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Properties of Solutions. Solution Concentration Solubility Colligative Properties. I. Solution Process. A. Solutions: homogeneous mixture Solid, liquid, or gaseous solutions The solute : the substance present in the smaller amount I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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1. Properties of Solutions • Solution • Concentration • Solubility • Colligative Properties

2. I. Solution Process • A. Solutions: homogeneous mixture • Solid, liquid, or gaseous solutions • The solute: the substance present in the smaller amount • The solvent : the substance present in the larger amount

3. B. Solution Process • Can be exothermic or endothermic • Entropy increases during solution process • Dynamical equilibrium DHsoln= DH1+ DH2+DH3

4. C. Solubility • Solubility • Max amount of solute that can be dissolved in given amount of solvents • Saturated solution • contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. • Unsaturated solution • contains less solute • Supersaturated solution • contains more solute

5. moles of A xA = sum of moles of all components x 100% mass of solute x 100% = mass of solution mass of solute mass of solute + mass of solvent II. Concentration Unit A. Percent by Mass % by mass = B. Mole Fraction(x)

6. moles of solute liters of solution moles of solute m = mass of solvent (kg) M = II. Concentration Unit C. Molarity(M): depend on temperature D. Molality(m) * M ~ m in dilute aqueous solutions.

7. 2.00 g % mass= x 100% = 1.96 % 102.0 g 0.0434 moles C2H5OH m = = 0.434 m 0.100 kg solvent 0.0434 moles C2H5OH M = = 0.426 M 0.1020 L solution 0.0434 mol C2H5OH x = = 0.00776 0.0434 mol + 5.549 mol Q: 2.00 g of ethanol is mixed with 100.0 g of H2O to make 102 mL of solution. What is the % mass, m, M, and mole fraction of this solution? The density of the solution is 1.00 g/cm3. Final solution: m = 100.0g + 2.00g =102.0g, V = m/d = 102.0 g /(1.00 g/cm3) =102 mL 100.0g H2O x 1 mol H2O/ 18.02 g H2O = 5.549 mol H2O 2.00 g ethanol x 1 mol eth/ 46.07 g eth = 0.0434 mol ethanol

8. moles of solute moles of solute m= m= moles of solute M = mass of solvent (kg) mass of solvent (kg) liters of solution 5.86 moles C2H5OH = 0.657 kg solvent Q: What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? • Assume 1 L of solution: • 5.86 moles ethanol = 270 g ethanol • 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg = 8.92 m

9. III. Factors affecting solubility A. Solute-solvent interaction • The stronger solvent-solute interaction, the greater the solubility • Like dissolves like (similar intermolecular forces) • non-polar molecules are soluble in non-polar solvents ex. CCl4 in C6H6 • polar molecules are soluble in polar solvents ex. C2H5OH in H2O • ionic compounds are more soluble in polar solvents ex. NaCl in H2O Q: which of the following compounds will be miscible with CH3(CH2)4CH3? (a) H2O ______ (b) CCl4 _______ No Yes

10. Solid solubility gas solubility Solid solute: solubility generally as T , but not always • Gas solute: solubility as T increases B. Temperature Effect

11. = mass of NaOH 130 g x g mass of saturated soln (130+100) g 65.0 g Q: The solubility of NaOH in water is 130 g/100g H2O at 40oC; 65.0 g of saturated NaOH solution at 40oC contains how many gram of NaOH? is constant at 40oC x = 36.7 g

12. low P high P low c high c C. Pressure effect • Solids and liquids: little effects • Gases: solubility in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). C: concentration (M) of the dissolved gas c = kP P: pressure of the gas over the solution k: constant (mol/L•atm), depends on T and gas

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14. IV. Colligative Properties Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. • Vapor pressure (VP) lowering • Boiling point (BP) elevation • Freezing point (FP) depression • Osmotic pressure (p)

15. A. Vapor pressure lowering • Raoult’s law: P1=x1P1o P1o: vapor pressure of pure solvent, x1 : mole fraction of solvent • If the solution contains only one solute • when solute is nonvolatile (no measurable VP) P1=(1-x2) P1o x2 : mole fraction of solute DP= P1o - P1 = x2P1o

16. A. Vapor pressure lowering When solute is volatile PA=xAPAo PB=xBPBo PT= PA+PB

17. Q: The VP of pure ethanol (C2H5OH) and 1-propanol (C3H7OH) at 36oC are 108 mmHg and 40.0 mmHg, respectively. A mixture of 14.3 g of ethanol and 11.4 g of 1-propanol is prepared. (a) Determine the VP above the solution. (b) What is the mole fraction of ethanol in the vapor above this solution? Mole fraction of ethanol and 1-propanol in solution: mole ethanol = 14.3 g x (1mol C2H5OH/46.07 C2H5OH) = 0.310 mol mole 1-propanol = 11.4 g x (1 mol C3H7OH/60.1 g C3H7OH) = 0.190 mol mole fraction of ethanol xeth = 0.310 /(0.310 +0.190) = 0.620 mole fraction of 1-propanol xpro= 1 - xeth =1 -0.620 = 0.380 • VP of the solution: (Raoult’s law) Peth=xethx Petho= 0.620 x 108 mmHg = 67.0 mmHg Ppro=xprox Pproo= 0.380 x 40.0 mmHg = 15.2 mmHg PT = Peth+ Ppro = (67.0 + 15.2 )mmHg = 82.2 mmHg (b) Mole fraction in the vapor: (Dalton’s law) xeth= Peth/PT = 67.0 mmHg/82.2 mmHg = 0.815

18. DTb = Kbm B. Boiling point elevation • When nonvolatile solutes are added, the VP of the solution is reduced =>BP is increased DTb = Tb- Tb0 > 0 T bo: bp of pure solvent T b : bp of solution m : molality of the solution Kb: the molal boiling-point elevation constant (0C/m)

19. DTf = Kfm C. Freezing point depression DTf = Tf0 - Tf > 0 T fo: fp of pure solvent Tf : fp of solution m : molality of the solution Kf is the molal freezing-point depression constant (0C/m)

20. moles of solute m= mass of solvent (kg) = 3.202 kg solvent 1 mol 62.01 g 478 g x Tf = Tb0 + DTb Tf = Tf0 – DTf Q: What is the freezing point and boiling point of a solution that contains 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g. DTf = Kf m, DTb = Kbm Kf = 1.86 0C/m, Kb = 0.52 0C/m = 2.41 m = 1.86 0C/m x 2.41 m = 4.48 0C DTf = T f0 – Tf = Kfm = 0.00 0C – 4.48 0C = -4.48 0C = 0.52 0C/m x 2.41 m = 1.3 0C DTb = Tb – Tb0= Kbm = 100.0 0C + 1.3 0C = 101.3 0C

21. D. Osmotic Pressure Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more conc. solution. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (p) is the pressure required to stop osmosis.

22. D. Osmotic Pressure (p) High P Low P M: molarity of the solution p = MRT R : gas constant T : temperature (in K)

23. Example of Osmosis A cell in an: isotonic solution hypotonic solution hypertonic solution

24. Vapor-Pressure Lowering Boiling-Point Elevation DTb = Kbm 0 P1 = X1 P1 Freezing-Point Depression DTf = Kfm p = MRT Osmotic Pressure (p) Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

25. actual number of particles in soln after dissociation van’t Hoff factor (i) = number of formula units initially dissolved in soln V. Colligative Properties of Electrolyte Solutions 0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution 0.2 m ions in solution i should be 1 nonelectrolytes 2 NaCl CaCl2 3

26. Boiling-Point Elevation DTb = iKbm Freezing-Point Depression DTf = i Kfm p= iMRT Osmotic Pressure (p) V. Colligative Properties of Electrolyte Solutions

27. DTf = Kf m 0.4 m ions in solution 0.9 m ions in solution 0.8 m ions in solution 0.5 m urea in solution Q: If all ionic substances are completely ionized, the aqueous solution with the highest boiling point would be (1) 0.2 m NH4NO3 (2) 0.3 m MgBr2 (3) 0.4 m LiCl (4) 0.5 m urea (1) 0.2 m NH4NO3 solution (2) 0.3 m MgBr2 solution (3) 0.4 m LiCl solution (4) 0.5 m urea solution highest boiling point: (2) MgBr2

28. p = MRT p M= = RT 0.0821 atm.L/mol.K x 298 K mass B-12 0.500 g 1 atm = 3.68 x10-4 mol mol B-12 760 mmHg 85.6 mmHg x Q: At 25oC, the osmotic pressure of 80.0 mL of solution containing 0.500 g of vitamin B-12 is 85.6 mmHg, what is the molar mass of vitamin B-12? =0.00460 M mol B-12 = 0.00460 M x 0.0800 L = 3.68 x10-4 mol Molar mass B-12= = 1.36 x 103 g/mol