Transcript Chapter 15

Chapter 15: Solutions
Solution
• Solution – a homogeneous mixture of two
or more substances in a single physical
state.
• Properties of solutions –
• The particles are very small (atoms,
molecules or ions)
• the particles in a solution are evenly
distributed or uniformly mixed (a spoonful
of lemonade tastes the same as the whole
glass)
Parts of solutions
• Solute – the thing that gets dissolved
• Solvent – the thing that does the
dissolving
• Soluble – something that can be
dissolved (salt and sugar)
• Insoluble – a substance that cannot
be dissolved (Hg and oil)
Types of solutions
• Solid Solutions – alloys – made by melting
different metals and then cooling them
• Makes material stronger, higher melting
points and greater resistance to corrosion
• Ex. Dental filling – mercury in silver
• Sterling silver – copper in silver
• 16 karat gold – gold, copper and silver
• Coinage bronze – copper, tin and zinc
Types of solutions
• Gaseous solutions – air we
breathe. Nitrogen with oxygen in
it
• If gas molecules mix they
become a solution. Anytime gases
are near each other they will mix,
therefore any gas mixture is a
solution.
Types of solutions
• Liquid Solutions – MOST COMMON! the
solvent and the solution are both liquids.
The solute can be a gas, liquid or solid.
• Miscible – liquids that can be mixed in any
amount (water and ethanol)
• Immiscible – liquids that cannot mix in any
proportion (oil and water)
Types of solutions
• Aqueous solutions – solutions with water as the
•
•
•
solvent
2 types
1. When ionic compounds dissolve, ions are
present and make an Electrolyte. (NaCl)
2. When molecular compounds dissolve, no
ions are present and make a nonelectrolyte.
(Sugar)
15-2 Concentration of Solutions
• Concentration – the amount of solute in a
given amount of solvent.
• Molarity = moles of solute
•
Liters of solution
• What is the Molarity of a NaOH solution if
10.0g of NaOH is dissolved in enough
solvent to make 0.100L of solution?
You try it!!!
• Find the Molarity of a solution formed by
mixing 10.0g of H2SO4 with enough water
to make 100.0mL of solution
• M = 10.0g H2SO4 x 1 mol H2SO4 = 1.02 mol/L
•
0.100L sol
98.1 g H2SO4
• Molality = moles of solute
•
Kg of Solvent
• If 18.0g of C6H12O6 is dissolved in 1Kg of
water.
• 18.0g C6H12O6 x 1mol C6H12O6 = 0.100 mol C6H12O6 / 1Kg H2O
• 1Kg H2O
180g C6H12O6
• Mole Fraction = moles of component
•
Total moles of solution
• What is the mole fraction of SO2 in a gas
containing 128.0g of SO2 dissolved in
every 1500.g of CO2
• 1st change grams to moles
• 128g SO2 x 1 mole SO2 = 1.999 moles SO2
64.04 g SO2
• 1500 g CO2 x 1 mole CO2 = 34.09 moles
CO2
•
43.99 g CO2
• XSO2 = moles of SO2
•
Total moles of solution
• 1.999moles SO2
• (1.999 moles SO2 + 34.09 moles CO2)
• = 0.05539
Example.
• A gas mixture contains 50.4 g of N2O and
65.2g of O2 what is the mole fraction of
N2O? (0.360)
15-3 The formation of Solutions
• How they form
• Why does salt dissolve? The particles are
attracted to the water. The water attaches
itself on the face of NaCl and pulls it
apart. Then the water surrounds the ions.
• solvation - the interaction between solute
and solvent
• hydration – when the solvent is water
• energy is needed when bonds are broken
between the solute and solvent
• when the solvent and solute are attracted
energy is released.
• Cold packs = NH4OH
• Hot packs = Na2S2O3
• Solubility – the amount of solute that will
dissolve in a specific solvent under given
conditions.
• Depends on…
• 1. Nature of solute and solvent – polar
compounds dissolve polar compounds. IE
“likes dissolve likes” cholesterol is
nonpolar and fat is nonpolar
• 2. Temperature – As temperature
increases, gas particles get more energy
and solubility of a gas decreases (better to
keep soda cold!)
• Solubility of a solid – as temp increases,
solubility increases (sugar to iced tea or
hot tea)
• Pressure – solubility of a gas is increases
as the pressure increases because when
the pressure is increases, the gas particles
hit the solution faster.
• Scuba divers – nitrogen is dissolved in
your blood because of the extra pressure,
if they come up too fast the nitrogen will
bubble out of your blood. (bends)
Factors affecting the Rate of
Dissolving
• surface area – the greater the surface
area, the faster the solid can be dissolved.
(big ice cubes or little ones)
• Stirring – allows the solute to be exposed
to all of the solvent not just the top
• Temperature – warmer is faster
• Saturation – if a solution contains as
much solute as can possibly be
dissolved under the existing
conditions of temperature and
pressure
• Saturated and concentrated are not
the same things. A solution can be
saturated with only a little amount of
solute.
• Unsaturated – a solution that has
less than the maximum amount of
solute
• Supersaturated – contains a
greater amount of solute than
needed to form a saturated
solution. The extra solute will
eventually go back into solution.
(rock candy)
• 15-4 Colligative Properties – depend on
the collective effort f the solute properties
and not their identity.
• 1. Vapor pressure reduction – when a
nonvolatile solute is added to a solvent,
the solute takes up space at the surface
which prevents some of the solvent from
leaving. Gases are still returning to the
liquid at the same rate. This reduces the
vapor pressure of the solution.
• 2. Boiling point Elevation – antifreeze is
added to a car to make the water not boil.
Antifreeze is a nonvolatile substance so it
reduces the vapor pressure and increases
the boiling point because it takes longer to
reach atmospheric pressure
• Eq. ∆Tb = Kbm
• Where ∆Tb = boiling point elevation
• Kb = molal boilin point constant (varies
with solvent)
• m= molality
• Problem. Water with sugar added to it will
boil at a higher temperature than pure
water. By how much will the boiling point
of water be elevated if 100g of C12H22O11
is added to 500 g of water. Kb = 0.52 C/m
• Molality = mol solute
•
Kg solution
•
= 100. g sucrose x 1 mole sucrose = 0.584mol/kg
•
0.500kG H2O
342.3g sucrose
• ∆Tb = Kbm = 0.52C/m x 0.584 m = 0.30
C
• Freezing Point Depression – the temp at
which the vapor pressure of the solid
and liquid are the same. If the solute is
nonvolatile (rock salt)
• Eq. ∆Tf = Kfm
• Problem. Calculate the freezing point
depression of a solution of 100.g of
C12H6O2 in 0.500Kg of water. The Kf is
1.86C/m
• Molality = mol solute / Kg solution
• 100.g C2H6O2 x 1 mol
• 0.500Kg
62.0g
• =3.23m
• ∆Tf = Kfm = 1.86C/m x 3.23m = 6.01C