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Honors Chemistry, Chapter 13

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Title: Honors Chemistry, Chapter 13


1
Chapter 13 Solutions
2
Solutions
  • If you put sugar in water, it is soluble and will
    form a solution. Soluble means that a substance
    is capable of being dissolved in another
    substance.
  • Water and sugar form a solution. A solution is a
    homogeneous mixture of two or more substances in
    a single phase.

3
Components of a Solution
  • The simplest type of solution, like sugar in
    water consists of
  • The dissolving medium is called the solvent.
  • The substance being dissolved is called the
    solute.

4
Solute-Solvent Combinations
5
Suspensions and Colloids
  • If particles in a solute are so large that they
    settle out unless the mixture is constantly
    stirred or agitated, the mixture is called a
    suspension. Particles over 1000 nm in diameter
    (1000 X the size of atoms or molecules) form
    suspensions.
  • Particles that are intermediate in size between
    those in solution and suspensions form mixtures
    known as colloidal dispersions, or simply
    colloids. Particles between 1 nm and 1000 nm in
    diameter may form colloids.

6
Classes of Colloids
7
Solutions, Colloids, Suspensions
8
Solutes Electrolytes vs. Nonelectrolytes
  • A substance that dissolves in water to make a
    solution that conducts electric current is called
    an electrolyte.
  • A substance that dissolves in water to make a
    solution that does not conduct an electric
    current is called a nonelectrolyte.

9
Chapter 13, Section 1 Review
  • Distinguish between heterogeneous and homogeneous
    mixtures.
  • List three different solute-solvent combinations.
  • Compare the properties of suspensions, colloids,
    and solutions.
  • Distinguish between electrolytes and
    non-electrolytes.

10
Factors Affecting the Rate of Dissolution
  • What can be done to increase the rate of
    dissolution?
  • Increasing the surface area of the solute
  • Agitating the solution
  • Heating the solvent

11
Solubility Characteristics
  • Solution equilibrium is the physical state in
    which the opposing process of dissolution and
    crystallization of a solute occur at equal rates.
  • A solution that contains the maximum amount of
    dissolved solute is described as a saturated
    solution.

12
Solubility Characteristics
  • A solution that contains less solute than a
    saturated solution under the existing conditions
    is an unsaturated solution.
  • The solubility of a substance is the amount of
    that substance required to form a saturated
    solution with a specific amount of solvent at a
    specified temperature.

13
Dissolving Ionic Compounds in Aqueous Solution
  • Water tends to surround charged ions. This
    solution process is referred to as hydration.
  • When a crystalline compound incorporates water
    into the crystal structure, the compound is known
    as a hydrate. (Example CuSO4.5H2O)

14
Nonpolar Solvents
  • Nonpolar solvents are covalently bonded solvents
    without strong molecular dipoles.
  • Ionic compounds are generally not soluble in
    nonpolar solvents such as carbon tetrachloride
    and toluene because they differ widely in
    bonding, polarity, and intermolecular forces.

15
Liquid Solutes and Solvents
  • Liquid solutes and solvents that are not soluble
    in each other are immiscible.
  • Liquids that dissolve freely in one another in
    any proportion are said to be completely
    miscible. (There is no partially miscible.)
    Examples Water and Ethanol Hexane and
    Octane Benzene and CCl4

16
Effects of Pressure on Solubility
  • Pressure has very little effect on the
    solubilities of liquids and solids in liquid
    solvents. However, increases in pressure
    increase gas solubilities in liquids.
  • Gas solvent solution

17
Henrys Law
  • Henrys Law states that the solubility of a gas
    in a liquid is directly proportional to the
    partial pressure of that gas on the surface of
    the liquid.
  • When the pressure over a liquid with dissolved
    gas is reduced, bubbles of gas begin to form in
    the solution. The rapid escape of gas from a
    liquid in which it is dissolved is known as
    effervescence.

18
Three Steps In Dissolving
  • Solute particles become separated from the solid.
    (Requires addition of energy.)
  • Solvent particles must move aside to make room
    for the solute particles. (Requires addition of
    energy.)
  • Solvent particles are attracted to and solvate
    solute particles. (Releases energy.)

19
Heat of Solution
  • The net amount of heat energy absorbed or
    released when a specific amount of solute
    dissolves in a solvent is the heat of solution.
  • Heats of solution may be either negative
    (exothermic, heat is released) or positive
    (endothermic, heat is required).

20
Chapter 13, Section 2 Review
  • List and explain three factors that affect the
    rate at which a solid solute dissolves in a
    liquid solvent.
  • Explain solution equilibrium, and distinguish
    among saturated, unsaturated, and supersaturated
    solutions.
  • Explain the meaning of like dissolves like in
    terms of polar and non-polar substances.

21
Chapter 13, Section 2 Review
  • List three interactions that contribute to the
    heat of solution, and explain what causes
    dissolution to be exothermic or endothermic.
  • Compare the effects of temperature and pressure
    on solubility.

22
Concentration of Solutions
  • The concentration of a solution is a measure of
    the amount of solute in a given amount of solvent
    (molality) or solution (molarity).

23
Molarity
  • Molarity is the number of moles of solute in one
    liter of solution.
  • Example 0.5 mole of NaOH in 1 liter of solution
    is 0.5 molar NaOH
  • To prepare 0.5 molar NaOH, 20.0 g (0.5 mole) of
    NaOH is dissolved in less than a liter of water,
    then the volume is carefully brought up to 1
    liter.

24
Sample Problem 13-1
  • You have 3.50 L of solution that contains 90.0 g.
    of sodium chloride, NaCl. What is the molarity
    of the solution?
  • Given 90.0 g of NaCl
  • 3.5 liter of solution
  • 90.0 g of NaCl x 1 mole of NaCl 1.54 mol
    58.44 g NaCl

25
Sample Problem 13-1
1.54 mol NaCl 0.440 M NaCl 3.50 L of
solution
26
Sample Problem 13-2
You have 0.8 L of 0.5 M HCl solution. How many
moles of HCl does this solution contain? Given
0.8L of HCl Solution 0.5 M HCl is the
solution Concent. 0.5 mole of HCl x 0.8 L
0.4 mole HCl 1 L of Solution
27
Sample Problem 13-3
  • To produce 40 g. of Ag2CrO4, you will need at
    least 23.4 grams of K2CrO4 in solution as a
    reactant. All you have on hand in the stock room
    is 5 L of a 6 M K2CrO4 solution. What volume of
    the solution is needed to give you the 23.4 g of
    K2CrO4 needed for the reaction?

28
Sample Problem 13-3
Given 5 L of 6.0 M K2CrO4 Mass needed 23.4
g of K2CrO4 Mass of product 40.0 g
Ag2CrO4 1 mol K2CrO4 194.2 g. 23.4 g K2CrO4
x 1 mol K2CrO4 1.20 mol 194.2 g K2CrO4
29
Sample Problem 13-3
1.20 mol K2CrO4 6.0 M K2CrO4 x L of K2CrO4 x
0.020 L K2CrO4 Solution
30
Molality
  • Molality is the concentration of a solution
    expressed in moles of solute per kilogram of
    solvent.
  • Example 40.0 g of NaOH (1 mole) dissolved in 1
    kg of water would make a 1 m NaOH solution

31
Sample Problem 13-4
  • A solution was prepared by dissolving 17.1 g of
    sucrose (C12H22O11) in 125 g of water. Find the
    molal concentration of this solution.
  • Given 17.1 g of C12H22O11
  • 125 g of H2O
  • 1 mol of C12H22O11 342.34 g

32
Sample Problem 13-4
342.34 g C12H22O11 x 1 mol C12H22O11
342.34 g C12H22O11 0.0500
mol C12H22O11 125 g H2O x 1 kg H2O 0.125
kg H2O 1000 g H2O
33
Sample Problem 13-4
0.0500 mol C12H22O11 0.400 m C12H22O11
0.125 kg H2O
34
Sample Problem 13-5
  • A solution of iodine, I2, in CCl4 is used when
    iodine is needed for a certain chemical test.
    How much iodine must be added to prepare a 0.480
    m solution of iodine in CCl4 if 100.0 g of CCl4
    is used?
  • Given 0.480 m I2
  • 100 g of CCl4
  • 253.8 g of I2 1 mole

35
Sample Problem 13-5
100.0 g CCl4 x 1 kg CCl4 0.100 kg CCl4
1000 g CCl4 0.48 m x mol I2
0.100 kg CCl4 x 0.0480 mol I2
36
Sample Problem 13-5
0.0480 mol I2 x 253.8 g I2 12.2 g I2
1 mol I2
37
Chapter 13, Section 3 Review
  • Given the mass of solute and volume of solvent,
    calculate the concentration of a solution.
  • Given the concentration of a solution, determine
    the amount of solute in a given amount of
    solution.
  • Given the concentration of a solution, determine
    the amount of solution that contains a given
    amount of solute.
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