Chapter 16 Clligative prperties f slutins In the last chapter we dealt with pure slids and liquids. In this chapter we will deal with what happens t a liquid when there is sme material disslved in it. While the emphasis will be n aqueus slutins, the cncepts and equatin yu learn will apply t any slutin aqueus r nnaqueus. In the last chapter yu learned abut hw all liquids have a vapr pressure and hw this vapr pressure varies with temperature. Mst if nt all f the prperties we deal with in this chapter will be based n the vapr pressure f the slvent 16-1 Mlality and Mle Fractin S, we just finished with cllids, let s get back t what happens in an aqueus slutin Ins interact strngly with water, s get hydrated ins, these hydrated ins are unifrmly dispersed at the mlecular level, s there are n clumps r particles like we had fr cllids Plar nnelectrlytes als interact strngly with water, s again they get unifrmly dispersed thrughut the slutins It is these strng interactins between the slute and the slvent that affect many f the bulk prperties f the slvent. Interestingly, it is nt the interactin per se that has the effect, but simply the rati f slute t slvent mlecules The rati f slvent t slute mlecules is a measure f cncentratin s let s review ur measures f cncentratin: Mlarity (M) = mles slute/liter slutin mass % = g slute/g slutin x 100% mle fractin = n 1/(n 1 + n 2 +...) Nte: we had mle fractin in gases, but the same idea applies t slutins While yu are the mst used t using mlarity as a measure f cncentratin, there are sme prblems with it. 1.) Yu need the special vlumetric glassware t make it in the lab. And 2.) As the temperature changes, the vlume f the slutin will expand and cntract, s as T V and M Fr the average chemist ding simple stichimetry these changes are s small that they are insignificant, but they can be imprtant in sme f these clligative prperties we will talk abut here.
Since clligative prperties depend n the rati f slvent t slute mlecules, mle fractin is a better way t measure this kind f cncentratin. The nly prblem is that mst f the slutins we will be dealing with are mstly water and very little slute, s we will have water smething like.999999 and slute maybe.000001 mle fractin. And this is nt really cnvenient S I will intrduce yet anther measure f cncentratin Key equatin/definitin: Mlality (m) = mles slute/kg slvent Practice prblems EXAMPLE Let s disslve 5 grams f acetic acid (CH3COOH) in 100 grams f water. If the resulting slutin has a ttal vlume f 105 mls, calculate the fllwing: Mlarity (M) Mles/liter Need mle f acetic acid = 5/((12.01x2)+(16.00x2)+(1.008x4) = 60.05 =.0833 Need liters 105 ml x 1L/1000mL =.105 L M = ml/l =.0833ml/.105L =0.793M Mass Percent = mass slute/mass/slutin x 100% = 5/(100+5) x100% = 4.76% Mle fractin Mle acetic acid/ttal number f mles Ml acetic acid (frm abve) = 5/60.05 =.0833 Ml water =100 g/ 18 = 5.56 ml =.0833/(5.56+.0833) =.0148 Mlality(m) = ml slute / kg f slvent ml slute (frm abve) =.0833 kg f slvent = 100 g x 1kg/1000g =.1 kg =.0833ml/.1kg =.833m Nte mlality and mlarity nt the same!
One additinal detail The clligative prperties we will deal with depend n the number f slute particles in the slutin. Hw many mles f particles d yu get frm 1 mle f CH CH OH? 1 mle f NaCl? 1 mle f FeCl? 3 2 3 T take int accunt the number f particles(ins) released frm an inic cmpund we will use the term clligatve mlality Key term Clligative mlality m c = im where m is mles f the uninized electrlyte and i is the number f particles (r ins) prduced where than electrlyte disslves in slutins Practice prblem (Clicker questin) What is the clligative mlality f a.1m slutin Magnesium phsphate 16-2 Rault s Law When a nnvlatile slute is disslved in a vlatile slvent, the vapr pressure f the slvent decreases Last chapter we saw that the vapr pressure was an equilibrium between the rate f evapratin f the slvent and the rate f cndensatin f the slvent frm the vapr Hw des this change when yu add a slute? The rate f evapratin f the slvent is directly prprtinal t the number f slvent mlecules n the surface f the liquid. When yu add slute mlecules yu lwer the cncentratin f slvent mlecules at the surface f the liquid, and this decreases the rate f evapratin. By decreasing the number f mlecules getting int the vapr, yu lwer the vapr pressure f the slvent (Figure 16.1) We can accunt fr this decrease in pressure using an equatin called Rault s law Key equatin: Rault s Law: VP = VP slutin slvent slvent The amunt that the vapr pressure f the slvent has decreased is called the 0 vapr pressure lwering = VP - VP pure slvent slutin =VP pure slvent - pure slvent x VP pure slvent = VP Pure slvent (1- pure slvent) but, fr a 2 cmpnent system + = 1; s 1- = s: slvent slute slvent slute
Key equatin: Vapr pressure lwering = VP slute slvent Practice prblem: The vapr pressure f water at 80 C is 355 Trr Calculate the vapr pressure f an aqueus slutin made by disslving 50 grams f ethylene glycl (C2H6O 2) in 50 grams f water. What is the vapr pressure lwering f water in this slutin? VP slitin = slvent x VP slvent slvent = n slvent /(n slvent + n slute) n slvent = 50 g water /18g/mle = 2.78 mles n slute = 50 g /[(2(12)+6(1) + 2(16)] = 50/62 =.806 mles slvent = 2.78/(2.78+.806) =.775 VP =.775(355) = 275 Trr slutin Pressure lwering = slute VP slvent slute =.806/(.806+2.78) =.225 VP lwering =.225 x 355 Trr = 80 Trr Check 355-275 = 80 Trr Answer checks 16-3 Biling Pint Elevatin In the last chapter we fund that the biling pint f a slutin is the pint temperature at which the vapr pressure f the slutin = the atmspheric pressure. S what happens t the biling pint f the slutin we just calculated in the previus prblem? Since we have lwered the VP f the slutin, yu will have t heat the slutin t a higher temperature t get it t bil. Key definitin: The biling pint elevatin is the amunt by which the biling pint f a slutin exceeds the biling pint f the pure slvent = T -T Sketch right half f Figure 16.4 n bard b b In slutins where m c 1.0 m, the mle fractin f the slute is directly prprtinal t the f the slute S if VP lwering is prprtinal t, then it is als prprtinal t m c Figure 16.6 At 1 atm the vapr pressure lwering is directly prprtinal t the change in the biling pint s
T b - T b is prprtinal t m c slute as well. S Key equatin Biling pint elevatin equatin: T b - T b = Kbmc slute where K is called the biling pint elevatin cnstant b K b fr varius slvents are shwn in table 16.2 Practice prblem Let s calculate the biling pint fr that ethylene glycl slutin we wrked with in the previus example T b -T b = Kbmc slute X-100 =.513 C kg/ml x m c slute m c slute = mles slute /kg slvent In the previus example we had 50 grams f ethylene glycl (C2H6O 2) in 50 grams f water, s kg f slvent =.05 kg water EG had a mlar mass f 62 s 50/62 =.806 mles m =.806/.05 = 16.1 ml/kg i=1 (nnelectrlyte), s m c = 1(16.1) = 16.1 X-100 =.513 C kg/ml x 16.1 ml/kg = 8.27 C X-100 = 8.27 X = 108.27 C 16-4 Freezing Pint Depressin S if the biling pint ges up, why des the freezing pint g dwn? Lets think a little mre abut the vapr pressure f a slid and a liquid and what happens when smething melts r freezes. Sketch like 10.45 f Zumdahl n bard Let s lk at the VP f water and ice just abve and belw the MP, 0 Ntice that the VP f slid is lwer <0 and the VP f the liquid is lwer >0. And they equal each ther at zer s ur definitin f nrmal melting pint where slid and liquid have equal vapr pressures makes sense. Nw try this thught experiment with a set up like 10.46 frm Zumdahl if Temp < this pint, then slid has lwer vp, and liquid will mve r slid and freeze. If Temp > this pint, then liquid has lwer VP and slid will mve t liquid and melt if Temp = this pint, then in equilibrium and liquid and slid can cexist. Nw return t sketch like 10.45 -Z What happens when the VP f the slutin get lwer? The liquid curve drps Where is the intersectin f slid and liquid s they can be in equilibrium? At a lwer T!
If yu want yu can nw return t Mcquarry figure 16.4 and pint ut regin at bttm f liquid phase yu have been discussin. If yu d this be sure t emphasize that the line between slid and liquid in this figure has a visible slpe t the left, which is crrect, but the line is mre nearly vertical s the ÄT yu get at the bttm tip f the liquid phase is ~ same as the ÄT @ 1 atm nrmal Using the same chain f lgic we used fr the biling pint elevatin, yu can find that the freezing pint depressin is als prprtinal t the m c f the slute s: Key Equatin Freezing pint elevatin equatin: where K f is called the Freezing pint depressin cnstant Ntice that this equatin clsely resembles the biling pint elevatin equatin, but the tw T s have been switched arund. Practice prblem 1 Let s calculate the biling pint fr that ethylene glycl slutin we wrked with in the previus examples 0-X = 1.86 C kg/ml x m c slute m c slute = mles slute /kg slvent In the previus example we had 50 grams f ethylene glycl (C2H6O 2) in 50 grams f water, s kg f slvent =.05 kg water EG had a mlar mass f 62 s 50/62 =.806 mles m =.806/.05 = 16.1 ml/kg i=1 (nnelectrlyte), s m c = 1(16.1) = 16.1 0-X = 1.86 C kg/ml x 16.1 ml/kg = 29.9 C 0-29.9=X X = -29.9 C Practice prblem 2 Which slutin will have the lwer freezing pint, 10 g f NaCl, 10 g f FeCl 3 r 10 g f Ethylene glycl in 100 mls f water? All three slutes use the same equatin and same K f, s yu need t find the slutin with the largest m c 10 g f EG is 10/62 =.16 m NaCl has a mlar mass f 58.44 s 10/58.44 =.171 m, but NaCl has tw ins s the I r van t Hff factr is 2 s.171 x 2 = m =.342 c
FeCl 3 has a mlar mass f 55.84+3(35.45) = 162.19 mles f FeCl 3 = 10/162.19 =.0617 but I = 4 s m = 4(.0617) =.247 c Lking at the three m c, NaCl will be have the lwest melting pint Clicker questin Which slutin will have the highest biling pint 20 g f Ethylene Glycl, 20 g f NaCl r 20 g f FeCl in 100 ml f water? 3 The bk has an interesting discussin n antifreeze and using salt t melt ice, I will leave that fr yu t read n yur wn I will spend a little mre time n the van t Hff factr. When I first intrduced it a few pages ag I said the I factr was the number f ins yu get frm a strng electrlyte when it ges int slutin. But what abut a weak electrlyte, ne that desn t cmpletely inize? Well here yu can t knw what the I factr is withut a few experiments, but the experiments are useful because with can turn the clligative prperties equatins arund t experimentally determine hw much a weak electrlyte inizes Practice questin: If a.0500m slutin f acetic acid has a freezing pint f -0.095 C,what is the van t Hff factr (I) fr Acetic Acid at this temperature 0-(-.095) = 1.86 m c.095/1.86 = m c =.0512 m c = i m.0512 = X(.0500) X=i=.0512/.0500) = 1.024 If Acetic acid was did nt inize, i=1, if it was a strng acid and inized cmpletely i=2. The % that it has inized = (1.024-1)/1 x 100 = 2.4 % 16-5 Osmtic Pressure Has anybdy ever had a bag f salt that yu use fr thrwing n the sidewalk site in the garage fr a few years? What happens t it? It picks up water. Why? The bag f salt in an pen garage is nt a well defined clsed system s let s simplify it Figure 16.11
Why des the water level f the pure water g dwn and the salt water g up? 2 ways t explain. Bk way:as cme t equilibrium rate f cndensatin f water frm air ver bth is the same. But rate f evapratin f the salt water is lwer because there are less mlecules f water /unit surface area. S desn t lse water as fast s net effect is t gain the water that the pure water is lsing. My way: VP f Pure slvent > VP f slutin. Gas ges frm high pressure t lw pressure, s mves int the salt water with the lwer pressure This wrked because the vapr phase presented a barrier between the tw slutins that the salt culd nt pass thrugh. Rather than using a vaprbarrier, let s use a physical barrier, a semi-permeable membrane Key definitins: A semipermeable membrane is a material that will let slvent mlecules pass between tw slutins, but will nt allw slute mlecules t pass. Osmsis is the name fr the spntaneus passage f a slvent thrugh a semipermeable membrane frm a dilute slutin int a mre cncentrated slutin Let s use a semipermeable membrane t separate these tw slutins Figure 16.12 Nw the same thing happens water passes frm the pure water side slutin side. But the way we have it set up here, ntice that we begin t get a pressure build up because there is a difference in the heights f the tw sides. Water will cntinue t mve frm the pure water side t the slutin side until the water pressure frm the height difference equals the vapr pressure difference between the slvent and slutin sides Key defintin: Osmtic pressure is the pressure that is required t prevent the passage f slvent frm the dilute side f a semipermeable membrane t the cncentrated side f that membrane van t Hff fund that Key Equatin Ð = RTM c Where R is the gas cnstant in L atm/k ml r L bar/k ml T is in K and M is the clligative Mlarity f the slutin = im c
Practice Prblem 1. What is the smtic pressure f.55m NaCl at 25 C in units f atm? Ð=RTM c =.08206 L atm/k ml x 298 x 1.1 ml/l (i=2 fr NaCl) = 26.9atm Practice Prblem 2 Bichemists use smsis t calculate the mlecular mass f an unknwn. Say we just purified a new prtein, say that makes hair grw in men. We want t knw it s mlecular weight. First we disslve exactly 1.00 mg in 1.00 ml f slutin. At this pint we dn t knw the mlarity f the slutin because we dn t knw the MM f the prtein. We can, hwever, put this slutin n ne side f a semipermeable membrane and measure its smtic press and experimentally fine the mlarity, and hence the MW f the prtein. Let s try it. We make ur slutin up and find it has an smtic pressure f 1.35 trr -3 Our smtic pressure is then 1.35 trr x 1 atm/760 trr = 1.78x10 atm when the experiment is dne at rt (298 K) Using the equatin: ð = M c RT -3 1.78x10 =?x.08206 Latm/K ml x 298K -3-6? = 1.78x10 /.08206/298 = 72.8x10 ml/l We will assume ur prtein desn t inize s i=1 and M C =M Our slutin was 1 mg/ml = 1 g/l therefre -6 72.8x10 ml = 1g/MW -6 MW = 1g/72.8x10 ml MW = 13,700 MW Nte this is a small prtein!
Reverse Osmsis S far we have talked abut the pressure that builds up n the cncentrated side f the membrane spntaneusly as the water mves frm the lw cncentratin side t the high cncentratin side. But what if we put a + pressure n the high cncentratin side? Nw we are supplying energy t d wrk n the system, the result is that we can push pure water ut f the ther side f the membrane. This is hw reverse smsis water purificatin wrks. Osmtic pressure in cells The membrane arund a cell is a semipermeable membrane. The inside f a cell has high cncentratins f lts f prteins, ins and small mlecules that cannt pass thrugh this membrane, s there is always an smtic pressure inside a cell. When a bilgist suspends cells in a slutin he r she has t be careful f the salt cncentratin f that slutin (Figure 16.18) t try t match the smtic pressure f the cells. If the salt is t dilute, yu make a hyptnic slutin, and the cells begin t swell. If yu add t much salt, yu make a hypertnic slutin and the cells begin t cllapse. An istnic slutin is ne in which the smtic pressure f the medium matches the smtic pressure f the cell s the cell is stable. If the salt cncentratin is t lw yu can actually make cells explde. Mst animal cells will burst if Ð > 7.5 bar. Plant cells, with a rigid cell wall can sustain higher pressures 16-6 Ideal Slutins Slutins where bth slute and slvent have a vapr pressure S far we have dealt with slutins where the slvent was vlatile and the slute was nnvlatile. Let s up the ante. What happens when bth the slvent and the slute have a vapr pressure (are vlatile). Fr this discussin we will nly deal with Rault s law, and will nt lk at BP r FP In applying Rault s law befre I assumed that the slute was nnvlatile, s nly s slute cntributed t the VP ver the slutin. This might apply t smething like NaCl, but definitely desn t apply t smething like alchl which als has a vapr pressure. When the slute is vlatile we use a mdified frm f Rault s Law: Key Equatin: Rault s Law fr an ideal slutin when bth slute and slvent are vlatile P = P + P = P + P ttal A B A A B B And a plt f P vs mle fractin lks like figure 16.20
Ideal and Nn-ideal slutins Just as we had ideal and nn-ideal gases, we can have ideal and nn-ideal slutins. A nn-ideal slutin is a slutin that des nt bey Rault s law, an ideal slutin des. Yu usually bserve ideal slutins when the tw liquids in the slutin are similar in structure, like Benzene and Tluene. Slutins can act in a nn-ideal manner fr many reasns, let s lk at tw 1. Negative deviatins frm Rault s law If the slute and the slvent have strng interactins, like they frm hydrgen bnds, then the slvent will hld n t the slute mre tightly, and the slute will have a lwer vapr pressure than expected, s the net pressure will be lwer than expected and have a negative deviatin frm Rault s Law (Make sketch n bard + figure 16.19a ) What ther prperty can we assciate with a strng slute-slvent interactin? 2. Psitive deviatins frm Rault s law (Sketch n bard +16.19b) On the ther hand if yu mix tw liquids and they give yu a higher vapr pressure than expected, this indicates that the slute-slvent interactins are weaker than the slute-slute and slvent-slvent interactin. This means that the slute mlecules are bund less well t the slutin than they are t the pure state s they clump tgethr and try t escape, and the VP f the slute and the ttal VP will be higher. This is a Psitive deviatin frm Rault s Law An example is a mixture f Ethanl and Hexane. Plar ethanl is fine by itself, nn-plar hexane fine by itself. Plar and nn-plar wn t interact s dn t pick up any new interactins when they mix, and, in fact, yu lse sme f the H- bnd interactins in the pure ethanl as it makes rm fr the hexane. Practice calculatins 1. Calculate the vapr pressure f a slutin that has an equal number f mles f benzene and tluene, given that the VP f Benzene is 183 Trr and that f Tluene is 59.2 If mles f Benz = mles f Tl, then = =.5 Benz Tl VP Slutin = VP Bez + VP Tl VP Benz =.5(183) = 91.5 Trr VP Tl =.5(59.2) = 29.6 Trr VP = 91.5 + 29.6 = 121.1 Trr Slutin
2. What is the mle fractin f Benzene and Tluene in this vapr? Ging back t the chapter n gases Benz = VP benz/(vp Benz + VP tl) = 91.5/121.1 =.756 Tl = VP Tl/(VP Benz + Vptl) = 29.6/121.1 =.244 Distillatin Ntice in the abve example. The f benzene n the liquid was.5, but it the vapr it is nw.756. Yu have significantly increased the purity f the benzene This is hw a still wrks Yu start with a wine slutin that is say 6% alchl. The VP f alchl is almst twice that f water s, when yu bil the slutin the vapr is enriched in alchl. When yu cndense the vapr with a still, the resulting slutin has a higher alchl cntent (and has becme Brandy r Cgnac) Yu can make the still mre efficient by packing with with inert materials that will add t the number f times the vapr is cndensed and re-evaprated befre the final vapr is taken frm the still head. This is a lab yu will prbably t with Dr. Dixsn if yu take rganic chem next year. Even with the best still in the wrlds, the best purity yu will ver get fr yur ethanl water mixture is 95.6% alchl This is because the water alchl mixture is nt an ideal slutin, and the slvent-slute interactins make the f vapr and liquid equal t each ther when yu reach 95.6% alchl. Key definitin: Azetrpe: A slutin that distills withut a change in cmpstin If yu want t find ut hw t get yur hch up t 200 prf alchl yu will have t read yur bk! 16-7 Henry s Law Nw let s lk at the slubility f gases in liquids Let s d a thught experiment draw n bard similar t fig 11.5-Zumdahl At start have a clsed cylinder with disslved gas in equilibrium with gas phase. What happens if press dwn n cylinder? Smaller vlume ver liquid, higher pressure, sme f the gas mlecules are frces back int the liquid, s slubility must have increased!
Thus we find that the cncentratin f a gas in a liquid is directly prprtinal t the partial pressure f the gas ver the liquid. This is summarized in Henry s Law Key Equatin: Henry s Law P gas = k h Mgas Where P gas is the partial pressure f the gas M gas is the mlar cncentratin f the gas in the slutin and k h is the Henry s Law cnstant fr that gas The Henry s law cnstant is different fr each gas, each slvent and every temperature Practice Prblems: Part 1. A sft drink is canned s that the bttle at 25 C cntain CO 2 gas at a pressure f 5 atm ver the liquid. When yu pen the can the pressure naturally drps t 1 atm, but further, because the air is nt made f CO2the partial 4 pressure f CO 2 is nly 4.0 x 10 atm (.0004 atm). Calculate the cncentratin f CO 2 in the liquid befre and after the bttle is pened. Needed fact, the k fr Henry s law fr CO is 29 L atm/ml at 25 C (table 16.3) Henry s law P gas = k h Mgas In sealed can, P=5, k=29l atm/ml s C = 5=29(X) X=5 atm /29Latm/ml =.172 ml/l =.172M 2 In the pen can we have P =.0004= 29(X) X=.0004/29Latm/ml = 1.38x10-5 M Part 2. Hw big is yur belch frm a can f sda? a can f sda is 355 ml =.355 L; s in full can yu have.156 ml/l x.355 L =.055 mles CO 2 After yu pen the cab yu have: -6.0000138 x.335 = 4.6x10 mles f CO Remaining 2 That means yu have released.055 -.0000046 ~.055 mles f CO 2 Using Gd ld PV=nRT; V = nrt/p =.055 (.08206)298/1 = a 1.3 L belch, r serval smaller belches Agin the chapter ends with sme interesting discussin n the Bends and hw He/ O gas mixtures are used in deep diving t avid the bends. 2