Problem: What is the percent ionization of a 2 M solution of acetic acid at 25 °C? a) 0.6% b) 0.3% c) 0.0036% d) 0.0018%

FREE Expert Solution

Recall that the percent ionization is given by:


% ionization=[H+][HA]initial×100


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:


Ka=productsreactants=[H3O+][CH3COO-][CH3COOH]


Ka=[H3O+][CH3COO-][CH3COOH]


1.8×10-5=(x)(x)2-x


[CH3COOH]initialKa500


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

The equation simplifies to:

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Problem Details

What is the percent ionization of a 2 M solution of acetic acid at 25 °C? 

a) 0.6% 

b) 0.3% 

c) 0.0036% 

d) 0.0018%

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