Question

Calculate the pH of a mixture that contains 0.23 M of HCOOH and 0.20 M of HC6H5O. The Ka of HCOOH is 1.8

Answer 1

Make an I.C.E. table of the stronger acid (HCOOH ) first.
You'll need to use the H3O+ concentration from this part of the work to use on the 2nd I.C.E table for HC6H5O

                                            HCOOH<---->H+   + COO-

initial                                         0.23          0           0

at equilibrium                       0.23-x         x              x

Ka= [H+][COO-]/[HCOOH]

1.8*10^-4 = x*x / (0.23-x)

4.14*10^-5 - 1.8*10^-4x =x^2

x^2 +1.8*10^-4x - 4.14*10^-5 = 0

x=0.006 or x= - 0.006

x can't be negative. so x = 0.006

Now use 2nd acid:

                            HC6H5O <---> H+ + C6H5O-

initial:                         0.2               0.006     0

at equilibrium:    0.2-x                  0.006+x   x

Ka= [H+][C6H5O-]/[HC6H5O]

1.3*10^-10 = (0.006+x)*x /(0.2-x)

since Ka is very small, x will be small. so x can be ignored as compared to 0.006 and 0.2,so

since x will be small as compared to 0.006

[H+]=0.006+x = 0.006

pH= -log[H+]

   =-log(0.006)

= 2.2