Chapter Contents Solution Composition Concentrations H solution Hess’s Law undersea Solubilities Henry’s Law: Gases and Raoult’s Law Temperature Effects Colligative Properties T BP Elevation T FP Depression Osmotic Pressure van’t Hoff Factor Colloids and Emulsions
Solution Composition Molarity, M = moles solute / liter sol’n. Cannot be accurately predicted for mixtures because partial molar volumes vary. If volumes don’t add, masses and moles do! Molality, m = moles solute / kg solvent Not useful in titration unless density known. Useful in colligative effects. Mole fraction, X A = moles A / total moles
Conc. of 50% by wt. NaOH Density at 20°C is 1.5253 g cm 3 each liter of solution weighs 1525.3 g ½ that mass is NaOH, or 762.6 5 g n NaOH = 762.6 5 g [ 1 mol/39.99 8 g ] = 19.06 7 [NaOH] = 19.06 7 M but also 19.06 7 mol / 0.7626 5 kg H 2 O = 25.00 1 m and n water = 762.6 5 g [ 1 mol/18.01 6 g ] = 42.33 2 X NaOH =19.06 7 /(19.06 7 +42.33 2 )=0.3105 4
100 cc ea. H 2 O & C 2 H 5 OH Want Proof? 50% by Volume 100 proof Want Volume? Need densities! At 20°C, = 0.9982 3 & 0.7907 4 g/cc, resp. sol’n. mass = 99.82 3 +79.07 4 = 178.89 7 g By mass: 100%(79.07 4 / 178.89 7 ) = 44.20 1 % From tables: = 0.9265 0 g/cc V = 178.89 7 g/0.9265 0 g/cc = 193.0 9 cc It’s really 2 100cc / 193.0 9 cc = 103.5 8 proof
But even if it requires heat, mixing may well happen since entropy favors it! Conceptual Mixing Enthalpies 1.Expand both solvent and solute at the expense of H 1 and H 2 in lost intermolecular interactions. 2.Merge the expanded liquids together recovering H 3 from the new interactions. 3. If the exothermic mixing exceeds the endothermic expansion, there will be a net exo- thermic heat of solution.
Underwater Hess’s Law Unrelated to basket weaving. Since solutions are fluid, they need not expand then mix, requiring “upfront” $$. Instead they acquire AB interaction as they lose AA and BB ones; pay as you go. Hess doesn’t care; the overall enthalpy change$ will be the same.
Solubilities It’s true that which of A or B is the solute or solvent is mere naming convention … Which was the solute in that 50% cocktail? Still solutes with low solubility are surely in the mole fraction minority. And it is worthwhile asking what state parameters influence their solubility?
Gas Solubilities No doubt about it: pressure influences solubility. And directly. CO 2 in soft drinks splatter you with dissolution as you release the pressure above the liquid. Henry’s Law codifies the relationship: P A = k H[A(aq)] (k H is Henry’s constant) It applies only at low concentrations; so It applies not at all to strongly soluble gases!
Raoult’s and Henry’s Laws Apply at opposite extremes. Raoult when X~1 Henry when X~0 So Raoult to solvent and Henry to solute. When X B is small, X B =[B]/55.51M for [water]=55.51M Henry’s OK with X. XAXA 01 P P°BP°B P°AP°A k’ H;B k’ H;A P = P° X P k H ’ X
Raoult vs. Henry Difference When X~1, the solvent is not perturbed by miniscule quantities of solute. Solvent vaporization is proportional to solvent molecules at solution’s surface. Raoult When X~0, solute is in an utterly foreign environment, surrounded only by solvent. k H reflects the absence of A-A interaction, and Henry applies.
Solubility and Temperature Sometimes the AB interactions are so much weaker than AA or BB that A and B won’t mix even though entropy favors it. Since T emphasizes entropy, some of the immiscible solutions mix at higher T. Solid solubilities normally rise with T. Exceptions are known … like alkali sulfates.
Gases Flee Hot Solutions You boiled lab water to drive out its dissolved gases, especially CO 2. That’s why boiled water tastes “flat.” Genghis Khan invented tea (cha) to flavor the water his warriors refused to boil for their health as they conquered Asia and Eastern Europe. Increased T expands V gas, making it more favored by entropy vs. dissolved gas. This time, no exceptions!
Changed Phase Changes The Phase Diagram Mixing in a solute lowers solvent P vapor So T BP must rise. Since the solvent’s solid suffers no P vapor change, T FP must fall. Liquid span must increase in solution. P T
Elevating Depressions Both colligative properties arise from the same source: Raoult’s Law. Thermo. derivations of resulting T give: Freezing Point Depression: T FP = –K f m solute where K f ~ R T FP 2 / H fusion Boiling Point Elevation: T BP = +K b m solute where K b ~ R T BP 2 / H vap K f > K b since H fusion < H vap
Practical Phase Changes Antifreeze / Summer Coolant are the same Ethylene glycol (1,2- Ethanediol) is soluble in the radiator water, non- corrosive, nonscaling, and raises the boiling point in summer heat while lower- ing freezing point in winter. “Road salt” is CaCl 2 now since NaCl corrodes cars.
Colligative Utility Ligare means “to bind.” These features are bound up with just numbers of moles. NOT the identity of the molecules! Indeed, K f and K b are seen not to depend on solute properties but on solvent ones. So they’re used to count solute moles to convert weights to molar weights! Not sensitive enough for proteins, MW~10 kg
Exquisite Sensitivity To count protein moles, we need Osmotic Pressure that is very sensitive to [solute]. Solvent will diffuse across a membrane to dilute a concentrated solute solution. If the solute is too large (protein!) to diffuse back, the volume must increase. Rising solution creates (osmotic) pressure to an equilibrium against further diffusion.
MW by Osmotic Pressure, Thermodynamic derivation of the balance between & diffusion on the equilibrium gives: V = n R T (!) or = M R T E.g., 0.5 g in 50 cc yields 10 cm of pressure at 25°C (so R T = 24.5 atm L /mol) 10 cm (1 ft/30.5 cm) (1 atm/33 ft) =.010 atm [protein] = / R T = 0.00041 mol/L Wt = 0.5 g/0.05 L = 10 g/L MW = 24 kg
Moles of What? Doesn’t matter if property’s colligative. Counts moles of ions if solute dissociates. van’t Hoff Factor, i, measures ionization. i multiplies molality in any of the colligative expressions to show apparent moles present. It’s a stand-in for non-idealities too; pity. So in 0.001m K 3 PO 4, i should be nearly 4, and colligative properties see 0.004m? NO!
Weak Electrolyte Corrections PO 4 3– is a conjugate base of HPO 4 2– K a3 = 4.8 10 –13 so K b1 = K w /K a3 = 0.021 for PO 4 3– + H 2 O HPO 4 2– + OH – Equilibrium lies to left, so start with [OH – ] = [HPO 4 2– ] = 0.001–x and [PO 4 3– ] = x (0.001–x) 2 / x = 0.021 or x ~ 4.8 10 –5 ~ 0 Counting K +, total moles ~ 0.003+2(0.001) So i ~ 0.005/0.001 = 5 not 4. (4.95 with care)
Reverse Osmosis If dilution across a semipermeable (keeps out solute) membrane builds pressure, Pressure should be able to squeeze water back out of a solution! …if the membrane survives. Desalination plants are critical in desert nations like the Gulf States & N. Africa. Waste water is much more (salt) concentrated, an environmental hazard to local sea life unless ocean currents are swift enough to dilute it.
When is a Solution Not a Solution? When it’s a problem? Insoluble materials precipitate out of a solution at a rate that increases with their mass. So small particles stay suspended. With particle sizes of 1 m to 1 nm such suspensions are called colloids. Since visible ~ 0.5 m, the larger colloids scatter visible light efficiently! (Tyndall effect)
Taxonomy of Suspensions SolidLiquidGas Solid Solid Suspension e.g., pigmented plastics Solid Emulsion e.g., opal or pearl or butter Solid Foam e.g.,styrofoam coffee cups Liquid Sol or paste e.g., toothpaste Emulsion e.g., milk or Sauce Bernaise Foam e.g., suds; fire extinguisher Gas Solid Aerosol e.g., smoke, dust Liquid Aerosol e.g. fog, atomizer spray fahgedaboudit Dispersed Material Phase Dispersing Medium Phase
Aqueous Colloids Particles might be charged and stabilized (kept from coagulating) by electrostatics. Even neutral ones will favor adjacency of one charge which develops double layer (an oppositely charged ionic shell) to stabilize the colloid. “Salting out” destroys the colloid by over- whelming the repulsions with ionic strength. Small, highly charged ions work best, of course. + + + + + + + + + – – – – – – – – –
Surface Chemistry (liquids) Colloid study, a subset of surface science. Colloid molecules must be insoluble in the dispersing medium. Solubility governed by “like dissolves like.” But surface tensions play a role as well since solutes display surface excess concentration. Interfaces between phases are not simply at the bulk concentrations; influences segregation.
Surface Chemistry (solids) Industrial catalysts for many processes are solids. Atoms and molecules adhere, dissociate, migrate, reassociate, and desorb. Efficiency scales with catalyst surface area. Area measured by adsorbing monolayers of gas (N 2 ) and observing discontinuities as monolayer is covered.