Recall that the percent ionization is given by:
The dissociation of acetic acid (CH3COOH) is as follows:
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq); Ka = 1.8 × 10–5
From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and Ka expression.
The Ka expression for CH3COOH is:
Since the ratio is greater than 500, we can remove (–x) from the equation.
The equation simplifies to:
What is the percent ionization of a 2 M solution of acetic acid at 25 °C?
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