Question

The $ {K_a} $ for $ C{H_3}COOH $ is $ 1.8 \times {10^{ - 5}} $ . Find out the percentage dissociation of $ 0.2M{\text{ }}C{H_3}COOH $ in $ 0.1M{\text{ HCl}} $ solution.
A) 0.018
B) 0.36
C) 18
D) 36

Answer 1

Hint :We are given two acids, one is the acetic acid which is a weak acid and will partially dissociate in here. Hence the dissociation constant $ {K_a} $ is given. Another is strong Hydrochloric Acid which dissociates completely in water.

Complete Step By Step Answer:
Given that the concentration of HCl solution is $ 0.1M $ and that HCl is a strong acid which dissociation coefficient $ \alpha = 1 $ , the dissociation of HCl can be shown as:
 $ HCl \to {H^ + } + C{l^ - } $
0.1 0.1 0.1
The concentration of acetic acid is given as $ 0.2M $ . Acetic acid is a weak acid and will dissociate partially. Consider the amount dissociated to be ‘x’. The dissociation can be shown as:
 $ C{H_3}COOH \rightleftharpoons C{H_3}CO{O^ - } + {H^ + } $

T=0	0.2	-	-
T=equilibrium	$ 0.2 - x $	$ x $	$ x $

Here, $ {H^ + } $ is the common ion, and will exert a common ion effect. The total concentration of $ {H^ + } $ ions now will be $ = x + 0.1 $ (both from acetic acid and Hydrochloric acid). The dissociation Constant $ {K_a} $ of acetic acid can be given as:
  $ {K_a} = \dfrac{{[C{H_3}CO{O^ - }][{H^ + }]}}{{[C{H_3}COOH]}} $
Substituting the values we get,
  $ 1.8 \times {10^{ - 5}} = \dfrac{{(x)(0.1 + x)}}{{0.2}} $
  $ 0.36 \times {10^{ - 5}} = (x) \times (x + 0.1) $
  $ {x^2} + 0.1x - 0.36 \times {10^{ - 5}} = 0 $
  $ x = 0.35 \times {10^{ - 4}} $
To find the percentage dissociation the formula would be: $ \dfrac{{Dissociated{\text{ }}Amount}}{{Initial{\text{ }}Amount}} \times 100 $
Therefore, the percentage dissociation of acetic acid will be $ = \dfrac{{0.35 \times {{10}^{ - 4}}}}{{0.2}} \times 100 = 0.0175\% $
This value is approximately equal to $ 0.018\% $ .
Hence the correct answer is Option A).

Note :
Common Ion effect refers to the suppression of the Dissociation constant of any weak electrolyte by addition of a small amount of strong electrolyte having a common ion. In the given problem the $ {H^ + } $ is the common ion, hence its overall concentration increases. Also, remember that if $ {K_a} $ of any electrolyte is given, then that electrolyte is always a weak electrolyte. Strong electrolytes have unity as the dissociation constant.