EXAM #2 H2CO3   HCO H ; Ka1 = 4.3 x 10-7

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Presentation transcript:

EXAM #2 H2CO3   HCO3 - + H+ ; Ka1 = 4.3 x 10-7 HCO3-   CO3 -2 + 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 - ] 1.0-X

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

pH = pHa + log [A-] ; pH = 10.7 + log (0.03)/(0.05) = 10.48 [HA] EXAM #3 0.03 mole A- and 0.05 mole HA; THIS IS A BUFFER!!!!!!!! pH = pHa + log [A-] ; pH = 10.7 + log (0.03)/(0.05) = 10.48 [HA] pH =pHa + log [A-] ; pH = 10.7 + log (0.03+0.001)/(0.05-0.001)=10.50 [HA] EVERY MOLE OF OH- PRODUCES A MOLE OF A-- EVERY MOLE OF OH- NEUTRALIZES A MOLE OF HA-

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)2 ; X= 2.9 x 10-5 = [OH-] [C5H5N ] (0.5-X) 2.9 x 10-5 = [OH-] ; pOH = 4.53 ; pH = 9.47

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)