1 Solutions Why does a raw egg swell or shrink when placed in different solutions?
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Transcript of 1 © 2006 Brooks/Cole - Thomson Solutions Why does a raw egg swell or shrink when placed in...
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© 2006 Brooks/Cole - Thomson
Solutions
Why does a raw egg swell or shrink when placed in different solutions?
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© 2006 Brooks/Cole - Thomson
Some Definitions
A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase.
One constituent is usually regarded as the SOLVENT and the others as SOLUTES.
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Solutions can be classified as saturated or unsaturated.
A saturated solution contains the maximum quantity of solute that dissolves at that temperature.
Definitions
SUPERSATURATED SOLUTIONS contain more than is possible and are unstable.
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Dissolving An Ionic Solid
Active Figure 14.9
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Energetics of the Solution
Process
Figure 14.8
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Energetics of the Solution
Process
If the enthalpy of formation of the solution is more negative that that of the solvent and solute, the enthalpy of solution is negative.
The solution process is exothermic!
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SupersaturatedSodium
Acetate
• One application of a supersaturated solution is the sodium acetate “heat pack.”
• Sodium acetate has an ENDOthermic heat of solution.
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Supersaturated Sodium Acetate
Sodium acetate has an ENDOthermic heat of solution.
NaCH3CO2 (s) + heat ----> Na+(aq) + CH3CO2
-(aq)
Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC.
Na+(aq) + CH3CO2-(aq) ---> NaCH3CO2 (s) + heat
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Colligative PropertiesOn adding a solute to a solvent, the properties
of the solvent are modified.
• Vapor pressure decreases
• Melting point decreases
• Boiling point increases
• Osmosis is possible (osmotic pressure)
These changes are called COLLIGATIVE PROPERTIES.
They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.
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An IDEAL SOLUTION is one where the properties depend only on the concentration of solute.
Need concentration units to tell us the number of solute particles per solvent particle.
The unit “molarity” does not do this!
Concentration Units
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Concentration UnitsMOLE FRACTION, X
For a mixture of A, B, and C
XA mol fraction A = mol A
mol A + mol B + mol C
m of solute = mol solute
kilograms solvent
WEIGHT % = grams solute per 100 g solution(also ppm, ppb, which is like %, which is parts per hundred)
MOLALITY, m
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Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate mol fraction, molality, and weight % of ethylene glycol.
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Calculating Concentrations
250. g H2O = 13.9 mol
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol.
Xglycol = 1.00 mol glycol
1.00 mol glycol + 13.9 mol H 2O
X glycol = 0.0672
Calculate mole fraction
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Calculating Concentrations
Calculate molality
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g
of H2O. Calculate X, m, and % of glycol.
conc (molality) = 1.00 mol glycol0.250 kg H 2O
4.00 molal
%glycol = 62.1 g
62.1 g + 250. g x 100% = 19.9%
Calculate weight %
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Dissolving Gases & Henry’s Law
Sg Gas solubility (mol/L) = kH • Pgas
kH for O2 = 1.66 x 10-6 M/mmHg
When Pgas drops, solubility drops.
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Understanding Colligative Properties
To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.
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Psolvent = Xsolvent • Posolvent
Understanding Colligative Properties
Vapor pressure of H2O over a solution
depends on the number of H2O molecules
per solute molecule.
Psolvent proportional to Xsolvent
VP of solvent over solution = (Mol frac solvent)•(VP pure solvent)
RAOULT’S LAW
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PA = XA • PoA
Raoult’s LawAn ideal solution is one that obeys Raoult’s
law.
Because mole fraction of solvent, XA, is always
less than 1, then PA is always less than PoA.
The vapor pressure of solvent over a solution is
always LOWERED!
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Vapor Pressur
e Lowerin
g
Figure 14.14
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Raoult’s LawAssume the solution containing 62.1 g of
glycol in 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg)
Solution
Xglycol = 0.0672 and so Xwater = ?
Because Xglycol + Xwater = 1
Xwater = 1.000 - 0.0672 = 0.9328
Pwater = Xwater • Powater = (0.9382)(31.8 mm Hg)
Pwater = 29.7 mm Hg
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Elevation of Boiling Point
Elevation in BP = ∆TBP =
KBP•m
(where KBP is characteristic of solvent)
V P so lv enta fter a ddingso lute
V P Pure so lv ent
BP pureso lv ent
BP so lutio n
1 a tm
P
T
The boiling point of a solution is higher than that of the pure solvent.
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Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g
of water. What is the BP of the solution?
KBP = +0.512 oC/molal for water (see Table 14.3).
Solution
1. Calculate solution molality = 4.00 m
2. ∆TBP = KBP • m
∆TBP = +0.512 oC/molal (4.00 molal)
∆TBP = +2.05 oC
BP = 102.05 oC
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Change in Freezing Point
The freezing point of a solution is LOWER than that of the pure
solvent.
FP depression = ∆TFP = KFP•m
Pure water Ethylene glycol/water solution
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Lowering the Freezing Point
Water with and without antifreeze When a solution freezes, the solid phase is pure water. The solution becomes more concentrated.
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How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?.
Solution
Calc. required molality (*This one is a little
different because the solute is ionic, not
molecular.)
∆TFP = KFP • m
-10.00 oC = (-1.86 oC/molal) • Conc
Conc = 5.38 molal
Freezing Point Depression
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How much NaCl must be dissolved in 4.00 kg of water to lower
FP to -10.00 oC?.
Solution
Conc req’d = 5.38 molal
This means we need 5.38 mol of dissolved particles per kg of solvent.
*Recognize that m represents the total concentration of all dissolved particles.
Recall that 1 mol NaCl(aq)--> 1 mol Na+(aq) + 1 mol Cl-(aq)
Freezing Point Depression
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How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 oC?.
SolutionConc req’d = 5.38 molalWe need 5.38 mol of dissolved particles per
kg of solvent. Each mol of NaCl contributes 2 mol of dissolved particles.
NaCl(aq) --> Na+(aq) + Cl-(aq)To get 5.38 mol/kg of particles we need
5.38 mol / 2 = 2.69 mol NaCl / kg2.69 mol NaCl / kg ---> 157 g NaCl / kg(157 g NaCl / kg)•(4.00 kg) = 629 g NaCl
Freezing Point Depression
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Boiling Point Elevation and Freezing Point
Depression ∆T = K•m•iA generally useful equation
i = van’t Hoff factor = number of particles produced per formula unit.
Compound Theoretical Value of i
glycol 1
NaCl 2
CaCl2 3
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Osmosis
Dissolving the shell in vinegar
Egg in corn syrupEgg in pure water
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Osmosis
The semipermeable membrane allows only the movement of solvent molecules.
Solvent Solution
Semipermeable membrane
Solvent molecules move from pure solvent to solution in an attempt to make both have the
same concentration of solute.
Driving force is entropy
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Process of Osmosis
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Osmosis at the Particulate Level
Figure 14.17
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Osmosis at the Particulate Level
Figure 14.17
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Osmosis
• Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration.
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Osmosis and Living Cells
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Reverse OsmosisWater Desalination
Water desalination plant in Tampa
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Osmosis Calculating a Molar Mass
Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin.
Solution
(a) Calc. ∏ in atmospheres
∏ = 10.0 mmHg • (1 atm / 760 mmHg)
= 0.0132 atm
(b) Calculate molarity
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Osmosis Calculating a Molar Mass
Conc = 5.39 x 10-4 mol/L
(c) Calc. molar mass
Molar mass = 35.0 g / 5.39 x 10-4 mol/L
Molar mass = 65,100 g/mol
Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass of hemoglobin.
Solution
(b)Calc. molarity from ∏ = iMRT
Conc = 0.0132 atm
(0.0821 L • atm/K • mol)(298K)