Transcript Slide 1

Chapter 15: Solutions
15.1 Solubility
Objectives: To understand the process of
dissolving.
To learn why certain substances dissolve in
water.
Chapter 15.1: Solutions
Solution: Homogeneous mixture in which the
components are uniformly intermingled.
Examples: air, shampoo, orange soda, coffee,
gasoline, cough syrup, etccc…
Solvent: substance present in the largest amount.
Solute: the other substance.
Aqueous solution: water as the solvent.
Figure 15.1: Dissolving of solid
sodium chloride.
Figure 15.2: Polar water molecules
interacting with positive and negative
ions of a salt.
Chapter 15.1: Solutions
Solubility of Polar
Substances:
Ethanol
Can form hydrogen bonds.
Makes it compatible with water.
Water
molecule interacts strongly with
the polar O—H bond in ethanol.
Figure 15.4: Structure of
common table sugar.
Contains many polar O-H bonds each of which can
hydrogen-bond to a water molecule.
Figure 15.5: A molecule typical of those
found in petroleum.
Similar
electronegativity
Share electrons ==
Nonpolar
Like dissolves Like
• Nonpolar solutes dissolve in nonpolar
solvents.
15.2: Solution Composition: An
Introduction
• Objective: To learn qualitative terms
associated with concentration of a
solution.
15.2: Solution Composition: An
Introduction
• Saturated solution: contains as much solute as will
dissolve at that temperature.
• Unsaturated: solution that has not reached the limit of
solute.
• Supersaturated solution: contains more dissolved solid
than a saturated solution will hold at that temperature. A
supersaturated solution is very unstable.
• Concentrated solution: relatively large amount of solute
• Dilute Solution: small amount of solute.
15.3: Factors Affecting the Rate of
Dissolution
• Objectives:
1) To understand the factors that affect the
rate at which a solute dissolves.
15.3: Factors Affecting the Rate of
Dissolution
• Three factors affect the speed of the
dissolving process.
– Surface area
– Stirring
– Temperature
15.4: Solution Composition: Mass
Percent
• Objective:
To understand the concentration term mass
percent and learn how to calculate it.
15.4: Solution Composition: Mass
Percent
Mass percent = mass of solute x 100%
mass of solution
Mass percent = grams of solute x 100%
grams of solute + grams of solvent
15.4: Solution Composition: Mass
Percent
A solution is prepared by mixing 2.5 g of
calcium chloride w/ 50.0g of water.
Calculate the mass percent of calcium
chloride in this solution.
4.76%
15.4: Solution Composition: Mass
Percent
A solution is prepared by mixing 1.00 g of
ethanol with 100.0 g of water. Calculate
the mass percent of ethanol in this
solution.
0.990%
15.4: Solution Composition: Mass
Percent
A solution of milk contains 4.5% by mass of
lactose. Calculate the mass of lactose
present in 175 g of milk.
7.9 g
15.5: Solution Composition:
Molarity
Objectives:
1) To understand molarity.
2) To learn to use molarity to calculate the
number of moles of solute present.
15.5: Solution Composition:
Molarity
To measure the concentration by volume.
concentration is defined as the amount of
solute in a given volume of solution.
Molarity (M) describes the amount of solute
in moles and the volume of the solution
in liters.
Molarity is the number of moles of solute per
volume of solution in liters.
15.5: Solution Composition:
Molarity
Molarity is the number of moles of solute per
volume of solution in liters.
M=molarity= moles of solute = mol
liters of solution L
15.5: Solution Composition:
Molarity
Calculate the molarity of a solution prepared
by dissolving 11.5 g of solid NaOH in
enough water to make 1.5L of solution.
0.192 M
15.5: Solution Composition:
Molarity
Calculate the molarity of a solution prepared
by dissolving 1.56 g of gaseous HCl into
enough water to make 26.8 ml
1.59 M
15.5: Solution Composition:
Molarity
Give the concentrations of all the ions in
each of the following solutions.
a. 0.50 M Co(NO3)2
b. 1 M FeCl3
0.50 M Co, 1.0M NO3
1 M Fe, 3.0 M Cl
15.5: Solution Composition:
Molarity
To analyze the alcohol content of a certain
wine, a chemist needs 1.00 L of an
aqueous 0.200 M K2Cr2O7. How much
solid K2Cr2O7 (molar mass =294.2 g)
must be weighed out to make this
solution?
0.200 mol, 58.8 g
Standard solution: is a solution whose
concentration is accurately known.
Figure 15.7: Steps involved in the
preparation of a standard aqueous solution.
15.5: Solution Composition:
Molarity
How many moles of Ag+ ions are present in
25 ml of a 0.75 M AgNO3 solution?
0.50 M Co, 1.0M NO3
1 M Fe, 3.0 M Cl
15.6: Dilution
Objective: To learn to calculate the
concentration of a solution made by
diluting a stock solution.
15.6: Dilution
Objective: To learn to calculate the
concentration of a solution made by
diluting a stock solution.
15.6: Dilution
Stock solution is purchased. Often need to add
water or another solvent to get to the desired
concentration.
Moles of solute after dilution=moles of solute before dilution
# of moles of solute stays the same, but more water is added
increasing the volume, so the molarity decreases.
M= moles of solute
volume (L)
15.6: Dilution
Volume of dilute molarity of moles of solute
Solution (liters) x dilute soln. = present
15.6: Dilution
What volume of 16M sulfuric acid must be
used to prepare 1.5 L of a 0.10M H2SO4
solution?
9.4 ml
15.6: Dilution
What volume of 19M sodium hydroxide must
be used to prepare 1.0L of a 0.15M
NaOH solution?
7.9 ml
15.6: Dilution
What volume of water is needed to prepare
500.0 ml of a 0.250 M Ca(NO3)2 solution
from a 5.00 M Ca(NO3)2 solution?
475 ml
15.7: Stoichiometry of Solution
Reactions
Objectives: To understand the strategy for
solving stoichiometric problems for
solution reactions
15.7: Stoichiometry of Solution
Reactions
Step 1: Write the balanced equation for the
reaction. For reactions involving ion, it is
best to write the net ionic equation.
Step 2: Calculate the moles of reactants.
Step 3: Determine which reactant is limiting.
Step 4: Calculate the moles of other
reactants or products, as required.
Step 5: Convert to grams or other units, if
required.
15.7: Stoichiometry of Solution
Reactions
Calculate the mass of solid NaCl that must
be added to 1.50 L of a 0.100 M AgNO3
solution to precipitate all of the Ag+ ions
in the form of AgCl. Calculate the mass
of AgCl formed.
21.5 g AgCl
15.7: Stoichiometry of Solution
Reactions
Calculate the mass of solid sodium sulfate
that must be added to 250.0 ml of a 0.2
M solution of barium nitrate to precipitate
all of the barium ions in the form of
barium sulfate. Also calculate the mass
of barium sulfate forms.
7.10 g Na2SO4; 11.7 g BaSO4
15.7: Stoichiometry of Solution
Reactions
Calculate the mass of sodium iodide that
must be added to 425.0 ml of a 0.100 M
Pb(NO3)2 solution to precipitate all of the
Pb2+ ions as PbI2. Also calculate the
mass of PbI2 forms.
12.7 g NaI; 19.6g PbI2
15.7: Stoichiometry of Solution
Reactions
Calculate the mass of the white solid CaCO3
that forms when 25.0 ml of a 0.100 M
Ca(NO3)2 solution is mixed with 20.0 ml
of a 0.150M Na2CO3 solution.
0.250 g CaCO3
15.7: Stoichiometry of Solution
Reactions
Calculate the mass of the blood-red solid
Ag2CrO4 that forms when 50.0 ml of a
0.250M AgNO3 solution is mixed with
30.0 ml of a 0.250M K2CrO4 solution.
2.07 g Ag2CrO4
15.8: Neutralization Reactions
Objective: To learn how to do calculations
involved in acid-base reactions.
Neutralization reaction: acid-base reaction
When just enough strong acid is added to
react exactly with the strong base, the
acid is neutralized. One product is water.
15.8: Neutralization Reactions
What volume of a 0.100 M HCl solution is
needed to neutralize 25.0 ml of a 0.350
M NaOH solution?
8.75 x 10-2 L
15.8: Neutralization Reactions
What volume of a 0.150 M HNO3 solution is
needed to neutralize 45.0 ml of a 0.550
M KOH solution?
165 ml
15.8: Neutralization Reactions
What volume of a 1.00 x 10-2 M HCl solution
is needed to neutralize 35.0 ml of a 5.00
x 10-3 M Ba(OH)2 solution?
35.0 ml
15.9: Normality
Objectives: To learn about normality and
equivalent weight.
To learn to use these concepts in
stoichiometric calculations.
15.9: Normality
Normality:
Equivalent of an acid: amt. of acid that can
furnish 1 mol of H+ ions.
Equivalent of a base: amt of base that can
furnish 1 mol of OH- ions.
Equivalent weight: mass in grams of 1
equivalent of that acid or base.
15.9: Normality
Normality: is defined as the number of equivalents
of solute per liter of solution.
Normality=N= # of equivalents= equivalents
1 liter of soln.
L
HCl
H2SO4
H3PO4
Molar Mass
36.5 g
98.0
98.0
Equivalent Wt (g)
36.5 g
49.0g
32.7 g
15.9: Normality
A solution of sulfuric acid contains 86 g of
H2SO4 per liter of solultion. Calculate the
normality of this solution.
1.8 N H2SO4
15.9: Normality
Arsenic acid, H3AsO4, can furnish three H+
ions per molecule. Calculate the
equivalent weight of H3AsO4.
47.31 g
15.9: Normality
Calculate the equivalent weight of HBr.
80.91 g
15.9: Normality
A solution of phosphoric acid contains 50.0g
of H3PO4 per liter of solution. Calculate
the normality of this solution.
1.53 N
15.9: Normality
A solution of hydrochloric acid contains 25.0
g of HCl per liter of solution. Calculate
the normality of this solution.
0.686N
15.10: The Properties of Solutions:
Boiling Point and Freezing Point
Objective: To understand the effect of a
solute on solution properties.
1.0M NaCl solution freezes at about -1oC.
Boils at 104oC.
Figure 15.9: A bubble in the interior of
liquid water surrounded by solute
particles and water molecules.
Figure 15.10: Pure water.
Colligative property: depends on the number
of solute particles.