Question

Calculate the pH of a mixture that contains 0.20 M of HCOOH and 0.18 M of HClO. The Ka of HCOOH is 1.8 × 10-4 and the Ka of HClO is 4.0 × 10-8.

Answer 1

both formic acid and HCIO are weak acids

so 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 HCOOH since the Ka values are relatively close to one another, the "stronger" weak acid will affect the "weaker" weak acid.

Remember that H2O(l) is not part of the equilibrium

.......HCOOH(aq) + H2O(l) <--> HCOO^- (aq) + H3O^+ (aq)
I...........0.20.....O...........O
C...........-x.......+x............+x
E.........0.20-x...x.................x

Ka = ([HCOO2^-][H3O^+])/ [HCOOH]
1.8*10^-4 = (x^2)/(0.20-x)

x^2-1.8 x 10^-4 x 0.36 x 10^-4 ( solve quadratic equation using calculator)

Solving for "x" through the quadratic formula, you should end up with about x = 6*10^-3

Now moving on to the "weaker" weak acid. Remember that both weak acids are in the solution. The stronger weak acid will affect the weaker weak acid.

.......HCIO(aq) + H2O(l) <--> CIO- (aq) + H3O^+ (aq)

I.......0.18......0........6*10^-3
C.............-x....+x.............+x
E.........0.18-x..x.....6*10^-3 + x

Ka = ([CIO-][H3O^+])/ [HCIO]
4*10^-8 = (x)(6*10^-3 +x)/(0.18-x)
x^2 + 2.4 x 10 ^-10 x + 7.2 X 10^-8 = 0
x = 2.6x10^-4

Now the total [H3O+] = 6*10^-3 + 2.6x10^-4
= 6.26*10^-3

pH = - log [H3O+]
= 2.20

see, there may be some mistakes in the calculations, but the process is correct.