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occurs when neutral combinations of particles separate into ions while in aqueous solution.

occurs when neutral combinations of particles separate into ions while in aqueous solution. Dissociation. NaCl  Na 1+ + Cl 1–. sodium chloride. sodium hydroxide. NaOH  Na 1+ + OH 1–. hydrochloric acid. HCl  H 1+ + Cl 1–. sulfuric acid.

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occurs when neutral combinations of particles separate into ions while in aqueous solution.

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  1. occurs when neutral combinations of particles separate into ions while in aqueous solution. Dissociation NaCl Na1+ + Cl1– sodium chloride sodium hydroxide NaOH Na1+ + OH1– hydrochloric acid HCl H1+ + Cl1– sulfuric acid H2SO4 2 H1+ + SO42– CH3COOH CH3COO1– + H1+ acetic acid ? In general, acids yield hydrogen ions (H1+) ? in aqueous solution; bases yield hydroxide ions. (OH1–)

  2. NaCl Na1+ + Cl1– CH3COOH CH3COO1– + H1+ Strong electrolytes exhibit nearly 100% dissociation. NOT in water: 1000 0 0 in aq. solution: 1 999 999 Weak electrolytes exhibit little dissociation. NOT in water: 1000 0 0 in aq. solution: 980 20 20 “Strong” or “weak” is a property of the substance. We can’t change one into the other.

  3. 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

  4. 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 FREEZING PT. DEPRESSION …normal boiling point (NBP) …higher BP BOILING PT. ELEVATION

  5. Applications (NOTE: Data are fictitious.) 1. salting roads in winter water + a little salt –11oC 103oC water + more salt –18oC 105oC 2. antifreeze (AF) /coolant

  6. 3. law enforcement

  7. Effect of Pressure on Boiling Point

  8. Calculations for Colligative Properties The change in FP or BP is found using… DTx = Kx m i DTx = change in To (below NFP or above NBP) Kx = 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  Na1+ + Cl1–………………i = 2 barium bromide, BaBr2  Ba2+ + 2 Br1–……i = 3

  9. Freezing Point Depression Boiling Point Elevation DTf = Kf m i DTb = Kb m i Then use these in conjunction with the NFP and NBP to find the FP and BP of the mixture. (Kf = cryoscopic constant, which is 1.86 K kg/mol for the freezing point of water) (Kb = ebullioscopic constant, which is 0.51 K kg/mol for the boiling point of water)

  10. 168 g glucose (C6H12O6) are mixed w/2.50 kg H2O. Find BP and FP of mixture. For H2O, Kb = 0.512, Kf = –1.86. i = 1 (NONELECTROLYTE) DTb = Kb m i = 0.512 (0.373) (1) = 0.19oC BP = (100 + 0.19)oC = 100.19oC DTf = Kf m i = –1.86 (0.373) (1) = –0.69oC FP = (0 + –0.69)oC = –0.69oC

  11. 168 g cesium bromide are mixed w/2.50 kg H2O. Find BP and FP of mixture. For H2O, Kb = 0.512, Kf = –1.86. Cs1+ Br1– i = 2 CsBr Cs1+ + Br1– DTb = Kb m i = 0.512 (0.316) (2) = 0.32oC BP = (100 + 0.32)oC = 100.32oC DTf = Kf m i = –1.86 (0.316) (2) = –1.18oC FP = (0 + –1.18)oC = –1.18oC

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