Effect of Temperature and Pressure on Solubility

1. Effect of Temperature and Pressure on Solubility College Chemistry

2. Solubility: Temperature Dependence • All solubilities are temperature dependent; must report temperatures with solubilities. • Most solids are more soluble at higher temperatures. Exceptions exist. • All gases are less soluble at higher temperatures. • Temperature related to sign of Hsoln; • negative means less soluble at high temperatures • positive means more soluble • E.g. Predict the temperature dependence of the solubility of Li2SO4, Na2SO4 and K2SO4 if their Hsoln are 29.8 kJ/mol, 2.4 kJ/mol and +23.8 kJ/mol, respectively.

3. Fractional Crystallization – Using Solubility at Temperatures to Our Advantage • Fractional crystallization – separation of a mixture of substances into pure compounds on the basis of their differing solubilities • Suppose we have impure KNO3 (10% is NaCl) • We gradually cool the sample and KNO3 will crystallize out of the solution before NaCl b/c KNO3 is less soluble at this temperature

4. Gas Solubility • Solubility of gases almost always decreases with increasing temperature • Ex: when you boil water, bubbles of air try to escape and “boil out” long before the solution begins to bubble • Fishing and Gas solubility – most fishers like fishing early in the morning or pick a deep spot in the river • Why? Fish need oxygen and the temperature is less there  more oxygen

5. Gas Solubility

6. Solubility: Pressure Dependence • Pressure has little effect on the solubility of a liquid or solid, but has dramatic effect on gas solubility in a liquid. • Henry’s law c = kHP. Allows us to predict the solubility of a gas at any pressure. • States that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution • c = molar concentration (molarity) P = pressure • kH = constant that depends only on temperature • Because of higher concentration, there are more gas particles moving around and hitting each other  increases pressure

7. Solubility of Gases • Ex: soft drinks are pressurized with a mixture of air and CO2 • Because of high partial pressure of CO2, the amount dissolved in the soft drink is many times the amount that would dissolve in the atmosphere • When the cap is removed, the pressurized gases escape and the pressure falls to the atmospheric pressure

8. Exceptions of Henry’s Law • If gases dissolved in water end up reacting with the water, solubility will be much higher than you would expect • An example is oxygen in blood • Normally, O2 isn’t very soluble in water • However, its solubility in blood is great because of the high content of hemoglobin (Hb) that can bind oxygen) • Hb + 4 O2⇌Hb(O2)4

9. Physical Behavior of Solutions: Colligative Properties • Compared with the pure solvent the solution’s: • Vapor pressure is lower • Boiling point is elevated • Freezing point is lower • Osmosis occurs from solvent to solution when separated by a membrane. • All the above are considered colligative properties – properties that depend only on the # of solute particles in solution and not on the nature of the solute particles (no difference between NaCl and KCl)

10. Vapor-Pressure Lowering of Solutions: Raoult’s Law • Raoult’s Law: Psoln = PsolvxXsolv • Non–volatile solute: vapor pressure decreases upon addition of solute. • Non-volatile solute: does NOT have a measurable vapor pressure (Ex: salt water) • Linear for dilute solutions • Vapor pressure lowering : P = Po P = Po(1Xsolv). • Decrease in vapor pressure, DP, is proportional to the solute concentration (measured in mole fraction)

11. Example 12.7 • Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (molar mass = 180.2 g/mol) in 460 mL of water at 30°C. What is the vapor-pressure lowering? The vapor pressure of pure water at 30°C is 31.82 mmHg. Assume the density of 1.00 g/mL. • We know: P1 = X1P1° (trying to find P1, need to find X1) • 1. # of moles of glucose and water • n1(water) = 460 mL x (1.00 g/1 mL) x (1 mol/18.02 g) = 25.5 mol • n2(glucose) = 218 g x (1 mol/180.2 g) = 1.21 mol • 2. mole fraction of water • X1 = n1 / total = 25.5 mol/ (25.5 + 1.21) = 0.955 • 3. Find vapor pressure of water • P1 = (0.955) x (31.82 mmHg) • P1 = 30.44 mmHg • 4. Vapor pressure lowering • 31.82 mmHg – 30.44 mmHg = 1.4 mmHg

12. Vapor-Pressure Lowering of Solutions: Volatile Solute • In ideal solutions each substance in the solution obeys Raoult’s law. Thus, • PA = XAP°o and Ptotal = PA + PB. • Combining gives: . • Notice the equation predicts the linear variation in total vapor pressure with variations in the mole fraction of one of the components in the liquid phase.

13. Fractional Distillation • Fractional distillation – separates compounds based on their different boiling points • Similar to fractional crystallization • I used this a LOT in the lab In grad school

14. Colligative Properties • Properties of a solvent that depend only on the amount of solute particles present (i.e.: freezing point) • Why is vapor pressure decreased and boiling point raised? • The solute particles take away space from the solvent and block the solvent from being evaporated • This means less solvent particles near the surface and therefore less “pressure” • Boiling point raises because more energy is needed http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch15/graphics/15_13.gif

15. Colligative Properties • This same argument can be used to explain why freezing point is lowered • The solute particles get in the way of the pure solvent • It takes a lower temperature to get(stronger “push”) the solvent to freeze

16. Colligative Properties • We will be dealing with freezing point depression today • Can be found by the equation • DTf=Kfm • DTf= freezing point of pure solvent – f.p. of solution • Kf= freezing point constant for the solvent (Table 12.1) • m= molality • m= moles of solute/ kg of solvent

17. BP Elevation and FP Depression of Solutions The magnitude of the change in FP and BP is directly proportional to the concentration of the solute (molality) – expressed in terms of the total number of particles in the solution. • BP Elevation The magnitude of the BP increase is given by the equation: where Kb has units of °Ckg/mol or °C/m • FP Depression: linear variation with composition and given by: where the units for this constant are the same as for Kb

18. Boiling Point Elevation and Freezing Point Depression • Freezing point is lowered when a solute is added • Ex: salt is added to icy sidewalks to get the freezing point to lower and not have ice • Boiling point of a solution is raised when a solute is added • For example, you add salt to water to help it boil • Contrary to popular belief, the salt does NOT help the water boil at a lower temperature (or faster)

19. BP Elevation and FP Depression

20. Example 12.8 • Ethylene glycol (EG) is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p 197C). Calculate the freezing point of a solution containing 651 g of this substance in 2505 g of water. Would you keep this substance in your car radiator during the summer. MM = 62.01 g • 651 g EG x (1 mol EG / 62.07 g EG) = 10.5 mol EG • Molal = moles solute / mass solvent (kg) • m = 10.5 mol EG / 2.505 kg H2O = 4.19 m • Remember, DTf = Kfm • DTf = (1.86°C/m)(4.19 m) • DTf = 7.79°C • Since water freezes at 0°C, the solution will freeze at -7.79°C

21. Example 12.8 Cont. • To find the boiling point elevation: • DTb = Kbm • = (0.52 °C/m)(4.19 m) • = 2.2°C • Since water normally boils at 100°C, it would be 102.2°C that this boils • It would be preferable to leave the antifreeze in your car to keep the water from boiling • Kb and Kf will either be given OR you will need to find it (and have all the information necessary)

22. Osmosis and Osmotic Pressure Osmosis: the passage of solvent through a membrane from the less concentrated side to the more concentrated side. • Osmotic pressure: the amount of pressure necessary to stop Osmosis. • Small molecules such as water can move through certain types of materials (membranes). . where M = is molarity of solute particles, R = gas constant, T = temperature, and p = osmotic pressure

23. Example 12.9 • The average osmotic pressure of seawater, is about 30.0 atm at 25°C. Calculate the molar concentration of an aqueous solution of sucrose (C12H22O11) that is isotonic (same osmotic pressure) with seawater. • p = MRT • Rearrange to M = p/RT • M = 30.0 atm / (0.0821 L*atm/K*mol)(298 K) • M = 1.23 M

24. Using Colligative Properties to Find Molar Mass • Let’s say you add an unknown substance to the solvent, how can you find what it is? Find molar mass! • You have a solution of 15.00 g glucose in 100g water. Freezes at -1.53°C. What is the molar mass? 1.86°C is Kf • Start by finding molality • m=DTf/Kf • m=1.53°C/1.86°C • m=0.823

25. Helpful example • Remember molality is equal to moles/kg solvent and molar mass is in grams • Convert from moles/kg to grams • So we have 0.823 molality or .823 moles per 1kg (1000g) of water • We only have 100g (0.1 kg)in this case with 15g of glucose • (0.100 kg) x (0.823 mol / 1 kg) = 0.0823 mol • Now we can solve for molar mass • M= 15g / 0.0823 m • M=182 g/mol

26. Colligative Properties of Ionic Solutions • Colligative properties of solutions depends upon the total concentration of particles. • Each equation describing colligative properties must be modified to account for this with ionic solutions since each ionic compound gives more than one mole of ions for every mole of compound. • BP elevation: • Freezing point depression: • Osmotic pressure: Where i = van’t Hoff factor. • van’t Hoff factor can be something other than integer under certain circumstances, but for completely ionic solutions is equal to the number of ions/ionic compound to be found in solution: E.g. NaCl: i = 2; Na2SO4: i = 3;

27. Dissociation Equations and the Determination of i i = 2 NaCl(s)  Na+(aq) + Cl-(aq) i = 2 AgNO3(s)  Ag+(aq) + NO3-(aq) i = 3 MgCl2(s)  Mg2+(aq) + 2 Cl-(aq) i = 3 Na2SO4(s)  2 Na+(aq) + SO42-(aq) AlCl3(s)  Al3+(aq) + 3 Cl-(aq) i = 4

28. Ideal vs. Real van’t Hoff Factor The ideal van’t Hoff Factor is only achieved in VERY DILUTE solution.

29. Example 12.12 • Calculate the osmotic pressure of a 0.010 M potassium iodide (KI) solution at 25°C. • KI dissolves into 2 ions, K+ and I- • p = iMRT • p = (2.0)(0.010 M)(0.0821 L*atm/mol*K)(298 K) • p = 0.489 atm

30. Colloids • Colloids are not true solutions • Colloids – dispersion of particles of one substance throughout a dispersion medium made of another substance • Particle size is on the order of 10 to 200 nm • Might be super-sized molecules (e.g., proteins) or aggregates of ions • Colloidal particles cannot be filtered and do not settle out of solution • Colloids exhibit the Tyndall effect (whereas solutions don’t) • The particles in a colloid are large enough to scatter light passing through) • Examples: • blood, milk, jelly, propofol

32. polar head greasy tail Soaps and Detergents • A soap is prepared by hydrolyzing fat with alkali • A detergent is a synthetic chemical with a structure similar to soap • Both have a polar (hydrophilic) head and a non-polar (hydrophobic, greasy) tail

33. Monolayers • The greasy tails stick out of the surface of water • This breaks down the surface tension of water