Pka = -logKa
        =
-log4.2*10-7
        = 6.3767
PH = Pka + log[KHCO3]/[H2CO3]
        = 6.3767 +
log0.337/0.337
        = 6.3767
2. naOH -------> Na+ + OH-
      no of moles of H2C2O4 =
0.317
     no of moles of KHC2O4 Â
= 0.317
     Ka = 5.9*10-2
    PKa = -logka
           Â
= -log0.059Â Â = 1.229
    PH  = Pka +
log[KHC2O4]/[H2C2O4]
   PH  = 1.229 + log0.317/0.317
   PH = 1.229
HBr +H2O ------> H3O+ + Br-
0.071
MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0.071M
          Â
= 1.229 + log