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Solutions

Solutions. Solution Definitions. Solution: homogeneous mixture, evenly mixed at the particle level (like salt water). Solute: substance being dissolved. Solvent: present in greater amount that can dissolve the solute. Classes of Solutions. aqueous solution :. solvent = water.

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Solutions

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  1. Solutions

  2. Solution Definitions • Solution: homogeneous mixture, evenly mixed at the particle level (like salt water). Solute: substance being dissolved Solvent: present in greater amount that can dissolve the solute

  3. Classes of Solutions aqueous solution: solvent = water water = “the universal solvent” amalgam: solvent = Hg e.g., dental amalgam tincture: solvent = alcohol e.g., tincture of iodine (for cuts) organic solution: solvent contains carbon e.g., gasoline, benzene, toluene, hexane

  4. water salt Solution Definitions alloy: a solid solution of metals -- e.g., bronze = Cu + Sn; brass = Cu + Zn solvent: the substance that dissolves the solute soluble: “will dissolve in” miscible: refers to two gases or two liquids that form a solution; more specific than the term “soluble”

  5. Solubility • Solubility - maximum grams of solute that will dissolve in 100 g of solvent at a given temperature • varies with temperature

  6. UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form Solubility increasing concentration

  7. 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. SOLUBILITY CURVE unsaturated: solution could hold more solute; below line saturated: solution has “just right” amt. of solute; on line supersaturated: solution has “too much” solute dissolved in it; above the line

  8. gases solids Solubility vs. Temperature for Solids Solubility Table 140 KI 130 120 NaNO3 110 100 KNO3 90 80 HCl NH4Cl • shows the dependence • of solubility on temperature 70 Solubility (grams of solute/100 g H2O) 60 NH3 KCl 50 40 30 NaCl KClO3 20 10 SO2 0 10 20 30 40 50 60 70 80 90 100 Temperature (°C) LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517

  9. Classify as unsaturated, saturated, or supersaturated. per 100 g H2O 80 g NaNO3 @ 30oC unsaturated 60 g KCl @ 60oC saturated 50 g NH3 @ 10oC unsaturated 70 g NH4Cl @ 70oC supersaturated Per 500 g H2O, 120 g KNO3 @ 40oC saturation point @ 40oC for 100 g H2O = 66 g KNO3 So sat. pt. @ 40oC for 500 g H2O = 5 x 66 g = 330 g 120 g < 330 g unsaturated

  10. Solubility • Solids are more soluble at... • high temperatures. • Gases are more soluble at... • low temperatures

  11. Describe each situation below. (A) Per 100 g H2O, 100 g Unsaturated; all solute NaNO3 @ 50oC. dissolves; clear solution. (B) Cool solution (A) very Supersaturated; extra slowly to 10oC. solute remains in solution; still clear. (C) Quench solution (A) in Saturated; extra solute an ice bath to 10oC. (20 g) can’t remain in solution, becomes visible.

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

  13. AFTER Water COLD Water HOT A B Solubility Experiment 1: Add 1 drop of red food coloring Before Miscible – “mixable” two gases or two liquids that mix evenly Water HOT Water COLD A B

  14. Solubility Experiment 2: Add oil to water and shake AFTER Before Immiscible – “does not mix” two liquids or two gases that DO NOT MIX Oil Water Water T0 sec T30 sec

  15. Solutions • What the solute and the solvent are determines: • whether a substance will dissolve. • how much will dissolve.

  16. Factors Affecting the Rate of Dissolution As T , rate 1. temperature As size , rate 2. particle size More mixing, rate 3. mixing 4. nature of solvent or solute

  17. 200 180 160 140 120 100 80 60 40 20 0 Solid Solubility KI KNO3 Solubility (g solute / 100 g H2O) NaNO3 Na3PO4 NaCl 20 40 60 80 100 Temperature (oC) Timberlake, Chemistry 7th Edition, page 297

  18. Gas Solubility CH4 2.0 O2 Higher Temperature …Gas is LESS Soluble CO Solubility (mM) 1.0 He 0 10 20 30 40 50 Temperature (oC)

  19. Particle Size and Mixing • In order to dissolve - the solvent molecules must come in contact with the solute. • Stirring moves fresh solvent next to the solute. • The solvent touches the surface of the solute. • Smaller pieces increase the amount of surface of the solute.

  20. Molecular Polarity • Nonpolar molecules: electrons are shared equally • Molecule has no charge • e.g. Ethane

  21. Polar Molecules • Electrons are not • equally shared • Molecule has • slight charge • “Like dissolves like” • (Polarity) H+ d+ O2- H+ d-

  22. Solubility Rules General Rules for Solubility of Ionic Compounds (Salts) in Water at 25 oC • Most nitrate (NO31-) salts are soluble. • Most salts of Na+, K+, and NH4+ are soluble. • Most chloride salts are soluble. Notable exceptions • are AgCl, PbCl2, and Hg2Cl2. • Most sulfate salts are soluble. Notable exceptions • are BaSO4, PbSO4, and CaSO4. • Most hydroxide compounds are only slightly • soluble.* The important exceptions are NaOH and • KOH, Ba(OH)2 and Ca(OH)2 are only moderately • soluble. • Most sulfide (S2-), carbonate (CO32-), and phos- • phate (PO43-) salts are only slightly soluble. • *The terms insoluble and slightly soluble really mean the same • thing. Such a tiny amount dissolves that it is not possible to • detect it with the naked eye. Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 218

  23. Oil and Water Don’t Mix • Oil is nonpolar • Water is polar • “Like dissolves like” • polar + polar = solution • nonpolar + nonpolar = solution • polar + nonpolar = suspension (won’t mix evenly) Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 470

  24. Solvation • Soap / Detergent • polar“head” with long nonpolar“tail” • dissolves nonpolar grease in polar water micelle

  25. Solubility of Common Compounds Soluble compounds NO3- salts Na+, K+, NH4+ salts Except for those containing Ag+, Hg22+, Pb2+ Cl-, Br-, I- salts Except for those containing SO42- salts Ba2+, Pb2+, Ca2+ Insoluble compounds S2-, CO32-, PO43- salts Except for those containing OH1- salts Na+, K+, Ca2+ Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 218

  26. Making Molar Solutions …from liquids (More accurately, from stock solutions)

  27. Concentration…a measure of solute-to-solvent ratio concentrated vs. dilute “lots of solute”“not much solute” “watery” Add water to dilute a solution; boil water off to concentrate it.

  28. mol D. molality (m) = moles of solute mol L M L kg of solvent M = Concentration“The amount of solute in a solution” A. mass % = mass of solute mass of sol’n B. parts per million (ppm) also, ppb and ppt – commonly used for minerals or contaminants in water supplies C. molarity (M) = moles of solute L of sol’n – used most often in this class % by mass – medicated creams % by volume – rubbing alcohol

  29. # of moles Molarity (M) = volume (L) V = 250 mL n = 8 moles [ ] = 32 molar V = 1000 mL V = 1000 mL V = 5000 mL n = 2 moles n = 4 moles n = 20 moles Concentration = 2 molar [ ] = 4 molar [ ] = 4 molar

  30. Glassware

  31. 1000 mL + 0.30 mL Glassware – Precision and Cost beaker vs. volumetric flask When filled to 1000 mL line, how much liquid is present? beaker 5% of 1000 mL = volumetric flask 50 mL Range: Range: 999.70 mL– 1000.30 mL 950 mL – 1050 mL precise; expensive; good for making solutions imprecise; cheap

  32. Wash bottle Volume marker (calibration mark) Weighed amount of solute How to mix a Standard Solution Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 480

  33. mol L ? mol 1 L M = 3 M = How to mix solid chemicals How many grams of sodium hydroxide would we need to make 100. mL of 3.0 M NaOH? How much will this weigh? 1 Na @ 23g/mol + 1O @ 16g/mol + 1 H @ 1 g/mol MMNaOH = 40g/mol 40.0 g NaOH X g NaOH = 3.0 molNaOH = 120 g NaOH 1 mol NaOH To mix this: add 120 g NaOH into 1L volumetric flask with~750 mL cold H2O. Mix, allow to return to room temperature – bring volume to 1 L.

  34. mol M L How many moles of solute are required to make 1.35 L of 2.50 M solution? = mol = M L 3.38 mol = 2.50 M (1.35 L) A. What mass of sodium hydroxide is this? 40.0 g NaOH X g NaOH = 3.38 mol NaOH = 135 g NaOH 1 mol NaOH B. What mass of magnesium phosphate is this? 262.9 g Mg3(PO4)2 X g Mg3(PO4)2 = 3.38 mol Mg3(PO4)2 = 889 g Mg3(PO4)2 1 mol Mg3(PO4)2

  35. mol M = L Find molarity if 58.6 g of barium hydroxide are in 5.65 L solution. Step 1). How many moles barium hydroxide is this? 1 mol Ba(OH)2 X mol Ba(OH)2 = 58.6 g Ba(OH)2 = 0.342 mol Ba(OH)2 171.3 g Ba(OH)2 Step 2). What is the molarity of a 5.65 L solution containing 0.342 mol solute? 0.342 mol M = = 0.0605 M Ba(OH)2 5.65 L

  36. You have 10.8 g potassium nitrate. How many mL of solution will make this a 0.14 M solution? convert to mL

  37. mass of solvent only 1 kg water = 1 L water Molality

  38. Molality • Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 kg water = 3.2m MgCl2

  39. Molality • How many grams of NaCl are required to make a 1.54 msolution using 0.500 kg of water? 0.500 kg water 1.54 mol NaCl 1 kg water 58.44 g NaCl 1 mol NaCl = 45.0 g NaCl

  40. 500 mL of 1.54M NaCl Preparing Solutions molarity molality 1.54mNaCl in 0.500 kg of water • mass 45.0 g of NaCl • add 0.500 kg of water (volume will be just over 500 total mL) • mass 45.0 g of NaCl • add water until total volume is 500 mL 500 mLwater 500 mL volumetric flask 500 mL mark 45.0 gNaCl

  41. Dilution • Preparation of a desired solution by adding water to a concentrate. • Moles of solute remain the same.

  42. Dilution • What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M)V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

  43. 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

  44. 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.

  45. 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

  46. 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

  47. 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

  48. 3. law enforcement

  49. 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

  50. 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)

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