chapter 11 solutions copyright © 2005 by pearson education, inc. publishing as benjamin cummings
TRANSCRIPT
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Chapter 11 Solutions
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
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Solute and SolventSolutions• Are homogeneous mixtures of two or more substances.• Consist of a solvent and one or more solutes.
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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.
Nature of Solutes in Solutions
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Examples of Solutions
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WaterWater• 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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(Intermolecular Forces)
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Formation of a Solution
Na+ and Cl- ions,• On the surface of a NaCl
crystal are attracted to polar water molecules.
• In solution are hydrated as several H2O molecules surround each.
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When NaCl(s) dissolves in water, the reaction can be written as:
H2O
NaCl(s) Na+(aq) + Cl- (aq)
solid separation of ions
Equations for Solution Formation
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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.
Learning Check
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Two 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.
Like Dissolves Like
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Which of the following solutes will dissolve in water? Why?
1) Na2SO4
2) gasoline (nonpolar)
3) I2
4) HCl
Learning Check
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In water, • Strong electrolytes produce ions and conduct an
electric current. • Weak electrolytes produce a few ions. • Nonelectrolytes do not produce ions.
Solutes and Ionic Charge
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Strong electrolytes, • Dissociate in water producing positive and negative ions.• Produce 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)
Strong Electrolytes
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A weak electrolyte,• Dissociates only slightly in water.• In water forms a solution of only a few ions and
mostly undissociated molecules.
HF(g) + H2O(l) H3O+(aq) + F- (aq)
NH3(g) + H2O(l) NH4+(aq) + OH- (aq)
Weak Electrolytes
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Solubility • Is the maximum amount of solute that dissolves in a
specific amount of solvent. • Can be expressed as grams of solute in 100 grams of
solvent, usually water.
g of solute
100 g water
Solubility
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Unsaturated solutions • Contain less than the maximum
amount of solute. • Can dissolve more solute.
Saturated solutions • Contain the maximum amount of solute that can dissolve. • Have undissolved solute at the bottom of the container.
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Soluble and Insoluble SaltsIonic compounds that• Dissolve in water are
soluble salts.• Do NOT dissolve in water
are insoluble salts.
***
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Equations for Forming SolidsA molecular equation shows the formulas of the
compounds.
Pb(NO3)(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq)
An ionic equation shows the ions of the compounds.
Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + 2Cl−(aq)
PbCl2(s) + 2Na+(aq) + 2NO3−(aq)
NOTE: the “spectator ions”
A net ionic equation shows only the ions that form a solid. Pb2+(aq) + 2Cl−(aq) PbCl2(s)
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Learning Check A precipitate forms in the following reaction.
Write the molecular, ionic and net ionic equations for the reaction.
BaCl2(aq) + Na2SO4(aq)
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The mass percent (%m/m)
• Concentration is the percent by mass of solute in a solution.
mass percent (%m/m)
= g of solute x 100
g of solute + g of solvent
• Is the g of solute in 100 g of solution.
mass percent = g of solute x 100
100 g of solution
Mass Percent
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Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution).
g of KCl = 8.00 g
g of solvent (water) = 42.00 g
g of KCl solution = 50.00 g
8.00 g KCl (solute) x 100 = 16.0% (m/m)
50.00 g KCl solution
Calculating Mass Percent
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The 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
Volume Percent
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Percent Conversion Factors
• Two conversion factors can be written for each type of % value.
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Write two conversion factors for each solutions:
A. 8.50%(m/m) NaOH
B. 5.75%(v/v) ethanol
Learning Check
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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 NaClSTEP 2 g solution g NaClSTEP 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.225 g solution x 10.0 g NaCl = 22.5 g NaCl
100 g solution
Using Percent Factors
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Molarity (M)
Molarity (M) is:
• A concentration term for solutions.
• Gives the moles of solute in 1 L solution.
• moles of soluteliter of solution
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Preparing a 1.0 Molar Solution
A 1.00 M NaCl solution is prepared:• By weighing out 58.5 g NaCl (1.00 mol) and• Adding water to make 1.00 liter of solution.
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What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3?
1) 0.557 M
2) 1.44 M
3) 1.71 M
Learning Check
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Molarity Conversion FactorsThe units of molarity are used as conversion factors in calculations with solutions.
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How many grams of AlCl3 are needed to prepare
125 mL of a 0.150 M solution?
1) 20.0 g AlCl3
2) 16.7g AlCl3
3) 2.50 g AlCl3
Learning Check
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How many milliliters of 2.00 M HNO3 contain 24.0 g HNO3?
1) 12.0 mL
2) 83.3 mL
3) 190. mL
Learning Check
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DilutionIn a dilution,
• Water is added
• Volume increases
• Concentration decreases
M1V1 = M2V2
initial diluted
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Learning Check
What is the final volume (mL) of 15.0 mL of a 1.80 M
KOH diluted to give a 0.300 M solution?
1) 27.0 mL
2) 60.0 mL
3) 90.0 mL
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Using Molarity of Reactants
How many mL of 3.00 M HCl are needed to completely
react with 4.85 g CaCO3?
2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)
STEP 1 Given: 3.00 M HCl; 4.85 g CaCO3
Need: volume in mL
STEP 2 Plan:
g CaCO3 mol CaCO3 mol HCl mL HCl
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Using Molarity of Reactants (cont.)
2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)
STEP 3 Equalities
1 mol CaCO3 = 100.09 g; 1 mol CaCO3 = 2 mol HCl
1000 mL HCl = 3.00 mol HCl
STEP 4 Setup
4.85 g CaCO3 x 1 mol CaCO3 x 2 mol HCl x 1000 mL HCl
100.09 g CaCO3 1 mol CaCO3 3.00 mol HCl
= 32.3 mL HCl required
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Learning Check
If 22.8 mL of 0.100 M MgCl2 is needed to completely
react 15.0 mL of AgNO3 solution, what is the molarity of
the AgNO3 solution?
MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq)
1) 0.0760 M
2) 0.152 M
3) 0.304 M