by Steven S. Zumdahl & Donald J. DeCoste
University of Illinois
Introductory Chemistry: A Foundation, 6th Ed.
Introductory Chemistry, 6th Ed.
Basic Chemistry, 6th Ed.
Chapter 15
Solutions
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Solutions
• Solutions are homogeneous mixtures.
• Solvent: the substance present in the highest percentage
• Solute: the dissolved substance, which is present in lesser amount
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Solutions (cont.)
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The Solution Process: Ionic Compounds
• When ionic compounds dissolve in water they dissociate into ions and become hydrated.
• When solute particles are surrounded by solvent molecules we say they are solvated.
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The Solution Process: Covalent Molecules
• Covalent molecules that are small and have “polar” groups tend to be soluble in water.
• The ability to H-bond with water enhances solubility.OH
H
C O
H
HH H
O
H
H
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Solubility
• When one substance (solute) dissolves in another (solvent), it is said to be soluble.– Salt is soluble in water– Bromine is soluble in methylene chloride
• When one substance does not dissolve in another, it is said to be insoluble.– Oil is insoluble in water
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Solubility (cont.)
• There is usually a limit to the solubility of one substance in another.– Gases are always soluble in each other– Some liquids are always mutually soluble
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Solutions & Solubility
• Molecules that are similar in structure tend to form solutions: “like dissolves like”
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Solutions & Solubility (cont.)
• The solubility of the solute in the solvent depends on the temperature.– Higher temp = greater solubility of solid in liquid– Lower temp = greater solubility of gas in liquid
• The solubility of gases depends on the pressure.– Higher pressure = greater solubility
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Describing Solutions Qualitatively
• A concentrated solution has a high proportion of solute to solution.
• A dilute solution has a low proportion of solute to solution.
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Describing Solutions Qualitatively (cont.)
• A saturated solution has the maximum amount of solute that will dissolve in the solvent.– Depends on temp
• An unsaturated solution has less than the saturation limit.
• A supersaturated solution has more than the saturation limit.– Unstable
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Describing Solutions Quantitatively (cont.)
• Solutions have variable composition.
• To describe a solution accurately, you need to describe the components and their relative amounts.
• Concentration: the amount of solute in a given amount of solution– Occasionally amount of solvent
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Solution Concentration Percentage
• Mass percent = grams of solute per 100 g of solution– 5.0% NaCl has 5.0 g of NaCl in every 100 g of
solution
• Mass of solution = mass of Solute + mass of solvent
• Divide the mass of solute by the mass of solution and multiply by 100% to get mass percent.
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Solution Concentration Molarity
• Moles of solute per 1 liter of solution
• Used because it describes how many moles of solute in each liter of solution
• If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar, etc.
molarity = moles of soluteliters of solution
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Molarity & Dissociation
• The molarity of the ionic compound allows you to determine the molarity of the dissolved ions.
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Molarity & Dissociation (cont.)
• CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq)
• A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution
• Because each CaCl2 dissociates to one Ca+2, 1.0 M CaCl2 = 1.0 M Ca+2
• Because each CaCl2 dissociates to 2 Cl-1, 1.0 M CaCl2 = 2.0 M Cl-1
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Dilution
• Dilution: adding solvent to decrease the concentration of a solution
• The amount of solute stays the same, but the concentration decreases.
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Dilution (cont.)
• Dilution Formula
M1 x V1 = M2 x V2
# Moles/L · # L = # moles– In dilution we take a certain number of moles
of solute and dilute to a bigger volume.
• Concentrations and volumes can be most units as long as they are consistent.
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Solution Stoichiometry
• Many reactions occur in solution. Therefore you need to be able to predict amounts of reactants and products in terms of concentrations and volumes as well as masses.
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Solution Stoichiometry (cont.)
• Basic strategy is the same:1. Balance the equation
2. Change given amounts to moles (M x V = #moles)
3. Determine limiting reactant
4. Calculate moles of required substance
5. Convert moles of the required substance into the desired unit
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Example #1:
Calculate the mass of solid NaCl required to precipitate all the Ag+1 ions from 1.50 L of a 0.100 M AgNO3 solution.
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Example #1 (cont.)
• Write and balance the reaction:– The reaction is a precipitation reaction. It
involves Cl-1 ions from NaCl reacting with Ag+1 ions from AgNO3 to form AgCl(s). Therefore we get:
Ag+1(aq) + Cl-1(aq) AgCl(s) (balanced)
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• Change the given amounts to moles:– We are given 1.50 L of 0.100 M AgNO3 . Since 1
AgNO3 dissociates into 1 Ag+1
11
Ag mol 0.150 Solution L 1
Ag mol 0.100 x L 1.50
Example #1 (cont.)
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Example #2:
Calculate the mass of solid NaCl required to precipitate all the Ag+1 ions from 1.50 L of a 0.100 M AgNO3 solution.
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Example #2 (cont.)
• Determine the limiting reactant:– Since we are going to precipitate all the Ag+1 by
adding Cl-1 , the Ag+1 is the limiting reactant
• Determine the number of moles of the required substance:– We need to calculate the moles of Cl-1 required
to precipitate 0.150 moles of Ag+1
11
11 Cl mol 0.150
Ag mol 1
Cl mol 1 x Ag mol 0.150
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• Convert moles of the required substance into the desired unit:– We need 0.150 moles of Cl-1. Since 1 NaCl
dissociates into 1 Cl-1, the moles of NaCl needed = 0.150 moles. 1 mol NaCl = 58.44 g NaCl
NaCl g 8.76 NaCl mol 1
NaCl g 58.44 x NaCl mol 0.150
Example #2 (cont.)
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Neutralization Reactions
• Acid-Base reactions are also called neutralization reactions.
• Often we use neutralization reactions to determine the concentration of an unknown acid or base.
• The procedure is called a titration. With this procedure we can add just enough acid solution to neutralize a known volume of a base solution.– Or vice-versa
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Normality
• Normality: concentration unit used mainly for acids and bases
• One equivalent of an acid is the amount of acid that can furnish 1 mol of H+1
• One equivalent of a base is the amount of base that can furnish 1 mol of OH-1
• Equivalent weight: the mass of 1 equivalent of an acid or base
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Equivalents
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Solution Concentration Normality
• Equivalents of solute per 1 liter of solution
• Used because it describes how many H+ or OH- in each liter of solution
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Solution Concentration Normality (cont.)
• If an acid solution concentration is 2.0 N, 1 liter of solution contains 2.0 equiv. of acid - which means 2 mol H+1
– 2 liters = 4.0 equiv acid = 4.0 mol H+1 – 0.5 liters = 1.0 equiv acid = 1.0 mol H+1
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Normality
Normality = equivalents of solute
liters of solution
Liters x normality = equivalents of solute
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Normality and Neutralization
• One equivalent of acid exactly neutralizes one equivalent of base.
• Can be used to simplify neutralization stoichiometry problems to the equation
Nacid x Vacid = Nbase x Vbase
# equiv/L x # L = # equiv
# eqivacid = # equivbase
• The volumes can be most any unit, as long as they are consistent.
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Example #3:
What volume of 0.075 N KOH is required to neutralize 0.135 L of 0.45 N H3PO4?
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Example #3 (cont.)
• Determine the quantities and units in the problem
Acid Solution Base Solution
Normality 0.45 N 0.075 N
Volume 0.135 L ? L• Solve the formula for the unknown quantity
base
acid x acid base
basebaseacidacid
NVNV
V x N V x N
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• Plug the values into the equation and solveAcid Solution Base
Solution
Normality 0.45 N 0.075 N
Volume 0.135 L ? L
L 0.81 N 0.075
L 0.135N 0.45V
NVNV
x
base
base
acid x acid base
Example #3 (cont.)