HF(aq) + H2O(l) ⇌ F–(aq) + H3O+(aq); Ka = 3.5 × 10–4
From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and Ka expression.
The Ka expression for HF is:
Plugging in the equilibrium concentrations from the ICE table into the Ka expression:
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 Ka:
Since the ratio is greater than 500, we can remove (–x) from the equation.
The equation simplifies to:
Find the percent ionization of a 0.337 M HF solution. The Ka for HF is 3.5 x 10-4.
C) 1.2 x 10-2%
D) 3.5 x 10-2%
E) 1.1 %
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