This is a VERY complicated problem but here goes.
H2SO4 is a strong acid for the first H that ionizes but a relatively weak acid for the second H. HCOOH is a weak acid.
First, determine the H^+ from H2SO4.
H2SO4 ==> H^+ + HSO4^-
HSO4^- ==> H^+ + SO4^2-
k2 for HSO4^- is (H^+)(SO4^-)/(HSO4^-).
Substitute into k2. For H^+ you substitute 0.62+x. For SO4^2- you substitute x and for (HSO4^-) you substitute 0.62-x. Solve the quadratic for x, then obtain 0.62+x and that will give you the H^+ from the H2SO4. If I didn't make a mistake you should get about 0.08M.
Then HCOOH is a weak acid and it ionizes as
HCOOH ==> H^+ + HCOO^-
Set up an ICE chart and look up Ka for HCOOH. The H^+ will be decreased by Le Chatelier's Principle (due to the H^+ from H2SO4). Substitute H^+ from H2SO4 for H^+ an solve for HCOO^-. Then add
H^+ from H2SO4 + H^+ from HCOOH to find the total H^+ and divide by liters soln to find M. Convert to pH.
What is the pH of a solution that is 0.62M H2SO4(aq) and 1.7M HCOOH (formic acid).
1 answer