SOLUTIONS. General Chemistry I CHAPTER

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1 11 CHAPTER SOLUTIONS 11.1 Composition of Solutions 11.2 Nature of Dissolved Species 11.3 Reaction Stoichiometry in Solutions: Acid-Base Titrations 11.4 Reaction Stoichiometry in Solutions: Oxidation-Reduction Titrations 11.5 Phase Equilibrium in Solutions: Nonvolatile Solutes 11.6 Phase Equilibrium in Solutions: Volatile Solutes 11.7 Colloidal Suspensions

2 473 Dissolution of sugar in water

3 COMPOSITION OF SOLUTIONS Percent composition mass of solute Mass (or Weight) % = 100 mass of solution vol of solute Vol % = 100 vol of solution Parts per million & parts per billion mass of solute ppm = 10 O(mg L mass of solution 6 1 mass of solute ppb = 10 O( µ g L mass of solution 9 1 ) )

4 474 Mole Fraction n n X =, X = = 1 X (for a binary mixture) n1+ n2 n1+ n2 Molarity & Molality Molarity Molality moles solute liters solution ( M) = = mol L moles solute kilograms solvent ( m) = = mol kg 1 1 Molality is independent of temperature!

5 477 Preparing a solution of NiCl 2 using a volumetric flask

6 477 Dilution of solution Chemical amount conserved. n = cv = i i cv moles solute cv i i c f f f = = final solution volume V n: number of moles of solute c i(f) : initial (final) concentration in molarity V i(f) : initial (final) solution volume in liters f

7 NATURE OF DISSOLVED SPECIES Aqueous Solutions of Molecular Species Polar molecules dissolved by water ~ Like dissolves like Sugars: C m (H 2 O) n ~ Sucrose (C 12 H 22 O 11 ), Fructose (C 6 H 12 O 6 ), Ribose (C 5 H 10 O 5 ) ~ Do not contain water molecules ~ Include polar OH groups ~ Dipole-dipole interaction between OH groups and water molecules hydrogen bonds

8 479 Fig Electrostatic potential energy surface of a fructose molecule and its hydrated form in aqueous solution. Four water molecules are bonded with hydrogen bondings.

9 479 Aqueous Solutions of Ionic Species (Electrolytes) Solubility: Maximum mass dissolved in 1L at 25 o C K 2 SO 4 (s) 2 K + (aq) + SO 2 4 (aq), Solubility of K 2 SO 4 : 120 g L 1 at 25 o C Dissolution of ionic species Ion-dipole forces Each ion is surrounded by an intact solvation shell of water molecules. ~ Good conductor, strong electrolyte ~ Electrophoresis under an electric field

10 480 Insoluble salts BaSO 4 (s) Ba 2+ (aq) + SO 2 4 (aq), Solubility of BaSO 4 : g L 1 at 25 o C Precipitation reaction BaCl 2 (aq) + K 2 SO 4 (aq) BaSO 4 (s) + 2 KCl(aq) Ba 2+ (aq) + 2 Cl (aq) + 2 K + (aq) + SO 2 4 (aq) BaSO 4 (s) + 2 K + (aq) + 2 Cl (aq) Net ionic equation Ba 2+ (aq) + SO 2 4 (aq) BaSO 4 (s) Spectator ions: Cl and K +

11 Potassium sulfate (K 2 SO 4 ) 480 Fig Dissolves in water. Conducts electricity. Fig 11.4 Reacts with BaCl 2 BaSO 4 Fig 11.5

12 11.3 REACTION STOICHIOMETRY IN SOLUTIONS: ACID-BASE TITRATIONS Reactions in Solution Ex. 2 Br (aq) + Cl 2 (aq) 2 Cl (aq) + Br 2 (aq) We have 50.0 ml of M solution of NaBr. Q. What is the volume of M Cl 2 to react completely with Br? n = ( L)( mol/l) = mol n = mol (1/2) = mol V = ( mol) / ( mol/l) = L Q. What is the concentration of Cl? V = V + V = + = L L L Br ( aq) 2 aq total Cl ( ) 3 [Cl ] = n / V Cl total = mol / L = M n Cl = n Br 481

13 Background on Acid-Base Reactions 483

14 483 Names of common acids Binary acids Oxoacids Organic acids HF, HCl, H 2 S H 2 CO 3, HNO 3 HCOOH, CH 3 COOH(HAc) HCN H 2 SO 4, H 3 PO 4 C 6 H 5 COOH HCl HNO 3 H 2 SO 4 HAc Ionization of pure water H 2 O(l) H + (aq) + OH (aq) [ 2 H 2 O(l) H 3 O + (aq) + OH (aq) ] hydronium ion

15 483 Arrhenius s definition of Acids and Bases Acid ~ produces H + in aqueous solution (> [H + ] water ) HCl(g) H + (aq) + Cl (aq) Base ~ produces OH in aqueous solution (> [OH ] water ) NaOH(s) Na + (aq) + OH (aq) NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH (aq) Titration Determination of unknown amount of a sample Titrant (in buret, known concentration, volume to be measured) Titrate (in flask, unknown concentration, known volume) End point, Indicator

16 484 Acid-Base Titration CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H 2 O(l) Indicator ~ Phenolphthalein Ex ml of M NaOH ml of a vinegar sample Calculate [CH 3 COOH(aq)] in the sample. At the end point, n(naoh) = n(ch 3 COOH) = (1.306 mol/l)( L) = mol [CH 3 COOH] = ( mol)/( L) = M

11.4 REACTION STOICHIOMETRY IN SOLUTIONS: OXIDATION-REDUCTION TITRATIONS 485 Background on Oxidation-Reduction (Redox) Reactions Oxidation ~ increase in the oxidation number, donate electrons

17 11.4 REACTION STOICHIOMETRY IN SOLUTIONS: OXIDATION-REDUCTION TITRATIONS 485 Background on Oxidation-Reduction (Redox) Reactions Oxidation ~ increase in the oxidation number, donate electrons Reduction ~ decrease in the oxidation number, accept electrons Ca(s) + Cl 2 (g) CaCl 2 2e = 2 1e Mg(s) + O 2 (g) 2 MgO 2 2e = 2 2e Fig Mg burning in air. Flash powder for photograph fireworks

18 Balancing Oxidation-Reduction Equations 486 CuS(s) + NO 3 (aq) Cu 2+ (aq) + SO 2 4 (aq) + NO(g) in aqueous nitric acid Step 1. Divide into two unbalanced half-reactions. CuS Cu 2+ + SO 2 4 NO 3 NO Step 2. Balance all elements except oxygen and hydrogen. Already done. Step 3. Balance oxygen by adding H 2 O. CuS + 4 H 2 O Cu 2+ + SO 2 4 NO 3 NO + 2 H 2 O

19 Step 4. Balance hydrogen. 487 ~ Add H 3 O + / H 2 O (acidic) or H 2 O / OH (basic) CuS + 8 H 2 O + 4 H 2 O Cu 2+ + SO H 3 O + NO H 3 O + NO + 2 H 2 O + 4 H 2 O Step 5. Balance charge using e. CuS + 12 H 2 O Cu 2+ + SO H 3 O e NO H 3 O e NO + 6 H 2 O Step 6. Combine the two half-reactions canceling e. 3 ( CuS + 12 H 2 O Cu 2+ + SO H 3 O e ) 8 ( NO H 3 O e NO + 6 H 2 O ) CuS + 8 NO H 3 O + 3 Cu SO NO + 12 H 2 O

20 489 Disproportionation ~ Redox reaction in which a single substance is both oxidized and reduced. Redox Titration Titration of Fe 2+ in an iron ore with KMnO 4 MnO 4 (aq) + 5 Fe 2+ (aq) + 8H + (aq) Mn 2+ (aq) + 5 Fe 3+ (aq) + 4H 2 O(l) Volume of M KMnO 4 at the end point ~ ml n (MnO 4 ) = ( L)( mol/l) = mol n (Fe 2+ ) = n(mno 4 ) 5 = mol

21 491 Titration of Ca 2+ in an iron ore with KMnO 4 ~ Indirect method Ca 2+ (aq) + C 2 O 2 4 (aq) CaC 2 O 4 (s) oxalate calcium oxalate CaC 2 O 4 (s) + 2H + (aq) Ca 2+ (aq) + H 2 C 2 O 4 (aq) oxalic acid 2 MnO 4 (aq) + 5 H 2 C 2 O 4 (aq) + 6 H 3 O + (aq) 2 Mn 2+ (aq) + 10 CO 2 (g) + 14 H 2 O(l) Oxalic acid, H 2 C 2 O 4

$11.5 PHASE EQUILIBRIUM IN SOLUTIONS: NONVOLATILE SOLUTES 491 Raoult s law ~ for Ideal solutions Vapor pressure of solvent in solution P 1 its mole fraction X$

22 11.5 PHASE EQUILIBRIUM IN SOLUTIONS: NONVOLATILE SOLUTES 491 Raoult s law ~ for Ideal solutions Vapor pressure of solvent in solution P 1 its mole fraction X 1 P 1 = X 1 P 1 o P 1o : vapor press of pure solvent Plot of P 1 vs. X 1 ~ a straight line Fig Vapor pressure of solvent in ideal and nonideal solutions.

23 Real solutions 492 ~ Positive or negative deviations from the linear behavior of Raoult s law Real solutions ideal solutions as X 1 1 (dilute) Negative deviations ~ solute-solvent attractions > solvent-solvent attractions Positive deviations ~ solute-solvent attractions < solvent-solvent attractions

24 Colligative properties 492 Collective properties depending on the number of dissolved particles rather than the nature of the particular particles involved Determination of Molar mass Let us consider Nonvolatile solute, Nonelectrolyte, Binary solution Vapor pressure lowering Boiling-point elevation Freezing-point depression Osmosis

25 492 Vapor-Pressure Lowering P= P P = XP P = XP o o o o P 1 (above a dilute solution) < P 1o (above a pure solvent) Boiling-Point Elevation Let P 1o = 1 atm and n 1 >>n 2 (dilute solution). Slope of P 1 vs T curve: S = P 1 / T b TT bb = PP 1 SS = XX oo 2PP 1 SS = 1 SS nn 2 nn 1 + nn 2 = 1 SS nn 2 nn 1 = 1 SS mm 2 /MM 2 mm 1 /MM 1

26 493 Boiling-point elevation constant (or Ebullioscopic constant) K = M 1 b 1 ( 1000 g kg ) S m/ 2 M2 T = K m/ 1 g kg ( 1000 ) b b 1 T = b Kbm K b = K kg mol 1 for water Fig Boiling point elevation.

27 494

28 494 EXAMPLE When 6.30 g of an unknown hydrocarbon is dissolved in g of Benzene, the boiling point of the solution increases by o C. What is the molar mass of the unknown substance? mm = TT bb KK bb = KK = mmmmmm kkkk KK kkkk mmmmmm 1 nn = mmmmmm kkkk kkkk = mmmmmm mmmmmmmmmm mmmmmmmm = mmmmmmmm nn = 6.30 gg mmmmmm = 178 gg mmmmmm 1 Hydrocarbon: anthracene (C 14 H 10 )

29 495 Freezing-Point Depression T = T T = K f f f f m K f : cryoscopic constant K f = 1.86 K kg mol -1 for water For dissociating solutes, count total molalities. Ca(NO 3 ) 2 total 3 moles (Ca 2+, 2 NO 3 s) Fig. Freezing-point depression

30 497 Fig Freezing-point depression of ionic substances. HAc(M), NaCl(MX), FeCl 3 (MX 3 )

31 Osmosis 498 Fig Measurement of the osmotic pressure.

32 Osmosis 499 Osmosis is the movement of solvent particles from a region of lower to a region of higher concentration through a semi-permeable membrane. (Larger solute molecules can not pass through the membrane.) Osmotic pressure, π, is the pressure that must be applied to prevent the net flow of solvent. π =crt van t Hoff equation c = molarity R = atm L/mol K T = Kelvin temperature

33 Reverse Osmosis 499

34 499 EXAMPLE A chemist dissolves 2.00 g of a protein in L water. The osmotic pressure is atm at 25 o C. What is the approximate molar mass of the protein? cc = ππ RRRR = aaaaaa LL aaaaaa mmmmmm 1 KK KK = mmmmmm LL 1 MM = 20.0 gg mmmmmm = 23,000 gg mmmmmm 1 Osmotic pressure is useful for large molecules with low solubilities. Under the identical conditions, Osmotic pressure with h = 22 cm Vapor-pressure lowering = atm Boiling-point elevation = K Freezing-point depression = K

11.6 PHASE EQUILIBRIUM IN SOLUTIONS: VOLATILE SOLUTES 499 Ideal mixture of volatile substances P 1 = X 1 P 1o, P 2 = X 2 P 2o = (1-X 1 )P o 2 Henry s Law For X 2 <<1, P 2 = k 2 X 2 k 2 : Henry s law

35 11.6 PHASE EQUILIBRIUM IN SOLUTIONS: VOLATILE SOLUTES 499 Ideal mixture of volatile substances P 1 = X 1 P 1o, P 2 = X 2 P 2o = (1-X 1 )P o 2 Henry s Law For X 2 <<1, P 2 = k 2 X 2 k 2 : Henry s law constant For X 1 <<1, P 1 = k 1 X 1 ~ Henry s law applies to a volatile solute in a dilute solution. ~ Carbonation of soft drinks ~ O 2 dissolved in blood Fig Vapor pressures above a mixture of two volatile liquids.

36 Distillation P 1o = atm for hexane (C 6 H 14 ) at 25 o C. P 2o = atm for heptane (C 7 H 16 ) at 25 o C. A solution of n 1 = 4.00 mol and n 2 = 6.00 mol. 501 Mole fractions in the solution X 1 = 0.400, X 2 = P 1 = X 1 P 1o = (0.400)(0.198 atm) = atm P 2 = X 2 P 2o = (0.600)( atm) = atm P tot = P 1 + P 2 = atm Mole fractions in the vapor (Dalton s law) P 1 = Y 1 P tot, P 2 = Y 2 P tot Y 1 = atm / atm = Y 2 = 1 Y 1 = = 0.312

condensation on the basis of their different boiling points

37 Fractional Distillation ~ Separation of two or more components of a liquid solution by successive evaporation and condensation on the basis of their different boiling points 502 Fig Vapor pressure vs. X 2. Fig Boiling temperature vs. X 2.

38 Fig An apparatus for fractional distillation. 503

39 503 Azeotropes Maximum-boiling azeotrope Fig 11.19a HCl / H 2 O ~ large negative deviation, strongly attractive ~ T b = o C at X HCl = Minimum-boiling azeotrope Fig 11.19b C 2 H 5 OH / H 2 O ~ large positive deviation ~ T b = o C at X H2O = ~ the last 4% of water can not be removed by distillation An azeotrope behaves like a single-component fluid.

Negative deviation Positive deviation 503 Fig. 11.

40 Negative deviation Positive deviation 503 Fig (a) Maximum-boiling azeotrope. (b) Minimum-boiling azeotrope.

41 The Strongest Alcoholic Drink in the World v% 80 v% 72 v% 76 v%

42 COLLOIDAL SUSPENSIONS Colloids A dispersion of large particles (1 nm ~ 1 μm) in a solvent Intermediate between a solution and a heterogeneous mixture Homogenous appearance but scatters light

43 Flocculation ~ Acceleration of the settling out of a colloid through the addition of salts. ~ Salts reduce the electrostatic repulsions between suspended particles. (ex. river deltas, paints) 504 Fig Process of flocculation: Dispersion Aggregation Sedimentation

505 Precipitation ~ flocculation, centrifugation, membrane filtration Fig. 11.

44 505 Precipitation ~ flocculation, centrifugation, membrane filtration Fig (a) This colloidal suspension of PbCrO 4 appears cloudy. (b) After flocculation, the precipitate settles to the bottom.

45 Problem Sets For Chapter 11, 3, 13, 18, 25, 34, 43, 55, 62, 72, 83

SOLUTIONS. Dissolution of sugar in water. General Chemistry I. General Chemistry I CHAPTER

SOLUTIONS. Dissolution of sugar in water. General Chemistry I. General Chemistry I CHAPTER 11 CHAPTER SOLUTIONS 11.1 Composition of Solutions 11.2 Nature of Dissolved Species 11.3 Reaction Stoichiometry in Solutions: Acid-Base Titrations 11.4 Reaction Stoichiometry in Solutions: Oxidation-Reduction