lecture 20lecture 20 colligative...

Chemistry 433Chemistry 433

Lecture 20Lecture 20Colligative Properties

Freezing Point DepressionBoiling Point Elevation

Osmosis

NC State University

Colligative propertiesg p pThere are a number of properties of a dilute solution that depend only on the number of particles and not on their kinddepend only on the number of particles and not on their kind. Colligative properties include the lowering of the vapor pressure of a solvent and elevation of the boiling temperature by the addition of a nonvolatile solute, the depression of the freezing point of a solution by a solute, and osmotic pressure.

They all have a common treatment using the chemical potential of the pure substance compared to the chemicalpotential of the mixture In this case the mixture consists ofpotential of the mixture. In this case the mixture consists ofa non-volatile (solid) solute.

Freezing point depression g p pAt the freezing point of a solution, solid solvent is in equilibrium with the solvent in solution. As we have seen,this equilibrium implies that the chemical potential of each phase is equal to the other:

μ1solid(Tfus) = μ1

soln(Tfus)

where subscript 1 denotes solvent and T is the freezingwhere subscript 1 denotes solvent and Tfus is the freezing point of the solution.

μ1soln(Tfus) = μ1

* + RT ln x1 = μ1liq + RT ln x1

So that we can equate μ for the liquid and solidSo t at e ca equate μ o t e qu d a d so d

μ1solid = μ1

liq + RT ln x1

Freezing point depression g p pSolving for x1, we getln x = (μ solid μ liq)/RT = Δ G/RTln x1 = (μ1

solid - μ1liq)/RT = - ΔfusG/RT

ΔfusG = ΔfusH - TΔfusSln x1 = (μ1

solid - μ1liq)/RT = - ΔfusH/RT + ΔfusS/R

For pure solvent x1 = 1 andln 1 = - ΔfusH/RT* + ΔfusS/RFor an activity a2 of solute, ln x1 = ln(1 - x2) ≈ - x2y 2 , 1 ( 2) 2- x2 = - ΔfusH/RT + ΔfusS/RIn the above we present the fusion temperature of pure solvent as T* and the fusion temperature of the solution as Tsolvent as T and the fusion temperature of the solution as T. Noting that ln(1) = 0 we can eliminate ΔfusS/R from the two equations.

*ΔfusS/R = ΔfusH/RT*

Freezing point depression g p px2 = ΔfusH/RT - ΔfusH/RT*

x = Δ H/R(1/T 1/T*)x2 = ΔfusH/R(1/T - 1/T )The above expression can be further approximated using(1/T - 1/T*) = (T* - T)/TT* ≈ ΔT/T*2

*2x2 = ΔfusH/RT*2 ΔTThe above formula can be compared with the formula for freezing point depressiong p pΔTfus = Kfm (m is molality)For a dilute solution x2 ≈ M1m/1000g kg-1 for small values of m Thereforevalues of m. Therefore,Kf = (RT*2)(M1/1000g kg-1)/ΔfusHis called the freezing point depression constant.

Justification: ln(1-x) ~ -x

f(x) = dn f(x)dxn

′

xn

n!Σn = 1

∞

The derivative is:dx x = x′n!n 1

dln (1 – x) 1

and we are expanding about 0 so that we will

( )dx = – 1

1 – x

need to substitute 0 in for x in the derivative.The first term in the Taylor expansion is:

-1/(1-0)x = -x

Boiling point elevation We can determine the value of Kb for water.The phenomenon of boiling point elevation can be derived i l t l l f hi I b th thin a completely analogous fashion. In both cases the physics behind the effect is a lowering of the chemical potential of the solution relative to the pure substance. This can be seen in the diagram below where we plot the chemical potential as a function of the temperature.

QuestionWhich mathematical term explains why the chemical potential of a solution lowered relative to pure liquid?A Δ H/RT*2ΔTA. ΔvapH/RT*2 ΔTB. -ΔvapH/RT*2 ΔTC. RT ln x11D. -RT ln x1

QuestionWhich mathematical term explains why the chemical potential of a solution lowered relative to pure liquid?A Δ H/RT*2ΔTA. ΔvapH/RT*2 ΔTB. -ΔvapH/RT*2 ΔTC. RT ln x1 μ1

soln = μ1* + RT ln x11 μ1 μ1 1

D. -RT ln x1

QuestionWhich is an accurate expression of the boiling point elevationin terms of the activity of the solvent?A l Δ H/R(1/T 1/T*)A. ln x1 = ΔvapH/R(1/T - 1/T*)B. x2 = ΔvapH/R(1/T - 1/T*)C. RT*2/ΔvapH (M1/1000g kg-1)vap 1D. μ1

soln = μ1* + RT ln x1

QuestionWhich is an accurate expression of the boiling point elevationin terms of the activity of the solvent?A l Δ H/R(1/T 1/T*)A. ln x1 = ΔvapH/R(1/T - 1/T*)B. x2 = ΔvapH/R(1/T - 1/T*) Activity of the soluteC. RT*2/ΔvapH (M1/1000g kg-1) Molality of the solutevap 1D. μ1

soln = μ1* + RT ln x1 Chemical potential (not

boiling point elevation)

Boiling point elevation g pIn this plot notice that the slope increases as the phase changes from solid to liquid and then to vapor The slopechanges from solid to liquid and then to vapor. The slope is proportional to - S (since ∂μ/∂T = -S) and the entropy increases in the same order. Notice that the violet line representing the chemical potential as a function of temperature is shifted down by the addition of solute. Mathematically this is due toyμ1

soln = μ1liq + RT ln x1

Because the chemical potential of the solid and vapor are not shifted by the addition of solute the intersection pointnot shifted by the addition of solute the intersection point (i.e. temperature of phase transition) goes down for fusion, but goes up for vaporization.

QuestionWhat is the slope of the chemical potential with temperature?

A. free energyB. enthalpyC. entropyD. none of the above

QuestionWhat is the slope of the chemical potential with temperature?

A. free energyB. enthalpyC. entropyD. none of the above

QuestionWhat is Kf for water?

K = RT*2/Δ H (M /1000 g kg-1)Kf = RT 2/ΔfusH (M1/1000 g kg 1)A. 3.46 K-1 mol-1 kgB. 1.86 K-1 mol-1 kgC. 0.67 K-1 mol-1 kgD. 0.095 K-1 mol-1 kg

QuestionWhat is Kf for water?

K = RT*2/Δ H (M /1000 g kg-1)Kf = RT 2/ΔfusH (M1/1000 g kg 1)A. 3.46 K-1 mol-1 kgB. 1.86 K-1 mol-1 kgC. 0.67 K-1 mol-1 kgD. 0.095 K-1 mol-1 kg

QuestionWhat concentration of salt must be achieved to cause a 1 oCdecrease in the melting temperature of the ice on a road?decrease in the melting temperature of the ice on a road?

A. 1.6 mol kg -1

B. 0.16 mol kg -1g

C. 5.4 mol kg -1

D. 0.54 mol kg -1

QuestionWhat concentration of salt must be achieved to cause a 1 oCdecrease in the melting temperature of the ice on a road?decrease in the melting temperature of the ice on a road?

A. 1.6 mol kg -1

B. 0.16 mol kg -1g

C. 5.4 mol kg -1

D. 0.54 mol kg -1

QuestionWhich is larger?A K = RT*2/Δ H (M /1000 g kg-1)A. Kf = RT 2/ΔfusH (M1/1000 g kg 1)B. Kb = RT*2/ΔvapH (M1/1000 g kg-1)

QuestionWhich is larger?A K = RT*2/Δ H (M /1000 g kg-1)A. Kf = RT 2/ΔfusH (M1/1000 g kg 1)B. Kb = RT*2/ΔvapH (M1/1000 g kg-1)

Osmotic pressureOsmotic pressure arises from requirement that the chemical potential of a pure liquid and its solution must be the same if th i t t th h i bl bthey are in contact through a semi-permeable membrane. Osmotic pressure is particularly applied to aqueous solutions where a semi-permeable membrane allows water to pass back and forth from pure water to the solution, but the solute cannot diffuse into the pure water. The point here is that the solute lowers the chemical potential on the solution side of thelowers the chemical potential on the solution side of the membrane and therefore there will be a tendency for water to move across the membrane to the solution side. Ultimately, th ill b b l f f if b ild ththere will be a balance of forces if a pressure builds up on the solution side of the membrane. This pressure can arise due to an increase in the hydrostatic pressure due to a rise in a column of solution or due to pressure inside a closed membrane. The easiest to visualize is a column of water.

Osmotic pressure arises from an imbalance in chemical potentialimbalance in chemical potential

when solute is addedμ* >μsoln

μsoln μ*

H2O + Solute Pure H2O

The flow of solvent leads to an increase in hydrostatic pressure

P + Π

h

H2O + solute P

Pure H2O

Osmotic pressurepRecall that the pressure at the bottom of a column of a fluid is given by P = ρ g h. If water flows into the solution the height of the column of solution increases and the hydrostatic pressure also increases. At some point the chemical potential due to the concentration difference is exactly opposed to thedue to the concentration difference is exactly opposed to the chemical potential due to the pressure difference. We express this asμ *(T P) = μ soln(T P+Π x )μ1 (T,P) = μ1

soln(T,P+Π,x1) The chemical potential of the solution isμ1

soln(T,P) = μ1*(T,P) + RT ln x1

μ1*(T,P) = μ1

soln(T,P+ Π,x1) = μ1*(T,P+ Π) + RT ln x1

Recall that ∂μ/∂T = Vm (subscript for molar volume) soP + Π

μ1* T,P + Π – μ1

* T,P = VmdPP

Osmotic pressurepThus,

VmdPP+Π

+ RT ln a1 = 0

assuming Vm does not vary with applied pressure, we can writeΠVm + RT ln x1 = 0

VmdP+P

RT ln a1 0

ΠVm RT ln x1 0For a dilute solution and ln x1 = ln(1-x2) ≈ -x2we have thatΠV RT x = 0ΠVm - RT x2 = 0which be expressed asΠV = n2RT.The above expression bears a surprising similarity to the ideal gas law. Keep in mind, however, that Π is the osmotic pressure and n2 is the number of moles of solute. p essu e a d 2 s t e u be o o es o so ute

Osmotic pressurepThus, we can compute the osmotic pressure fromΠ = n2RT/VorΠ = cRTwhere c is the molarity, n2/V, of the solution.where c is the molarity, n2/V, of the solution.This equation is called the van't Hoff equation for osmotic pressure. The osmotic pressure can be used to determine the molecular masses of solutes particularly solutes withthe molecular masses of solutes, particularly solutes with large molecular masses such as polymers and proteins.

QuestionWhat is the height of a column of water that will result from addition of enough NaCl to make a 0.1 M solution.

A. 25 mB. 2.5 mB. 2.5 mC 0.25 mD 0.025 m

QuestionWhat is the height of a column of water that will result from addition of enough NaCl to make a 0.1 M solution.

A. 25 mB. 2.5 mB. 2.5 mC 0.25 mD 0.025 m

Π = cRT = ρgh

h = cRT/ρg = (100 mol/m3)(8.31 J/mol–K)(298 K)(1000 kg/m3)(9.8 m/s2)

= 25 m

Use of osmotic pressure to determine molar mass

The van’t Hoff equation can be modified to form used

Π RT Π = w RT

The van t Hoff equation can be modified to form used for the determination of molar mass by osmometry.

Π = cRT Π = MRT

Here we related to the concentration c in moles/liter toHere we related to the concentration c in moles/liter tothe concentration w in grams/liter and the molar massM in grams/mole.

The experimental configuration uses the measurementof height as an estimate of the osmotic pressure. Theequation Π = ρgh is used (h = Π/ρg).

Use of osmotic pressure to determine molar mass

Pure H2O Pure H2O

Use of osmotic pressure to determine molar mass

h

H2O + unknown

Pure H2O

Use of osmotic pressure to determine molar mass

A sample of 1 5 mg of a protein of unknown molarA sample of 1.5 mg. of a protein of unknown molar mass is added to an osmometer. The solution volume is 1 mL. The solution height increases by 1 cm. Themeasurement temperature is 298 K. What is the molarmass of the protein?

A. 37,900

B 39 700B. 39,700

C. 79,300

D. 97,300

Use of osmotic pressure to determine molar mass

A sample of 1 5 mg of a protein of unknown molarA sample of 1.5 mg. of a protein of unknown molar mass is added to an osmometer. The solution volume is 1 mL. The solution height increases by 1 cm. Themeasurement temperature is 298 K. What is the molarmass of the protein?

A. 37,900

B 39 700

M = wRTΠ

= wRTρgh

= (1.5 kg/m3)(8.31 J/mol–K)(298 K)(1000 kg/m3)(9.8 m/s2)(0.01 m)

= 37 9 kg / molB. 39,700

C. 79,300

= 37.9 kg / mol= 37,900 g / mol

D. 97,300