Answer
The concentration of hydrofluoric acid is equal to $2.9 \times 10^{-4} \space M$
Work Step by Step
1. Find the $[H_3O^+]$ value:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 3.65}$
$[H_3O^+] = 2.2 \times 10^{-4}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
$HF(aq) + H_2O(l) \lt -- \gt F^-(aq) + H_3O^+(aq) $
\begin{matrix} & [HF] & [H_3O^+] & [F^-] \\ Initial & [Initial \space HF] & 0 & 0 \\ Change & -x & +x & +x\\ Equil & [Initial \space HF] -x & x & x \end{matrix}
3. Now, use the $K_a$ and x values and equation to find the initial concentration value.
$Ka = \frac{[H_3O^+][F^-]}{ [Initial \space HF] - x}$
$ 6.8\times 10^{- 4}= \frac{[x^2]}{ [Initial \space HF] - x}$
$ 6.8\times 10^{- 4}= \frac{( 2.2\times 10^{- 4})^2}{[Initial \space HF] - 2.2\times 10^{- 4}}$
$[Initial \space HF] - 2.2\times 10^{- 4} = \frac{ ( 2.2\times 10^{- 4})^2}{ 6.8\times 10^{- 4}}$
$[Initial \space HF] = \frac{ ( 2.2\times 10^{- 4})^2}{ 6.8\times 10^{- 4}} + 2.2\times 10^{- 4}$
$[Initial \space HF] = 2.9 \times 10^{-4}$
