1. Previously in Chem104: plant pigments do acid/base chemistry it’s just equilibrium new names: K a, K b for same K expressions Today in Chem104: the concept of K w the concept of the K w circle p-functions (pH, pK a, pK w ) pH scale P-functions in calcs

2. Dual Personality of Water: both an acid and a base Amphoteric (or amphiprotic) The Autoionization Reaction of water. 2 H 2 0 H 3 O+ + OH- K eq =

3. KwKw Because K w = [H 3 O+][OH-] = 10 -14 and K w is a constant: then the solution is neutral when [H 3 O+] = [OH-] = 10 -7 M [H 3 O+] [OH-]

4. KwKw But, if [H 3 O+] > [OH-], [H 3 O+] > 10 -7 M, the solution is acidic [OH-] < 10 -7 M [H 3 O+] > 10 -7 M

5. KwKw Or if [OH-] > [H 3 O+], [OH-] > 10 -7 M, the solution is basic Because K w = [H 3 O+] x [OH-] must be 10 -14 [OH-] > 10 -7 M [H 3 O+] < 10 -7 M

6. Because all these exponential numbers are a pain: i.e.K w = [H 3 O+][OH-] = 10 -14 [H 3 O+] = [OH-] = 10 -7 M (in neutral water) K a = 10 -10 we will use: The P-Function p of (a number) = -log 10 (a number) so “pH” means p of [H 3 O+], or -log 10 [H 3 O+] so the pH of neutral water, where [H 3 O+] = 10 -7 M is pH 7

7. The P-Function and K’s p of (K eq ) = -log 10 K eq = pK eq p of (K a ) = -log 10 K a = pK a p of (K w ) = -log 10 K w = pK w Examples: K eq = 10 -10 the pK eq = 10 K a = 1.20 x 10 -5 the pK eq = 4.92 K w = 10 -14 the pK w = 14

8. The P-Function simplifies exponential numbers K a = 1.20 x 10 -5, pK eq = 4.92 [H 3 O+] = 4.20 x 10 -5 M, pH = 4.376 [OH-] = 6.66 x 10 -6 M, pOH = 5.176 ( note number of sig figs) Examples converting from p-function: if pH = 7.47, [H 3 O+] = 10 -7.47 M= 3.39 x 10 -8 M, if pK a = 4.92, K a = 10 -4.92 = 1.20 x 10 -5

9. P-Function simplifies a large range of numbers: graphically 10 1 10 -1 10 -3 10 -5 10 -7 10 -9 10 -11 10 -13 10 -14 [H 3 O+], M -1 0 1 3 5 7 91113pH Note that on a p-scale, the smaller the p-number, the larger the actual number converts to a simpler scale

10. Apply the P-function to each side p of K w = p of [H 3 O+][OH-] = p of 10 -14 Working in P-Functions can simplify problems Recall K w = [H 3 O+][OH-] = 10 -14 -log K w = -log ( [H 3 O+][OH-] )= -log 10 -14 pK w = pH + pOH = 14 -log K w = -log [H 3 O+] + ( -log [OH-] ) = -log 10 -14

11. K w = 10 -14 Now apply this equation: pK w = pH + pOH = 14 to this picture [H 3 O+] [OH-]

12. pK w = 14 Now apply this equation: pK w = pH + pOH = 14 to this picture pH pOH

13. pK w When the solution is acidic [H 3 O+] > 10 -7 M, pH < 7 : pH is small Because pK w = pH + pOH must be 14 pH < 7 pOH > 7

14. pK w When the solution is ____________ [H 3 O+] __10 -7 M, pH ___ 7 : pH is ________ pOH is _______ pH is _______ Fill in the blanks!

15. Let’s do some problems !!