Lecture 10 Acid-Base Equilibria -II

K a = C = 0.01 M [H + ] = (10 -4  ) ½ = M pH=3 90% acid, 10% base 2 C = M ??

K b = 2.6  C=10 -3 M [OH - ]=5  [H + ]=2  pH=9.7

Sodium lactate C=0.01 M K a = 1.4  = K b = [OH - ] =(  ) ½ = [H + ] = pH=7.92  8

Polyprotic acids: Zwitter-ion Aminoacids:

pK 1 =2.35 pK 2 =9.87 H 2 A  HA + H + HA  A + H +

H2AH2AHAA pH of H 2 CO 3 -same as any weak acid; use K 1 pH of Na 2 CO 3 Na 2 CO 3  2 Na + + CO 3 2- CO 3 2- same as any weak base, use K 2

H2AH2A HAA NaHCO 3 NaHCO 3  Na + + HCO 3 - In HA, [A]=[H 2 A] because 2HA  A + H 2 A [H + ] 2 = K 1  K 2

NaHCO 3 pK 1 =6.35 pK 2 = 10.33

H 3 PO 4 H 2 PO 4 - HPO 4 2- PO