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EXAM #2 H 2 CO 3   HCO 3 - + H+ ; Ka1 = 4.3 x 10 -7

EXAM #2 H 2 CO 3   HCO 3 - + H+ ; Ka1 = 4.3 x 10 -7 HCO 3 -   CO 3 -2 + H+ ; Ka2 = 5.6 x 10 -11. Ka1 = [ H 2 CO 3 ][ H+ ] ; 4.3 x 10 -7 = (X) 2 ; X = 6.5 x10 -4 = H+ = HCO 3 - [HCO 3 - ] 1.0-X.

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EXAM #2 H 2 CO 3   HCO 3 - + H+ ; Ka1 = 4.3 x 10 -7

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  1. EXAM #2 H2CO3   HCO3 - + H+ ; Ka1 = 4.3 x 10-7 HCO3-   CO3 -2 + H+ ; Ka2 = 5.6 x 10-11 Ka1 = [H2CO3][H+] ; 4.3 x 10-7 = (X)2 ; X = 6.5 x10-4 = H+ = HCO3- [HCO3 - ] 1.0-X

  2. 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 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)

  3. 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-

  4. EXAM #4 WEAK BASE IONIZATION 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

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

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