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Chem 300 - Ch 25/#2 Today’s To Do List
• Binary Solid-Liquid Phase Diagrams Continued (not in text…)
• Colligative Properties
Stable Compound Formation
K/Na with incongruent MP & Unstable Compound Formation
Colligative Properties
Depends upon only the number of (nonvolatile) solute particles
Independent of solute identity From colligatus: “depending upon the
collection”• Vapor pressure lowering• Boiling point elevation• Freezing point depression• Osmotic pressure
Basis for Colligativity
Solvent chem potential (μ1) is reduced when solute is added:• μ*
1 μ*1 + RT ln x1 (“1” is solvent)
• Since x1 < 1 ln x1 < 0
• Thus μ1 (solution) < μ1 (pure solvent)
Chemical Potential
Vapor Pressure lowering
Freezing Point Depression: ΔTfus = Kf m
Thermo Condition:• At fp: solid solvent in equilib with solvent
that’s in soln
• μsolid1(Tfus) = μsoln
1(Tfus)
• μsolid 1 = μ*
1 + RT ln a1 = μliq1 + RT ln a1
Rearranging:• ln a1 = (μsolid
1 - μliq1)/RT
ln a1 = (μsolid 1 - μliq
1)/RT
Take derivative:• ( ln a1/ T)P, x1 = [(μsolid
1 - μliq1)/RT]/ T
• Recall Gibbs-Helmholtz equation:• [ (μ/T)/ T]P, x1 = - H1/T2
• Substitute in above:
• ( ln a1/ T)P, x1 = (Hliq1 – Hsol
1)/RT2 = ΔfusH/RT2
( ln a1/ T)P, x1 = ΔfusH/RT2
Integrate between T*fus and Tfus :
• ln a1 = ƒ(ΔfusH/RT2)d T
Since it’s a dilute solution:• a1 ~ x1 = 1- x2
• ln (1 – x2) ~ - x2
Substitute above:• - x2 = (ΔfusH/R)(1/T*
fus – 1/Tfus)
- x2 = (ΔfusH/R)(Tfus - T*fus)/T*
fus Tfus
Solute lowers the freezing point:• Tfus < T*
fus
Express in molality:• x2 = n2/(n1 + n2) = m/(1000/M1 + m)
• But m << 1000/M1
• x2 ~ M1m/1000 (substitute above for x2)
Note: T*fus ~ Tfus
• (Tfus - T*fus)/T*
fus Tfus ~ (Tfus - T*fus)/T*2
fus
= - Δ T/T*2fus
Substitute!
Δ Tfus = Kf m
• Where Kf = M1 R(T*fus)2 /(1000ΔfusH)
• Kf is function of solvent only
Similar expression obtained for bp elevation: Δ Tvap = Kb m
• Where Kb = M1 R(T*vap)2 /(1000ΔvapH)
Compare terms
Example Comparison
Calc. fp and bp change of 25.0 mass % soln of ethylene glycol (M1 = 62.1) in H2O.
m = nGly/kg H2O = (250/62.1)/(750/103) = 5.37
Δ Tfus = Kf m = (1.86)(5.37) = 10.0 OC
Δ Tvap = Kb m = (0.52)(5.37) = 2.8 OC
Osmotic Pressure
Example
Calc. Osmotic pressure of previous example at 298 K.
Π = c2RT
• c2 = 4.0 R = 0.0821 L-atm/mol-K
• Π = c2RT = (4.0)(0.0821)(298) = 97 atm
Debye-Hückel Model of Electrolyte Solutions
The Model: An electrically charged ion (q) immersed in a solvent of dielectric constant ε
Experimental Observations:• All salt (electrolyte) solutions are nonideal
even at low concentrations• Equilibrium of any ionic solute is affected by
conc. of all ions present.
Next Time
How to explain the experimental evidence:Debye-Huckel Model of electrolyte solutions