Solutions
Solutions
• A solution is a homogeneous mixture.
• A solution is composed of a solute dissolved in a
solvent.
• Solutions exist in all three physical states:
Gases in Solution
• Temperature effects the solubility of gases.
• The higher the temperature, the lower the
solubility of a gas in solution.
• An example is carbon dioxide in soda:
– Less CO2 escapes when you open a cold soda than
when you open the soda warm
Polar Molecules
• When two liquids make a solution, the solute is
the lesser quantity, and the solvent is the greater
quantity.
• Recall, that a net dipole is present in a polar
molecule.
• Water is a polar molecule.
Polar & Nonpolar Solvents
• A liquid composed of polar molecules is a polar
solvent. Water and ethanol are polar solvents.
• A liquid composed of nonpolar molecules is a
nonpolar solvent. Hexane is a nonpolar solvent.
Like Dissolves Like
• Polar solvents dissolve in one another.
• Nonpolar solvents dissolve in one another.
• This it the like dissolves like rule.
• Methanol dissolves in water but hexane does not
dissolve in water.
• Hexane dissolves in toluene, but water does not
dissolve in toluene.
Miscible & Immiscible
• Two liquids that completely
dissolve in each other are
miscible liquids.
• Two liquids that are not miscible
in each other are immiscible
liquids.
• Polar water and nonpolar oil are
immiscible liquids and do not
mix to form a solution.
Solids in Solution
• When a solid substance dissolves in a liquid, the
solute particles are attracted to the solvent
particles.
• When a solution forms, the solute particles are
more strongly attracted to the solvent particles
than other solute particles.
• We can also predict whether a solid will dissolve
in a liquid by applying the like dissolves like rule.
Like Dissolves Like for Solids
• Ionic compounds, like sodium chloride, are
soluble in polar solvents and insoluble in nonpolar
solvents.
• Polar compounds, like table sugar (C12H22O11),
are soluble in polar solvents and insoluble in
nonpolar solvents.
• Nonpolar compounds, like naphthalene (C10H8),
are soluble in nonpolar solvents and insoluble in
polar solvents.
The Dissolving Process
• When a soluble crystal is placed into a solvent, it
begins to dissolve.
• When a sugar crystal is placed
in water, the water molecules
attack the crystal and begin
pulling part of it away and into
solution.
• The sugar molecules are held
within a cluster of water
molecules called a solvent cage.
Dissolving of Ionic Compounds
• When a sodium chloride crystal is place in water,
the water molecules attack the edge of the crystal.
• In an ionic compound, the water
molecules pull individual ions
off of the crystal.
• The anions are surrounded by
the positively charged hydrogens
on water.
• The cations are surrounded by
the negatively charged oxygen
on water.
Rate of Dissolving
• There are three ways we can speed up the rate of
dissolving for a solid compound.
• Heating the solution:
– This increases the kinetic energy of the solvent and the
solute is attacked faster by the solvent molecules.
• Stirring the solution:
– This increases the interaction between solvent and
solute molecules.
• Grinding the solid solute:
– There is more surface area for the solvent to attack.
Solubility and Temperature
• The solubility of a compound is the maximum
amount of solute that can dissolve in 100 g of
water at a given temperature.
• In general, a
compound
becomes more
soluble as the
temperature
increases.
Saturated Solutions
• A solution containing exactly the maximum
amount of solute at a given temperature is a
saturated solution.
• A solution that contains less than the maximum
amount of solute is an unsaturated solution.
• Under certain conditions, it is possible to exceed
the maximum solubility of a compound. A
solution with greater than the maximum amount
of solute is a supersaturated solution.
Supersaturated Solutions
• At 55C, the solubility of NaC2H3O2 is 100 g per
100 g water.
• If a saturated solution at 55C is cooled to 20C,
the solution is supersaturated.
• Supersaturated
solutions are
unstable. The
excess solute can
readily be
precipitated.
Supersaturation
• A single crystal of sodium acetate added to a
supersaturated solution of sodium acetate in water
causes the excess solute to rapidly crystallize from
the solution.
Concentration of Solutions
• The concentration of a solution tells us how much
solute is dissolved in a given quantity of solution.
• We often hear imprecise terms such as a “dilute
solution” or a “concentrated solution”.
• There are two precise ways to express the
concentration of a solution:
– mass/mass percent
– molarity
Mass Percent Concentration
• Mass percent concentration compares the mass of
solute to the mass of solvent.
• The mass/mass percent (m/m %) concentration is
the mass of solute dissolved in 100 g of solution.
mass of solute
× 100% = m/m %
mass of solution
g solute
× 100% = m/m %
g solute + g solvent
Calculating Mass/Mass Percent
• A student prepares a solution from 5.00 g NaCl
dissolved in 97.0 g of water. What is the
concentration in m/m %?
5.50 g NaCl
× 100% = m/m %
5.00 g NaCl + 97.0 g H2O
5.00 g NaCl
× 100% = 4.90 %
102 g solution
Mass Percent Unit Factors
• We can write several unit factors based on the
concentration 4.90 m/m% NaCl:
4.90 g NaCl
100 g solution
100 g solution
4.90 g NaCl
4.90 g NaCl
95.1 g water
95.1 g water
4.90 g NaCl
95.1 g water
100 g solution
100 g solution
95.1 g water
Mass Percent Calculation
• What mass of a 5.00 m/m% solution of sucrose
contains 25.0 grams of sucrose?
• We want grams solution, we have grams sucrose.
100 g solution
25.0 g sucrose ×
= 500 g solution
5.00 g sucrose
Molar Concentration
• The molar concentration, or molarity (M), is the
number of moles of solute per liter of solution, is
expressed as moles/liter.
moles of solute
=M
liters of solution
• Molarity is the most commonly used unit of
concentration.
Calculating Molarity
• What is the molarity of a solution containing
18.0 g of NaOH in 0.100 L of solution?
• We also need to convert grams NaOH to moles
NaOH (MM = 40.00 g/mol).
18.0 g NaOH
1 mol NaOH
×
= 4.50 M NaOH
0.100 L solution 40.00 g NaOH
Molarity Unit Factors
• We can write several unit factors based on the
concentration 4.50 M NaOH:
4.50 mol NaOH
1 L solution
1 L solution
4.50 mol NaOH
4.50 mol NaOH
1000 mL solution
1000 mL solution
4.50 mol NaOH
Molar Concentration Problem
• How many grams of K2Cr2O7 are in 250.0 mL of
0.100 M K2Cr2O7?
• We want mass K2Cr2O7, we have mL solution.
0.100 mol K2Cr2O7 294.2 g K2Cr2O7
×
250.0 mL solution ×
1 mol K2Cr2O7
1000 mL solution
= 7.36 g K2Cr2O7
Molar Concentration Problem
• What volume of 12.0 M HCl contains 7.30 g of
HCl solute (MM = 36.46 g/mol)?
• We want volume, we have grams HCl.
1 mol HCl
1000 mL solution
7.30 g HCl ×
×
36.46 g HCl
12.0 mol HCl
= 16.7 mL solution
Dilution of a Solution
• Rather than prepare a solution by dissolving a
solid in water, we can prepare a solution by
diluting a more concentrated solution.
• When performing a dilution, the amount of solute
does not change, only the amount of solvent.
• The equation we use is: M1 × V1 = M2 × V2
– M1 and V1 are the initial molarity and volume and
M2 and V2 are the new molarity and volume
Dilution Problem
• What volume of 6.0 M NaOH needs to be diluted
to prepare 5.00 L if 0.10 M NaOH?
• We want final volume and we have our final
volume and concentration.
M1 × V1 = M2 × V2
(6.0 M) × V1 = (0.10 M) × (5.00 L)
(0.10 M) × (5.00 L)
V1 =
= 0.083 L
6.0 M
Solution Stoichiometry
• In Chapter 10, we performed mole calculations
involving chemical equations, stoichiometry
problems.
• We can also apply stoichiometry calculations to
solutions.
solution
concentration
balanced
equation
molarity known  moles known 
moles unknown  mass unknown
molar mass
Solution Stoichiometry Problem
• What mass of silver bromide is produced from the
reaction of 37.5 mL of 0.100 M aluminum
bromide with excess silver nitrate solution?
AlBr3(aq) + 3 AgNO3(aq) → 3 AgBr(s) + Al(NO3)3(aq)
• We want g AgBr, we have volume of AlBr3
0.100 mol AlBr3 3 mol AgBr 187.77 g AgBr
37.5 mL soln ×
×
×
1 mol AgBr
1 mol AlBr3
1000 mL soln
= 2.11 g AgBr
Conclusions
• Gas solubility decreases as the temperature
increases.
• Gas solubility increases as the pressure increases.
• When determining whether a substance will be
soluble in a given solvent, apply the like dissolves
like rule.
– Polar molecules dissolve in polar solvents.
– Nonpolar molecules dissolved in nonpolar solvents.
Conclusions Continued
• Three factors can increase the rate of dissolving
for a solute:
– Heating the solution
– Stirring the solution
– Grinding the solid solute
• In general, the solubility of a solid solute
increases as the temperature increases.
• A saturated solution contains the maximum
amount of solute at a given temperature.
Conclusions Continued
• The mass/mass percent concentration is the mass
of solute per 100 grams of solution:
mass of solute
× 100% = m/m %
mass of solution
• The molarity of a solution is the moles of solute
per liter of solution.
moles of solute
=M
liters of solution
Conclusions Continued
• You can make a solution by diluting a more
concentrated solution:
M1 × V1 = M2 × V2
• We can apply stoichiometry to reactions involving
solutions using the molarity as a unit factor to
convert between moles and volume.