Answer - We are given,
[HCrO4-] = 0.109 M ,
We know Ka2 for the H2CrO4 =
3.2*10-7
we need to put ICE table for calculating the
[H3O+]
   HCrO4- + H2O
------> H3O+ +
CrO42-
I
0.109Â 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 0
C
  -x                  Â
      Â
+x         +x
E
0.109-x                 Â
  Â
+x         +x
Ka = [H3O+] [CrO42-]
/ [HCrO4-]
3.2*10-7 = x*x /(0.005-x)
3.2*10-7 *(0.109-x) = x2
The Ka value is too small, so we can neglect the x in the
0.109-x.
3.2*10-7*0.109 =x2
x = 0.000187 M
so, x = [H3O+] = 0.000187 M
so, pH = -log [H3O+]
          =
-log 0.000187 M
          =
3.73
So answer for this question is 3.73
