1. © University of South Carolina Board of Trustees Spaghetti Science What happens somewhat before a pot of water boils? ●Bubbles form on sides of the pot.

2. © University of South Carolina Board of Trustees Chapt. 12 Solutions Sec. 3 Does Higher Temperature Increase Solubility?

3. © University of South Carolina Board of Trustees Solubility Experiments

4. HeatX Heat + X (s) + solvent X (sol) © University of South Carolina Board of Trustees Effect of Temperature on Solubility X Heat + X (s) + solvent X (sol)  H pos. (endothermic)  Temp.  Temp.,  Solubility    Temp. = add heat X Heat + X (s) + solvent X (sol)    

5. XHeat X (s) + solvent X (sol) + Heat © University of South Carolina Board of Trustees Effect of Temperature on Solubility  H neg. (exothermic)  Temp.  Temp.,  Solubility    Temp. = add heat X X (s) + solvent X (sol) + Heat    

6. © University of South Carolina Board of Trustees Example Problem What happens somewhat before a pot of water boils? ●Bubbles form on sides of the pot. ●N 2 /O 2(gas)  N 2 /O 2(sol) + heat ●Nitrogen and oxygen have a negative enthalpy of solution (exothermic). ●Gas solubility is decreasing. Draw a conclusion about the enthalpy of solution of air.

7. © University of South Carolina Board of Trustees Chapt. 12 Solutions Sec. 3 Does Higher Pressure Increase Solubility?

8. © University of South Carolina Board of Trustees Pressure and Solubility ●Gases  Pressure,  Solubility ●Liquids little effect ●Solidslittle effect

9. © University of South Carolina Board of Trustees Pressure and the Solubility of Gases Doubling the pressure ______ the solubility of gases. doubles Henry’s Law m = k P Conc. in solution Pressure of gas Henry’s Law Constant (table) (partial) (in a mixture) (William) (1803)

10. © University of South Carolina Board of Trustees Example Problem If the atmospheric CO 2 level is 0.03 mole %, how many grams of CO 2 will be dissolved in 1 L of water at 20  C?

11. © University of South Carolina Board of Trustees Henry’s Law Constants in Water (molal/atm) Gas0 o C20 o C40 o C60 o C CO 2 7.60 x10 -2 3.91 x10 -2 2.44 x10 -2 1.63 x10 -2 C 2 H 4 1.14 x10 -2 5.60 x10 -3 3.43 x10 -3 --- He4.22 x10 -4 3.87 x10 -4 3.87 x10 -4 4.10 x10 -4 N 2 1.03 x10 -3 7.34 x10 -4 5.55 x10 -4 4.85 x10 -4 O 2 2.21 x10 -3 1.43 x10 -3 1.02 x10 -3 8.71 x10 -4

12. © University of South Carolina Board of Trustees Chapt. 12 Solutions Sec. 4 Colligative Properties

13. © University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes: lower vapor pressure higher boiling point lower freezing point osmotic pressure Colligative Properties

14. © University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes:  lower vapor pressure  P = P  x  sol  lower vapor pressure  P = - P  x  sol higher boiling point lower freezing point osmotic pressure Colligative Properties

15. © University of South Carolina Board of Trustees Vapor Pressure Lowering Benzene pure 0.02 mol frac 0.04 mol frac 0.06 mol frac 0.08 mol frac

16. © University of South Carolina Board of Trustees Boiling Point Elevation Benzene pure 0.02 mol frac 0.04 mol frac 0.06 mol frac 0.08 mol frac

17. © University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes: lower vapor pressure  P = - P  x  sol  higher boiling point  T = + k b x m sol lower freezing point osmotic pressure Colligative Properties

18. © University of South Carolina Board of Trustees Example Problem Find the molar mass of a solute, if a solution of 1.33 g of the compound dissolved in 25.0 g of benzene has a boiling point of 81.22 o C.

19. © University of South Carolina Board of Trustees Adding a nonvolatile solute to a pure solvent causes: lower vapor pressure  P = - P  x  sol  higher boiling point  T = + k b x m sol lower freezing point osmotic pressure Colligative Properties Mole-Meter