1. EXAM #2
H2CO3   HCO H ; Ka1 = 4.3 x 10-7
HCO3-   CO H ; Ka2 = 5.6 x 10-11
H2CO3 
 HCO3 - H+
1.0 M -X X
1.0-X
Ka1 = [H2CO3][H+] ; 4.3 x 10-7 = (X)2 ; X = 6.5 x10-4 = H+ = HCO3-
[HCO3 - ] X

2.  H+ -X X EXAM #2 – calculate all ions in the following ;
H2CO3   HCO H ; Ka1 = 4.3 x 10-3
HCO3-   CO H ; Ka2 = 5.6 x 10-11
HCO3- 
 H+
CO3 -2
6.5 x10-4 -X X
6.5 x X
6.5 x10-4 +X
APPROXIMATE X IN BOTH TERMS
Ka2 = [CO3 -2 ][H+] ; 5.6 x = (X)(6.5 x10-4) ; X= 5.6 x =[CO3-2]
[HCO3 - ] (6.5 x10-4)

3. pH = pHa + log [A-] ; pH = 10.7 + log (0.03)/(0.05) = 10.48 [HA]
EXAM # mole A- and 0.05 mole HA; THIS IS A BUFFER!!!!!!!!
pH = pHa + log [A-] ; pH = log (0.03)/(0.05) = 10.48
[HA]
pH =pHa + log [A-] ; pH = log ( )/( )=10.50
[HA]
EVERY MOLE OF OH- PRODUCES A MOLE OF A--
EVERY MOLE OF OH- NEUTRALIZES A MOLE OF HA-

4. C5H5N H2O   OH- C5H5NH+ 0.5 -X X 0.5-X
EXAM #4 WEAK BASE IONIZATION
C5H5N
H2O 
 OH-
C5H5NH+
0.5 -X X
0.5-X
Kb = [C5H5NH+][OH-] ; 1.7 x 10-9 = (X) ; X= 2.9 x 10-5 = [OH-]
[C5H5N ] (0.5-X)
2.9 x 10-5 = [OH-] ; pOH = 4.53 ; pH = 9.47

5. C5H5N H2O   OH- C5H5NH+ 0.5 0.001 -X X 0.5-X
EXAM #4a WEAK BASE IONIZATION WITH COMMON ION (OH-)
C5H5N
H2O 
 OH-
C5H5NH+
0.5
0.001 -X X
0.5-X
Kb =[C5H5NH+][OH-]; 1.7 x 10-9 = (X)(0.001+X) ;X= 8.9x10-7 = [C5H5NH+]
[C5H5N ] (0.5-X)