FeSO4 ------> Fe+2 +SO42-
    SO42- + H2O ----->
HSO4- + OH_
IÂ Â Â
0.1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â
-x                        Â
+XÂ Â Â Â Â Â Â Â Â +x
EÂ Â 0.1-x
                    Â
+x         +x
ka of HSO4- is 0.012
Kb = Kw/ka
    = 1*10-14/0.012
   = 83.3*10-14
Kb = [HSO4-][OH-]/[SO42-]
83.3*10-14 = x*x/0.1-x
x2Â Â Â Â =
83.3*10-14
x = 9.12*10-7
[OH-] = x =9.12*10-7 M
POH = -log9.12*10-7
      = -log9.12+7log10
    = -0.9604 +7 = 6.0396
PH = 14-POH
    = 14-6.0396
   = 7.9604
