the rate of solvation.

Types of solutions

 Unsaturated:  A solution that more solute could be dissolved in.

 Saturated:  In a saturated solution the rate of solvation and the rate of crystallization are in equilibrium.  Supersaturated solution:  A super saturated solution is a solution that has more solute dissolved in it that it would normally allow.

Solubility of Gases

 Unlike solid solutes the solubility of gases (CO 2 , or O 2 for example) increases as the temperature of the solvent goes down.

 The solubility of gases is also affected by pressure. How?

 Think about soda.

Henry’s Law

 Henry’s Law states that at a constant temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.  Henry’s Law   S 1 /P 1 = S 2 /P 2 Example:  If 0.85g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25 o C, how much will dissolve in 1.0 L at 1.0 atm at the same temperature?

 0.21 g

Decompression Sickness

 Decompression sickness is a term used to describe the effects of a drop in external pressure on a persons body.

 One common symptom of decompression sickness is AGE, Arterial Gas Embolism.  The bubbles that form in the blood stream from gas leaving solution can restrict blood flow.

Boiling Point Elevation

 Dissolving a solute in a solution raises the boiling point of the solution.

 This is called boiling point elevation.

 We can calculate the difference in boiling point (Δ T b ) by multiplying a solutions molality by a constant we look up for our particular solvent (K b ).

 Δ T b = K b m

 What is the boiling point of a 0.625m aqueous solution of a nonelectrolyte?

 What is the boiling point of a 0.4m aqueous solution of Ca(OH) 2 (A strong electrolyte) K b = 0.512 o C/m

Molal Boiling Point Elevation Constants.

Solvent

Water Benzene Chloroform Ethanol

Boiling Point ( o C)

100.0

80.1

61.7

78.5

K b ( o C/m)

0.512

2.53

3.63

1.22

Freezing Point Depression

 Adding a solute to a solvent also lowers the solvents freezing point.

 The equation is very similar to boiling point elevation.

 ΔT f = K f m

Solvent

Water Benzene Ethanol

Freezing Point ( o C)

0.00

5.5

-114.1

K f ( o C/m)

1.86

5.12

1.99

Example

 Sodium chloride is often used to prevent icy roads and to freeze ice cream. What are the boiling point and freezing point of a 0.029m AQUEOUS solution of sodium chloride?

 BP: 100.03 0 C  FP: -0.11 o C

   Calculate the freezing and boiling point of a solution containing 6.42 g of sucrose (a nonelectrolyte) in 100.0 g of water.

T f T b = -0.350 o C = 100.096 o C    Calculate the freezing and boiling point of a solution containing 23.7 g of Copper (II) sulfate (A strong electrolyte) in 250 g of water T f T b = -2.02 o C = 100.606 o C    Calculate the freezing and boiling point of a solution containing 0.15 moles of the compound naphthalene, a nonelectrolyte in 175 g of benzene (Normal F.P. = 5.5, Normal B.P. = 80.1, K f = 5.12, K b = 2.53) T f T b = 1.1 o C = 82.3 o C