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SOLUTIONS

SOLUTIONS. SOLUTION – A homogeneous mixture SOLVENT – The major component of a solution SOLUTE – The minor component(s) of a solution. 3G-1 (of 15). Polar. Nonpolar. LIKES DISSOLVE LIKES. Polar Component with a Nonpolar Component . Polar molecules attract with LDF, DDA

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SOLUTIONS

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  1. SOLUTIONS SOLUTION – A homogeneous mixture SOLVENT – The major component of a solution SOLUTE – The minor component(s) of a solution 3G-1 (of 15)

  2. Polar Nonpolar LIKES DISSOLVE LIKES Polar Component with a Nonpolar Component Polar molecules attract with LDF, DDA Nonpolar molecules attract with LDF To make a solution, polar and nonpolar molecules attract with LDF These are too weak to overcome the polar molecules’ LDF, DDA 3G-2 (of 15)

  3. LIKES DISSOLVE LIKES Two Nonpolar Components Nonpolar molecules attract with LDF To make a solution, different nonpolar molecules attract with LDF All molecules equally attract each other a solution forms 3G-3 (of 15)

  4. LIKES DISSOLVE LIKES Two Polar Components Polar molecules attract with LDF, DDA To make a solution, different polar molecules attract with LDF, DDA All molecules equally attract each other a solution forms Unless… 3G-4 (of 15)

  5. LIKES DISSOLVE LIKES Hydrogen-Bonding Solvent The solute must be able to H-Bond with the solvent to dissolve S  Cl  Cl No H-Bonding H  O H O H O  H H H H-Bonding 3G-5 (of 15)

  6. LIKES DISSOLVE LIKES Hydrogen-Bonding Solvent The solute must be able to H-Bond with the solvent to dissolve H  H  N  H H  N  H  H H-Bonding H-Bonding H  O H O H O  H H H H-Bonding 3G-6 (of 15)

  7. LIKES DISSOLVE LIKES Hydrogen-Bonding Solvent The solute must be able to H-Bond with the solvent to dissolve H  C  H  O H-Bonding H  O H O H O  H H H H-Bonding 3G-7 (of 15)

  8. Solvent Solute Solubility C6H6 C8H18 C6H6 CH2Br2 CH2Cl2 C8H18 CH2Cl2 CH2Br2 H2O C8H18 H2O CH2Br2 H2O CH3OH H2O CH3CCH3 ║ O High Low Low High Low Low (a little) High High 3G-8 (of 15)

  9. When water-soluble polar molecules dissolve in water, the MOLECULES separate from each other, and exist as intact, neutral molecules These solutions do not conduct electricity because no ions are formed NONELECTROLYTE – A water-soluble compound whose solution does not conduct electricity 3G-9 (of 15)

  10. Alcohols (CxHyOH) and sugars (Cx(H2O)y) are nonelectrolytes Solute molecules surrounded by water molecules are said to be HYDRATED H2O C2H5OH (l)→ C2H5OH (aq) H2O C6H12O6(s)→ C6H12O6(aq) 3G-10 (of 15)

  11. When acid molecules dissolve in water, the waters rip the acid molecules into IONS These solutions conduct electricity because ions are formed ELECTROLYTE – A water-soluble compound whose solution conducts electricity Acids are electrolytes 3G-11 (of 15)

  12. IONIZATION – The formation of ions during the dissolving process H2O HCl(g)→ H+(aq) + Cl- (aq) Strong acids produce many ions in solution, and their solutions are good conductors HCl, HBr, HI, and acids with at least 2 more O’s than H’s are strong acids Strong acids are STRONG ELECTROLYTES 3G-12 (of 15)

  13. Weak acids produce few ions in solution, and their solutions are poor conductors All other acids are weak acids Weak acids are WEAK ELECTROLYTES H2O HF (g)→ HF(aq) 3G-13 (of 15)

  14. When water-soluble ionic compounds dissolve in water, the IONS separate from each other ION-DIPOLE ATTRACTIONS between the ions and the water molecules pull the ions into solution These solutions conduct electricity because ions are formed Soluble ionic compounds are strong electrolytes 3G-14 (of 15)

  15. DISSOCIATION – The separation of ions from an ionic crystal during the dissolving process H2O NaCl (s)→ Na+(aq) + Cl- (aq) 3G-15 (of 15)

  16. SOLUBILITY – The maximum amount of solute that will dissolve in a specific amount of solvent SATURATED – A solution that contains as much dissolved solute as possible Dissolving Rate Crystallizing Rate SOLUTION EQUILIBRIUM – When the rates of dissolving and crystallizing are equal A saturated solution is in solution equilibrium 3H-1 (of 8)

  17. FACTORS AFFECTING SOLUBILITY (1) Temperature Increasing the temperature generally increases a solid’s solubility Increasing the temperature decreases a gas’s solubility (2) Pressure No effect on a solid’s solubility Increasing the pressure increases a gas’s solubility HENRY’S LAW – The amount of gas dissolved in a liquid is directly proportional to the pressure of the gas in contact with the liquid pgas = kCgas(Cgasis concentration of dissolved gas) 3H-2 (of 8)

  18. Mass Percent = Mass Solute x 100 ___________________ Mass Solution CONCENTRATION UNITS (1) MASS PERCENT – The mass of solute per mass of solution, times 100 Find the mass percent of KCl in a solution made with 20.0 g KCl dissolved in 80.0 g H2O. 20.0 g KCl x 100 _____________________ 100.0 g Solution = 20.0 % 3H-3 (of 8)

  19. (2) MOLE FRACTION (X) – The moles of solute per moles of solution Find the mole fraction of KCl in a solution made with 20.0 g KCl dissolved in 80.0 g H2O. = 0.2683 mol KCl 20.0 g KCl x mol KCl _______________ 74.55 g KCl = 4.440 mol H2O 80.0 g H2O x mol H2O _______________ 18.02 g H2O 0.2683molKCl ____________________________________ 0.2683 + 4.440mol Solution = 0.0570 3H-4 (of 8)

  20. (3) MOLARITY (M) – The moles of solute per liter of solution Used to easily calculate moles of solute by only measuring the volume of solution M = n ____ V n = quantity of matter (moles), V = volume (liters) Find the molarity of a solution with 6.24 grams of magnesium chloride dissolved in enough water to make 500.0 mL of solution. = 0.06554 mol MgCl2 6.24 g MgCl2 x mol MgCl2 _________________ 95.21 g MgCl2 0.06554 mol MgCl2 _________________________ 0.5000 L Solution = 0.131 M MgCl2 3H-5 (of 8)

  21. Electrolyte solutions really consist of individual ions Each MgCl2 formula unit has 1 Mg2+ ion and 2 Cl- ions  0.131 M x 1 = 0.131 M Mg2+ 0.131 M x 2 = 0.262 M Cl- Find the molarities of each ion in a 0.10 M Al2(SO4)3 solution 3H-6 (of 8)

  22. Find the mass of potassium nitrate needed to prepare 250. mL of a 0.200 M potassium nitrate solution. M = n ___ V MV = n 0.200 mol KNO3 _____________________ L solution = 0.05000mol KNO3 x 0.250 L solution = 5.06 g KNO3 x 101.11 g KNO3 ___________________ mol KNO3 0.05000mol KNO3 3H-7 (of 8)

  23. (4) MOLALITY (m) – The moles of solute per kilogram of solvent Used because it does not change when the temperature changes m = mol solute _______________ kg solvent Find the molality of a solution with 11.8 grams of glucose (C6H12O6, m = 180.16 g/mol) dissolved in 150.0 grams of water. _ = 0.06550 mol C6H12O6 11.8 g C6H12O6 x mol C6H12O6 ______________________ 180.16 g C6H12O6 0.06550 mol C6H12O6 ___________________________ 0.1500 kg H2O = 0.437 m C6H12O6 3H-8 (of 8)

  24. COLLIGATIVE PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES – Properties that depend only on the number of dissolved solute particles, not what they are VOLATILE – A substance (or solute) that evaporates easily NONVOLATILE – A substance (or solute) that does not evaporate 3I-1 (of 11)

  25. (1) Vapor Pressure Lowering A liquid’s equilibrium vapor pressure is lowered by having a nonvolatile solute dissolved in it The pressure exerted by a vapor in equilibrium with its liquid is called the EQUILIBRIUM VAPOR PRESSURE With solute particles dissolved in the liquid, there are less solvent molecules on the surface, therefore less solvent molecules can evaporate, so the rate of evaporation decreases Now vapor molecules condense faster than liquid molecules evaporate, so the amount of vapor decreases When the two rates are again equal, there are less vapor molecules than before, so a lower equilibrium vapor pressure 3I-2 (of 11)

  26. Temp (ºC) EVP of pure H2O (torr) EVP of a dilute salt water solution (torr) 10 20 30 9.2 17.5 31.8 9.1 17.4 31.6 EVP Pure Water Water Solution (Temp °C) 3I-3 (of 11)

  27. 1882 FRANÇOIS-MARIE RAOULT RAOULT’S LAW – The equilibrium vapor pressure of a solution is proportional to the mole fraction of the solvent in the solution psolution= Xsolventpºsolvent psolution = equilibrium vapor pressure of the solution Xsolvent= mole fraction of the solvent in the solution pºsolvent= equilibrium vapor pressure of the pure solvent A solution will obey Raoult’s Law if the solution is dilute the solvent and solute particles are about the same size the solvent and solute particles have about the same attractive forces 3I-4 (of 11)

  28. Find the equilibrium vapor pressure of a solution at 20ºC that is prepared with 0.500 moles of sucrose dissolved in 35.00 moles of water, if the equilibrium vapor pressure of water at 20ºC is 17.5 torr. 35.00 mol H2O ____________________________________ 35.00 + 0.500 mol solution = 0.9859 psoln= Xsolvpºsolv = (0.9859)(17.5 torr) = 17.3 torr 3I-5 (of 11)

  29. When a solute is an electrolyte (a salt or acid), the fact that it dissociates or ionizes must be considered Find the EVP of a solution at 20ºC that is prepared with 0.250 moles of Na2SO4dissolved in 10.00 moles of water, if the EVP of water at 20ºC is 17.5 torr. nsolute = 3 x 0.250 mol = 0.750 mol 10.00 mol H2O __________________________________ 10.00 + 0.750 mol solution = 0.9302 psoln = Xsolvpºsolv = (0.9302)(17.5 torr) = 16.3 torr 3I-6 (of 11)

  30. Find the EVP of a solution at 23ºC that is prepared with 0.300 moles of KCldissolved in 15.00 moles of water, if the EVP of water at 23ºC is 21.1 torr. nsolute = 2 x 0.300 mol = 0.600 mol 15.00 mol H2O __________________________________ 15.00 + 0.600 mol solution = 0.9615 psoln = Xsolvpºsolv = (0.9615)(21.1 torr) = 20.3 torr 3I-7 (of 11)

  31. In a solution, if both the solvent and solute are volatile, vapor pressure is produced from both components in the solution psolution= psolvent + psolute psolution= Xsolventpºsolvent + Xsolutepºsolute 3I-8 (of 11)

  32. psolution= Xsolventpºsolvent + Xsolutepºsolute Find the EVP of a solution prepared with 0.300 moles of acetone and 0.100 moles of chloroform if the EVP of acetone is 293 torr and the EVP of chloroform is 345 torr. 0.300 mol acetone ________________________ 0.400 mol solution = 0.7500 0.100 mol chloroform ___________________________ 0.400 mol solution = 0.2500 (0.7500)(293 torr) (0.2500)(345 torr) + 219.8torr + 86.25torr = 306.05torr = 306 torr 3I-9 (of 11)

  33. If the EVP of a solution equals the sum of the EVP’s of the 2 components, the solution obeys Raoult’s Law IDEAL SOLUTION – A solution that obeys Raoult’s Law 3I-10 (of 11)

  34. POSITIVE DEVIATION – Occurs when the attractions between solvent and solute molecules are WEAKER than the attractions in either pure component NEGATIVE DEVIATION – Occurs when the attractions between solvent and solute molecules are STRONGER than the attractions in either pure component Solution formation is endothermic Solution formation is exothermic 3I-11 (of 11)

  35. (2) Boiling Point Elevation A liquid’s boiling point is raised by having a nonvolatile solute dissolved in it Atmospheric Pressure A solvent boils when its EVP equals the prevailing atmospheric pressure With solute particles dissolved in the solvent, its EVP is lowered Only when the temperature increases will the EVP again equal atmospheric pressure  the boiling point of the solution is now higher than the boiling point of the solvent 102ºC 100ºC 101ºC 3J-1 (of 16)

  36. EVP Pure Water Water Solution 100 (Temp °C) 3J-2 (of 16)

  37. To determine the increase in boiling point from the solvent to the solution: ΔTb = Kbmi ΔTb = increase in the boiling point Kb = molal boiling point constant of the solvent m= molality of the solution i = van’t Hoff factor of the solute = moles of particles in solution _____________________________________ moles of dissolved solute C6H12O6 MgBr2 HCl Al2(SO4)3 1 i = 3 2 5 3J-3 (of 16)

  38. Find the boiling point of a solution that is prepared with 0.150 moles of sodium chloride dissolved in 90.0 grams of water. The boiling point of pure water is 100.00ºC and molal boiling point constant for water is 0.51 Cºkg H2O/mol solute. ΔTb = Kbmi = 1.667m NaCl 0.150 mol NaCl ____________________ 0.0900 kg H2O ΔTb = (0.51 Cºkg H2O/mol solute)(1.667 mol solute/kg H2O)(2) = 1.7 Cº 100.00ºC + 1.7 Cº = 101.7ºC 3J-4 (of 16)

  39. All colligative properties can be used to calculate the molar mass of a nonvolatile solute 3J-5 (of 16)

  40. Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with 4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon disulfide has a boiling point of 47.5ºC. molar mass X = grams X _____________ moles X molar mass N.E. = 4.80 grams N.E. ____________________ ? moles N.E. 3J-6 (of 16)

  41. Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with 4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon disulfide has a boiling point of 47.5ºC. Pure carbon disulfide has a boiling point of 46.3ºC and molal boiling point constant of 2.37 Cºkg CS2/mol solute. ΔTb = Kbmi ΔTb = 47.5ºC – 46.3ºC = 1.2 Cº ΔTb = Kb (mol solute)i _______________ (kg solvent) (kg solvent) ΔTb = (mol solute) _____________________ Kb i = (0.1500 kg CS2)(1.2 Cº) _____________________________________ (2.37 Cºkg CS2/mol solute)(1) = 0.0759mol solute 3J-7 (of 16)

  42. Find the molar mass of a nonelectrolyte (i = 1) if a solution prepared with 4.80 grams of the nonelectrolyte dissolved in 150.0 grams of carbon disulfide has a boiling point of 47.5ºC. Pure carbon disulfide has a boiling point of 46.3ºC and molal boiling point constant of 2.37 Cºkg CS2/mol solute. 4.80 grams N.E. _______________________ 0.0759 moles N.E. = 63 g/mol 3J-8 (of 16)

  43. (3) Freezing Point Depression A liquid’s freezing point is lowered by having a nonvolatile solute dissolved in it A solvent freezes when the EVP of its liquid equals the EVP of its solid With solute particles dissolved in the liquid solvent, its EVP is lowered Only when the temperature decreases will the EVP of the liquid solution equal the EVP of the solid solvent 0ºC -1ºC -2ºC  the freezing point of the solution is now lower than the freezing point of the solvent 3J-9 (of 16)

  44. EVP Pure Water Ice Water Solution 0 (Temp °C) 3J-10 (of 16)

  45. To determine the decrease in freezing point from the solvent to the solution: ΔTf = Kfmi ΔTf = decrease in the freezing point Kf= molal freezing point constant of the solvent m= molality of the solution i = van’t Hoff factor of the solute 3J-11 (of 16)

  46. Find the freezing point of a solution that is prepared with 0.150 moles of calcium chloride dissolved in 97.0 grams of water. The freezing point of pure water is 0.00ºC and molal freezing point constant for water is 1.86 Cºkg H2O/mol solute. ΔTf = Kfmi = 1.546m CaCl2 0.150 mol CaCl2 _______________________ 0.09700 kg H2O ΔTf = (1.86 Cºkg H2O/mol solute)(1.546 mol solute/kg H2O)(3) = 8.63 Cº 0.00ºC - 2.88 Cº = -8.83ºC 3J-12 (of 16)

  47. Find the molar mass of a nonelectrolyte if a solution prepared with 10.0 grams of the nonelectrolyte dissolved in 100.0 mL of benzene has a freezing point of 3.51ºC. molar mass X = grams X _____________ moles X molar mass N.E. = 10.00 grams N.E. ____________________ ? moles N.E. 3J-13 (of 16)

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