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Unit 10: Solutions

Chemistry. Unit 10: Solutions. 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 .

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Unit 10: Solutions

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  1. Chemistry Unit 10: Solutions

  2. 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

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

  4. 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

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

  6. 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

  7. 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)

  8. 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.

  9. 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

  10. lecithin e.g., soap eggs detergent MODEL OF A SOAP MOLECULE Na1+ NONPOLAR HYDROCARBON TAIL POLAR HEAD emulsifying agent (emulsifier): molecules w/both a polar AND a nonpolar end -- -- allows polar and nonpolar substances to mix

  11. Na1+ 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 Na1+ 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

  12. 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 hold more solute; below the line saturated: sol’n has “just right” amt. of solute; on the line supersaturated: sol’n has “too much” solute dissolved in it; above the line

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

  14. per 100 g H2O Using an available solubility curve, classify as unsaturated, saturated, or supersaturated. 80 g NaNO3 @ 30oC unsaturated 45 g KCl @ 60oC saturated 30 g KClO3 @ 30oC supersaturated 70 g Pb(NO3)2 @ 60oC unsaturated

  15. (Unsaturated, saturated, or supersaturated?) Per 500 g H2O, 100 g KNO3 @ 40oC saturation point @ 40oC for 100 g H2O = 63 g KNO3 So saturation pt. @ 40oC for 500 g H2O = 5 x 63 g = 315 g 100 g < 315 g unsaturated

  16. Describe each situation below. (A) Per 100 g H2O, 100 g NaNO3 @ 50oC. unsaturated; all solute dissolves; clear sol’n. (B) Cool sol’n (A) very slowly to 10oC. supersaturated; extra solute remains in sol’n; still clear (C) Quench sol’n (A) in an ice bath to 10oC. saturated; extra solute (20 g) can’t remain in sol’n and becomes visible

  17. 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

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

  19. 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.

  20. 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

  21. 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? Na1+ OH1– 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

  22. mol M L (convert to mL) Find molarity if 58.6 g barium hydroxide are in 5.65 L sol’n. Ba2+ OH1– 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? K1+ NO31– KNO3 10.8 g = 0.1068 mol = 0.763 L = 763 mL

  23. m m mol mol V V V of gases at STP P M L M L P V of sol’ns Molarity and Stoichiometry PbI2 KI 1 2 1 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.

  24. mol M L M L M L 1 2 1 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. PbI2 KI Based on (1), find volume of 4.0 M KI sol’n. 89 g PbI2 = 0.39 mol KI = 0.098 L of 4.0 M KI

  25. mol M L M L M L Cu CuSO4 How many mL of a 0.500 M CuSO4 sol’n will react w/excess Al to produce 11.0 g Cu? Al3+ SO42– 3 2 3 1 __CuSO4(aq) + __Al(s)  __Cu(s) + __Al2(SO4)3(aq) 11.0 g Cu = 0.173 mol CuSO4 = 0.346 L = 346 mL of 0.500 M CuSO4

  26. 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:

  27. 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.

  28. 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 0.500 M (VD) Moral: Buy the concentrate and mix it yourself to any desired concentration. VD = 200 L @ $6.35 ea. = $2,540

  29. 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.

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

  31. Strong electrolytes exhibit nearly 100% dissociation. NaCl Na1+ + Cl1– Weak electrolytes exhibit little dissociation. CH3COOH CH3COO1– + H1+ 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.

  32. 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.

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

  34. 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)

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

  36. 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

  37. 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

  38. h h

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