Ion-Dipole Interaction

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1 제 14 장 : 용액의물리적성질 용액 : 두종류이상의물질이균일하게섞여있는혼합물 ( 단일상 ) 용매 : 용액중더많이존재하는성분 ( 예외 : 물은적게존재해도항상용매 ) 용질 : 용액중더조금존재하는성분 혼합물의특성이균일하지않으면불균일 (heterogeneous) 혼합물이다. 상이두개이상. 혼합물전체가균일한특성을보이면균일 (homogeneous) 혼합물혹은용액이라한다

2 용액의형태 용질용매용액 gas O 2 in a gas N 2 Air Gas CO 2 in a Liquid H 2 O Carbonated Water liquid H 2 O in a Gas Air Fog liquid EtOH in a Liquid H 2 O Wine Liquid Hg in a Solid Ag Dental-filling Solid NaCl in a Liquid H 2 O 식염수 Solid Ag in a Solid Au 14 Karat gold

3 용해의원리 Like Dissolves Like 극성이상이할때 : 불용 극성이유사할때 : 가용 Immiscible 두용액이서로석이지않음 Miscible 두용액이서로가용성

4 용해의원리 - 용질과용매의분자구조가비슷하거나같은원자단이있으면잘용해됨. ( 예 ) 설탕 (-OH 있음 ), 에탄올 (C 2 H 5 -OH) 등은물 (H-OH) 에잘용해 - 전기적성질 ( 극성및비극성 ) 이비슷하면잘용해 좌 : 물 ( 극성 )- 나프탈렌 ( 비극성 ) : 안섞임우 : 나프탈렌 ( 비극성 )- 벤젠 ( 비극성 ) : 섞임 극성은극성에비극성은비극성에섞인다.

5 . 1 이온결정은물과같이극성이큰용매에잘녹는다 2 나프탈렌과같은무극성분자는벤젠과같은무극성용매에잘녹는다 3 물과에탄올은 -OH 에의한수소결합때문에서로잘섞인다 4 극성분자는극성용매에, 무극성분자는무극성용매에잘녹는다.... 용해의원리 용해도곡선고체는온도상승, 대부분용해도상승기체는온도상승시, 용해도감소

6 용해도 (Solubility): 결합의관점 물질의용해결합력 : 용질 - 용매 > 용매 - 용매 + 용질 - 용질

7 용해도 : 열역학적관점 반응의엔탈피변화 : DH = DH1 + DH2 + DH3 DH > 0 ( 흡열 ); DH <0 ( 발열 )

8 NaCl 의용해의예 ( 고체 / 액체 )

9 용해도에영향을주는인자 1. 용질 / 용매극성 - Like dissolves like 2. 온도 - (Inter Molecular Forces) i) 고체 / 액체 - 온도증가시주로증가운동에너지증가 충돌 ( 용매 / 용질 ) 증가 ii) 기체 온도증가시감소 운동에너지증가로기체탈출 ] ( 기체가용해될때발열반응이진행되기때문에온도가낮을수록기체의용해도는증가한다.) 3. 압력 - i) 고체 / 액체 거의영향없음 ii) 기체 압력증가시용해도증가 ( 헨리의법칙

10 용액속의기체 온도 (Temperature) 효과 : 기체의용해도는온도가높아지면감소한다 ( 운동에너지증가 ). 온도가높으면용매, 용질 ( 기체 ) 모두운동에너지가높아지나기체는액체 ( 용매 ) 의인력을끊고나갈수있게된다. 고온 저온 기체의부피는압력에반비례하고녹는기체의양은압력과비례 ( 낮은압력에서적용 ) 헨리의법칙이잘적용되는기체 : H 2, O 2, N 2, CO 2 => 무극성분자헨리의법칙이잘적용되지않는기체 : HCl, NH 3, SO 2 => 극성분자

잠수마다 1 기압씩증가 : 34 ft : 2 기압 ; 68 ft : 3 기압 만약 0 o C 1 기압에서 CO 2

11 용액속의기체 압력 (pressure) 효과 : 기체의용해도는압력이높아지면증가한다 ( 헨리의법칙, Henry s law). c = k P 심해에서 N 2 의용해도증가와급상승시 N 2 의부피증가로인한혈관파손병--잠수병 ( 케숑병 ) 34ft 잠수마다 1 기압씩증가 : 34 ft : 2 기압 ; 68 ft : 3 기압 만약 0 o C 1 기압에서 CO 2 의용해도가 0.348g/100 ml 라면, 2.5 기압에서의용해도는? 0.348g/100ml * 2.5 = 0.870g/100ml

12 Disaster: (1700 dead) from Gas Solubility 기체의용해도는온도가높아지면감소. Lake Nyos in Cameroon ( 아프리카 ), the site of a natural disaster. In 1986 a huge bubble of CO 2 escaped from the lake and asphyxiated more than 1700 people. CO 2 는공기밀도보다높고그때 CO 2 층은 20 m 이상을덮었다.

13 불포화, 포화, 과포화용액 용해 (Solubility): 물질간분자수준에서서로섞이는것. Unsaturated Saturated supersaturated 용질을더녹일수있는용액. 용질이녹을수있는최대량을포함한용액. 용해한계보다더많은용질이녹아있는용액 매우불안정하여용질의작은결정을넣어주면용질이석출됨 온도차에의한재결정

14 용액의농도 - 몰농도 (Molarity, M) : 용질의몰수 / 용액 Liter 수 ( 부피 ) - 몰랄농도 (Molality, m): 용질의몰수 / 용매의 Kg 수 노르말농도 (Normality,N) 용질의당량 / 용액의 Liter 수 - 질량백분율농도 (Mass percent (% m)): ( 용질의질량 ) / 용액전체질량 ) * 100

15 몰농도 (Molarity) 계산의예 : Conc. Sulfuric acid ( 진한황산,Mw = 98) is a solution that has a density of 1.84 g/ml and contains 98.3% H 2 SO 4 by mass. What is the molarity of this acid (M)? D=m/v 1.84 g/ml=1840 g / 1000mL (Molarity=moles/L) 1840 g x.983=1810 g H 2 SO 4 (at 98.3%) 1810 g H 2 SO 4 x (1 mol H 2 SO 4 /98g)=18.5 moles M=moles/liter= 18.5 moles/1l = 18.5 M

16 노르말농도 ( 당량 ) 당량 : 산혹은염기가 1 몰의하전 (H +, OH - ) 을줄수있는능력 예 HCl : 1mol HCl = 1eq H + H 2 SO 4 : 1 mol H 2 SO 4 = 2 eq H + H 3 SO 4 : 1 mol H 3 PO 4 = 3 eq H + NaOH: 1 mol NaOH = 1 eq OH M HCl = 5mol HCl /L = 5.0 N HCl 5.0 M H 2 SO 4 = 5mol /L H 2 SO 4 2 eq/1 mol = 10 eq / L = 10 N

몰랄농도 (molality,m) 예 : 3.5 g of CoCl 2 (Mw = 129.3) is dissolved in 100mL solution. Assume the density of the solution is 1.0 g/ml, what is concentration of the solution in molality(m)?

17 몰랄농도 (molality,m) 예 : 3.5 g of CoCl 2 (Mw = 129.3) is dissolved in 100mL solution. Assume the density of the solution is 1.0 g/ml, what is concentration of the solution in molality(m)? mol CoCl2 = 3.5 g CoCl2 mol CoCl g = m = m mass H2O = (100ml 1 g) g 1 ml = 96.5 g m = 3.5 g CoCl2 mol CoCl g.0965kg = m = 0.28 molality (m)

18 용액의묽힘 M1 x V1 = M2 x V2 ( M: 용액의몰농도, V: 용액의부피 ) 진한염산 ( 농도 = 12M) 50 ml 를 2.50L 로희석시최종용액의농도는? M1 x V1 = M2 x V2 에서 12 M x 0.05L = M2 x 2.5 L M2 = (12 M x 0.05 L) / 2.5 L = 0.24 M HCl

19 총괄성질 (colligative properties) : 용액중에녹아있는용질의농도에의해용액의물성이달라지는정도를결정하는성질을말함. 이것은용질의종류에관계없이용액에녹아있는용질의분자수에비례하는성질이다. ( 보통분자간상호인력을무시할수있을정도의낮은농도의용액에적용된다.) 1. Vapor pressure ( 증기압 ) Pv = x_solvent * Po 2. Freezing point depression ( 어는점내림 ) ΔTf = Kf * Σmi mi: 몰랄농도 3. Boiling point elevation ( 끓는점오름 ) ΔTb = Kb * Σmi 4. Osmotic pressure (π) π = (nrt)/v

20 Colligative properties ( 총괄성질 ) Vapour pressure is always lower Boiling point is always higher Freezing point is always lower Osmotic pressure drives solvent from lower concentration to higher concentration

21 비휘발성용질과 Raoult 의법칙 The vapor pressure of the solvent in a solution containing a non-volatile solute is always lower than the vapor pressure of pure solvent at the same temperature The solute molecules occupy volume reducing the number of solvent molecules at the surface The rate of evaporation decreases and so the vapor pressure above the solution must decrease to recover the equilibrium

22 Vapor pressure: Raoult s law solution of a nonvolatile (p vap 0) solute in a solvent: 묽은용액에서그증기압력내림의크기는용액중의용질의몰분율에비례한다는관계. ( 용매의증기압은용매의몰분율에비례 ) p A º p A = X A p A º (A: 용매 ) Raoult s law solvent A solvent A + nonvolatile solute B

23 Freezing point depression? The well known phenomena of freezing point depression and boiling point elevation are both consequences of vapor pressure lowering

24 freezing point depression and boiling point elevation ( 빙점강하 & 비점상승 ) Nonvolatile solutes ( 비휘발성용질 ) : raise the boiling point and lower the freezing point for H 2 O: DT b = K b m solute boiling point elevation K b = 0.51ºC/m DT f = K f m solute freezing point depression K f = 1.86ºC/m e.g., We saw that a 50/50 (v/v) solution of ethylene glycol in water had a molality of 18.0 m. What are its freezing and boiling points? 18*0.51 ~ oC 18*1.86 ~ oC

25 Osmotic pressure The solvent passes into the solution increasing its volume The passage of the solvent can be prevented by application of a pressure The pressure to prevent transport is the osmotic pressure

26 Osmotic pressure ( 삼투압 ) osmotic pressure, P solvent solution equilibrium solvent solution semipermeable membrane ( 반투막 ) -allows solvent molecules to pass, but not solute tries to equalize concentration, creates pressure Find: P = mrt In dilute solution, m M P = MRT ( 몰랄농도 ~ 몰농도 ) (n/v) (similar to ideal gas equation: P = (n/v)rt or PV = nrt)

27 Determining molar mass A solution contains 20.0 mg insulin in 5.00 ml develops an osmotic pressure of 12.5 mm Hg at 300 K. 인슐린의분자량은? atm M P RT M 12.5mmHg 1 760mmHg L atm K mol K moles insulin = M (6.68*10-4M) x V (5.0*10-3L) (M=mole/V)*(V) =mole = 3.34x10-6 mol Molar mass = mass of insulin/moles of insulin (mole = mass/mw) = g/3.34x10-6 mol = 5990 g/mol 4 M

2) 용해도곡선 어떤물질의용해도가온도에따라변하는정도를나타내는그래프 KNO saq K aqno aq 에서 KNO 가용해되는과정이흡열이므 로온도를높이면평형이오른쪽으로이동하여더욱많이용해된다. Ce SO 는발열적용해이므로온도를높이면용해도가떨어진다. 3) 재결정 ( 분별결정 )

2) 용해도곡선 어떤물질의용해도가온도에따라변하는정도를나타내는그래프 KNO saq K aqno aq 에서 KNO 가용해되는과정이흡열이므 로온도를높이면평형이오른쪽으로이동하여더욱많이용해된다. Ce SO 는발열적용해이므로온도를높이면용해도가떨어진다. 3) 재결정 ( 분별결정 ) 1) 용해 두종류이상의순물질이균일하게섞이는현상을용해라고하며, 이때의균 일한혼합물을용액이라한다. 2) 용해의조건 용매 + 용질 용해 석출 용액 가 ) 용매 - 용질사이의인력 용매 용매사이의인력용질 용질사이의인력일경우 나 ) 용질과용매의분자구조가비슷하거나같은원자단 ( 작용기 ) 을가진경우 용매와용질일반적으로용액의성분중에서액체성분을용매 (solvent), 고체나기체성분을용질