Chemistry 101
Prof. Hammon
CHAPTER 12 – SOLUTIONS
Solution – A mixture of 2 or more soluble pure substances,
physically combined (particles in mixture do not alter identity,
i.e. salt water).
Solute – the substance that is dissolved.
Solvent – the substance that does the dissolving or the substance
present in the larger amount.
SOLUBILITY
Solubility – the maximum amount of a solute that will dissolve
in a solvent (at constant temp).
i.e. solubility of salt is 36.2 g/100 g of H2O at 25oC
Saturated solution – solvent contains all the solute it can.
Unsaturated solution – solvent contains less solute than it can.
Supersaturated solution – solvent contains more solute than
can normally hold, unstable.
Miscible – when 2 liquids dissolve in each other in all
proportions. i.e. ethanol and water
(solvent – solute interaction stronger than solvent-solvent and
solute – solute interaction)
Chemistry 101
Prof. Hammon
Effect of Temperature on Solubility
For solids & liquids: solubility in liquids increases with
increasing temperature.
i.e. glycine solubility is 33.2 g/100 g of H2O at 30oC, and 52.8
g/100 ml at 80oC
60
Solubility
(g/100g H2O) 40
20
0
100
o
Temp. ( C)
For gases: solubility in liquids decreases with increasing
temperature.
i.e. thermal pollution: heating water decreases solubility of
oxygen in water, so concentration of O2 in water may not be
enough for fish to survive.
0.045
Solubility
(g/1 liter H2O)
0.025
0
20
35
o
Temp. ( C)
Chemistry 101
Prof. Hammon
The Effect of Pressure on Solubility
For solids & liquids: little effect.
For gases: solubility in liquids increases, with increasing
pressure (Henry’s Law).
i.e. Scuba diving: as pressure on body from water increases as
descend, the solubility of supplied air gases (oxygen, nitrogen)
in the blood increases.
When diver ascends to the surface, pressure on body decreases
and solubility of gases in blood decreases (gas bubbles form).
0.075
Solubility
(g/1 liter H2O)
0.050
0.025
1
1.5
Pressure of Gas (atm)
Chemistry 101
Prof. Hammon
Concentration Units
Concentration of solution – the amount of solute dissolved in a
given quantity of solvent.
Percent by Mass
Mass % = mass solute / mass solution x 100
Molarity (M)
M = moles of solute/ liters of sol’n
Molality (m)
Molality (m) = moles solute / kg solvent
Example: Calculate the molality of a solution prepared by
adding 0.20 mLs of H2O to 18.5 grams of t-butanol.
Mole Fraction (xsolute)
Mole Fraction = moles of solute / moles solution
Chemistry 101
Prof. Hammon
Example: A solution is prepared by dissolving 73.6 grams of
H2SO4 in 972 grams of water, having a total volume of 1.00 liter
of solution. Calculate % by mass, molarity, molality and mole
fraction.
Chemistry 101
Prof. Hammon
Colligative Properties: properties that depend on the number of
solute particles, not the type of solute particles.
Examples: FP Depression, BP elevation, VP Lowering, Osmotic
Pressure (pressure required to stop osmotic flow: flow of solvent
from lower concentration to a higher solute concentration)
Freezing Point (FP) Depression: Temp difference between FP
of solvent & FP of solution ( TFP = FPsolvent - FPsolution)
FPsol’n < FPpure solvent (due to solute disrupting crystalline struct.)
TFP = KFP * msolute
TFP = FP temp pure solvent – FP temp sol’n
KFP = FP constant for solvent
msolute = molality sol’n (moles solute/kg solvent)
Example: What is the freezing point of a 0.0290 molal ethanol in
water solution? (KFP water = 1.86 oC/m)
Chemistry 101
Prof. Hammon
Example
Calculate the molecular weight of an unknown. 1.00 gram of
the unknown was added to 50.0 g of benzene. The freezing
point of benzene is 5.50oC and the freezing point of the solution
is 3.69oC. The KFP for benzene is 4.90oC/m.
Chemistry 101
Prof. Hammon
Boiling Point (BP) Elevation: Temp difference between BP of
solution & BP of solvent
( TFBP = BPsolution - BPsolvent)
BPsol’n > BPpure solvent
(due to attraction of solvent to solute, increasing boiling point)
TBP = KBP * msolution
TBP = BP temp solution – BP temp solvent
KBP = BP constant for solvent
msolution = molality sol’n (moles solute/kg solvent)
Example What is the BP of a 0.0290 molal aqueous ethanol
solution? (KBP water = 0.512 oC/m)
Chemistry 101
Prof. Hammon
Example
Calculate the molality of an aqueous glucose solution if the
boiling point of the solution is 101.27oC at 1 atm. KBP for H2O
is 0.512oC*kg/mol.
Chemistry 101
Prof. Hammon
Vapor Pressure (VP) Lowering
VPsol’n < VPpure solvent
VPsolution = Xsolvent VPosolvent (Raoult’s Law)
VPsolution = vapor pressure sol’n
Xsolvent = mole fraction for solvent
VPosolvent = VP of pure solvent at same temp
Example
Calculate the molar mass of urea. Given 20.0 g urea in 125 g
water at 25.0oC. VPH2O = 23.76 mm Hg and observed VP of
solution = 22.67 mm Hg.
Chemistry 101
Prof. Hammon
For Ionic Solutes: # of solute particle (colligative property)
affects BP elevation, FP depression, VP lowering, etc
Example of ionic solutes:
# of particles
NaCl(aq)
Na+(aq) +
Cl-(aq)
H2SO4(aq)
2 H+(aq) +
SO42-(aq)
van’t Hoff factor (i) – measures particle disassociation in
solvent
i = moles of particles in sol’n / moles of formula units dissolved
What is i for NaCl? ______
Ionic solutes FP Depression:
TFP = KFP * msolute * i
What is the FP for a 0.150 molal aq sodium chloride solution?
(KFP water = 1.86 oC/m)
Chemistry 101
Prof. Hammon
van’t Hoff factor (i) (cont)
NaCl(aq)
Na+(aq) +
Cl-(aq)
MgSO4(aq)
Mg2+(aq) +
SO42-(aq)
Solute(0.05 m soln)
NaCl
MgSO4
i expected
2
2
i observed
1.9
1.3
Ion Pairing – ions may get close enough to stick together,
lowering the concentration of expected particles.
*Greater charges more ion pairing.
Na+(aq) Cl-(aq)
Mg2+(aq) SO42-(aq)
*Less concentrated solutions less ion pairing, so i closer to
expected.