HCOOH + H2O ------------->
H3O+ + HCOO-
  Â
0.21Â Â Â Â Â Â Â Â Â Â Â Â 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 Â
0
  0.21 -
x                              Â
x            Â
x
Ka = [ H3O+][HCOO-]/[HCOOH]
1.8 x 10-4 = x X x/0.21 - x
  x2 + 1.8 x 10-4x - 0.378 x
10-4 = 0
x = - 1.8 x 10-4+ sqrt [(1.8 x 10-4
)2+ 4 x 1 x 0.378 x 10-4]/2
x = 0.6 x 10-2
concentration of species
[HCOOH] = 0.21 - x
               Â
= 0.21 - 0.006
               Â
= 0.204M
[HCOO-] = 0.006M
[H3O+]Â Â =
0.006M
[OH-] =
Kw/[H3O+] = 10-14/0.006
         = 0.166 x
10-11M
PH = -log[H3O+]
    = -log[6 x 10-3]
PHÂ Â = 2.22
