chapter 11 solutions copyright © 2005 by pearson education, inc. publishing as benjamin cummings
TRANSCRIPT
Chapter 11 Solutions
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
Solute and SolventSolutions• Are homogeneous mixtures of two or more substances.• Consist of a solvent and one or more solutes.
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
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
Examples of Solutions
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.
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
(Intermolecular Forces)
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.
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
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
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
Which of the following solutes will dissolve in water? Why?
1) Na2SO4
2) gasoline (nonpolar)
3) I2
4) HCl
Learning Check
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
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
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
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
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.
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)
Learning Check A precipitate forms in the following reaction.
Write the molecular, ionic and net ionic equations for the reaction.
BaCl2(aq) + Na2SO4(aq)
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
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
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
Percent Conversion Factors
• Two conversion factors can be written for each type of % value.
Write two conversion factors for each solutions:
A. 8.50%(m/m) NaOH
B. 5.75%(v/v) ethanol
Learning Check
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
Molarity (M)
Molarity (M) is:
• A concentration term for solutions.
• Gives the moles of solute in 1 L solution.
• moles of soluteliter of solution
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.
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
Molarity Conversion FactorsThe units of molarity are used as conversion factors in calculations with solutions.
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
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
DilutionIn a dilution,
• Water is added
• Volume increases
• Concentration decreases
M1V1 = M2V2
initial diluted
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
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
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
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