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Ions in Aqueous Solution

NO 3 –. NO 3 –. Co 2 +. Co 2 +. NO 3 –. NO 3 –. OH –. Na +. Na +. OH –. Ions in Aqueous Solution. Co(NO 3 ) 2 ( aq ). Co(NO 3 ) 2 (s ). Pb 2+ (aq) + 2 NO 3 – (aq). add water. dissociation :. “splitting into ions”. NaOH ( aq ). NaOH (s ). Na + ( aq ) + OH – ( aq ).

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Ions in Aqueous Solution

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  1. NO3– NO3– Co2+ Co2+ NO3– NO3– OH– Na+ Na+ OH– Ions in Aqueous Solution Co(NO3)2 (aq) Co(NO3)2 (s) Pb2+(aq) + 2 NO3–(aq) add water dissociation: “splitting into ions” NaOH (aq) NaOH (s) Na+ (aq) + OH– (aq) add water

  2. yields 1 2 __Co2+(aq) + __OH–(aq) 1 __Co(OH)2(s) OH– OH– OH– OH– OH– OH– OH– OH– NO3– NO3– Na+ Na+ Na+ Na+ Co2+ Co2+ Co2+ Co2+ NO3– NO3– Mix them and get the overall ionic equation… reactants 1 2 2 2 __Co2+(aq) + __NO3–(aq) + __Na+(aq) + __OH–(aq) 1 __Co(OH)2(s) + __NO3–(aq) + __Na+(aq) 2 2 products Cancel spectator ions to get net ionic equation…

  3. OH– NO3– clear Zn(NO3)2 solution clear Ba(OH)2 solution Ba2+ Zn2+ OH– NO3– Mix together Zn(NO3)2(aq) and Ba(OH)2(aq): Zn(NO3)2(aq) Ba(OH)2(aq) Zn2+(aq) + 2 NO3–(aq) Ba2+(aq) + 2 OH–(aq)

  4. OH– OH– OH– OH– NO3– yields OH– OH– OH– OH– NO3– Ba2+ Zn2+ Ba2+ Zn2+ Zn2+ Zn2+ NO3– NO3– Mix them and get the overall ionic equation… reactants 2 2 Zn2+(aq) + NO3–(aq) + Ba2+(aq) + OH–(aq) 2 + NO3–(aq) + Ba2+(aq) Zn(OH)2(s) products Cancel spectator ions to get net ionic equation… 2 Zn2+(aq) + OH–(aq) Zn(OH)2(s)

  5. Chemistry II Unit 5: Solutions

  6. e.g., water e.g., salt Solution Definitions solution: a homogeneous mixture -- evenly mixed at the particle level -- e.g., salt water alloy: a solid solution of metals -- e.g., bronze = Cu + Sn brass = Cu + Zn solvent: the substance that dissolves the solute

  7. soluble: “will dissolve in” miscible: refers to two liquids that mix evenly in all proportions -- e.g., food coloring and water

  8. As temp. , rate As size , rate Factors Affecting the Rate of Dissolution 1. temperature 2. particle size 3. mixing With more mixing, rate We can’t control this factor 4. nature of solvent or solute

  9. Classes of Solutions aqueous solution: solvent = water water = “the universal solvent” amalgam: solvent = mercury e.g., dental amalgam for cavities tincture: solvent = alcohol e.g., tincture of iodine (for cuts) organic solution: solvent contains ________ carbon Organic solvents include benzene, toluene, hexane, etc.

  10. Non-Solution Definitions insoluble: “will NOT dissolve in” e.g., sand and water immiscible: refers to two liquids that will NOT form a solution e.g., water and oil suspension: appears uniform while being stirred, but settles over time e.g., liquid medications, Italian dressing

  11. H H–C–H H–C–H H–C–H H–C–H H H H O Molecular Polarity nonpolar molecules: -- e– are shared equally -- tend to be symmetric fats and oils e.g., polar molecules: -- e– NOT shared equally water e.g., “Like dissolves like.” polar + polar = solution nonpolar + nonpolar = solution polar + nonpolar = suspension (won’t mix evenly)

  12. Using Solubility Principles Chemicals used by body obey solubility principles. -- water-soluble vitamins: e.g., vitamin C vitamins A & D -- fat-soluble vitamins: e.g., Anabolic steroids and HGH are fat-soluble, synthetic hormones.

  13. Cl Cl C=C Cl Cl Using Solubility Principles (cont.) Dry cleaning employs nonpolar liquids -- polar liquids damage wool, silk -- also, dry clean for stubborn stains (ink, rust, grease) -- tetrachloroethylene was in longtime use

  14. lecithin e.g., soap eggs detergent MODEL OF A SOAP MOLECULE Na+ NONPOLAR HYDROCARBON TAIL POLAR HEAD emulsifying agent (emulsifier): molecules w/both a polar AND a nonpolar end -- -- allows polar and nonpolar substances to mix

  15. Na+ NONPOLAR HYDROCARBON TAIL POLAR HEAD H2O H2O oil H2O H2O soap vs. detergent -- -- made from petroleum made from animal and vegetable fats -- works better in hard water Hard water contains minerals w/ions like Ca2+, Mg2+, and Fe3+ that replace Na+ at polar end of soap molecule. Soap is changed into an insoluble precipitate (i.e., soap scum). micelle: a liquid droplet covered w/soap or detergent molecules

  16. SOLUBILITY CURVE sudden stress causes this much ppt KNO3 (s) KCl (s) Solubility (g/100 g H2O) HCl (g) Temp. (oC) Solubility how much solute dissolves in a given amt. of solvent at a given temp. unsaturated: sol’n could dissolve more solute; below the line saturated: sol’n dissolves “just right” amt. of solute; on the line supersaturated: sol’n has “too much” solute dissolved in it; above the line

  17. Sol. Sol. As To , solubility ___ As To , solubility ___ To To Solids dissolved Gases dissolved in liquids in liquids [O2]

  18. gases solids per 100 g H2O 150 Using an available solubility curve, classify as unsaturated, saturated,or supersaturated. KI 140 130 120 NaNO3 110 100 80 g NaNO3 @ 30oC KNO3 90 unsaturated 80 HCl NH4Cl 70 45 g KCl @ 60oC Solubility (grams of solute/100 g H2O) KCl NH3 60 saturated 50 50 g NH3 @ 10oC 40 30 NaCl unsaturated 20 70 g NH4Cl @ 70oC KClO3 10 SO2 supersaturated 0 10 20 30 40 50 60 70 80 90 100

  19. Solubility vs. Temperature for Solids gases solids (Unsaturated, saturated, or supersaturated?) 150 KI 140 Per 500 g H2O, 120 g KNO3 @ 40oC 130 120 NaNO3 110 saturation point @ 40oC for 100 g H2O = 70 g KNO3 100 KNO3 90 80 HCl NH4Cl 70 Solubility (grams of solute/100 g H2O) So saturation pt. @ 40oC for 500 g H2O = 5 x 70 g KCl NH3 60 50 40 30 NaCl = 350 g 20 120 g < 350 g KClO3 10 SO2 unsaturated 0 10 20 30 40 50 60 70 80 90 100

  20. gases solids Describe each situation below. 150 KI 140 (A) Per 100 g H2O, 100 g NaNO3 @ 50oC. 130 120 unsaturated; all solute dissolves; clear sol’n. NaNO3 110 100 KNO3 90 (B) Cool sol’n (A) very slowly to 10oC. 80 HCl NH4Cl 70 Solubility (grams of solute/100 g H2O) supersaturated; extra solute remains in sol’n; still clear KCl NH3 60 50 40 30 NaCl (C) Quench sol’n (A) in an ice bath to 10oC. 20 KClO3 10 SO2 saturated; extra solute (20 g) can’t remain in sol’n and becomes visible 0 10 20 30 40 50 60 70 80 90 100

  21. gases solids Solubility Curve 150 KI 140 130 120 NaNO3 110 100 KNO3 90 80 HCl NH4Cl 70 Solubility (grams of solute/100 g H2O) KCl NH3 60 50 40 30 NaCl 20 KClO3 10 SO2 0 10 20 30 40 50 60 70 80 90 100

  22. beaker vs. volumetric flask 1000 mL + 5% 1000 mL + 0.30 mL min: min: Range: Range: max: max: Glassware – Precision and Cost When filled to 1000 mL line, how much liquid is present? WE DON’T KNOW. 5% of 1000 mL = 50 mL 999.70 mL 950 mL 1050 mL 1000.30 mL precise; expensive imprecise; cheap

  23. water in grad. cyl. mercury in grad. cyl. measure to bottom measure to top ** Measure to part of meniscusw/zero slope.

  24. Concentration…a measure of solute-to-solvent ratio concentrated dilute “lots of solute” “not much solute” “not much solvent” “watery” Add water to dilute a sol’n; boil water off to concentrate it.

  25. Selected units of concentration A. mass % = mass of solute x 100 mass of sol’n B. parts per million (ppm) = mass of solute x 106 mass of sol’n also, ppb and ppt (Use 109 or 1012 here) mol -- commonly used for minerals or contaminants in water supplies M L molarity (M) = moles of solute L of sol’n C. -- used most often in this class

  26. mol M L How many mol solute are req’d to make 1.35 L of 2.50 M sol’n? mol = M L = 2.50 M (1.35 L ) = 3.38 mol A. What mass sodium hydroxide is this? Na+ OH– NaOH 3.38 mol = 135 g NaOH B. What mass magnesium phosphate is this? Mg2+ PO43– Mg3(PO4)2 3.38 mol = 889 g Mg3(PO4)2

  27. mol M L (convert to mL) Find molarity if 58.6 g barium hydroxide are Dissolved in 5.65 L sol’n. Ba2+ OH– Ba(OH)2 58.6 g = 0.342 mol = 0.061 M Ba(OH)2 You have 10.8 g potassium nitrate. How many mL of sol’n will make this a 0.14 M sol’n? K+ NO3– KNO3 10.8 g = 0.1068 mol = 0.763 L = 763 mL

  28. Molarity and Stoichiometry PbI2 KI V of sol’ns 2 2 Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq) What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2? Find mol KI needed to yield 89 g PbI2. Strategy: (1) (2) Based on (1), find volume of 4.0 M KI sol’n. mass mass mol mol M L M L part. part.

  29. 2 2 Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq) What volume of 4.0 M KI sol’n is req’d to yield 89 g PbI2? Strategy: (1) (2) Find mol KI needed to yield 89 g PbI2. Based on (1), find volume of 4.0 M KI sol’n. PbI2 KI 89 g PbI2 = 0.39 mol KI = 0.098 L of 4.0 M KI

  30. How many mL of a 0.500 M CuSO4 sol’n will react w/excess Al to produce 11.0 g Cu? Cu CuSO4 3 2 3 CuSO4(aq) + Al(s)  Cu(s) + Al2(SO4)3(aq) 11.0 g Cu Cu CuSO4 = 0.173 mol CuSO4 = 0.346 L = 346 mL of 0.500 M CuSO4

  31. Dilutions of Solutions Acids (and sometimes bases) are purchased in concentrated form (“concentrate”) and are easily diluted to any desired concentration. **Safety Tip: When diluting, add acid (or base) to water. C = conc. D = dilute MC VC = MD VD Dilution Equation:

  32. 14.8 14.8 Conc. H3PO4 is 14.8 M. What volume of concentrate is req’d to make 25.00 L of 0.500 M H3PO4? MC VC = MD VD (VC) = 14.8 0.500 (25) VC = 0.845 L How would you mix the above sol’n? 0.845 1. Measure out _____ L of conc. H3PO4. ~20 2. In separate container, obtain ____ L of cold H2O. 3. In fume hood, slowly pour H3PO4 into cold H2O. 4. Add enough H2O until 25.00 L of sol’n is obtained.

  33. 400 samples Cost Analysis with Dilutions 2.5 L of 12 M HCl (i.e., “concentrate”) 0.500 L of 0.15 M HCl Cost: $25.71 Cost: $6.35 How many 0.500 L-samples of 0.15 m HCl can be made from the bottle of concentrate? MC VC = MD VD (Expensive water!) (2.5 L) = 12 M 0.150 M (VD) Moral: Buy the concentrate and mix it yourself to any desired concentration. VD = 200 L @ $6.35 ea. = $2,540

  34. You have 75 mL of conc. HF (28.9 M); you need 15.0 L of 0.100 M HF. Do you have enough to do the experiment? MC VC = MD VD (0.075 L) 28.9 M 0.100 M (15 L) = Calc. how much conc. you need… (VC) 28.9 M 0.100 M (15 L) = VC = 0.052 L = 52 mL needed Yes; we’re OK.

  35. Dissociation occurs when neutral combinations of particles separate into ions while in aqueous solution. sodium chloride NaCl  Na+ + Cl– sodium hydroxide NaOH  Na+ + OH– H+ + Cl– hydrochloric acid HCl  sulfuric acid H2SO4 2 H+ + SO42– acetic acid CH3COOH  CH3COO– + H+ acids In general, _____ yield hydrogen (H+) ions in aque- ous solution; _____ yield hydroxide (OH–) ions. bases

  36. Strong electrolytes exhibit nearly 100% dissociation. NaCl Na+ + Cl– Weak electrolytes exhibit little dissociation. CH3COOH CH3COO– + H+ 1000 0 0 NOT in water: in aq. sol’n: 1 999 999 1000 0 0 NOT in water: in aq. sol’n: 980 20 20 “Strong” or “weak” is a property of the substance. We can’t change one into the other.

  37. electrolytes: solutes that dissociate in sol’n -- conduct elec. 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.

  38. nonelectrolytes: solutes that DO NOT dissociate -- DO NOT conduct elec. current (not enough ions) -- e.g., any type of sugar

  39. Colligative Properties properties thatdepend on the conc. of a sol’n Compared to solvent’s… a sol’n w/that solvent has a… …normal freezing point (NFP) …lower FP (freezing point depression) …normal boiling point (NBP) …higher BP (boiling point elevation)

  40. FP BP water water + a little salt water + more salt Applications of Colligative Properties (NOTE: Data is fictitious.) 1. salting roads in winter 0oC (NFP) 100oC (NBP) –11oC 103oC –18oC 105oC

  41. FP BP water water + a little AF 50% water + 50% AF Applications of Colligative Properties (cont.) 2. Antifreeze (AF) (a.k.a., “coolant”) 0oC (NFP) 100oC (NBP) –10oC 110oC –35oC 130oC

  42. Applications of Colligative Properties (cont.) 3. law enforcement starts melting at… finishes melting at… penalty, if convicted white powder comm. service A 120oC 150oC B 130oC 140oC 2 yrs. C 20 yrs. 134oC 136oC

  43. Calculations with Colligative Properties

  44. Adding a nonvolatile solute to a solvent decreases the solution’s freezing point (FP) and increases its boiling point (BP). The freezing point depression and boiling point elevation are given by: DTx = Kx m i DTx = FP depression or BP elevation Kx = Kf (molal FP depression constant) or Kb (molal BP elevation constant) -- they depend on the solvent -- for water: Kf= 1.86oC/m, Kb = 0.52oC/m m = molality of solute

  45. i = van’t Hoff factor (accounts for # of particles in solution) In aq. soln., assume that… 1 -- i = __ for nonelectrolytes 2 -- i = __ for KBr, NaCl, etc. Jacobus Henricus van’t Hoff (1852 – 1911) 3 -- i = __ for CaCl2, etc. In reality, the van’t Hoff factor isn’t always an integer. Use the guidelines unless given information to the contrary.

  46. Kf = 1.86oC/m Kb = 0.52oC/m 0.690 m Find the FP and BP of a soln. containing 360 g barium chloride and 2.50 kg of water. DTx = Kx m i 360 g BaCl2 = 1.725 mol BaCl2 i = 3 DTf = Kf m i = 1.86(0.690)(3) = 3.85oC FP = –3.85oC DTb = Kb m i = 0.52(0.690)(3) = 1.08oC BP = 101.08oC

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