1. Outline:3/12/07 è Chem. Dept. Seminar Wed @ 4pm è 2 more lectures until Exam 2… è Chemistry Advising – Today @ 4pm Today: è More Chapter 18 Polyprotic acid titrations Solubility Product (K sp )

2. Worksheet #8 practice… #1a. pH = 2.5 [H 2 A] = 0.997 M [HA - ] = 0.00316 M [A 2- ] = 1  10 -8 M #1b. pH = 10.5 #2a. pH = 9.74 #2b. pH = 11.1

3. Quiz # 7 Please put away all books/papers If you don’t have a calculator, just set up the problems fully…

4. Quiz # 7 Please turn your papers over and pass them to the right…

5. Quiz #7 : Buffers #1 pH = pK a + log ([base]/[acid]) 4.00 = 4.75 + log ([base]/[1.0M]) [base] = 10  0.75 = 0.178 M [base] = 0.178 mol/1.0L = 14.6 g CH 3 COONa / 1.0L = 14.6 g CH 3 COONa / 1.0L

6. Quiz #7 : Weak Base #2 Pyr + H 2 O = pyrH + + OH  K b = 10  8.72 = 1.91  10  9 0.015  x +x +x 1.91  10  9 = x 2 / 0.015 x = 5.34  10  6 = [OH  ] pH = 14 – log( 5.34  10  6 ) = 8.73

7. Quiz #7 : Titration #3 pH = pK a + log ([base]/[acid]) pH = 3.74 + log ([0.0005]/[0.0020]) = 3.14 = 3.14 HA + OH   A  + H 2 O (titration) 0.0025 + 0.0005  0 + 0 (init) 0.0020 + 0  0.0005 (equil)

8. Titration of Polyprotic Acids Weak acid: K a1 = x 2 /[H 2 A]

9. Titration of Polyprotic Acids Buffer: pK a2 +log[A]/[HA] Buffer: pK a1 +log[HA]/[H 2 A]

10. Titration of Polyprotic Acids log((K a1 × K a2 ) 0.5 ) See page 785

11. Try example 18 – 12 (page 787) Sulfurous Acid, H 2 SO 3, has two acidic hydrogen atoms, with pK a values of 1.85 and 7.20. Construct a titration curve for the titration of 125 mL of 0.150 M sulfurous acid with 0.800 M NaOH. Sulfurous Acid, H 2 SO 3, has two acidic hydrogen atoms, with pK a values of 1.85 and 7.20. Construct a titration curve for the titration of 125 mL of 0.150 M sulfurous acid with 0.800 M NaOH.

12. n changes color as the titration passes the stoichiometric point if : pK in ≈ pH stoichiometric point

13. The Solubility-Product Constant, K sp Consider for which K sp is the “solubility product”. (BaSO 4 is ignored because it is a pure solid so its concentration is constant.)

14. The solubility product is another example of equilibrium calculations Solubility product calcs depend on the common ion effect (LeChâtelier). They have particular applications with metal ions and pH calculations (environmental applications). Solubility Equilibria

15. Types of Equilibrium Constants: Lots of different names…. K eq, K H, K sp, K a, K b, K f, K c, K p … All the same idea!

16. Solubility Equilibria n Insoluble compounds: solubility is less than 0.01 mol of dissolved material per liter of solution, K sp << 1 n Slightly soluble: 10 -5 < K sp < 10 -2 n Soluble: K sp > 10 -2

17. The solubility product is another example of equilibrium calculations Solubility product calcs depend on the common ion effect (LeChâtelier). They have particular applications with metal ions and pH calculations (environmental applications). Solubility Equilibria

18. The Common Ion Effect Solubility is decreased when a common ion is added (Le Châtelier again) as F - (from NaF, say) is added, the equilibrium shifts left, therefore CaF 2 (s) is formed (precipitation occurs). As NaF is added to the system, the solubility of CaF 2 decreases. Factors that Affect Solubility