Colligative Properties. occurs when neutral combinations of particles separate into ions while in aqueous solution. sodium chloride sodium hydroxide hydrochloric.

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Presentation transcript:

Colligative Properties

occurs when neutral combinations of particles separate into ions while in aqueous solution. sodium chloride sodium hydroxide hydrochloric acid sulfuric acid acetic acid Dissociation In general, acids yield hydrogen ions in aqueous solution; bases yield hydroxide ions. NaCl  Na 1+ + Cl 1– NaOH  Na 1+ + OH 1– HCl  H 1+ + Cl 1– H 2 SO 4  2 H 1+ + SO 4 2– CH 3 COOH  CH 3 COO 1– + H 1+ (H 1+ ) (OH 1– ) ? ?

NaCl Na 1+ + Cl 1– CH 3 COOH CH 3 COO 1– + H 1+ Weak electrolytes exhibit little dissociation. “Strong” or “weak” is a property of the substance. We can’t change one into the other. Strong electrolytes exhibit nearly 100% dissociation. NOT in water: in aq. solution: NOT in water: in aq. solution:

electrolytes: solutes that dissociate in solution -- conduct electric current because of free-moving ions e.g., acids, bases, most ionic compounds -- are crucial for many cellular processes -- obtained in a healthy diet -- For sustained exercise or a bout of the flu, sports drinks ensure adequate electrolytes. nonelectrolytes: solutes that DO NOT dissociate -- DO NOT conduct electric current (not enough ions) e.g., any type of sugar

…normal boiling point (NBP)…higher BP FREEZING PT. DEPRESSION BOILING PT. ELEVATION Colligative Properties  depend on concentration of a solution Compared to solvent’s… a solution w/that solvent has a… …normal freezing point (NFP)…lower FP

1. salting roads in winter FP BP water0 o C (NFP) 100 o C (NBP) 2. antifreeze (AF) /coolant FP BP water0 o C (NFP) 100 o C (NBP) water + a little AF–10 o C110 o C 50% water + 50% AF–35 o C130 o C water + a little salt water + more salt –11 o C 103 o C –18 o C105 o C Applications (NOTE: Data are fictitious.)

3. law enforcement white powder starts melting at… finishes melting at… penalty, if convicted A109 o C175 o Ccomm. service B150 o C180 o C2 years C194 o C196 o C20 years

Effect of Pressure on Boiling Point Boiling Point of Water at Various Locations Location Feet above sea level P atm (kPa) Boiling Point (  C) Top of Mt. Everest, Tibet29, Top of Mt. Denali, Alaska20, Top of Mt. Whitney, California14, Leadville, Colorado10, Top of Mt. Washington, N.H.6, Boulder, Colorado5, Madison, Wisconsin New York City, New York Death Valley, California

Elevation = 869 feet Normal, Illinois

Calculations for Colligative Properties The change in FP or BP is found using…  T x = K x m i  T x = change in T o (below NFP or above NBP) K x = constant depending on… (A) solvent (B) freezing or boiling m = molality of solute = mol solute / kg solvent i = integer that accounts for any solute dissociation any sugar (all nonelectrolytes)……………...i = 1 table salt, NaCl  Na 1+ + Cl 1– ………………i = 2 barium bromide, BaBr 2  Ba Br 1– ……i = 3

Freezing Point DepressionBoiling Point Elevation  T f = K f m i  T b = K b m i Then use these in conjunction with the NFP and NBP to find the FP and BP of the mixture. (K b = ebullioscopic constant, which is 0.51 K kg/mol for the boiling point of water) (K f = cryoscopic constant, which is 1.86 K kg/mol for the freezing point of water)

(NONELECTROLYTE) 168 g glucose (C 6 H 12 O 6 ) are mixed w /2.50 kg H 2 O. Find BP and FP of mixture. For H 2 O, K b = 0.512, K f = –1.86. i = 1  T b = K b m i = (0.373) (1) = 0.19 o C BP = ( ) o C = o C  T f = K f m i = –1.86 (0.373) (1) = –0.69 o C FP = (0 + –0.69) o C = –0.69 o C

168 g cesium bromide are mixed w /2.50 kg H 2 O. Find BP and FP of mixture. For H 2 O, K b = 0.512, K f = –1.86.  T b = K b m i = (0.316) (2) = 0.32 o C BP = ( ) o C = o C  T f = K f m i = –1.86 (0.316) (2) = –1.18 o C FP = (0 + –1.18) o C = –1.18 o C Cs 1+ Br 1– i = 2 CsBr  Cs 1+ + Br 1–

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