1. CH 10 Solutions Principles of Solubility Nature of the solute and solvent Like likes like. Electrolytes will dissolve in a polar solvent (like water) Non-electrolytes will dissolve more easily in non polar or slightly polar solvents (like benzene, or toluene)

2. CH 10 Solutions Non electrolytes (molecular substances) dissolving in water. This ‘unusual’ case will involve Hydrogen bonding or otherwise polar molecules. Ex: alcohols (-OH group) Methanol, Ethanol Hydrogen peroxide H 2 O 2

3. Solubilities of Alcohols in Water Formula Name Solubility in Water (g/100 g) CH 3 OH methanol infinitely soluble CH 3 CH 2 OH ethanol infinitely soluble CH 3 (CH 2 ) 2 OH propanol infinitely soluble CH 3 (CH 2 ) 3 OH butanol 9 CH 3 (CH 2 ) 4 OH pentanol 2.7 CH 3 (CH 2 ) 5 OH hexanol 0.6 CH 3 (CH 2 ) 6 OH heptanol 0.18 CH 3 (CH 2 ) 7 OH octanol 0.054 CH 3 (CH 2 ) 9 OH decanol insoluble in water “greasy end” is larger proportion in large alcohols. CH 10 Solutions

4. DDT no more mosquitoes! Pesticide used widely in the 60’s Soluble in non polar solvents. Concentrates in tissues of fish, birds, and other wildlife. (biological ½ life of 8 years) Thins eggshells. CH 10 Solutions

5. Principles of Solubility Effect of Pressure/Henry’s Law For gases only Solubility is directly proportional to pressure. Soda is bottled under a pressure of 4atm. This drives more of the CO 2 into solution, when the cap is released, the pressure drops to 1 atm, and CO 2 bubbles and rapidly escapes solution. C g =k P g Concentration in solution = constant x Pressure of gas

6. CH 10 Solutions Principles of Solubility Effect of Temperature Solid + Water  Solution Usually + ΔH heat in Increased T favors an endothermic process So: Solubility of a solid usually increases with increased temperature Gas + Water  Soln. Usually –ΔH heat out Increased T is detrimental for an exothermic process So: Solubility of a gas usually decreases with higher temperature.

7. CH 10 Solutions Concentration Units Molarity (M) Mole Fraction (X) Mass % Solute Molality (m) Conversion between units Principles of solubility Effect of Temperature Effect of Pressure (Henry’s Law) Colligative Properties of Non-Electrolytes Vapor pressure lowering Boiling point elevation Freezing point lowering Osmosis/Osmotic Pressure (π) Colligative Properties of Electrolytes

8. CH 10 Solutions Concentration Units problem type 1. Molarity (M) Molarity = moles of solute/ liters of solution Units are moles/liter Tricks: change mass to moles change mL to Liters Typical Problems Making a volume of solution from the solid. Creation Preparing a volume of a solution from another (stock) solution. Dilution

9. CH 10 Solutions Problems Molarity (M) Prepare 35.0 mL of 0.200 M Aluminum nitrate from a solid sample. 1. How many moles of Al- nitrate would be in the solution? 2. Measure out that many grams and then add water to make 35.0 mL Prepare 35.0 mL of 0.20 M Aluminum Nitrate solution from a 0.50M stock solution. How many moles do you need? Remember M 1 V 1 =M 2 V 2 M c V c =M d V d Meausure out the volume of stock you need, then dilute Molar mass (gfm) of aluminum nitrate is 213.03 g/n

10. CH 10 Solutions Problem type 2. Mole Fraction (X) Similar to percent composition…you are interested in part of the whole Moles of substance/ total moles Typical problems will have you determine the mole fraction from % mass. determine the mole fraction of solute in solution

11. CH 10 Solutions Problems Mole Fraction (X) If you dissolve 12.0 g of methanol in 100.0 g of water. What is the mole fraction that is methanol? 1. Determine the moles of water 2. Determine the moles of methanol 3. Add together for the total moles, then take part/whole Methane is CH 4 Methanol is CH 3 OH

12. CH 10 Solutions Problem type 3. Mass % Solute Simply determine the percent composition by mass. Trick is to change the decimal to a percent (x100 + “%”) Solute is the solid dissolved in solution

13. CH 10 Solutions Problems mass % and ppb, ppm The federal drinking water standard allows 5 x10 -8 g arsenic per gram of water. How many parts per billion is this? 5.0 x 10^ -8 g/ 1.0 x 10^ 9 (1billion) =50 ppb,,,,parts per billion Very small impurities or contaminants in drinking water are measured in ppm or ppb NRDC: Arsenic in Drinking Water FAQ

14. CH 10 Solutions Problem Type 4. Molality (m) This is similar to molarity, but uses mass (kilograms) of the solvent, rather than liters of solvent. Typical problems will have you determine the molality from other concentration units. Molarity (M) = # moles solute/ Liter Solvent

15. CH 10 Solutions Problems : Molality (m) What is the molality of methanol if 12g of methanol are dissolved into 100g of water? 1. Find Moles of solute (methanol) 2. Find kilograms of solvent 3. Divide if the solvent is water, M and m are numerically equal 12g/32.0 g/n = 0.375 n 100g/ 1000g/kg = 0.100kg = 3.75m

16. CH 10 Solutions Conversion between units- decide on a fixed amount of solution When the original concentration [ ] is: Start with: Mass Percent100 g solution Molarity (M)1.00L of solution Molality (m)1000g of solvent (1Kg) Mole Fraction (X)1 mol (solute+solvent)

17. CH 10 Solutions Part III: Colligative Properties Vapor Pressure Lowering VPL = X 2 Pº 1 VPL= (vapor pressure lowering) X 2 = mole fraction of the solute Pº 1 = vapor pressure of the PURE solvent Each liquid has a vapor pressure. The pressure to escape the surface into the gas phase. Dissolved particles get in the way and block the escape, thereby reducing the vapor pressure. The higher the concentration (X 2 ), the more the VP is lowered.

18. CH 10 Solutions Raoult’s Law

19. CH 10 Solutions Part III: Colligative Properties *Boiling Point Elevation ΔT b *Freezing Point Depression ΔT f ΔT b = k b m ΔT f = k f m The change in temperature from the normal boiling/freezing points is a constant (k) times the molality(m). The higher the concentration of ‘contaminant’, the greater the change in freezing/boiling points Boiling Elevation: The particles lower the vapor pressure. K b for water is 0.52 °C/m Freezing Depression: The particles obstruct the formation of the solid crystal. There must be a lower temperature to form solid structure K f for water is 1.86 °C/m

20. CH 10 Solutions Problems: Colligative Properties freezing point depression ΔT f = k f m Change in freezing temp is (K f ) x (molality) Find the freezing point of a solution containing 20.0 g of ethanol in 50.0g of water. What is the molality of the solution? Grams to moles Moles / Kg of Solvent Molality x K f K f for water is 1.86 °C/m

21. CH 10 Solutions Problems: Colligative Properties boiling point elevation ΔT f = k b m Change in boiling temp is (K b ) x (molality) Find the boiling point of a solution containing 20.0 g of ethylene glycol in 50.0g of water. What is the molality of the solution? Grams to moles Moles / Kg of Solvent Molality x K b K b for water is 0.52 °C/m

22. CH 10 Solutions Colligative Properties Osmotic Pressure (π) π = M R T M = molarity R=.0821 L * atm/n *K T = Temperature Water moves through a semi- permeable membrane from an area of high vapor pressure to low vapor pressure.

23. CH 10 Solutions Problems: Using freezing point depression to find molar mass A solution is made using.0100g of a substance in 1.00g of water. The freezing point depresses 1.00 °C What is the molar mass? Solve for molarity