Answer
$[Initial HCOOH] = 0.02411M$
Work Step by Step
1. Find the $[H_3O^+]$ value:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.7}$
$[H_3O^+] = 1.995 \times 10^{- 3}$
2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [HCOO^-] = x = 1.995 \times 10^{-3}$
-$[HCOOH] = [HCOOH]_{initial} - x$
3. Now, use the Ka and x values and equation to find the initial concentration value.
$Ka = \frac{[H_3O^+][HCOO^-]}{ [Initial HCOOH] - x}$
$ 1.8\times 10^{- 4}= \frac{[x^2]}{ [Initial HCOOH] - x}$
$ 1.8\times 10^{- 4}= \frac{( 1.995\times 10^{- 3})^2}{[Initial HCOOH] - 1.995\times 10^{- 3}}$
$[Initial HCOOH] - 1.995\times 10^{- 3} = \frac{ 3.981\times 10^{- 6}}{ 1.8\times 10^{- 4}}$
$[Initial HCOOH] - 1.995\times 10^{- 3} = 0.02212$
$[Initial HCOOH] = 0.02212 + 1.995\times 10^{- 3}$
$[Initial HCOOH] = 0.02411$
