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Chapter 12 - Solutions

Chapter 12 - Solutions. A) Sapling HW 12 Due by 11:50 pm Monday, 11/4/2013 B) End-of-Chapter Problems. C) Final exam on Wednesday, December 11 from 8-10 am. Final exam includes all the chapters & lectures – including my lab reviews. D) Quiz #2: Chapters 11&12 on Monday, November 4, 2013

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Chapter 12 - Solutions

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  1. Chapter 12 - Solutions A) Sapling HW 12 Due by 11:50 pm Monday, 11/4/2013 B) End-of-Chapter Problems. C) Final exam on Wednesday, December 11 from 8-10 am. Final exam includes all the chapters & lectures – including my lab reviews. D) Quiz #2: Chapters 11&12 on Monday, November 4, 2013 Exam #2: Chapters 11&12 on Wednesday, November 6, 2013

  2. End-of-Chapter ProblemsPages 526 - 532 3 - 9 11 15 16 17 19 20 21 22 23 24 28 32 35 40 - 43 47 49 51 56 59 69 73 75 77 79 81 99

  3. I. Solutions A. Types Definitions: Solution – Homogeneous mixture of two or more substances. Solvent – Single solution component present in greatest amount. Solute(s) – Solution component(s) present in smaller amounts. Solubility – Maximum amount of solute that can dissolve at a given temperature (Note: Is a dynamic equilibrium) Solid Solute Dissolved Solute Saturated Solution – Solution with maximum amount of dissolved solute. Hydration - Attraction of solute ions for water molecules. Solvation - Attraction of solute ions for any solvent molecules. Miscible - Fluids that completely dissolve in each other.

  4. I. Solutions A. Types Types:SoluteSolvent Examples 1 Gas Gas Air (O2 Ar CO2 in N2 ) 2 Gas Liquid Soda Water (CO2 in H2O) 3 Liquid Liquid Ethyl Alcohol in H2O 4 Solid Liquid NaCl in H2O 5 Liquid Solid Hg in Ag – old dental filling 6 Solid Solid Brass – Zn, Sn, Pb, Fe in Cu

  5. I. Solutions B. Rules • General rule for solutions is “like dissolves like” in terms of polarity. Polar solvent can dissolve polar solutes and non-polar solvent can dissolve non-polar solutes. • Examples: • Polar CH3OH will dissolve as a molecule in polar H2O. • Ionic KI will dissolve as ions in polar H2O. • Non-polar C6H14 will not dissolve in polar H2O. • Non-polar C6H14 will dissolve in non-polar CCl4.

  6. I. Solutions B. Some Solubility Rules for IONIC compounds in water (Know These Chem 121 Rules) • All group IA (1), ammonium, acetate & nitrates are soluble. Examples: NaI NH4Cl Ca(C2H3O2)2 Pb(NO3)2 K3PO4 = soluble • All Chlorides - Iodides are soluble except silver, mercury & lead. Examples: CrCl3 = soluble PbBr2 = insoluble AgCl = insoluble • All sulfides are insoluble except for first rule above. Examples: (NH4)2S = soluble CdS= insoluble Ag2S = insoluble • All hydroxidesinsoluble except for first rule above & Ca, Sr, Ba. Examples: Ni(OH)2 = insoluble Al(OH)3 = insoluble Ba(OH)2 = soluble

  7. I. Solutions B. Solubility Rules for ionic compounds in H2O cont . - Why are some ionic compounds insoluble; why are some dissolutions in water endothermic & others exothermic? - Dissolving involves two factors: A - Break lattice structure = Energy required/added. B- Solvation of released parts = Energy released. • Dissolving most compounds is exothermic: B > A • Many ionic compounds are insoluble : A > B Examples: 1)CuSO4 orH2SO4(caution)in water = Exothermic (B >> A) 2) NH4NO3in water = Endothermic (A > B) 3) AgCl, BaSO4, CuS are insoluble ionic compounds (A >> B) Note: Heating the solvent for an insoluble compound usually increases its solubility; can be used for purification – recrystallization.

  8. I. Solutions B. General Rules • Polar CH3OH dissolves in H2O; energy released from H bonds formed. • AgCl is insoluble in water; hydration energy not enough to overcome high lattice energy. • Ionic LiF dissolves in H2O since polar H2O solvates ions (Hydration) which releases enough energy to pull ions out of LiF lattice. F-

  9. I. Solutions B. General Rules - Ivory Soap in water: CH3 - (CH2)16 - C – O- Na+ = O - 1) Water dissolves ionic/polar compounds & 2) Non-polar part of soap dissolves non-polar substances like grease. Soap forms micelles.

  10. I. Solutions B. General Rules – Le Chatelier’s Principle Le Chatelier’s Principle: A system at equilibrium will rearrange to relieve any stress. (Know this) A saturated gaseous solution is in equilibrium as illustrated: CO2 (aq) CO2 (gas) Apply pressure & above equilibrium will rearrange to remove extra pressure (dissolve gas; go to left).

  11. I. Solutions B. Rules - Henry’s Law • In 1803 William Henry studied the effect of pressure on gas solubility. • Henry’s Law: Solubility of a gas (S) α to the partial pressure of the gas above the solution: S1 = k x P1 or S2 = P2(Know these) S1 P1 • Important to a scuba diver; Henry's Law tells us how much N2(g) will be absorbed in our blood at various pressures. • Example: A diver is at 3.0 atm (P1) where solubility of N2 in blood is 53 ug/L (S1). What is the solubility (S2) of N2 in blood when the diver surfaces at 1.0 atm (P2)? S2=P2S2 = P2x S1 = 1.0 atm x 53 ug/L = 18 ug/L (18 ppb) S1 P1 P1 3.0 atm Note: 53 – 18 = 35 ug N2(g) released per liter of blood; get small N2 bubbles in blood.

  12. II. Solution Concentration Terms A. Review - (know all of these) 1. W / W % = g solute x 100 % (also have W / V & V / V %) g solution - 18.0 g NaCl is added to 200. g of water. What is the mass % NaCl? % = 18 g NaCl x 100 % = 8.26 % 218 g soln 2. ppm, ppb, ppt ppm = ug/mLppb = ng/mLppt = pg/mL (Note: g = mL for dilute, aqueous solutions; so, ppm also = ug/g, etc) - 4.2 x 10-8 g of Pb+2 is dissolved in 1000 mL H2O. What is ppt Pb+2 ? 4.2x10-8 g 1012 pg = 42 pg = 42 ppt 1000 mL g mL

  13. II. Solution Concentration Terms A. Review - know 3. Mole Fraction, X = Xa / XtotalNote: X is always ≤ 1.00 - 0.12 mol N2 is mixed with 0.33 mol O2 and 0.01 mol Ar. What is X of N2 XN2 = 0.12 mole (N2) = 0.26 (units cancel) 0.12 + 0.33 + 0.01 mole Total Note: Sum of all Xa’s in a solution = 1.0000 4.Molarity, M = moles solute / L solution M = m / L - 0.80 g of NaOH is dissolved in 200. mL. Calculate the M M = 0.80 g NaOH x 1 mol / 40. g NaOH = 0.10 mol = 0.10 M 0.200 L L Note: Use M1 x V1 = M2 x V2 only for dilution problems (moles initially = moles after)

  14. II. Solution Concentration Terms B. Molality, m 5. molality, m = moles Solute kg Solvent - This concentration term is strange, but useful for colligative property calculations; it’s the only one dealing with amount / “SOLVENT.” Example: 5.7 g of glucose (MW=180) is in 30.9 g of solution. Calculate m. - 5.7 g glucose x (1 mol / 180 g) = 0.032 mol glucose (solute) - 30.9 g soln - 5.7 g solute = 25.2 g solvent = 0.0252 kg solvent m = moles glucose = 0.032 mol glucose = 1.3 m kg solvent 0.0252 kg solvent

  15. III. Colligative Properties A. Vapor Pressure Lowering (VPL) - Colligative Property = Property that depends only on # of solute particles (not on the type of the solute particles). - Examples: VPL, BP elevation, FP depression, Osmosis - VPL = a nonvolatile solute lowers the vapor pressure of a solvent. - VP relationship = Raoult’s Law: Pa = Psa x Xa(Know this) Pa is VP of solvent(in the solution) Psa is VP of pure solvent Xa is the mole fraction of the solvent. - Note: are other forms of Raoult’s Law.

  16. III. Colligative Properties A. Vapor Pressure Lowering (VPL) Example: What is VP & change in VP of 1.00 mol H2O that contains 0.100 mole NaI at 20. oC? Given: VP of H2O (Psa) at 20. oC = 17.5 Torr Pa= Psa x Xa = 17.5 torr xXa Need mole fraction of water: **0.100 mol NaI yields 0.200 mols of particles (Na+ & I-) when it dissolves. **Total mols of particles = 1.00 mol H2O + 0.200 mol ions = 1.20 mols Xa = 1.00 mole H2O / 1.20 mole = 0.833[mole fraction of water] Pa = PsaxXa Pa = 17.5 Torr x 0.833 = 14.5 Torr ΔVP = 17.5 - 14.5 = 3.0 Torr

  17. III. Colligative Properties B. BP Elevation & FP Depression • an added nonvolatile solute lowers MP & raises BP of a solvent; the change in BP or MP depends on the type of solvent and the number of particles of solute dissolved (colligative property). For MP depression:ΔTf = Kfxm (know these) For BP elevation:ΔTb = Kbxm - ΔTf & ΔTb are changes in BP or MP & m = molality of SOLUTE particles. - Kf & Kb are constants in units of oC/m; are unique for each solvent: Kf = 1.86 oC/m for water Kf = 40.0 oC/m for camphor Kb = 0.512 oC/m for water Kb = 3.12 oC/m for acetic acid

  18. III. Colligative Properties B. BP Elevation & FP Depression Example: Calculate the freezing point of a 50.0 % (W/W) ethylene glycol (C2H6O2 or EG) solution in your car radiator? 50.0% - assume 100. g of ethylene glycol is added to 100. g (0.100 kg) of H2O. Kf = 1.86 oC/m for H2O ΔTf = Kfx m = (1.86 oC/m) x ?m 1) Need m of ethylene glycol (MW = 62.0 g/mol) in 50.0 % aqueous soln: - 100. g C2H6O2 x 1 mol EG / 62.0 g EG = 1.61 moles C2H6O2 - m = 1.61 moles C2H6O2 / 0.100 kg H2O = 16.1 mols EG/kg water = 16.1 m 2) ΔTf = Kfxm = (1.86 oC/m) x16.1 m= 30.0 oC . MP = (0.0-30.0)oC = -30.0oC. Car engine protected to -30 oC (-22 oF)

  19. III. Colligative Properties B. BP Elevation & FP Depression • FP or BP changes can be used to determine MW of low MW molecules. • Example: 0.99 g sorbitolin 0.010 kgwater has a MP of -1.0 oC. Calculate the MW of sorbitol. - Problem solving method: MW = g / mol. You are given g; get mol. ΔTf = KfxmGiven: Kf = 1.86 oC/m for water m = ΔTf / Kf = 1.0 oC / 1.86 oC/m = 0.54 m(mol sorbitol per kg water) - (0.54 mol sorbitol / kg water) x 0.010 kgwater = 0.0054 mol sorbitol - MW = g/mol = 0.99 g sorbitol / 0.0054 mol sorbitol = 180 g / mol (2 SF)

  20. III. Colligative Properties B. BP Elevation & FP Depression • Question: An aqueous 1.0 molal solution of ethanol (CH3-CH2-OH) lowers the mp of water by 1.86 oC, but an aqueous 1.0 molal solution of NaCl lowers the mp of water by 3.72 oC (twice that of ethanol). Why? • Answer: In water: 1.0 mol of ethanol yields 1.0 mols of particles; 1.0 mol of NaCl (ionic) yields 2.0 mols of particles (1.0 mol of Na+ & 1.0 mol of Cl-). • Better equation: ΔTf = ixKfxm(know) • i = van’t Hoff factor (pg 517): i= # moles of ions per mole of compound i = 1.0 for nonelectrolyte (CH3OH) i > 1.0 for electrolyte (2.0 for NaCl) Question:Why is i = 1.02 for 1.00 M acetic acid?

  21. III. Colligative Properties C. Osmosis - Jacobus Van’t Hoff received 1st nobel prize in chem. in 1901 for work on solns. Definition of Van’t Hoff Factor: i = # moles of ions per mole of compound. (Know this) - Summary of Colligative Property Equations using the Van’t Hoff factor: - Note: assume i = 1.00 unless problem dictates another value. ΔTf = i x Kfxm ΔTb = i x Kbxm π = i xMx R x T (for osmotic pressure) (Know These) Note: i is automatically included in Raoult’s Law.

  22. III. Colligative Properties C. Osmosis - Osmosis is the net flow of solvent molecules from a pure solvent through a semipermeable membrane into a solution. - Osmotic Pressure (π) is a colligative property & equals the pressure that, when applied to the solution, just stops osmosis.

  23. III. Colligative Properties Examples & Notes Note: During osmosis, water molecules go from the purest water to the lesspure water – from less concentrated to more concentrated. • Skin gets “pruney” when one swims in ocean or takes a long bath. • Red blood cells puff up & burst when placed in deionized water. • Osmotic pressure can be very large • Can use π to obtain MW of large molecules. • We can use reverse osmosis to purify sea water.

  24. III. Colligative Properties Osmosis & Colloids - π = i x M x R x T (Know this) M = Molarity (mols / Lsoln) R = Gas Constant = 0.0821 La/Kmol T in K - Example: Whatis the π of a 0.2 M (i=1.0) soln. at 300 K? • π = 1.0 x 0.2 mol/L x [0.0821 Latm/Kmol] x 300 K = 5 atm (5 atm can support a 1.0 in2 column of water over 150 feet high) - π can be used to determine M of dilute solutions due to large values of π. Useful for determining MW of dilute solutions of high MW molecules - proteins, carbohydrates, nucleic acids

  25. III. Colligative Properties Osmosis – MW example • Problem:Hemocyanin (Hy) is a protein found in crabs. If 1.5 g of Hy dissolved in 0.250 L of water has a π of 0.00370 atm at 277 K, then what is the MW of Hy? Given i = 1.1 for Hy. • π = i x M x R x T MW Hy = g/mol = 1.5g/molM = π / ( i x R x T) 1)M of Hy = (0.00370 a) = 1.48 x 10-4 mol/L 1.1 x (0.0821 La/Km) x 277 K 2)1.48 x 10-4 mol/L x 0.250 L= 3.70 x 10-5 mol Hy 3)MW = g / mole = 1.5 g / 3.70 x 10-5 mol Hy = 41000 g/mol Note: If used mp or bp, the ΔTof this same solution would have been ~10-4oC; much too small to measure accurately.

  26. IV - Colloids - Colloid = dispersion (not solution) of particles of one substance throughout another substance. (Know this) - Colloid differs from solutions in that dispersion particles are usually between 1 and 200 nm in diameter – they are not in solution, but they are small enough to stay in suspension. - Colloids will give Tyndall effect = the scattering of a light beam by the colloidal particles. (Know this) -Three names for colloids: 1) aerosols = liquid/solid dispersion in a gas; 2) emulsion = liquid dispersed in a liquid; 3) sol = solid dispersed in a liquid

  27. IV – Colloids; common names 1) Aerosols= liquids or solids dispersed in a gas Examples: Fog (liquid water in air) Dust in air (solid dirt in air) 2) Emulsions= liquids dispersed in a liquid Examples: Mayonnaise (oil in water) Milk (fat in water) 3) Sols= solids dispersed in a liquid Examples: Jell-O (collagens in water) Milk (casein in water)

  28. IV . Colloids - Tyndall Effect

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