## Chemistry: The Central Science (13th Edition)

The concentration of hydrofluoric acid is equal to $2.9 \times 10^{-4} \space M$
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. $Ka = \frac{[H_3O^+][F^-]}{ [Initial \space HF] - x}$ $6.8\times 10^{- 4}= \frac{[x^2]}{ [Initial \space HF] - x}$ $6.8\times 10^{- 4}= \frac{( 2.2\times 10^{- 4})^2}{[Initial \space HF] - 2.2\times 10^{- 4}}$ $[Initial \space HF] - 2.2\times 10^{- 4} = \frac{ ( 2.2\times 10^{- 4})^2}{ 6.8\times 10^{- 4}}$ $[Initial \space HF] = \frac{ ( 2.2\times 10^{- 4})^2}{ 6.8\times 10^{- 4}} + 2.2\times 10^{- 4}$ $[Initial \space HF] = 2.9 \times 10^{-4}$