Transcript CH 10 Solutions - Douglas County

Principles of Solubility
Nature of the solute and solvent

Like likes like.
Electrolytes will dissolve in a polar
solvent (like water)
Non-electrolytes will dissolve more
easily in non polar or slightly polar
solvents (like benzene, or toluene)
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Non electrolytes (molecular substances)
dissolving in water.

This ‘unusual’ case
will involve
Hydrogen bonding
or otherwise polar
molecules.

Ex:

alcohols (-OH
group)
Methanol, Ethanol

Hydrogen
peroxide H2O2
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Solubilities of Alcohols in
Water
Formula Name Solubility in
CH3OH
CH3CH2OH
CH3(CH2)2OH
CH3(CH2)3OH
CH3(CH2)4OH
CH3(CH2)5OH
CH3(CH2)6OH
CH3(CH2)7OH
CH3(CH2)9OH
Water (g/100 g)
methanol infinitely soluble
ethanol infinitely soluble
propanol infinitely soluble
butanol 9
pentanol 2.7
hexanol 0.6
heptanol 0.18
octanol 0.054
decanol insoluble in water
CH 10 Solutions

“greasy end” is
larger proportion
in large alcohols.
DDT no more mosquitoes!
Pesticide used widely in
the 60’s
 Soluble in non polar
solvents.
 Concentrates in tissues
of fish, birds, and other
wildlife. (biological ½ life
of 8 years)
 Thins eggshells.

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Principles of Solubility
Effect of Pressure/Henry’s Law
For gases only
 Solubility is directly proportional
to pressure.
Soda is bottled under a pressure of
4atm.
This drives more of the CO2 into
solution,
when the cap is released, the
pressure drops to 1 atm, and CO2
bubbles and rapidly escapes
solution.
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Cg
=k
Pg
Concentration in solution = constant x Pressure of gas
Principles of Solubility
Effect of Temperature


Solid + Water  Solution
 Usually + ΔH heat in
Increased T favors an
endothermic process



Gas + Water  Soln.
 Usually –ΔH heat out
Increased T is
detrimental for an
exothermic process
• So: Solubility of a gas
usually decreases with
higher temperature.
So: Solubility of a solid
usually increases with
increased temperature
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Concentration Units
problem type 1. Molarity (M)

Molarity = moles of solute/
liters of solution
Units are moles/liter
Tricks:
change mass to moles
change mL to Liters
CH 10 Solutions
Typical Problems
 Making a volume of
solution from the
solid. Creation
 Preparing a volume
of a solution from
another (stock)
solution. Dilution
Problems
Molar mass (gfm) of aluminum
nitrate is 213.03 g/n
Molarity (M)

1.
2.
Prepare 35.0 mL of 0.200 M
Aluminum nitrate from a
solid sample.
How many moles of Alnitrate would be in the
solution?
Measure out that many
grams and then add water
to make 35.0 mL

Prepare 35.0 mL of
0.20 M Aluminum
Nitrate solution from a
0.50M stock solution.

How many moles do
you need?



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Remember
M1V1=M2V2
McVc=MdVd
Meausure out the
volume of stock you
need, then dilute
Problem type 2.
Mole Fraction (X)

Similar to percent
composition…you
are interested in part
of the whole


Moles of substance/
total moles
Typical problems will have you
 determine the mole fraction
from % mass.

determine the mole fraction
of solute in solution
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Problems
Methane is CH4
Methanol is CH3OH
Mole Fraction (X)


1.
2.
3.
If you dissolve 12.0 g of
methanol in 100.0 g of water.
What is the mole fraction that
is methanol?
Determine the moles of water
Determine the moles of
methanol
Add together for the total
moles, then take part/whole
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Problem type 3.
Mass % Solute

Simply determine the
percent composition by
mass.
Trick is to change the
decimal to a percent
(x100 + “%”)

CH 10 Solutions
Solute is the solid
dissolved in
solution
.
Molality (m)

Problem Type 4


This is similar to molarity,
but uses mass (kilograms)
of the solvent, rather than
liters of solvent.
CH 10 Solutions
Molarity (M) = #
moles solute/
Liter Solvent
Typical problems will
have you determine the
molality from other
concentration units.
Problems : Molality (m)

What is the molality of methanol if
12g of methanol are dissolved
into 100g of water?



Find Moles of solute (methanol)
2. Find kilograms of solvent
3. Divide
if the solvent is water, M and m are
numerically equal
1.
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12g/32.0 g/n
= 0.375 n
100g/
1000g/kg

= 0.100kg
= 3.75m
Conversion between unitsdecide on a fixed amount of solution
When the original
concentration [ ] is:
Start with:
Mass Percent
100 g solution
Molarity (M)
1.00L of solution
Molality (m)
1000g of solvent (1Kg)
Mole Fraction (X)
1 mol (solute+solvent)
CH 10 Solutions
Part III: Colligative Properties
Vapor Pressure Lowering
VPL = X2Pº1



Each liquid has a vapor
pressure.
VPL= (vapor pressure lowering)
X2= mole fraction of the solute
Pº1= vapor pressure of the
PURE solvent
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The pressure to escape the
surface into the gas phase.
Dissolved particles get in the
way and block the
escape, thereby reducing
the vapor pressure.
The higher the concentration
(X2), the more the VP is
lowered.
Raoult’s Law
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Part III: Colligative Properties
*Boiling Point Elevation
*Freezing Point Depression
ΔTb = kbm
ΔTf = kfm
ΔTb
ΔTf
Boiling Elevation:
The particles lower the vapor
pressure.
The change in temperature
Kb for water is 0.52 °C/m
from the normal
Freezing Depression:
boiling/freezing points is a
The particles obstruct the formation
constant (k) times the
of the solid crystal. There must be
molality(m).
a lower temperature to form solid
The higher the concentration of
structure
‘contaminant’, the greater the
Kf for water is 1.86 °C/m
change in freezing/boiling
points
CH 10 Solutions
Problems: Colligative Properties
freezing point depression
ΔTf = kfm
 Change in freezing
temp is (Kf) x (molality)




Find the freezing point of
a solution containing
20.0 g of ethanol in
50.0g of water.
CH 10 Solutions


What is the molality
of the solution?
Grams to moles
Moles / Kg of
Solvent
Molality x Kf
Kf for water is
1.86 °C/m
Problems: Colligative Properties
boiling point elevation
ΔTf = kbm
 Change in boiling
 temp is (Kb) x (molality)




Find the boiling point of
a solution containing
20.0 g of ethylene glycol
in 50.0g of water.
CH 10 Solutions


What is the molality
of the solution?
Grams to moles
Moles / Kg of
Solvent
Molality x Kb
Kb for water is
0.52 °C/m
Colligative Properties
Osmotic Pressure (π)

π =MRT
M = molarity
R= .0821 L * atm/n *K
T = Temperature
CH 10 Solutions
Water moves
through a semipermeable
membrane from an
area of high vapor
pressure to low
vapor pressure.
Problems:
Using freezing point depression to find
molar mass



A solution is made using
.0100g of a substance in
1.00g of water.
The freezing point
depresses 1.00 °C
What is the molar mass?
CH 10 Solutions

Solve for molarity