Answer
Original acid molarity : $2.329 \times 10^{-3}M$
Work Step by Step
1. Find $[H_3O^+]:$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 3.26}$
$[H_3O^+] = 5.495 \times 10^{- 4}M$
2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [HCOO^-] = x$
-$[HCOOH] = [HCOOH]_{initial} - x$
3. Now, use the Ka and x values and equation to find the initial concentration value.
$Ka = \frac{[H_3O^+][HCOO^-]}{ [Initial HCOOH] - x}$
$ 1.7\times 10^{- 4}= \frac{[x^2]}{ [Initial HCOOH] - x}$
$ 1.7\times 10^{- 4}= \frac{( 5.5\times 10^{- 4})^2}{[Initial HCOOH] - 5.5\times 10^{- 4}}$
$[Initial HCOOH] - 5.5\times 10^{- 4} = \frac{ 3.025\times 10^{- 7}}{ 1.7\times 10^{- 4}}$
$[Initial HCOOH] - 5.5\times 10^{- 4} = 1.779\times 10^{- 3}$
$[Initial HCOOH] = 1.779\times 10^{- 3} + 5.5\times 10^{- 4}$
$[Initial HCOOH] = 2.329\times 10^{- 3}M$
