Chapter 13
Solutions
Homework
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Assigned Problems (odd numbers only)
“Problems” 25 to 59 (begins on page 478)
“Cumulative Problems” 109-129 (begins on page
482)
“Highlight Problems” 131, 133, page 484-485
Solutions
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A solution is a homogeneous mixture
of two or more substances
It is uniform, same composition
throughout
One substance is dissolved into
another
Requires the interaction of particles
of atomic or molecular size
Solutions
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Two parts: Solvent and solute
Solute: Substance being dissolved (present in a
smaller amount relative to the solvent)
Solvent: Substance that dissolves the solvent
(present in the greatest amount)
Most solutions are liquid but can be gaseous or
solid
Solutions
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“Like dissolves like”
Substances that are similar should
form a solution
Refers to the overall polarity of the
solvent (polar and nonpolar) and the
solute (polar, nonpolar, and ionic)
Must be an attraction between the
solute and solvent for a solution to
form
Solutions
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Nonpolar molecules
(no dipole, cannot
hydrogen bond)
Examples: oil, iodine
• Both do not
dissolve well in
water because it
(water) is polar
• Both dissolve well
in nonpolar solvents
such as carbon
tetrachloride
Ni(NO3)2
in H2O
H2O
I2 in
CCl4
CCl4
Solutions of Solids Dissolved in Water
(Water as a Solvent)
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Most common solutions (with a solid,
liquid, or gas) contain water as the
solvent
Water is a polar molecule due to its bent
shape
It also has the ability to hydrogen bond
Dissolves many polar and ionic
substances
Due to intermolecular interactions
(dipole-dipole or H-bonding) upon
mixing
Solutions of Solids Dissolved in Water
(Water as a Solvent)
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Polar compounds (a permanent dipole, can Hbond) and ionic compounds dissolve into polar
solvents
A polar molecule (with ionic bonding) dissolves
into water if attractions for water overcome the
attractions between the ions
As each ion enters the solution, it is immediately
surrounded by water molecules: hydration
(solvation)
Solutions of Solids Dissolved in Water
(Water as a Solvent)
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When sodium chloride crystals are placed in water, they
begin to dissolve
The attractive forces between the ions and water are
stronger than forces between the ions in the crystal
Water molecules surround each ion, keeping them apart
Solutions of Solids Dissolved in Water
(Water as a Solvent)
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When sodium chloride
crystals are placed in
water, they begin to
dissolve
The attractive forces
between the ions and
water are stronger
than forces between
the ions in the crystal
Water molecules
surround each ion,
keeping them apart
Electrolyte Solutions:
Dissolved Ionic Solids
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Compounds that ionize in water are
called electrolytes
• Electrolytes are solutes that exist as
ions in solution
• Formed from an ionic compound that
dissociates in water forming an
electroyte solution with cations and
anions
• These solutions conduct electricity
Ions In Solution (in Water)
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When ionic compounds dissolve in water
the ions dissociate
• Separate into the ions floating in water
• Potassium chloride dissociates in water into
potassium cations and chloride anions
KCl(aq) = K+ (aq) + Cl- (aq)
K
Cl
K+
Cl-
Ions In Solution (in Water)
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Copper(II) sulfate dissociates in water
into copper(II) cations and sulfate
anions
CuSO4(aq) = Cu+2(aq) + SO42-(aq)
Cu SO4
Cu+2
SO42-
Ions In Solution (in Water)
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Potassium sulfate dissociates in water
into potassium cations and sulfate
anions
K2SO4(aq)
K
=
SO4 K
2 K+ (aq) + SO42-(aq)
K+
SO42K+
Nonelectrolyte Solutions
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Compounds that do not ionize in
water are called nonelectrolytes
• Solute is a molecular substance
• Substance dispersed throughout the
solvent as individual molecules
• Each molecule is separated (dissolved)
by molecules of the solvent forming a
nonelectrolyte solution
• These solutions do not conduct
electricity
Electrolytes and Nonelectrolytes
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Strong electrolyte:
• Dissociates completely into ions
• Conduct electricity
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Weak electrolyte:
• Mainly whole molecules
• Very few separate (into ions)
• Conduct electricity less than strong electrolytes
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Nonelectrolyte:
• No dissociation into ions
• Do not conduct electricity
Electrolytes and Nonelectrolytes
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Strong electrolytes
are completely
ionized when
dissolved in water
Sodium chloride
dissociates to form
Na+ and ClGood conductor of
electricity
Electrolytes and Nonelectrolytes
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Weak electrolytes
are only partially
ionized when
dissolved in water
Hydrofluoric acid
only partially
dissociates to form
H+ and FPoor conductor of
electricity
Electrolytes and Nonelectrolytes
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Nonelectrolyes are
not ionized when
dissolved in water
e.g. sugar and
ethanol do not
dissociate into ions
in water
Do not conduct
electricity
Solubility and Saturation
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Solubility is the maximum amount of
solute that will dissolve into a given
amount of solvent
It is affected by
• Type of solute (solid, liquid, or gas)
• Type of solvent (and the solute
interaction)
• Temperature
Solubility and Saturation
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Unsaturated: Less solute than the
maximum amount possible is dissolved
into the solution
Saturated: Contains the maximum
amount of solute that can be dissolved
saturated
Solubility and Saturation
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A supersaturated solution contains more of the
dissolved particles than could be dissolved by the
solvent under normal circumstances
These solutions result from altering a condition of
the saturated solution such as T, V, or P
Solutions of Solids in Water:
Effect of Temperature on Solubility
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For most solids
(solute), solubility
increases with an
increase in
temperature
More sugar will
dissolve in hot
water than in cold
water
Solutions of Gases in Water:
Effect of Temperature on Solubility
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For gases, solubility
decreases with an
increase in
temperature
Henry’s Law: The
amount of a gas
dissolved in a solution
is directly proportional
to the pressure of the
gas above the solution
Solution Concentration
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Solution concentration
• The amount of solute (mass or moles) dissolved into a
certain amount of a solution or solvent
concentration of a solution amount of solute 100%
amount of solution
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Qualitative
• Dilute, concentrated, saturated, unsaturated
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Quantitative
• Mass to mass, volume to volume, and molarity
Solution Concentration
•Mass Percent
Mass of the solute divided by the total mass of solution
multiplied by 100
mass percent (% m/m) 
mass of solute (g)
 100%
mass of solute (g)  mass of solvent (g)
The mass of the solute and solution must be in the same
units
•Volume Percent
The volume of solute divided by the total volume of
solution multiplied by 100
volume percent (% v / v)

volume of solute 100 %
volume of solution
The solute and solution volumes must be in the same units
Mass Percent
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Grams of solute per grams of
solution
• Remember that the mass of solution
is grams of solute + grams of solvent
mass of solute
mass percent (% m/m) 
 100
mass of solution
Calculating Mass Percent: Example 1
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A 135 g sample of seawater is
evaporated to dryness, leaving 4.73 g
of solid residue. What is the mass
percent of the solute in the original sea
water?
mass percent (% m/m) 
mass of solute
100
mass of solution
4.73g solid
 100%  3.50%
135g solution
Mass Percent in Calculations: Example 2
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What mass of water must be added to 425 g
of formaldehyde to prepare a 40.0% (by
mass) solution of formaldehyde?
425 g formaldehyde
40% 
100%
? g total
40 g formaldehyde
40% 
100 g total
Mass Percent Conc. Example 2
40 g formaldehyde
40% 
100 g total
425 g formaldehyde
100 g total
1062.5 g total
40 g formaldehyde
1062.5 g total- 425 g formaldehyde = 637.5 g water
Solution Concentration: Molarity
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Molarity is the concentration
expression most commonly used in
the laboratory
The amount of solute is expressed in
moles
To obtain the molarity, we need to
know the solution volume in liters
and the number of moles of solute
present
Solution Concentration: Molarity
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Moles of solute per liters of solution
More useful than mass percent
• More common to measure liquids by volume, not mass
• Amount of solute expressed in moles (quantity of
particles)
• Chemical reactions occur between molecules and atoms
• Since it expressed in moles, you can do chemical
calculation (stoichiometry) problems
moles solute
Molarity(M) 
liters solution
Making Solutions of a Specific Molarity
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Make up in a volumetric flask
• Flask with a long, narrow neck that is marked
with a line indicating an exact volume
Method
• Add measured amount of solid (mass in grams)
• Add some water to dissolve the solid
• Fill with water up to the line (volume in mL or L)
Using Molarity in Calculations: Example 1
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Calculate the molarity of a solution made
by dissolving 15.0 g of NaOH (sodium
hydroxide) in enough water to give a final
volume of 100. mL.
Molar mass of NaOH  40.00g/mol
Convert volume to liters
1L
100. mL 
 0.100L
1000 mL
Using Molarity in Calculations: Example 1
Convert mass to moles
15.0g NaOH 1 mol NaOH
 0.375 mol NaOH
40.00g NaOH
moles solute
Molarity(M) 
liters solution
0.375 mol
m ol
 3.75
or 3.75 M
0.100L
liter
Using Molarity in Calculations: Example 2
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Formalin is an aqueous solution of
formaldehyde (H2CO). How many grams of
formaldehyde must be used to prepare 2.50 L
of 12.3 M formalin?
molarity × volume = moles
2.5 L 12.3 mol formaldehyde
 30.8 mol formaldehyde
1L
30.8 mol H 2CO 30.03 g H 2CO
 925 g H2CO
1 molH 2CO
Standard Solutions
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A solution whose concentration is exactly known
• A std. solution can be diluted to make up less
concentrated solutions
• A std. solution is like concentrated orange juice.
For example, one can of orange juice
concentrate is diluted with three cans of water.
Solution Dilution
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Dilution is the process in which more
solvent is added to a solution in order to
lower its concentration
A common laboratory routine is diluting a
solution of known concentration (stock
solution) to a lower concentration
A dilution always lowers the concentration
because the same amount of solute is
present in a larger amount of solvent
Molessolutestock solution  Molessolutediluted solution
Dilution
• Most often a solution of a specific molarity must be
prepared by adding a predetermined volume of solvent
to a specific volume of stock solution
• When solvent is added to dilute a solution, the number
of moles remains unchanged
• A relationship exists between the volumes and
molarities of the diluted and stock solutions
mol solute
M
L solution
Moles of solute
L solution  M  mol solute
=
(diluted solution)
(initial solution)
M1V1
Moles of solute
=
M2V2
Solution Dilution: Example 1
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Determine the volume required to prepare
0.75L of 0.10 M HCl from a 12 M HCl stock
solution.
• Initially we have 12 M HCl. Calculate
what volume of the stock solution will
contain the number of moles needed.
• How many moles of HCl do we
eventually want?
0.75 L 0.10 mol HCl
 0.075 mol HCl
L
Solution Dilution: Example 1
• How many liters of 12 M HCl contains
0.075 mol of HCl ?
0.075 mol HCl
1L
 6.2510 -3 L 12 M HCl
12 mol HCl
6.25 mL of 12 M HCl are
needed
Solution Dilution: Example 2
• What is the molarity of a solution prepared
when 25.0 mL of a 1.0 M CuSO4 is diluted to a
final volume of 250 mL.
Solution Dilution: Example 2
• What is the molarity of a solution prepared
when 25.0 mL of a 1.0 M CuSO4 is diluted to
a final volume of 250 mL.
Initial M1 = 1.0 M
Final M2 = ?
V1 = 0.025 L
Calculate the unknown molarity using the relationship
Moles of solute
=
(initial solution)
M1 V 1
Moles of solute
(diluted solution)
=
M2V2
V2 = 0.250 L
Solution Dilution: Example 2
• What is the molarity of a solution prepared
when 25.0 mL of a 1.0 M CuSO4 is diluted to
a final volume of 250 mL.
Set Up Problem by
solving for M2
1)
Moles of solute before dilution equals
moles of solute after dilution
M 1V1  M 2V2
M 1V1
 M2
V2
2) Calculate the unknown molarity by solving for M2
3) Set up problem
1.0
m ol
 0.025L
L
 M2
0.250 L
Solutions in Chemical Reactions:
Solution Stoichiometry
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Stoichiometry is the calculation of
quantitative relationships between
reactants and products
Calculations are based on balanced
chemical equations
The coefficients in the balanced equation
indicate the moles of products and
reactants
Solutions in Chemical Reactions:
Solution Stoichiometry
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Many reactions take place in solution and
the solution concentration (molarity)
directly relates the solution volume and
moles of solute present
Stoichiometric calculations are the same
as in chapter 8, but with the addition of
some molarity calculations
Quantitative Relationships Needed for
Solving Chemical Formula Based Problems
Grams B
Grams A
Molar mass
molarity
Liters A
Mole-mole
Factor
Moles A
M×V
molarity
Moles B
1 mol = 22.4 L
at STP
M×V
pV = nRT
PA , TA, VA
Liters B
P B , TB , V B
Solutions in Chemical Reactions:
Solution Stoichiometry
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When aqueous solutions of Na2SO4 and
Pb(NO3)2 are mixed, PbSO4 precipitates.
Solutions in Chemical Reactions:
Solution Stoichiometry
Na 2SO 4 (aq)  Pb(NO3 ) 2 (aq)  PbSO 4 (s)  2 NaNO3 (aq)
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Calculate the mass of PbSO4 that will be
formed when 1.25 L of 0.050 M Pb(NO3)2
reacts with 2.00 L of 0.025 M Na2SO4.
Given: 1.25 L of 0.050 M Pb(NO3)2 and 2.00 L of 0.025 M Na2SO4
Need: Mass (g) of PbSO4
Plan: Use volume and molarity to determine the moles of each reactant
Plan Na2SO4 : Msodium sulfate× Vsodium sulfate = mol Na2SO4
Plan Pb(NO3)2 : Mlead (II) nitrate× Vlead (II) nitrate = mol Pb(NO3)2
Solutions in Chemical Reactions:
Solution Stoichiometry
Na 2 SO4 (aq) 
Pb(NO3 )2 (aq) 
PbSO4 (s)  2 NaNO3 (aq)
Determine the moles of reactants
1.25 L 0.050 mol Pb(NO3 ) 2
 0.063molPb(NO3 ) 2
1L
2.00 L 0.025 mol Na 2 SO4
 0.050mol Na 2SO 4
1L
Solutions in Chemical Reactions:
Solution Stoichiometry
Na 2 SO4 (aq) 
Pb(NO3 )2 (aq) 
PbSO4 (s)  2 NaNO3 (aq)
Determine the limiting reactant
0.0625 mol Pb(NO3 ) 2 1 mol PbSO4
 0.0625 mol PbSO4
1 mol Pb(NO3 ) 2
0.050mol Na 2SO 4 1 mol PbSO 4
 0.050mol PbSO 4
1 mol Na 2SO 4
Limiting reactant
Solutions in Chemical Reactions:
Solution Stoichiometry
Calculate the mass of lead (II) sulfate that forms
0.05molPbSO 4 299.27g PbSO 4
 14.96 g PbSO4
1 molPbSO 4
Freezing Pt. Depression and
Boiling Point Elevation
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There are some physical properties of
solutions that are different from those of
the pure solvent
For example: Pure water freezes at T = 0 °C
Aqueous solutions freeze at lower temperatures
For example: The addition of antifreeze in water makes a solution with
a lower f.p. and higher b.p. than that of pure water
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The ability to lower a freezing point
or raise a boiling point of a solution
is called a colligative property.
Freezing Pt. Depression and
Boiling Point Elevation
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A colligative property of a solution is a
physical property that depends on the
quantity of solute particles present.
The ability to lower a freezing point or
raise a boiling point depends on the
quantity of solute particles present, not
the kind of particles.
For f.p. depression and b.p. elevation, the
concentration of the solution is expressed
in molality.
Freezing Pt. Depression and
Boiling Point Elevation
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The molality of a solution (m) is the
ratio that gives the number of moles
of solute per kilogram of solvent
Molality 
Moles solute
Kilogram s solvent
Freezing Pt. Depression and
Boiling Point Elevation
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The decrease in the freezing point relative to the
pure solvent is directly proportional to the
number of solute particles per mole of solvent
molecules
Kf is the freezing point
T f  m  K f
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depression constant for
the solvent
The increase in the boiling point relative to the
pure solvent is directly proportional to the
number of solute particles per mole of solvent
molecules
Kb is the boiling point
Tb  m  K b
elevation constant for the
solvent
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end