Chap 13A - Chem 210 WANG Chapter 13A Properties of Solution...

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Unformatted text preview: Chem 210 / WANG / Chapter 13A Properties of Solution Chapter 13 Properties of Solutions 13.1 The Solution Process 13.2 Saturated Solutions and Solubility 13.3 Factors Affecting Solubility 13.4 Ways of Expressing Concentration 13.5 Colligative Properties solvent – major component solute solute – minor component (may be more than one.) 13.6 Colloids Which Which of these pictures is: A “solution”? A “compound”? A B Mixture - (diatomic) element heterogeneous C A compound “Solution” vs. “Compound” “Compound” Definition of a Solution A solution A homogeneous mixture of two or more components. Like a compound, which has at least two compound, different elements; a solution also has at least two different components (solvent plus one or more solutes). Unlike compounds, in which atoms are joined through chemical bonds, solutes chemical and solvent stay together in solutions through intermolecular forces. intermolecular Compound has definite composition and definite properties; solution has variable solution variable composition, and its properties change properties depending on solute concentration. The major types of intermolecular forces in solutions. Physical states of solutions A solutiton = D homogeneous mixture Solutions can exist in any of the three states. Solid solutions dental fillings, 14K gold, sterling silver Liquid solutions saline, vodka, vinegar, sugar water Intermolecular forces keep keep solvent together. solute and solvent together. The The arrangement of atoms in two types of alloys. brass - a substitution alloy carbon steel -an interstitial alloy Alloys are “solid” solutions! Fall 2009 UM-SJTU JI Solution terms: dissolving dissolving, solvation and hydration All describe how solute disperses in solvent. Dissolving - Macroscopic description • Solute disappearing from sight. Solvation Solvation - Molecular description • Solvent molecules surrounding solute particles and stabilizing them in solution Hydration - Molecular description • = solvation when water is the solvent. solvation • H2O surrounding solute particles and stabilizing them in solution. Gaseous solutions the atmosphere, anesthesia gases Dissolving, solvation, hydration Figure 13.2 Figure 13.9 Hydrated ions Page 1 Chem 210 / WANG / Chapter 13A Properties of Solution Hydrates vs. anhydrous • Hydration shells around an aqueous ion. • A “hydrate” forms when water of hydration stays hydration in a crystalline solid obtained from aqueous solutions. When water is completely driven from the crystal, it becomes “anhydrous.” Chromium (III) chloride • Anhydrous CrCl3 (purple) • Hydrated CrCl3.6H2O (dark green) http:\\ Enthalpy of Solution = ∆H3 +∆H2 +∆H1 Instant cold packs looks like a plastic bag filled with liquid. You hit it, shake it up and it gets extremely cold. • The liquid inside the cold pack is water. In the water is another plastic bag or tube containing NH ammonium nitrate, NH4NO3(s) . • When you hit the cold pack, it breaks the tube so that the water mixes with solid NH4NO3. NH • The mixing process is an endothermic reaction, heat is absorbed. Figure 13.4 Heat of hydration Heat of hydration Used in “cold“coldpacks.” Temperature Temperature of the solution falls to about 35 F for 10 to 15 minutes. MgSO4 ∆Hsoln = -91.2 kJ/mol NH4 NO3 ∆Hsoln = +26.4 kJ/mol Dissolving of (a) NaCl in water (a) Na (b) CH3OH (methanol) in water methanol) in Like Like dissolves like: solubility of Energy Energy changes in the solution process Separation of solute molecules ∆H1 > 0 endothermic Separation of solvent molecules ∆H2 > 0 endothermic solvent & solute interact ∆ H3 < 0 Exothermic Exothermic Like – Dissolves – Like Rule • When solvent and solute molecules have similar intermolecular intermolecular forces, they are alike (polar alike (polar ↔ polar or nonpolar nonpolar) ↔ nonpolar). • Heat required to separate molecules ~ Heat released when they come together. • ∆Hsoln = ∆H1 + ∆H2 + ∆H3~ 0 • Mixing occurs spontaneously, since natural tendency is toward increasing disorder (entropy ↑). disorder O Vit. C C HO C HO H3C CH3 H C CH HC C C O C HO Vit. B C Fall 2009 UM-SJTU JI C A solution of methanol in water O CH2 CH2 N 2 H C H OH CH 2 C CH2 CH3 C CH CH O CH3 CH C CH CH CH3 CH3 C Vit. A C OH 2 Hint: Polar bonds such as O-H make polar molecules. H H HO C C CH C C C CH2 H 3C C O CH3 CH3 CH3 CH OH O HO CH3 1) Name the forces that must be overcome to break up each solute. 2) Name the forces that attract each solute to dissolve. HC C H C H H H H CH2 methanol Figure 13.6 Apply the like-dissolves-like rule to determine if each vitamin is water soluble. vitamin water water methanol methanol in water. Figure 13.3 C CH3 CH2 CH2 CH CH3 CH2 CH2 2 Vit. D CH2 CH CH3 CH2 CH 2 CH2 CH 2 2 CH3 Ref. page 540 Textbook Page 2 Chem 210 / WANG / Chapter 13A Properties of Solution LikeLike-dissolves like rule & Compds of double nature Compds double Figure 13.9 Lecithin – the phospholipid found in all cell membranes, has polar and nonpolar parts in tis structure. • One end of the molecule is nonpolar and the other end is polar. • It brings polar and nonpolar molecules together. Water solubility For Most SOLIDS… increases increases with temperature, temperature, but pressure pressure has little effect. Why? Figure 13.17 -OOC -OOC - COO COO -OOC COO -OOC COO - - Temperature and solubility 300 decreases decreases with temperature, temperature, but increases increases greatly with pressure. pressure. Why? Figure 13.18 250 SO2 KCl glycine NaBr KNO3 sucrose 200 150 100 50 0 0 Pressure compresses gas molecules into liquid solutions. Pressure has little effect on solids / liquids, which are not compressible. Nonpolar ‘tail’ (long chain of C and H) dissolves in grease/dirt. Polar ‘head’ (-COO-1) (stays in water. Micelles cause Micelles cause soapy water to appear cloudy, Colloidal Colloidal suspension. Rinse Rinse water carries micelles away, along with the grease/dirt trapped in them. For most GASES… Water solubility Higher T molecules have more kinetic energy to Stay in the gas phase. Higher T Molecules have more kinetic energy to •Break away from a solid. • Mix with water molecules. How soap works Solubility (g/100ml water) Explain the trend of solubility in Water and (b) Hexane. (Table 13.3) 20 40 60 80 100 Temperature (oC) Point out two major differences between SO2 and the rest. and Explain how KNO3 and sucrose differ from the others. Henry’s Law Gas solubility in a liquid is liquid directly directly proportional to the partial partial pressure of the gas above the liquid when temperature temperature is held constant. Sgas = kPgas S = Solubility of gas in solution Pgas = partial pressure of gas k = Henry’s constant at certain T. Fall 2009 UM-SJTU JI Gas Gas Solubility: Another way to see Henry’s Law Solubility (S) of a gas in a liquid is directly proportional to its partial pressure (P). S = constant P or S1 S =2 P1 P2 temperature must be held constant! The table below shows the solubility of several gases at P = 1 atm. Estimate the concentration of O2 in river water. Since air is composed of 20% O , if air = [O2] = 1.2 x 10-3 M 1.0 atm, partial pressure of O = 2P = O2 2 @ PO2 = 1 atm 0.20 atm S1 S =2 P1 P2 1.2x10-3 [O2] = 1.0 0.20 [O2] x 1.0 = 0.20 x 1.2 x10-3 [O2] = 2.4x10-4 M Page 3 Chem 210 / WANG / Chapter 13A Properties of Solution Solubility, saturated, unsaturated saturated unsaturated and supersaturated Solubility depends on a dynamic equilibrium dynamic between dissolving between dissolving and returning of solutes. of Ref. Fig 13.9 Differ in terms of the amount of solute in a solution. Unsaturated - Dissolving rate > Returning rate Supersaturated solution is not stable. Figure 13.10 Saturated – Dissolving rate = Returning rate. Supersaturated • Temporarily have dissolved too much. unstable • An unstable state – solute settles out when disturbed or over time. Solubility = Solute concentration in a Solute saturated saturated solution, a solution in equilibrium. solute (undissolved) (undissolved) Different concentration units: • Mass percentage • Parts per million • Parts per billion • Mole Fraction • Molarity • Molality solute (dissolved) Solution Concentration All units specify: • quantity of solute solute per • quantity of solvent solvent or solution. Calculate volume needed in dilution Prepare 350 mL of 4.5 M Hydrogen peroxide solution from 8.8 M concentrated solution. M1V1 = M2V2 Step 1: calculate M1 = 4.5 M M2 = 8.8 M V1 = 350 mL V2 = ? 4.5 x 350 = 8.8 x ? ? = 178.98 = 180 mL I need 180 mL of concentrated solution. Step 2: go to the lab and obtain a 350-mL 350volumetric flask. Step 3: Transfer the 180 mL concentrated solution into the flask with a burette (more precise than graduated cylinder). Step 4: Add distilled water to the etched mark. etched Fall 2009 UM-SJTU JI Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. Molarity M= moles solute = liters of solution mol L Molarity • Recognizes that compounds have different formula weights. • A 1 M solution of glucose contains the same number of molecules as 1 M ethanol. • [ ] - special symbol which means molar ( mol/L ) or M Molarity (a) Calculate the molarity of 1.33 g KOH in 2.5L solution MKOH = (1.33 gKOH ÷ 56.1 g/mol) / 2.5 L = 0.0095 M (b) Calculate the grams of solute in 500.0 mL of 0.0119 M NaCl gNaCl = 0.0119 MNaCl x (500.0 mL ÷ 1000) = 5.95 x10-3 mole x 58.5 g/mol = 0.348 g NaCl Mass Percentage (%), ppm, ppb ppm, % by mass = Mass solute 2 Mass of solution x 10 use g for both Parts per million (ppm)= Part (Solute) x 106 Total (solution) Use the same units (either mass or volume) for both Parts per billion (ppb)= Part (Solute) x 109 Total (solution) Use the same units (mass or volume) for both Metric prefixes Changing the prefix alters the size of a unit. Prefix mega kilo hecto deka base deci centi milli Micro Symbol M k h da d c m µ 106 103 102 101 100 10-1 10-2 10-3 10-6 Factor 1,000,000 1,000 100 10 1 0.1 0.01 0.001 0.000001 Page 4 Chem 210 / WANG / Chapter 13A Properties of Solution Express these aqueous solutions in ppb ppb and ppm NOTE: For dilute aqueous solutions, 1 g = 1 mL = 1 cm3 mL 1 kg = 1000 g = 1000 mL = 1 L (a) 5 µg solute per L solution L = kg kg = 5 µg / 1 kg x 109 = 5 ppb ppb -6g /103 g) = 5 x10-9 = (5 x 10 x 106 = 0.005 ppm 10 (b) 3.5 mg per kg = 3.5 mg / 1 kg x 109 = 3500 ppb 3500 = (3.5 x 10-3g /103g) = 3.5 x10-6 x 106 = 3.5 ppm 10 (c) 50 g per m3 cm3 cm g = 50 g / (102 cm)3 = 50 g / (106 cm3) 50,000 = (50 g / 106 g) = 50 x 10-6 x 109 = 50,000 ppb x 106 = 50 ppm 10 SAMPLE PROBLEM 2 Mass % versus Molarity What is the molarity of a 3.0% by mass H2O2 3.0% solution, density = 1.05 g/mL. 3% = 3.0 g H2O2 and 100 g solution 3.0 100 ? Mole H2O2 /1 L of solution Mole /1 3.0 g H2O2 100 g soln mole = 3.0 ÷ 34.0 = 0.088 mole H2O2 100g ÷ 1.05g/mL = 95.2 mL = 0.0952 L Molarity = moles ÷ Liters = 0.088 mole ÷ 0.0952 L = 0.92 M Molality (m) Molality, m = mole(Solute) Kg (solvent) SAMPLE PROBLEM 1(a) 40.5 mg Ca x Fall 2009 UM-SJTU JI g 103 mg x 106 3.5 g = 1.16x104 (b) The label on a 0.750-L bottle of Italian Chianti indicates “11.5% alcohol by volume”. How many milliliters of alcohol (C2H5OH) does the wine contain? 0.750 L wine x ppm Ca 11.5 L alcohol 100 L wine = 0.0862 L alcohol × Chianti wine in a traditional fiasco. 1000 mL 1L = 86.2 mL alcohol SAMPLE PROBLEM 3 Mole Fraction ( mf ) Compare M, %, ppm and ppb Which NaCl solution is most concentrated? 1x10-2 % 1 ppm 1000 ppb 1 x10-3 M 0.001 mole NaCl 1000 mL = 0.0585 g NaCl 1000 mL = 5.85 x 10-5 g/mL Since:1 g ~ 1mL = 5.85 x 10-5 0.01 g NaCl 100 g = 1 x 10-4 1 g NaCl 106 g = 1 x 10-6 1000 g NaCl 109 g = 1 x 10-6 This is the highest! These are the same! Both very low. SUMMARY: Concentration Definitions Concentration Term Ratio Molarity (M) Molality (m) moles of solute mass (kg) of solvent Mole fraction (mf): XA = Parts by mass Parts by volume Both of the same unit: mole SAMPLE PROBLEM 13.4 What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water? Mole fraction (m.f.) 32.0 g CaCl2 x mass of solute mass of solution Volume of solute volume of solution moles of solute mole solute + mole solvent mole(Solute) mole (solution) What is the mole fraction of each component in a solution of 36% HCl by mass? Sum of mf of all components must = 1 36% HCl = 36 g HCl and 64 g H2O 36 g HCl ÷ 36.5 g/mole = 0.986 mole HCl 64 g H2O ÷ 18.0 g/mole = 3.56 mole H2O XHCl = 0.986 ÷(0.986 + 3.56) = 0.22 XH2O = 1.00-0.22 = 0.78 moles of solute volume (L) of solution Moles of solute per Kg of solvent The ocean contains 19,000 ppm of chloride ion. What is the molality of this ion? 19000 ppm Cl-1 = 19000 g Cl-1 / 106 g solution Solute Cl-1= 19000g ÷ 35.5 g/mole = 535 mole Cl-1 mole Solvent H2O = 106 g – 19000 g Cl-1 = 9.8 x 105 g H2O = 9.8 x 102 kg H2O mCl- = 535 mole Cl-1 / 9.8 x 102 kg H2O = 0.55 m mole Cl- SAMPLE PROBLEM 1(b) (a) Find the concentration of calcium (in ppm) in a 3.50-g pill that contains 40.5 mg of Ca. molality = mole CaCl2 110.98 g CaCl2 = 0.288 mole CaCl2 0.288 mole CaCl2 kg 271 g H2O x 103 g = 1.06 m CaCl2 Page 5 Chem 210 / WANG / Chapter 13A Properties of Solution SAMPLE PROBLEM 13.5 SAMPLE PROBLEM 13.6 A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water? moles ethylene glycol = 142 g mole = 2.36 mol 60.09 g C3H8O mole moles water = 58.0g 18.02 g = 3.22 mol H2O Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution a a hair bleach. An aqueous solution H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate its: (a) Molality SAMPLE PROBLEM 13.7 Compare M, mf and m Which NaCl solution is most concentrated? 1 x10-2 M 1x10-2 mf 1x10-2 m 1000 ppm (b) Mole fraction of H2O2 2.36 mol C3H8O 2.36 mol C2H8O + 3.22 mol H2O C3H8O m.f. = 0.423 m.f. 0.423 3.22 mol H2O 2.36 mol C3H8O + 3.22 mol H2O (c) Molarity H2O m.f. = 0.577 m.f. 0.577 Fall 2009 UM-SJTU JI Page 6 ...
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