Answer
Original acid molarity: $6.358 \times 10^{-6}M$.
Work Step by Step
1. Find $[H_3O^+]$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 5.26}$
$[H_3O^+] = 5.495 \times 10^{- 6}$
2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [Conj. Base] = x$
-$[Acid] = [Acid]_{initial} - x$
3. Now, use the Ka and x values and equation to find the initial concentration value.
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Initial Acid] - x}$
$ 3.5\times 10^{- 5}= \frac{[x^2]}{ [Initial Acid] - x}$
$ 3.5\times 10^{- 5}= \frac{( 5.495\times 10^{- 6})^2}{[Initial Acid] - 5.495\times 10^{- 6}}$
$[Initial Acid] - 5.495\times 10^{- 6} = \frac{ 3.02\times 10^{- 11}}{ 3.5\times 10^{- 5}}$
$[Initial Acid] - 5.495\times 10^{- 6} = 8.628\times 10^{- 7}$
$[Initial Acid] = 8.628\times 10^{- 7} + 5.495\times 10^{- 6}$
$[Initial Acid] = 6.358\times 10^{- 6}$
