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Chapter 7 SolutionsChapter 7 Solutions
7.1 Solutions
1
Solute and SolventSolute and Solvent
Solutions
• are homogeneous mixtures of two or more substances.
• consist of a solvent and one or more solutes.
2
Nature of Solutes in SolutionsNature of Solutes in Solutions
Solutes• spread evenly
throughout the solution.
• cannot be separated by filtration.
• can be separated by evaporation.
• are not visible, but can give a color to the solution.
3
Examples of SolutionsExamples of Solutions
The solute and solvent in a solution can be a solid, liquid, and/or a gas.
4
ExampleExample
Identify the solute in each of the following solutions.
A. 2 g of sugar and 100 mL of water
B. 60.0 mL of ethyl alcohol and 30.0 mL of methyl alcohol
C. 55.0 mL of water and 1.50 g of NaCl
D. Air: 200 mL of O2 and 800 mL of N2
5
WaterWaterWater• is the most common
solvent.• is a polar molecule.• forms hydrogen
bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule.
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Formation of a SolutionFormation of a Solution
Na+ and Cl- ions• on the surface of a NaCl
crystal are attracted to polar water molecules.
• are hydrated in solution with many H2O molecules surrounding each ion.
7
When NaCl(s) dissolves in water, the reaction can be written as
H2O
NaCl(s) Na+(aq) + Cl-(aq)
solid separation of ions
ExampleExample
Solid LiCl is added to water. It dissolves because
A. the Li+ ions are attracted to the 1) oxygen atom ( -) of water.
2) hydrogen atom (+) of water.
B. the Cl- ions are attracted to the 1) oxygen atom ( -) of water. 2) hydrogen atom (+) of water.
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Like Dissolves LikeLike Dissolves LikeTwo substances form a solution
• when there is an attraction between the particles of the solute and solvent.
• when a polar solvent such as water dissolves polar solutes such as sugar, and ionic solutes such as NaCl.
• when a nonpolar solvent such as hexane (C6H14)
dissolves nonpolar solutes such as oil or grease.
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Like Dissolves LikeLike Dissolves Like
Solvents Solutes
Water (polar)
Ni(NO3)2
CH2Cl2(nonpolar) (polar)
I2 (nonpolar)
10
ExampleExample
Which of the following solutes will dissolve in water? Why?
1) Na2SO4
2) gasoline (nonpolar)
3) I2
4) HCl
11
7.27.2 Electrolytes and NonelectrolytesElectrolytes and Nonelectrolytes
12
In water, • strong electrolytes produce ions and conduct an
electric current. • weak electrolytes produce a few ions. • nonelectrolytes do not produce ions.
13
7.27.2 Electrolytes and Electrolytes and NonelectrolytesNonelectrolytes
Strong ElectrolytesStrong Electrolytes
Strong electrolytes • dissociate in water, producing positive and negative
ions.• conduct an electric current in water.• in equations show the formation of ions in
aqueous(aq) solutions. H2O 100% ions
NaCl(s) Na+(aq) + Cl−(aq) H2O
CaBr2(s) Ca2+(aq) + 2Br−(aq)
14
ExamplesExamples
Complete each for strong electrolytes in water. H2O
A. CaCl2(s)
1) CaCl2(s)
2) Ca2+(aq) + Cl2−(aq)
3) Ca2+(aq) + 2Cl−(aq)
H2O
B. K3PO4(s)
1) 3K+(aq) + PO43−(aq)
2) K3PO4(s)
3) K3+(aq) + P3−(aq) + O4
−(aq)
15
Weak ElectrolytesWeak Electrolytes
A weak electrolyte• dissociates only slightly in water.• in water forms a solution of a few ions and mostly
undissociated molecules.
HF(g) + H2O(l) H3O+(aq) + F-(aq)
NH3(g) + H2O(l) NH4+(aq) + OH-(aq)
16
NonelectrolytesNonelectrolytes
Nonelectrolytes • dissolve as molecules in
water. • do not produce ions in
water.• do not conduct an
electric current.
17
ExamplesExamples Classify each of the solute represented in the following
equations as a strong, weak, or nonelectrolyte
a. NH4OH(aq) NH4+(aq) + -OH(aq)
b.HCl (aq) H+(aq) + Cl-(aq)
EquivalentsEquivalents
An equivalent (Eq) is the amount of an electrolyte or an ion that provides 1 mole of electrical charge (+ or -).
1 mole of Na+ = 1 equivalent
1 mole of Cl− = 1 equivalent
1 mole of Ca2+ = 2 equivalents
1 mole of Fe3+ = 3 equivalents
19
Electrolytes in Body FluidsElectrolytes in Body Fluids
In replacement solutions for body fluids, the electrolytes
are given in milliequivalents per liter (mEq/L).
Ringer’s Solution
Na+ 147 mEq/L
Cl− 155 mEq/L
K+ 4 mEq/L
Ca2+ 4 mEq/L
The milliequivalents per liter of cations must equal the milliequivalents per liter of anions.
20
ExamplesExamples
A. In 1 mole of Fe3+, there are
1) 1 Eq. 2) 2 Eq. 3) 3 Eq.
B. In 2.5 moles of SO42−, there are
1) 2.5 Eq. 2) 5.0 Eq. 3) 1.0 Eq.
C. An IV bottle contains NaCl. If the Na+ is
34 mEq/L, the Cl− is
1) 34 mEq/L. 2) 0 mEq/L. 3) 68 mEq/L.
21
7.3 Solubility7.3 Solubility
Solubility is
• the maximum amount of solute that dissolves in a specific amount of solvent.
• expressed as grams of solute in 100 grams of solvent, usually water.
g of solute100 g water
22
Unsaturated SolutionsUnsaturated Solutions
Unsaturated solutions
• contain less than the maximum amount of solute.
• can dissolve more solute.
23
Saturated SolutionsSaturated Solutions
Saturated solutions
• contain the maximum amount of solute that can dissolve.
• have undissolved solute at the bottom of the container.
24
ExamplesExamples
At 40 C, the solubility of KBr is 80 g/100 g of H2O.Identify the following solutions as either
1) saturated or 2) unsaturated. Explain.
A. 60 g KBr added to 100 g of water at 40 C.
B. 200 g KBr added to 200 g of water at 40 C.
C. 25 g KBr added to 50 g of water at 40 C.
25
ExamplesExamples The solubility of NaNO3 in water at 50oC is 110/100mL.
In a laboratory, a student use 50.0 g of NaNO3 with 200 g of water at the same temperature
◦ How many grams of NaNO3 will dissovle?
◦ Is the solution saturated or unsaturated?
◦ What is the mass, in grams, of any solid NaNO3 on the bottom of the container?
Effect of Temperature on Effect of Temperature on SolubilitySolubility
Solubility• depends on temperature.• of most solids increases as temperature increases.
• Hot tea dissolves more sugar than does cold tea because the solubility of sugar is much greater in higher temperature
• When a saturated solution is carefully cooled, it becomes a supersaturated solution because it contains more solute than the solubility allows.
• of gases decreases as temperature increases.• At higher temperature, more gas molecules have the energy to
escape from the solution
• Leaving higher pressure inside the container
27
Solubility and PressureSolubility and Pressure
Henry’s law states • the solubility of a gas
in a liquid is directly related to the pressure of that gas above the liquid.
• at higher pressures, more gas molecules dissolve in the liquid.
28
29
A. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun?
B. Why do fish die in water that is too warm?
ExamplesExamples
Soluble and Insoluble SaltsSoluble and Insoluble Salts Ionic compounds that dissolve in water => soluble
salts
Ionic compounds that do not separate into ions in water => insoluble salt
How do we know which ionic compounds is soluble or insoluble?
Precipitation Reactions and Precipitation Reactions and Solubility GuidelinesSolubility Guidelines
ExamplesExamplesPredict whether the following
ionic salt is soluble in water◦FeCl3
◦Na2CO3
◦AgCl
◦BaSO4
7.4 Percent Concentration7.4 Percent Concentration
The concentration of a solution
is the amount of solute dissolved in a specific amount of solution.
amount of soluteamount of solution
33
Mass PercentMass Percent
Mass percent (% m/m) is the• concentration by mass of solute in a solution.
mass percent = g of solute x 100 g of solute + g of solvent
• amount in g of solute in 100 g of solution.
mass percent = g of solute a
100 g of solution
34
Calculating Mass PercentCalculating Mass PercentThe calculation of mass percent (% m/m) requires the• grams of solute (g KCl) and• grams of solution (g KCl solution).
g of KCl = 8.00 gg of solvent (water) = 42.00 gg of KCl solution = 50.00 g
35
ExampleExample
A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g of H2O. Calculate the mass percent (% m/m) of the solution.
1) 15.0% (m/m) Na2CO3
2) 6.38% (m/m) Na2CO3
3) 6.00% (m/m) Na2CO3
36
Volume PercentVolume PercentThe volume percent (% v/v) is
• percent volume (mL) of solute (liquid) to volume (mL) of solution.
volume % (v/v) = mL of solute x 100 mL of solution
• solute (mL) in 100 mL of solution.
volume % (v/v) = mL of solute 100 mL of solution
37
Mass/Volume PercentMass/Volume Percent
The mass/volume percent (% m/v) is
• percent mass (g) of solute to volume (mL) of solution.
mass/volume % (m/v) = g of solute x 100 mL of
solution
• solute (g) in 100 mL of solution.
mass/volume % (m/v) = g of solute
100 mL of solution
38
Percent Conversion FactorsPercent Conversion FactorsTwo conversion factors can be written for each type of % value.
39
Copyright © 2009 by Pearson Education, Inc.
Using Percent Concentration Using Percent Concentration (m/m) (m/m) as Conversion Factorsas Conversion Factors
How many grams of NaCl are needed to prepare225 g of a 10.0% (m/m) NaCl solution?
STEP 1: Given: 225 g solution; 10.0% (m/m) NaCl Need: g of NaCl
STEP 2: g solution g NaCl
STEP 3: Write the 10.0% (m/m) as conversion factors.10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl
STEP 4: Set up using the factor that cancels g solution.
40
ExampleExample
How many milliliters of a 5.75% (v/v) ethanol solution can be prepared from 2.25 mL of ethanol?
1) 2.56 mL
2) 12.9 mL
3) 39.1 mL
41
Using Percent Concentration (m/v) Using Percent Concentration (m/v)
as Conversion Factorsas Conversion FactorsHow many mL of a 4.20% (m/v) will contain 3.15 g of KCl?
STEP 1: Given: 3.15 g of KCl(solute); 4.20% (m/v) KCl Need: mL of KCl solution
STEP 2: Plan: g of KCl mL of KCl solution
STEP 3: Write conversion factors. 4.20 g KCl and 100 mL solution
100 mL solution 4.20 g KCl
STEP 4: Set up the problem
42
ExampleExampleHow many grams of NaOH are needed to prepare 125 mL
of a 8.80% (m/v) NaOH solution?
43
ExampleExample What is the % m/v of NaOH in a solution prepared by
dissovling 12.0 g of NaOH in enough water to make 220.0 mL solution?