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Chapter 11
• Suggested problems: 5, 9, 13, 15, 31, 33,
37, 39, 41, 43, 45, 53
Chapter 11
• Solutions
• Solution – solvent & solute.
• Solution – homogenous mixture – single
phase.
Chapter 11
• Types of solutions
• All nine combinations are possible:
• Solute Solvent
• Gas Gas
• Gas Liquid
• Gas Solid
• Liquid Gas
• Liquid Liquid
• Liquid Solid
• Solid Gas
• Solid Liquid
• Solid Solid
Chapter 11
• Effect of temperature on solubility
• If dissolving solute is exothermic then increase
in temperature makes solute less soluble
• If dissolving process is endothermic increasing
temperature will make solute more soluble.
• Dissolving ionic solid is usually endothermic
(takes energy to break up lattice) and so most
ionic solids are more soluble with increasing
temperature.
Chapter 11
• Dissolving a solid in a liquid is a three-step
process.
• 1) Breaking the bonds between the solid
particles - that takes energy - endothermic
• 2) Breaking the bonds between the liquid
molecules to make room for the solid particles -
that takes energy - endothermic
• 3) Making bonds between the solute and solvent
particles - that releases energy - exothermic
Chapter 11
• When a solute is soluble, steps 1 and 2
require less energy than step three
produces.
• When a solute is insoluble, the first two
steps require too much energy.
• Usually, it is the first step - breaking the
forces between solute particles - requires
too much energy.
Chapter 11
• NaCl is soluble in water because step 3,
where the Na+ and Cl- ions bind to the
water molecules, releases sufficent
energy.
Chapter 10
• ion-dipole.
Na+
H
HO δ+δ-
Chapter 11
• Hydration of NaCl(aq)
• NaCl(aq) ⇒ Na+(aq) + Cl-
Na ClNa
+
H H
O
Cl-
Chapter 11
• Oil won’t dissolve in water because,
though the oil molecules are easily
separated (step 1), there is no real
attraction for the oil molecules for the
water molecules (step 3).
Chapter 11
• Gases less soluble in liquids with
increasing temperature.
• Increasing temp increases molecular
vibrations
• Decreases intermolecular forces.
Chapter 11
• Gas-Liquid solutions
• Henry's law
• As the (partial) pressure of a gas is
increased above the liquid, the solubility of
the gas goes up.
• Carbonated beverages
Chapter 11
• Solutions
• Unsaturated: less solute dissolved than
possible
• Saturated: Maximum amount of solute
dissolved
• Supersaturated: More solute dissolved
than is “supposed” to dissolve.
Chapter 11
Molarity M = mols of solute
volume solution
% by weight (w/w) mass solute
mass solution
% by volume (v/v) volume solute
vol. solute + vol. solvent
Chapter 11
• molality m = mols solute
Kg of solvent
• mole fraction For a solution of A and B
χA = mols A
mols A + mols B
• χB = mols B
mols A + mols B
Chaper 11
• Calculate the wt of HCl in 5.00 mL of conc
HCl solution, of density = 1.19 g/mL and
37.23% HCl (w/w)
Chapter 11
• What is the M of H2SO4 solution, which
has a density of 1.2 g/mL and is 27%
(w/w)?
Chapter 11
• What is M of 16 g of CH3OH in 200 mL soln?
• M = mols solute/L solution
Chapter 11
• Find the M, m, and χacid and χwater of a solution of H2SO4 of density = 1.2 g/mL and 27.0% (w/w)
Chapter 11
• b) 1200 g solution minus 324 g H2SO4 =
876 g water
Chapter 11
• c) Mol Fraction:
Chapter 11
• Find the M, m, and χsolute and χwater of a solution of C6H12O6 of density = 2.45 g/mL and 34.5% (w/w)
Chapter 11
• b) 2450 g solution minus 845.3 g C6H12O6
= 1604.7 g water
•
Chapter 11
• c) Mol Fraction:
Chapter 11
• Those properties that depend on the
number, but not the nature, of the solute
particles
• Raoult's Law. P. 625 Accounts for vapor
pressure lowering.
Psoln = (χ solvent)(Po
pure solvent)
Chapter 11
Pure solvent Solution
Chapter 11
• The vapor pressure of water at 28oC is
28.35 Torr. Find the vapor pressure of a
solution of 68 g of C12H22O11 in 1000 g of
water at 28o C.
Chapter 11
• B.P. lowering and F.P. elevation760
ΔTf0 100 ΔTb
Chapter 11
• ΔTb = kb m and ΔTf = kf m
• Calculate the F.P. and B.P. of 2.60 g of
urea, CO(NH2)2 in 50 g of water.
Chapter 11
• Find the F.P. of 900 g ethylene glycol in 6
L water? Given: kf(H2O) = 1.86oC/m
• Ethylene glycol is (HO)CH2-CH2(OH)
Chapter 11
• 4.50 g of solute is dissolved in 125 g of
water. The solution freezes at -0.372oC.
What is the MW of the solute?
Chapter 11
• What is the MW of a solute if a solution of
0.510 g solute dissolved in 25 g of
benzene has a F.P. = 2.32oC?
Chapter 11
Chapter 11
Semi-permeablemembrane
Water moving from higher concentration(of water) to lowerconcentration of water.
Chapter 11
• ΠV = nRT or Π = MRT
• Because M = mols/V
• Π is osmotic pressure
Ocean pressure > Πbigcity
Chapter 11
• Electrolyte Solutions
• Colligative properties depend upon the
number of particles and not their nature
• So, ΔT = km is actually:
• ΔT = ikm where i is:
• actual number of particles after
dissolution
number of formula units initally dissolved
• Book calls i the van’t Hoff factor (p. 513)
Chapter 11
• Electrolyte Solutions
• Calculate the freezing point of 62 g of
H2SO4
in 500 g of water
• Remember, H2SO4 --> 2 H+ + SO42-
•
Chapter 11
• Electrolyte Solutions
• i actually usually less than the ideal value
• Due to formation of ion pairs
• That is, when ionic material dissolves
• Some cations and anions don’t separate
• Especially true when cation and anion
have high charge to size ratios
• Al2(SO4)3 --> Al3+ + SO42- --> Al3+---SO4
2-
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