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: 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

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

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

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

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)

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.

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

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.

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

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

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

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

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 106 mg ppb = 5.3 ppb

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

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

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

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

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

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

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

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

Concentration Solution: Answer: 1.15 M

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

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

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?

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.