Question

9) A formic acid, HCOOH (Ka = 1.7 x 10^-4), solution has a pH of 3.26. What is the original molarity of the HCOOH solution?

Answer 1

1) HCOOH(aq) <-----------> HCOO^-(aq) + H⁺(aq)

Initailly, [HCOOH] = x M & [H⁺] = [HCOO^-] = 0 M

Let at eqb., [HCOOH] = (x-y) M & [H⁺] = [HCOO^-] = y M

Now, pH = -log[H⁺] = -logy

Thus, 3.26 = -logy

or, y = 10^-3.26 = 5.5*10^-4 M

Hence , K_a = {[H⁺]*[HCOO^-]}/[HCOOH]

or, y²/(x-y) = 1.7*10^-4

or, x = 1.23*10^-3 M