1. Outline:3/7/07 è Pick up CAPA 15 & 16 - outside è 4 more lectures until Exam 2… Today: è Chapter 18 Buffers Buffer calculations Titrations

2. Lots of problems to practice: n Chapter 17: 17.9, 17.11, 17.13, 17.15, 17.17, 17.21, 17.23, 17.25, 17.27, 17.29, 17.31, 17.33, 17.35, 17.37, 17.39, 17.41, 17.43, 17.45, 17.49, 17.51, 17.53, 17.55, 17.57, 17.59, 17.61, 17.63, 17.67, 17.69, 17.71, 17.73, 17.81, 17.83

3. Lots of problems to practice: n Chapter 18: 18.1, 18.3, 18.5, 18.9, 18.11, 18.13, 18.1, 18.3, 18.5, 18.9, 18.11, 18.13, 18.15, 18.17, 18.19, 18.21, 18.23, 18.27, 18.29, 18.31, 18.33, 18.35, 18.37, 18.39, 18.41, 18.45, 18.47, 18.51, 18.55, 18.57, 18.59, 18.61, 18.63, 18.65, 18.67, 18.69, 18.71, 18.73, 18.75, 18.77, 18.79, 18.81, 18.83, 18.85, 18.87, 18.93, 18.101, 18.103

4. Summary of acid base rxns: weak acid + weak base weak acid + strong base strong acid + weak base strong acid + strong base K eq calculations solve for x pH + pOH = 14 Find K eq from K a K b = K w

5. Acid Base reactions (remember): è Add 0.10 mol NaOH to …. (water) No common ions: pH goes from 7.00 to 13.00 (HOAc/OAc  mix) Common ions: pH goes from 4.74 to 5.11 change = +0.37 change = +6.00 A Buffer!

6. Buffer calculations…. pH = pK a + log ([conj base]/[acid]) This one also exists: pOH = pK b + log ([conj acid]/[base]) Practice! n Definition: A buffer is usually a mixture of conjugate acid-base pairs in solution. A buffer resists strong changes in pH.

7. pH of a buffer… 0.3 L of 0.4 M acetic acid is added to 0.7 L of 0.2 M sodium acetate, what will be the pH of the resultant buffer? pH = 4.75 + log [OAc  ]/[HOAc] pH = pK a + log ([conj base] / [acid]) Do the dilutions: 0.3/1.0 × 0.4 = 0.12 M HA 0.7/1.0 × 0.2 = 0.14 M A 

8. pH = 4.75 + log (0.14/0.12) 0.3 L of 0.4 M acetic acid is added to 0.7 L of 0.2 M sodium acetate, what will be the pH of the resultant buffer? pH = pK a + log ([conj base] / [acid]) pH of a buffer… = 4.75 + 0.07 = 4.82

9. A buffer example: Human blood must be maintained at a pH of 7.40  0.05 for the rest of the body biochemistry to function. This is achieved with a buffer made from CO 2 :  CO 2 (aq) + H 2 O (l)  H 2 CO 3 (aq)  If the carbonic acid concentration of a healthy human is 1.35  10  2 M, what is the normal human concentration of HCO 3  ?

10.   7.40 = 6.25 + log ([HCO 3  ]/ 1.35  10  )   [HCO 3  ] = 1.52  10  M n Rather “twisted” addition to the problem: Inject 10 mL of 1.0 M HCl into a (previously healthy) human with a total blood volume of 8.0 L. What is the new pH? n “Buffer capacity” problem pH = pK a + log ([conj base]/[acid])

11. How many moles H + added? n 0.010 L 1.0 mol/L = 0.010 mol n 0.010 L  1.0 mol/L = 0.010 mol How many moles H + there already? [H + ] = 10  = 3.18  10  7 mol/L  8.0 L  = 2.54  10  6 mol = negligible! Initial [H + ]? [H + ]=(0.010 mol+2.54  10  6 mol)/ 8.0 L  = 1.25  10  3 M

12. Same old procedure….   H 2 CO 3  HCO 3  + H + K a =     calc in which direction   1.35  10  2 1.52  10  1.25  10  I    +1.25  10  3  1. 25  10   1.25  10   1.48  10  2 1.51  10  0   x + x + x   solve for x ; x = 4.2  10  M; pH = 7.38 or use: pH = pK a +log ([conj base] /[acid])  = 6.37+log(1.51  10  /1.48  10  2 ) = 7.38 Easier to use

13. More Buffer pH… n Try another one…. A buffer solution is prepared by mixing 0.300 L of 0.400 M (CH 3 ) 3 N and 0.700 L of 0.200 M (CH 3 ) 3 NHCl. What is the pH of this buffer solution? The pK b of (CH 3 ) 3 N is 4.19. pH= 9.74 This one also exists: pOH = pK b + log ([conj acid]/[base])

14. Addition of acid/base to a buffer n 1.0 L of a buffer solution is prepared with 0.025 M formic acid and 0.010 M sodium formate (pKa = 3.74). If 0.005 mols (0.2 g) of NaOH is added to the solution, what is the pH? pH= 3.74 + log (0.010+0.005/0.025  0.005) or pH = 3.74 – 0.12 = 3.62 pH = pK a + log ([conj base]/[acid]) HCOOH  HCOO  + H + OH 

15. Addition of acid/base to a buffer n What if you exceed the buffer capacity? If 0.050 mols (2.0 g) of NaOH is added to the solution, what is the pH? pH= 3.74 + log (0.010+0.050/0.025  0.050) pH = pK a + log ([conj base]/[acid]) Not defined! Simply neutralize 0.025 mols, and calculate the pOH of the remainder… = 0.025 M OH  Or pH = 12.4

16. weak acid of OH  added Buffer region Equivalence point Mid-point