Solutions Employee Web Site解决的方案工网站

1、solutionschapter 14types of solution a solution is a homogeneous mixture of two or more substances. solvent medium in which the solute dissolves usually a liquid usually the most abundant species 10 grams of h2o(l) in 25 grams of ch3oh(l) solute substance that is dissolveddiscuss (review) homogeneou

2、s and heterogeneous mixturesspontaneity of the dissolution process dissolving a substance in a solvent (usually a liquid) dissolving with a reaction 2na(s) + 2h2o(l) 2na+(aq) + 2oh-(aq) + h2(g) dissolving without a reactionwe will focus on the latter type of dissolution.)aq(ohc)s(ohc6126oh61262 spon

3、taneity of the dissolution process two major factors determine the dissolution of solutes change in energy, hsolution an _ process (a decrease in energy) favors dissolution. an _ process (an increase in energy) does not favor dissolution. change in disorder or randomness, smixing _ in disorder favor

4、s dissolution. _ in disorder does not favor dissolution. when a substance dissolves the disorder of the system almost always increases.spontaneity of the dissolution process heat of solution, hsolution, primarily depends on the strength of _ between solute and solvent particles. a negative value ind

5、icates that heat is _ favoring dissolution. the larger the magnitude of the value, the more dissolution is favored. factors affecting hsolution solute-solute attractions weak attractions favor solubility (why?) solvent-solvent attractions weak attractions favor solubility (why?) solvent-solute attra

6、ctions strong attractions favor solubility (why?)spontaneity of the dissolution processinput of energy is required to separate solute-solute (step a) and solvent-solvent (step b) attractions. energy is released due to solute-solvent attractions (step c). if the amount of energy released in step c is

7、 greater than the energy absorbed in steps a and b the process is exothermic and favored for dissolution. what if it is less?spontaneity of the dissolution process many solids, however, will dissolve in liquids by endothermic processes. the increase in disorder of the system upon mixing (magnitude o

8、f smixing) is enough to outweigh the endothermic process. the solute particles go from highly ordered in a crystalline solid to highly disordered in solution. most dissolving processes involve an overall increase in disorder. discuss mixing of gasesdissolution of solids in liquids the crystal lattic

9、e energy is the energy change accompanying the formation of one mole of formula units in the crystalline state from constituent particles in the gaseous state. the crystal lattice energy is always _ (i.e. the process is exothermic). m+(g) + x-(g) mx(s) + energy the amount of energy released increase

10、s with the strength of attraction between particles (e.g. ions in the solid) energy released increases with increasing charge density. why? dissolution of solids in liquids solute-solute attractive interactions present in the solid must be overcome for the solid to dissolve in the liquid. this can b

11、e related to the crystal lattice energy mx(s) + energy m+(g) + x-(g) if the magnitude of the crystal lattice energy is small only a small amount of energy is needed to start the dissolution process. this can be thought of as step a in figure 14-1. is this step exothermic or endothermic?dissolution o

12、f solids in liquids energy is also required to break up or expand the solvent molecules (i.e. solvent-solvent interactions). this can be fairly large if the solvent is water. why? this can be thought of as step b in figure 14-1. energy is released, however, due to attractive solute-solvent interacti

13、ons. this can be thought of as step c in figure 14-1. solvation is the process by which solvent molecules interact and surround solute particles or ions. hydration refers to when the solvent molecules are water.dissolution of solids in liquids solvation energy is the energy change involved in the so

14、lvation of one mole of gaseous ions. process is almost always exothermic. equivalent to the sum of steps b and c in figure 14-1. hydration energy is the energy change when water is the solvent.hydration is highly exothermic for ionic or polar covalent compounds. why?hsolution = (solvation energy) (c

15、rystal lattice energy)if hsolution is negative the dissolution process is exothermic.dissolution of solids in liquids a nonpolar solid such as naphthalene, c10h8, does not dissolve in polar solvents such as h2o. it does, however, dissolve in nonpolar solvents. why? “like dissolves like” lets look at

16、 the dissolving process of a soluble salt such as nacl (figure 14-2). the individual ions become solvated (hydrated if the solvent is water) due to electrostatic interactions between the ions and water molecules. most cations (e.g. na+) are surrounded by 4 to 9 h2o molecules.)aq(cl)aq(na)s(nacloh2 d

17、issolution of solids in liquids an increasing charge density (charge/radius ratio) increases the hydration energy (heat of hydration). table 14-1kj/mol 4750- 6.00 0.50 alkj/mol 2160- 2.78 0.72 cukj/mol 1650- 2.02 0.99 cakj/mol 351- 0.75 1.33 khydration ofheat chg/radius () radius ion+3+2+2+dissoluti

18、on of solids an increasing charge density, however, also increases the magnitude of the crystal lattice energy. for many salts that possess low-charge species (e.g. nacl) the hydration energy and crystal lattice energy nearly cancel each other. demo: the dissolution of nh4no3 is endothermic. what ca

19、n you say about the heat of solvation and crystal lattice energy? the dissolution of ca(ch3coo)2 is exothermic. with large charge densities the magnitude of the crystal lattice energy increases more than the hydration energy. this is the reason why the dissolution process for many solids that contai

20、n highly charged ions (e.g. cr2o3) is endothermic. many are insoluble. look at equation.dissolution of liquids in liquids miscibility is the ability of one liquid to dissolve in another. if two liquids are miscible one liquid completely dissolves in the other. types of attractive forces to consider

21、when determining miscibility solute-solute attractions (step a) solvent-solvent attractions (step b) solute-solvent attractions (step c)when will two liquids be the most miscible?dissolution of liquids in liquids like dissolves like determines miscibility polar liquids tend to dissolve in other pola

22、r liquids h2o and ch3ohdiscuss in terms of the types of interactions. what are some other liquids that will dissolve in water?dissolution of liquids in liquids will hexane, c6h14, dissolve in water, h2o? why or why not? explain in terms of types of interactions. will hexane dissolve in gasoline whic

23、h is nonpolar? why or why not? what are the strengths of attractive interactions?dissolution of gases in liquids like dissolves like holds fairly well for gases dissolving in liquids. polar gases dissolve in water polar gases can also react with water)aq(hf)g(hfoh2 )aq(br)aq(oh) l(oh)g(hbr32dissolut

24、ion of liquids in liquids nonpolar gases (e.g. o2) dissolves to a limited extent in h2o due to dispersion forces dissolved o2 is responsible for keeping fish alive co2, a nonpolar gas, dissolves in water appreciable because it reacts with h2oco2(g) + h2o(l) h2co3(aq)h2co3(aq) h+(aq) + hco3-(aq)hco3-

25、(aq) h+(aq) + co32-(aq)what happens to the acidity of water when co2(g) is dissolved?because gases have such weak solute-solute attractions, gases dissolve in liquids exothermically.rates of dissolution and saturation rate of dissolution of a solid increases if the size of the solid particles is dec

26、reased (e.g. ground to a powder). why? sugar cubes versus granulated sugar the amount of solid in a liquid will increase until the rate of dissolution equals the rate of crystallization. the opposing processes are in dynamic equilibrium. the solution is saturated. it is holding all the solute it can

27、 at a given temperature.demo: nacl crystals increasing in size in a saturated solution.particlesdissolvedsolidndissolutioationcrystallizrates of dissolution and saturation saturated solutions have an established equilibrium between dissolved and undissolved particles. nacl(s) na+(aq) + cl-(aq) the f

28、orward and reverse rates are equal. supersaturated solutions contain higher-than-saturated concentrations of solute. how is this possible? a supersaturated solution is in a metastable state. there needs to be mechanism to start the crystallization. hot packs, sodium acetate in h2oeffect of temperatu

29、re on solubility lechateliers principle states that a chemical system responds in a way that best relieves the stress or disturbance. exothermic dissolution endothermic dissolutionmol/kj8 .48)aq(br)aq(li)s(libroh2 )aq(mno)aq(kmol/kj6 .43)s(kmno4oh42 effect of temperature on solubility what will happ

30、en to the solubility if the temperature is changed? the system will respond according to lechateliers principle. adding heat to an exothermic process _ dissolution. why? adding heat to an endothermic process _ dissolution. why? for most solids dissolving in liquids the process is endothermic. most g

31、ases dissolve in liquids by exothermic processes. what does this mean if the temperature is increased?demo: dissolve nach3coo. is this process endothermic or exothermic? what will happen if the temp. is increased? cool the mixture. what type of solution results?effect of temperature on solubility in

32、creasing the temperature increases the solubility of most solids in h2o. are these processes exothermic or endothermic? na2so4 is the only exothermic processeffect of pressure on solubility changing the pressure has no appreciable effect on the solubilities of solids or liquids in liquids. pressure

33、changes have large effects on the solubilities of gases in liquids. carbonated beverages scuba divers get the bendseffect of pressure on solubility henrys law expresses that the concentration of dissolved gas is directly related to the pressure of the gas above the solution. pgas = kcgas pgas = pres

34、sure of the gas above the sollution k is a constant for a particular gas and solvent at a specific t cgas = the concentration of dissolved gas (molarity)molality and mole fraction previously, we covered molarity and weight precent to express concentration. what are they? molality number of moles of

35、solute per kilogram of solvent what is the molality of a solution that contains 142 grams of ch3oh in 315 grams of water?solventramslogkiofnumbersolutemolesofnumbermolality molality and mole fraction calculate the molality and the molarity of an aqueous solution that is 10.0% glucose, c6h12o6. the d

36、ensity of the solution is 1.04 g/ml. 10.0% glucose solution has several medical uses. 1 mol c6h12o6 = 180 g calculate the molality of a solution that contains 7.25 g of benzoic acid c6h5cooh, in 200 ml of benzene, c6h6. the density of benzene is 0.879 g/ml. 1 mol c6h5cooh = 122 gmolality and mole fr

37、action mole fraction is the number of moles of one component per moles of all the components. the sum of the mole fractions (xa + xb) = 1what are the mole fraction for methanol and water in the previous problem?what are the mole fractions of glucose and water in a 10.0% glucose solution?)componenstw

38、o(bmol.noamol.noamol.noxacolligative properties of solutions colligative properties depend solely on the number of particles dissolved in the solution and not the kinds of particles dissolved. physical property of solutions four kinds that will be discussed vapor pressure lowering freezing point dep

39、ression boiling point elevation osmotic pressurelowering of vapor pressure and raoults law addition of a nonvolatile solute to a solution lowers the vapor pressure of the solution fewer solvent molecules present at the surface since some solute molecules occupy the space. as a result, molecules evap

40、orate at a slower rate raoults law describes this effect in ideal solutions where xsolvent represents the mole fraction, is the vapor pressure of the pure solvent, and psolvent is the vapor pressure of the solvent in the solution. figure 14-90solventsolventsolventpxp0solventplowering of vapor pressu

41、re and raoults law the change in vapor pressure of the solvent in solution can be expressed in terms of the solute mole fraction.where psolvent is the change in vapor pressure of the solvent in the solution.this relationship assumes ideal solutions and that the solute is nonvolatile (i.e. has not va

42、por pressure). this is raoults law.0solventsolutesolventpxplowering of vapor pressure and raoults law determine the vapor pressure of a solution, at 25c, that is made by dissolving 5.00 grams of sucrose, c6h12o6, in 15.0 grams of water. the vapor pressure of pure water at 25c is 23.8 torr (appendix

43、e). determine the vapor pressure of a 5.25 molal aqueous sucrose solution at 47c.determining the vapor pressure of a two-component system both components are considered as volatile and contribute to the total vapor pressure. a solution of hexane and heptane each component behaves as if it were pure.

44、 therefore, the vapor pressure would be a sum of its components.ptotal = pa + pbfrom raoults lawtherefore, 0bbb0aaapxpandpxp0bb0aatotalpxpxpdetermining the vapor pressure when a two-component system the left-hand side corresponds to pure b (xb=1) and the right-hand side to pure a (xa=1). which is mo

45、re volatile? notice that the black line is always equal to the sum of pa and pb.determining the vapor pressure when a two-component system at 45c, the vapor pressure of pure heptane is 112 torr and the vapor pressure of pure octane is 36 torr. the solution contains two moles of heptane and three mol

46、es of octane. calculate the vapor pressure of each component (heptane and octane) and the total vapor pressure above the solution. what is the composition (in mole fractions) of the vapor above the solution? the volume above the solution is equal to 1.50 l. assuming that the gases behave ideally cal

47、culate the amount of heptane and octane in the gas phase.also, do example 14-5 for practice.boiling point elevation the boiling point of a liquid is the temperature at which its _ _ equals the external pressure. it was stated with raoults law that the addition of a nonvolatile solute decreases the v

48、apor pressure. therefore, a _ temperature must be acquired to cause the liquid to boil (vapor pressure equals atmospheric pressure). the amount of boiling point elevation depends on the number of moles of solute particles dissolved in the solvent.boiling point elevation the boiling point change can

49、be expressed astb = kbm tb is the change in boiling point (add) kb is the molal boiling point constant (table 14-2) kb corresponds to the change in boiling point by a one-molal solution m is the molality of the solutionwhat is the normal boiling point of a 2.50 m glucose, c6h12o6, solution?boiling p

50、oint elevation do this later. predict the boiling point elevation if 33.0 grams of nacl is added to 100 grams of water? note: calculate moles of solute particles not moles of solute.freezing point depression addition of a nonelectrolyte lowers the freezing point according to the following expression

51、tf=kfm tf is the change in the freezing point (subtract) kf is the the molal freezing point depression constant (table 14-2) m is the molality of the solutionupon freezing, the solvent solidifies as a pure substance. solute molecules make it more difficult for the solvent molecules to come together

52、and freeze. a lower temperature must be acquired to freeze the solvent.freezing point depression calculate the freezing point of a 2.50 m aqueous glucose solution. cacl2(s) is commonly added to melt snow on roads at lower temperatures. what is the freezing point when 20.0 grams of cacl2 is dissolved

53、 in 100 grams of h2o at 20c? for comparative purposes, calculate the lowering of the freezing point when 20.0 grams of nacl is dissolved in 100 grams of h2o at the same temperature. antifreeze (hoch2ch2oh) addition to water is another example of freezing point depression.changes in boiling and freez

54、ing point temperatures the relationships for calculating the change in freezing and boiling points are very similar. it is essentially the same effect. the difference is in the size or magnitude of the effect which is indicated by the constants, kf and kb.determination of molecular weight from freez

55、ing point depression or boiling point elevation these colligative properties (especially freezing point depression) can be used to determine molecular weight of a solute. kf and the amount of solvent in kilograms is usually known. the freezing point lowering was measured. therefore, the molality can

56、 be determined, and the moles of solute can be calculated. the molecular weight can then be determined.determination of molecular weight from freezing point depression or boiling point elevation a 37.0 g sample of a new covalent compound, a nonelectrolyte, was dissolved in 200 g of water. the result

57、ing solution froze at -5.580c. what is the molecular weight of the compound? either ethylene glycol (c2h6o2) or propylene glycol (c3h8o2) can be used to make an antifreeze solution. a 15.0 gram sample of either ethylene glycol or propylene glycol dissolved in 50.0 grams of water lowers the freezing

58、point to 7.33c. which glycol is added to the solution? the solutions behave ideally at these concentrations.colligative properties and dissociation of electrolytes since a colligative property only depends on the number of solute particles in a given mass, electrolytes have a larger effect on boilin

59、g point elevation and freezing point depression. how many moles of solute particles will be produced from one mole of sucrose (c6h12o6)? how many from 1 mole of mgcl2? as a result, the boiling point elevation and freezing point depression are larger for a given molar quantity of a typical electrolyt

60、e.colligative properties and dissociation of electrolytes solute particles, however, are not randomly distributed in an ionic solution. this causes the boiling point elevation and freezing point depression to be not as large. ionic solutions behave nonideally. the ions start to undergo association,