Question

Calculate the pH of a mixture that contains 0.25 M of HCOOH and 0.14 M of HClO. The Ka of HCOOH is 1.8 × 10-4 and the Ka of HClO is 4.0 × 10-8.

Answer 1

Since Ka of HClO is very small as compared to ka of HCOOH, Concentration of H+ from HClO can be ignored.

So all of H+ will come from HCOOH

HCOOH   <-----> HCOO- + H+

0.25                          0             0 (initial)

0.25-x                       x              x (at equilibrium)

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

1.8*10^-4 = x*x / (0.25-x)

Ka is small., so x will be small and it can be ignored as compared to 0.25.

So, above expression becomes:

1.8*10^-4 = x*x / 0.25

x= 6.71*10^-3 M

[H+] = x= 6.71*10^-3 M

pH = -loh [H+]

    = -log (6.71*10^-3)

   = 2.17