300 likes | 1.12k Vues
pH Change to 1M Acetic Acid/1M Acetate Ion Soln.. Moles H Added. 0. Moles OH- Added. Buffers. Buffer Solutions resist a change in pHBuffers contain relatively large concentrations of eitherAn acid, HA and its conjugate base A-A base, B, and its conjugate acid (BH ) . Buffers. NH4 to react with OH-.
                
                E N D
1. Buffer Example and Titration Calculations With your Chemistry 102 Host
Dr. Mike Daniel 
2. pH Change to 1M Acetic Acid/1M Acetate Ion Soln. 
3. Buffers Buffer Solutions resist a change in pH
Buffers contain relatively large concentrations of either
An acid, HA and its conjugate base A-
A base, B, and its conjugate acid (BH+)  
4. Buffers 
5. Buffers 
6. Buffers When H+ is added, it reacts essentially to completion with the weak base present
H+ + A- ? HA  or
H+ + B ? BH+ 
7. Buffers When OH- is added, it reacts essentially to completion with the weak acid present
OH- + HA ? H2O + A-
OH- + BH+ ? H2O +  B 
8. Buffers pH = pKa + log (base/acid)
Want pH ? pKa ? 1
pH determined by 
Ka of acid and ratio of acid/conjugate base
or 
Kb of base and ratio base/conjugate acid 
9. Buffer Choice Want pH ? pKa ? 1
How do I make a pH 4.0 buffer?
Choose a pKa near the desired pH 
10. Buffer Table 
11. Buffer Choice Choose a pKa near the desired pH
pH = pKa + log (base/acid)
4.0 = 3.74 + log (base/acid)
0.26 = log (base/acid)
10.26 = 1.8 = (Na formate / formic acid)
 
12. Basic Buffer Choice Ammonia pKb = 4.74 
pKa = 14.00  4.74 = 9.26
NH3 / NH4Cl used to buffer around pH 9.26
 
13. Buffer Capacity As long as ratio remains virtually constant, the pH will be virtually constant
This is true as long as concentrations of buffering materials (HA/A-) or (B/BH+) are large compared with H+  or OH- added. 
14. Follow Text page 714 (Brady & Holum)
25 mL of .2 M HCl titrated with .2 M NaOH
Equivalence Point  Where Stoichiometric amounts of acid and base have been added
End Point  Where indicator color change occurs Acid / Base Titrations  Strong A & B 
15.  VMHCl  = 5 X 10-3 mol
-VMNaOH
= moles of H+ leftover till equivalent point reached
At equivalence point, 
	5 X 10-3 mol NaCl/.050 L solution
pH = 7 Acid / Base Titrations  Strong A & B 
16. Acid / Base Titrations  Strong A & B 
17. Acid / Base Titrations  Strong A & B 
18. Acid / Base Titrations  Strong A & B 
19. 25 mL of .2 M Acetic Acid (HAc) titrated with .2 M NaOH
Initial pH calculated as before
 Acid / Base Titrations  Weak Acid with Strong Base 
20. During titration up to equivalence point 
	VMHAc  = 5 X 10-3 mol
	-VMNaOH
= moles of HAc leftover 
VMNaOH = moles OH- added = moles Ac- 
 made
Say 10. mL of .2 M NaOH added to 25 mL of .2 M HAc
 Acid / Base Titrations  Weak Acid with Strong Base 
21. 			HC2H3O2	? H+ + C2H3O- Acid / Base Titrations  Weak Acid with Strong Base 
22. Acid / Base Titrations  Weak Acid with Strong Base 
23. At Equivalence Point  all acetic acid converted to Acetate ion
At Equivalence Point you have a Sodium Acetate Solution   Acid / Base Titrations  Weak Acid with Strong Base 
24. At Equivalence Point you have a Sodium Acetate Solution  
To determine pH
Use Kb  and 
C2H3O2-  + H2O ? HC2H3O2 + OH-
To determine [OH-] and [H+] and pH Acid / Base Titrations  Weak Acid with Strong Base 
25. At Equivalence Point you have a Sodium Acetate Solution  
pH < or > 7 ?
pH = 8.88 Acid / Base Titrations  Weak Acid with Strong Base 
26. Acid / Base Titrations  Weak Acid with Strong Base 
27. After Equivalence Point  
VMNaOH  VMHAc(initial) = moles OH-  in total volume.
 From [OH-] determine [H+] and pH Acid / Base Titrations  Weak Acid with Strong Base 
28. Acid / Base Titrations  Weak Acid with Strong Base 
29. Acid / Base Titrations  Weak Base with Strong Acid 
30. Acid / Base Titrations  Weak Base with Strong Acid 
31. Acid / Base Titrations  Weak Base with Strong Acid