Chapter 13: Solubility
Chapter 13: Solubility CHEM 132
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Ch 13 LS Physical Properties of Solutions 131 Types of Solutions 1 A solution is a homogeneous mixture of two or more substances that are characterized by the original states of their components as well as their capacity to dissolve a solute a Saturated solution a solution that has the maximum amount of a solute contained in a solvent at a specific temperature Unsaturated solution a solution that has not reached the saturation point any time the amount of solute is less than the max capacity that a solution can hold c Supersaturated solution a solution that has more solute than it can hold not a very stable solution i Some solute comes out over time usually as crystals 1 Process called crystallization which forms from large particles 2 Precipitates form from smaller particles b ble 131 Types of Solutions Solute Solvent State of Resulting Solution Examples Gas Gas Gas Air Gas Liquid Liquid Soda waterSeltzer water C02 in water Gas Solid Solid H2 gas in palladium Liquid Liquid Liquid Ethanol in water Solid Liquid Liquid Table salt NaCl in water Solid Solid Solid Brass CuZn solder SnPb Palladium is a soft silverwhite metal that resembles platinum m A Molecular View of the Solution Process 1 2 3 4 5 When a substance or solute dissolves in another usually a solvent particles from the first substance disperse into the second The solute particles take over positions usually occupied by solvent particles The strength of solventsolvent solventsolute and solutesolute interactions determine the ease of the solute particle substitution in the solvent Heat of solution AHsom formula a AHsoln AH1 AHZ AH3 Application of Hess law i If solutesolvent interaction is greater than both solutesolute and solventsolvent interactions then the process is favorable 1 Exothermic heat is released AHSoln lt 0 ii If the interaction described above is less than the second interaction then the process is less favorable 1 Endothermic heat is absorbed AHsoln gt 0 Solution process accompanied by increase in disorder of molecules gt increase in disorder leads to favorable solubility Solubility is defined as the measure of how much solute will dissolve in a solvent at a given temperature Miscible describes two liquids that are completely soluble regardless of how much of each liquid the solution contains a Alcohols methanol ethanol etc are examples of miscible liquids in water 6 Solvation the process in which an ion or molecule is surrounded by solvent molecules arranged in a specific manner a When solvent is water process is called hydration 133 Concentration Units 1 Percent by masspercent by weightweight percent mass of solute a Percent by mass of solute mass of solutemass of solvent 100 mass of solute mass of solution 100 2 Molarity M moles of solute 3 MOIa ty liters of solution mom 1 3 Molality m moles of solute 3 mommy kilograms of solvent mmkg 4 Measuring volume of a solution is generally easier to do making molarity the preferred method of determination 5 Molality is independent of temperature because the concentration is expressed in number of moles of solute and mass of solvent Example 133 The density of a 486 M aqueous solution of methanol CH3OH is 0973 gmL What is the molality of the solution The molar mass of methanol is 3204 g moles of solute Formula we use m mass of solvent kg 1 First we calculate the mass of water in 1 L of solution using density as the conversion factor 1000 mL solution 0973 g 1 L solution 1 L solution 1mL 973 g water 2 Second since we know the solution contains 486 moles of methanol the amount of water solvent is Mass of H20 mass of solution mass of solute 3204 g CH3 OH Mass of H20 973 g 486 mol CH3OH 1 17101 CH 3 OH r Mass of H20 973 g 156 g 817 g 0817 kg 3 Finally we can calculate molality by converting to kg final step in part 2 and we find moles solute 486 mol CH 3 0H MOIahty mass solvent 0817 kg H 20 134 Effect of Temperature on Solubilitv In most cases the solubility of a solid substance increases with temperature There is no clear correlation between the sign of AHsoln and the variation of solubility with temperature The solubility of gases in water usually decreases as temperature increases Thermal pollutioi the heating of the environment usually waterways to temperatures which are harmful to its living inhabitants 0 Reduced solubility of molecular oxygen in hot water is one of the key reasons for why this happens 135 Effect of Pressure on the Solubilitv of Gamp External pressure has no in uence on the solubilities of liquids and solids Henry s Law states that the solubility of gas in a liquid is proportional to the pressure of the gas over the solution 0 c is proportional to P or c kP I c is the molar concentration of dissolved gas I P is the pressure in atm of the gas over the solution at equilibrium I k is the constant in molLatm that depends only on the temperature I ckwhenP1atm Most gases obey Henry s law with some exceptions 0 If a dissolved gas reacts with water higher solubility can result 0 Solubility of oxygen gas in blood is dramatically greater because of the high content of hemoglobin I Accounts for high solubility of molecular oxygen in blood 136 Colliggtive Properti Colligative propertiescollectjve propertiei important properties of solutions that depend on the number of particles of solute in solution 0 Not on the nature of the solute particles alone 0 Bound together by a common origin gt all depend on the number of solute particles present Four types of colligative properties 1 Vapor pressure lowering Raoult s law 2 Boiling point elevation affected by vapor pressure lowering 3 Freezing point depression disorder to order 4 Osmotic pressure osmosis and crenation 1 Vapor Pressure Lowering O Nonvolatile solute no measurable vapor pressure indicates the vapor pressure of its solution is ALWAYS less than the pure substance 0 Relationship between solution vapor pressure and solvent vapor pressure depends on solute concentration solute Raoult s Law the vapor pressure of a solvent over a solution P1 is given by the vapor pressure of the pure solvent P10 multiplied by the mole fraction of the solvent in the solution X1 Formula is P1 X1P1 O A decrease in vapor pressure shown as AP is directly proportional to the concentration of the solute Example 136 Calculate the vapor pressure of a solution made by dissolving 198 g of glucose molar mass is 1802 gmol in 435 mL 0435 L of water at 35 C What is the vapor pressure lowering The vapor pressure of pure water at 35 C is 4218 mmHg Assume the solvent density is 1 gmL We need to use the formula for Raoult s Law to solve 0 P1 X1P1 We know P10 4218 mmHg We need to know X1 mole fraction of water Need to calculate P1 vapor pressure of solution 0 Glucose is a nonvolatile solute indicating that the solution vapor pressure will be less than that of the pure substance VP solution lt VP pure substance 0 First we need to find the moles of each substance in the solution use 11 for moles of substance 1 g 1 mol 435 n1 water 435 mL 1mL 1802g 1802 1 mol glucose 198 112 glucose 198 g 1802gglucose 1802 Next we need to calculate the mole fraction of water X1 using the above information n 1 241 mol 241 X1 n1n2 241mol110 mol 52 mole fraction of water Now we know X1 and P10 so plug in the values and solve for P1 P1 X1P10 P1 0956 4218 mmHg P1 403 mmHg 3 sig figs Remember we are trying to find how much lower the vapor pressure will be so we want to find AP AP P10 P1 So AP 4218 mmHg 403 mmHg I this is what we set out to find If both components have a measurable vapor pressure in other words are volatile the vapor pressure of the solution becomes the sum of the individual partial pressures Raoult s law still works but is applied in components PA XAPAo and PB XBPBO Total pressure is given by Dalton s law of partial pressures more indepth on this in Ch 5 0 PT I sum of components is the total If a solution obeys Raoult s law it is known as an ideal solution 0 A main characteristic of an ideal solution is that both the solute and the solvent have equivalent IMFs which results in the heat of solution AHsoln always being zero Boiling Point Elevation Designgted as ATE Defined as ATb Tb Tbquot where O ATb is the boiling point elevation how much the boiling point increases 0 Tb is normal boiling point of the specific solution 0 Tb is the boiling point of the pure solvent Because ATb is proportional to the vapor pressure lowering it is ALSO proportional to the concentration or molality of the solution 0 ATb Kbm where Kb is the molal boiling point elevation constant in Cm I The temperature T here is NOT constant Freezin2 Point Depression Designgted as ATE Defined as ATf T Tf where 0 ATf is the freezing point depression O T is the freezing point of a pure solvent 0 Tf is the freezing point of the specific solution Similar to the boiling point ATf is proportional to the concentration of the solution so O ATf Kfm where Kf is the molal freezing point depression constant in Cm Table 132 Molal Boiling Point Elevation and Freezin2 Point Depression Cons tants of Several Common Liguids Solvent Normal Freezing Point Kf in Normal Boiling Point Kb in in C Cm in C Cm Water 0 186 100 052 Benzene 55 512 801 253 Ethanol 1173 199 784 122 Acetic 166 390 1179 293 acid Cyclohexa 66 200 807 279 ne measured at 1 atm Osmotic Pressure Semipermeable membrane allows solvent to pass through but blocks solute Osmosis the net movement of solvent through a semipermeable membrane from a pure solvent or from a dilute to a more concentrated solution 0 Osmotic pressure is the pressure required to stop osmosis Formula for osmotic pressure H MRT where O M is the molarity of the solution 0 R is the gas constant 00821 LatmKmol O T is the absolute temperature 0 H is the osmotic pressure expressed in atm Isotonic two solutions that have equivalent concentrations and osmotic pressures Hypertonic the solution that has the higher concentration of the two Hypotonic the solution that is more dilute that is the solution that is less concentrated Uses of osmotic pressure Preservation of jellyj am 0 When a bacterial cell is in a hypertonic solution the intracellular water tends to move out of the cell to a more concentrated solution causing the cell to shrink and die off a process known as crenation By killing off the bacterial cells the jam is able to be preserved longer than it would otherwise Using Colligative Properties to Determine Molar Mass Theoretically any of the four properties discussed above can be used but in practice only osmotic pressure and freezing point depression show pronounced changes Nonelectrolyte solutions 0 Don t conduct electricity in water or in their molten state 0 Don t produce ions in solution dissolve but do not separate into ions mmple 138 using empirical formul nd freezin2 point depression to nd molar mass and molecular formula Given 966 g sample of a compound with empirical formula C5H4 which is dissolved in 284 g of benzene C6H6 Freezing point of solution is 137 C below that of pure benzene so ATf 137 C We need to find molar mass and molecular formula of the compound after the combination 1 First we calculate molality m by using equation 139 and table 132 ATf Kfm 0 From table 132 we find Kf of benzene is 512 Cm Given this and the ATf we can find m Alf 137 O C 0 m Kf 512 Cm 2 Next we need to find the molar mass We know the solution contains 0268 mol of solute in 1 kg solvent found by calculating molality in step 1 and we know the number of kg in solvent we were given it in the problem 284 g 0284 kg so 0268 mol a 0284 kg 1 kg 00761 mol of solute b And the molar mass is g of compound 966 g C M013 mass mol of compound 00761mol 3 Finally we determine the ratio of molar mass to empirical molar mass to get the molecular formula molar mass 127 g mol 3 empirical molar mass 64 gmol 198 z 2 The empirical molar mass can be calculated using the periodic table b So C5H42 Clng which is naphthalene Colligative Properties of Electrolyte Solutions Electrolytes dissociate into ions in solution 0 Number of solute particles determine colligative properties I ie NaCl dissociates into Na and Cl39 ions so colligative properties of 01 m NaCl solution should be twice as great as those of a 01 m nonelectrolyte solution such as sucrose To account for the dissociation factor we need to modify three of the colligative property equations vapor pressure lowering is not affected by dissociation Boiling point elevation ATb iKbm Freezing point depression ATf inm Osmotic pressure H iMRT i is known as the van t Hoff factor or coefficient and is defined as actual number of particles E solution after dissociation 0 i number of formula units 1 initially dissolved E solution 0 1 formula units are the empirical formula of any ionic or covalent network solid compound used as an independent entity for stoichiometric calculations i is 1 for all nonelectrolytes i is 2 for strong electrolytes such as NaCl and KN03 i is 3 for strong electrolytes such as NaZSO4 and MgClz Ion pair a cation and an anion that are held together by electrostatic forces 0 Comes into play at high concentrations for electrolyte solutions Example 1310 The osmotic pressure of a 0010 M potassium iodide KI solution at 25 C is 0465 atm Calculate the van t Hoff factor for KI at this concentration K1 is a strong electrolyte dissociates into K and 139 ions so it should dissociate completely If so the osmotic pressure would be H iMRT 2 0010M 00821 LatmKmol 298 K 0489 atm Since the imeasured is lower we can say that the solution has some ionpair formation occurring compound does not completely dissociate H 0465atm Useti MRT 0010M00821Lat7mmol 298K slight1y less thanthe theoretical 2 Practice Exercise p 461 The freezing point depression of a 0100 m MgSO4 solution is 0225 C Calculate the van t Hoff factor of MgSO4 at this concentration MgSO4 dissociates into Mg2 and 804239 which makes it an electrolyte So the van t Hoff factor would be calculated using the given molality m 0100 and ATf 0225 C We find Kf in table 132 as 186 Cm We use the freezing point depression formula to get i O ATf inm 0 0225 i1860100 O 0225 0186i
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