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Solutions

Solutions. Solution : homogeneous mixture components are uniformly intermingled on a molecular level Solutions can be solid: brass (zinc in copper) liquid : salt water, sugar water, etc gas : Air (oxygen & others in nitrogen). Solutions. Unsaturated solution:

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Solutions

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  1. Solutions • Solution: • homogeneous mixture • components are uniformly intermingled on a molecular level • Solutions can be • solid: • brass (zinc in copper) • liquid: • salt water, sugar water, etc • gas: • Air (oxygen & others in nitrogen)

  2. Solutions • Unsaturated solution: • a solution that is capable of dissolving more solute • Saturated solution: • a solution that is in equilibrium with undissolved solid • Supersaturated solution: • a solution that contains more dissolved solute than is needed to form a saturated solution

  3. Solubility Example: Which of the following solutes would you expect to be soluble in water: CH3CH2CH3: CH3CH2OH: HCl: Vitamin A Vitamin C Remember: Substances with similar intermolecular forces tend to dissolve in each other

  4. Solubility of Gases • The solubility of a gas in a solvent depends on the nature of the solute and solvent, the temperature, and the pressure. • In general, the solubility of gases in water increases with increasing molar mass. • Larger LDF

  5. Solubility of Gases • The solubility of a gas in a solvent increases as the pressure of the gas over the solvent increases. • Henry’s Law:The solubility of a gas in a solvent is directly proportional to its partial pressure above the solution. Cg = kPg where Cg = solubility of the gas in the solution phase Pg = partial pressure of the gas k = proportionality constant (value depends on solute, solvent, and temperature

  6. Solubility Example: Calculate the concentration of CO2 in a soft drink that was bottled with a partial pressure of carbon dioxide of 3.5 atm over the liquid at 25oC. (k = 3.1 x 10-2 mol/L.atm)

  7. Solubility of Gases Example: Why does a bottle of soda bubble when the cap is first removed? • Carbonated beverages like soda are bottled under a carbon dioxide pressure slightly greater than 1 atm. • Opening the bottles, reduces the partial pressure of CO2 above the soda. • Solubility of CO2 decreases so CO2 bubbles out of the solution.

  8. Solubility of Gases • The solubility of solid solutes generally increases with increasing temperature.

  9. Solubility of Gases • The solubility of a gas in a solution decreases with increasing temperature. • Gas molecules have greater KE and can escape from the solution more easily.

  10. Solubility of Gases Example: A warm bottle of soda tends to taste “flat” compared to a cold bottle of soda. Explain why.

  11. Concentration • Several different units can be used to express the concentration of a solute in a solution: • mass (weight) percent • parts per million (ppm) • parts per billion (ppb) • mole fraction • Molality • Molarity • Varies with temperature Independent of temperature

  12. Concentration • Mass Percent = mass of component x 100 total mass of sol’n Example: A solution is prepared by dissolving 6.8 g of NaCl in 750.0 g of water. What is the mass percent of the solute? Mass % NaCl = ___g NaCl____ x 100 g NaCl + g H2O = ___6.8 g____ x 100 = 0.90 % 6.8 g + 750.0 g

  13. Concentration • ppm = mass of component x 106 total mass of sol’n Example: A 10.25 g sample of lake water contains 1.28 x 10-2 mg of arsenic. What is the concentration of arsenic in ppm? ppm As = 0.0128 mg As x __1 g As_ x 106 10.25 g water 103 mg As ppm As = 1.25 ppm

  14. Concentration • ppb = mass of component x 109 total mass of sol’n Example: A 225 g sample of lake water contains 1.2 mg of pesticide. What is the concentration of pestcide in ppb? ppb = 1.2 mg pest. x _1 g _ x 109 225 g water 106mg ppb = 5.3 ppb

  15. Concentration • Mole fraction = moles of component total moles of all components Example: Calculate the mole fraction HCl present in a solution prepared by dissolving 0.25 mol HCl in 9.50 mol H2O. XHCl = ___moles HCl____ mol HCl + mol H2O XHCl = ___0.25 mol ______ = 0.026 0.25 mol + 9.50 mol

  16. Concentration • Molality = m = moles of solute kg solvent Example: Calculate the molality of a solution prepared by dissolving 1.25 g of sodium hydroxide in 250 g of water. m = _moles NaCl_ kg H2O m = 1.25 g NaCl x 1 mol NaCl x 103 g H2O 250 g H2O 58.5 g NaCl 1 kg H2O m = 0.085 m

  17. Concentration • Molarity = M = moles of solute L solution Example: Calculate the molarity of a solution that contains 73.0 g of HCl per 250 mL of solution. M = _moles HCl_ L soln M = 73.0 g HCl x 1 mol HCl x 103 mL 250 mL soln 36.5 g HCl 1 L M = 8.0 M

  18. Concentration • Why does molarity vary with temperature?? • You must be able to interconvert between the different concentration units.

  19. Concentration Example: An aqueous solution of sodium hydroxide contains 4.4% NaOH by mass. Calculate the mole fraction and molality of the solution. To find mole fraction: Given: 4.4 g NaOH per 100.0 g solution Find: XNaOH XNaOH = mol NaOH total mol

  20. Concentration mol NaOH = 4.4 g NaOH x 1 mol = 0.11 mol 40.0 g grams H2O = 100.0 g – 4.4 g = 95.6 g mol H2O = 95.6 g H2O x 1 mol = 5.31 mol 18.0 g XNaOH = ___0.11 mol______ = 0.020 0.11 mol + 5.31 mol

  21. Concentration To find molality: Given: 4.4 g NaOH per 100.0 g solution Find: m m = mol NaOH kg solvent m = 0.11 mol NaOH x 1000 g H2O = 1.2 m 95.6 g H2O 1 kg H2O

  22. Concentration Example: Calculate the molarity of a 1.50 m solution of toluene (C7H8) in benzene if the solution has a density of 0.876 g/mL. Given: 1.50 mol toluene per 1 (exact) kg benzene d = 0.876 g/mL Find: M = mol toluene L solution

  23. Concentration Solution: Answer: 1.15 M

  24. Concentration • The dilution equation is used to calculate either • the new concentration of a solution prepared by diluting a stock solution Or • The volume of a stock solution needed to prepare a known volume of a more dilute solution C1V1 = C2V2

  25. Concentration Example: Calculate the molarity of a solution prepared by diluting 225 mL of 1.5 M KMnO4 to a total volume of 850.0 mL. Given: Find: Solution: Answer: 0.40 M

  26. Concentration Example: Describe in detail how you would prepare 500.0 mL of 0.60 M HCl from a 12.0 M HCl stock solution?

  27. Concentration Example: A solution is prepared by dissolving 1.50 g of sodium chloride in enough water to give 250.0 g of solution. A 25.0 g aliquot of this solution was then diluted with water to a total mass of 100.0 g. Calculate the weight percent sodium chloride present in the final solution.

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