13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects

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Week 3 Sections 13.3- 13.5

13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects

Temperature Effects 13.4 Ways of Expressing Concentration

Mass Percentage, ppm, and ppb Mole Fraction, Molarity, and Molality Conversion of Concentration Units

13.5 Colligative Properties

Lowering the Vapor Pressure Boiling-Point Elevation Freezing-Point Depression Osmosis

Determination of Molar Mass [13.6 Colloids]

Factors that FAVOR solubility:

1. Strong solute-solvent interactions 2. Weak solute-solute interactions 3. Weak solvent-solvent interactions

More often we’ll settle for the solute-solvent interactions being similar to the solute-solute and solvent-solvent interactions.

A general rule: Like dissolves like.

i.e. polar and polar non-polar and non-polar

Realize there is an inherent tendency for the two isolated materials to form solution, regardless of the energetics!!! This represents an entropy factor.

Consider the Solubilities (mol/100g water) of Alcohols

CH

₃

CH

₃

(CH

₂

)

₆

OH 0.0008

Fig 13.12 Structure of glucose—note red O atoms in OH groups

C₆H₁₂O₆ or C₅H₅OCH₂(OH)₅

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Fig. 25.26 Sucrose and Lactose Molecules.

Olestra has fatty acids at each of the OH in sucrose.

Fig 25.27 The Starch Molecule, note the blue atoms

Fig 25.28 Cellulose Molecule, note the blue atoms

Fig 25.29 Starch (a) and Cellulose (b) in 3-D representation

Table 13.2 Gas solubilities in water

at 20 ^oC with 1 atm gas pressure (Table 13.2) Solubility/Molarity He 0.40 x 10^-3 N₂ 0.69 x 10^-3 CO 1.04 x 10^-3 O₂ 1.38 x 10^-3 Ar 1.50 x 10^-3 Kr 2.79 x 10^-3 CO₂ 3.1 x 10^-2 NH₃ ~ 53

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Fig 13.14 Henry’s Law, C_g= k P_g See Table 13.2 for k in M/atm

Table 13.2 Gas solubilities in water

at 20 ^oC with 1 atm gas pressure (Table 13.2) Solubility/Molarity He 0.40 x 10^-3 N2 0.69 x 10^-3 CO 1.04 x 10^-3 O₂ 1.38 x 10^-3 Ar 1.50 x 10^-3 Kr 2.79 x 10^-3 CO₂ 3.1 x 10^-2 NH₃ ~ 53

Consider N2 dissolved in water at 4.0 atm.

Note k = 0.69 x 10^-3 mol/L-atm

Cg = k Pg

= (0.69 x 10^-3 mol/L-atm)(4.0 atm)

= 2.76 x 10^-3 mol/L

at normal atmospheric conditions, however, Pg = 0.78 atm Cg = (0.69 x 10^-3 mol/L-atm)(0.78 atm) = 0.538 x 10^-3 mol/L

Note that (2.76 - 0.54) x 10^-3 mol/L = 2.22 x 10^-3 mol/l Thus for 1.0 L of water, 0.0022 mol of nitrogen would be released = 0.0022 x 22.4L = 0.049 L = 49 mL !

To read about nitrogen narcosis, seehttp://www.gulftel.com/~scubadoc/narked.html and http://www.diversalertnetwork.org/medical/articles/index.asp

Also in Table 13.2

Fig 13.18 Temperature Effect on Solubility of Gases.

Fig 13.17 Effect of T on Solubilities of Ionic Cmpds.

Are these exothermic or endothermic processes?

The ammonia fountain

Add acetone to a sat’d CuSO4 sol’n Heat water and calcium acetate Heat water and potassium nitrate

Demonstrations

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Ways of expressing concentration:

a) percent, ppm, ppb usually m/m

b) mole fraction = X_A, X_B sum of X_i= 1 c) molarity = M or mol/L_solution

depends on T and density of soln preparation requires dilution d) molality = m or mol/kg_solvent

independent of T easily prepared

Prepare 0.500 L of a 0.100 M aqueous solution of KHCO₃(MW = 100.12 g)

Prepare 0.500 L of a 0.100 M aqueous solution of KHCO₃(MW = 100.12 g) (0.100 mol KHCO₃/L of soln)(0.500 L) = 0.0500 mol KHCO₃

Note: molarity = M = n/V therefore n = MV

Prepare 0.500 L of a 0.100 M aqueous solution of KHCO₃(MW = 100.12 g)

(0.100 mol KHCO₃/L of soln)(0.500 L) = 0.0500 mol KHCO₃ (0.0500 mol KHCO₃)(100.12g/mol KCHCO₃) = 5.01 g KHCO₃

(a) Prepare 0.500 L of a 0.100 M aqueous solution of KHCO₃(MW = 100.12 g) requires 5.01 g of KHCO₃dissolved and diluted to 0.500 L

(b) Use this solution as a ‘stock’ solution to prepare a final solution of 0.0400 M concentration. What is the final volume of this solution?

(a) Prepare 0.500 L of a 0.100 M aqueous solution of KHCO₃(MW = 100.12 g) requires 5.01 g of KHCO₃dissolved and diluted to 0.500 L

(b) Use this solution as a ‘stock’ solution to prepare a final solution of 0.0400 M concentration. What is the final volume of this solution?

Since n = MV, M₁V₁= M₂V₂

so, (0.100 M)(0.500 L) = n = (0.0400 M)( V₂)

or L

M L M M

V

V M 1.25

) 0400 . 0 (

) 500 . 0 )(

100 . 0 (

2 1 1

2= = =

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Consider a solution prepared by dissolving 22.4 g MgCl₂ in 0.200 L of water. Assume the density of water is 1.000 g/cm³and the density of the solution is 1.089 g/cm³.

Calculate

mole fraction molarity molality

A 9.386 M solution of H₂SO₄has a density of 1.509 g/cm³. Calculate

molality

% by mass

mole fraction of H₂SO₄

Week 3 Sections 13.3- 13.5 13.3 Factors Affecting Solubility

Solute-Solvent Interactions Pressure Effects Temperature Effects 13.4 Ways of Expressing Concentration

Mass Percentage, ppm, and ppb Mole Fraction, Molarity, and Molality Conversion of Concentration Units

13.5 Colligative Properties

Lowering the Vapor Pressure Boiling-Point Elevation Freezing-Point Depression Osmosis

Determination of Molar Mass

[13.6 Colloids]

Colligative Properties

Solution properties that depend only on the total # of ‘particles’ present.

Vapor Pressure Boiling Point Freezing Point Osmotic Pressure

Note that VP of solution is lower than that of pure solvent.

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Vapor Pressure lowering

Raoult’s Law P_A= X_AP_A^o

P_A = vapor pressure over solution X_A= mole fraction of component A (solvent) P_A^o= vapor pressure of pure

component A (solvent) also P_A= (1 – X_B) P_A^o

where X_B= mol fraction of B (solute)

At 25 ^oC, the vapor pressure of benzene is 0.1252 atm, i.e. P_A^o= 0.1252 atm. If 6.40 g of naphthalene (C₁₀H₈, 128.17 g/mol) is dissolved in 78.0 g of benzene (78.0 g/mol), calculate the vapor pressure of benzene over the solution

Raoult’s Law says P_A= X_AP_A^o therefore we need X_A

n_A= ? n_B= ?

At 25 ^oC, the vapor pressure of benzene is 0.1252 atm, i.e. P_A^o= 0.1252 atm. If 6.40 g of naphthalene (C₁₀H₈, 128.17 g/mol) is dissolved in 78.0 g of benzene (78.0 g/mol), calculate the vapor pressure of benzene over the solution

Raoult’s Law says P_A= X_AP_A^o therefore we need X_A let A = benzene

n_A= 78.0 g benzene (1 mol/78.0 g) = 1.00 mol benzene n_B= 6.40 g naph. (1 mol/128.17 g) = 0.0499 mol naph.

and P_A= P_benzene= (1.00/1.0499)(0.125) = 0.1193 atm note: ∆P = P_A– P_Aô= X_AP_Aô– P_Aô= P_Aô(X_a-1)

or ∆P = -X_BP_A^o

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Boiling Point Elevation and Freezing Point Depression Boiling Point Elevation

∆T_b= T_b’ – T_b= K_bm

where m is the molal concentration

Freezing Point Depression

∆T_f= T_f’ – T_f= - K_fm

where m is the molal concentration.

[note the definition and the negative sign!!!]

for H₂O, K_b= 0.052 ^oC/m and K_f= 1.86 ^oC/m

Consider a water solution which has 0.500 mol of sucrose in 1.000 kg of water. Therefore it has a concentration of 0.500 molal or 0.500 mol/kg.

recall K_b= 0.52 ^oC/m and K_f= 1.86 ^oC/m What is the boiling point and freezing point of this solution?

(a) When 5.50 g of biphenyl (C₁₂H₁₀, 154.2 g/mol) is dissolved in 100 g benzene (C₆H₆, 78.0 g/mol), the BP increases by 0.903 ^oC. Calculate K_bfor benzene.

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(a) When 5.50 g of biphenyl (C₁₂H₁₀, 154.2 g/mol) is dissolved in 100 g benzene (C₆H₆, 78.0 g/mol), the BP increases by 0.903 ^oC. Calculate K_bfor benzene.

Î K_b = 2.53 K kg/mol

(b) When 6.30 g of an unknown hydrocarbon is dissolved in 150.0 g of benzene, the BP of the solution increases by 0.597 ^oC.

What is the MW of the unknown substance?

A sample of sea water contains the following in 1.000 L of solution. Estimate the freezing point of this solution.

Na⁺ = 4.58 mol Cl- = 0.533 mol Mg²⁺ = 0.052 mol SO₄^2- = 0.028 mol Ca²⁺ = 0.010 mol HCO₃^- = 0.002 mol K⁺ = 0.010 Br^- = 0.001 mol neutral species = 0.001 mol

Sum of species = 1.095 mol

Consider Exercise 13.9

List the following aqueous solutions in increasing order of their expected freezing points.

0.050 m CaCl₂ 0.15 m NaCl 0.10 m HCl 0.050 m HOAc 0.10 m C12H₂₂O₁₁

Consider Exercise 13.9

List the following aqueous solutions in increasing order of their expected freezing points.

0.050 m CaCl₂ x 3 = 0.150 0.15 m NaCl x 2 = 0.30 0.10 m HCl x 2 = 0.20 0.050 m HOAc x 1 = 0.050 0.10 m C₁₂H₂₂O₁₁ x 1 = 0.10

These calculations assume total dissociation of the salts and zero dissociation of the last two.

This effect of the dissociation of electrolytes is usually taken into account through the van Hoff i factor.

∆T_b= i K_bm

where it is defined as

∆T_f(actual) K_fm_effective m_effective i = --- = --- = ---

∆Tf(ideal) K_fm_ideal m_ideal

In real systems, these i factors are NOT integers, but rather fractions whose values depend on concentration.

Osmosis (get started)

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Osmotic Pressure:

a fascinating behavior.

Yet it is the result of a very simple tendency to equalize the concentrations of solutions.

The critical part is the membrane!!!

Osmotic Pressure π V = n R T or π = (n/V) R T or π = M R T π = ρ g h, where ρ = density of solution

g = 9.807 m s^-2 h = height of column

π = ρ g h, where ρ = density of solution g = 9.807 m s^-2 h = height of column

If h = 0.17 m of a dilute aqueous soln with ρ = 1.00 g/cm³ π = [(1.00g/cm³)(10^-3kg/g)(106 cm³/m³)](9.807 m/s²)(0.17 m)

= 1.7 x 103 kg m^-1s^-1 = 1.7 x 10³Pa

or = (1.7 x 10³Pa) / (1.013 x 10⁵Pa/atm) = 0.016 atm

A chemist dissolves 2.00 g of protein in 0.100 L of water. The observed osmotic pressure is 0.021 atm at 25 ^oC. What is the MW of the protein?

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Fractional Distillation – How it’s done

Real solutions do not behave ideally!

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