Since Ka2 and Ka3 are very small as compared to
Ka1
We will use only Ka1
H3PO4Â Â Â -----> H +Â Â H2PO4
-
0.41Â 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
0Â Â Â Â Â Â (initial)
0.41-x                Â
x         Â
x       (at
equilibrium)
Ka1 = [H+] [H2PO4-]/[H3PO4]
7.5*10^-3 = x*x / (0.41-x)
3.075*10^-3 - 7.5*10^-3 x = x^2
x^2 + 7.5*10^-3 x - 3.075*10^-3 = 0
solving for x, the positive value of x is
0.052
So,
[H+] = 0.052 M
pH = -log [H+]
      = -log (0.052)
      = 1.28
Answer: 1.28
