1 / 11

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.

lada
Download Presentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related