Problem: Find the percent ionization of a 0.337 M HF solution. The Ka for HF is 3.5 x 10-4. A) 4.7% B) 3.2% C) 1.2 x 10-2% D) 3.5 x 10-2%E) 1.1 % 

$\boxed{\mathbf{\%}\mathbf{ }\mathbf{ionization}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}}{\mathbf{[}\mathbf{HA}{\mathbf{]}}_{\mathbf{initial}}}\mathbf{\times }\mathbf{100}}$

HF(aq) + H₂O(l) ⇌ F^–(aq) + H₃O⁺(aq); K_a = 3.5 × 10^–4

From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and K_a expression.

The K_a expression for HF is:

$\boxed{{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{F}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}\mathbf{HF}\mathbf{\right]}}}$

Plugging in the equilibrium concentrations from the ICE table into the K_a expression:

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{F}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}\mathbf{HF}\mathbf{\right]}}$

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{=}\frac{\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{337}\mathbf{-}\mathbf{x}}$

Now, we need to determine if we can remove (–x) from the equation. To do so, we need to determine the ratio of the initial concentration and K_a:

$\frac{\mathbf{[}\mathbf{HF}{\mathbf{]}}_{\mathbf{initial}}}{{\mathbf{K}}_{\mathbf{a}}}\mathbf{>}\mathbf{500}$

Since the ratio is greater than 500, we can remove (–x) from the equation. 

The equation simplifies to: