Answer – Given,
[C6H5COOH] = 0.45 M , Ka for the
C6H5COOH = 6.4*10-5
C6H5COOH + H2O ------>
H3O+ +
C6H5COO-
I
  0.45                     Â
        0      Â
  0
C
   -x                  Â
             +x        Â
+x
E
  0.45-x                  Â
        +x        Â
+x
Ka = [H3O+]
[C6H5COO-] /
[C6H5COOH]
6.4*10-5 = x*x /(0.45-x)
6.4*10-5 *(0.45-x) = x2
Now we need to set up quadratic equation
2.88*10-5 - 6.4*10-5 x = x2
x2 + 6.4*10-5 x - 2.88*10-5 =
0
a =1 , b = 6.4*10-5, c = -2.88*10-5
Using the quadratic equation
x = -b +/- √b2-4a*c / 2a
Plugging the value in this formula
x = 0.00533 M
so, x = [H3O+] = 0.00533 M
so, pH = -log [H3O+]
          =
-log 0.00533 M
          =
2.27
So, the pH of a 0.45M solution of Benzonic Acid is
2.27
