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Question: The \( {K_a} \) for \( C{H_3}COOH \) is \( 1.8 \times {10^{ - 5}} \) . Find out the percentage disso...

The Ka{K_a} for CH3COOHC{H_3}COOH is 1.8×10−51.8 \times {10^{ - 5}} . Find out the percentage dissociation of 0.2M CH3COOH0.2M{\text{ }}C{H_3}COOH in 0.1M HCl0.1M{\text{ HCl}} solution.
A) 0.018
B) 0.36
C) 18
D) 36

Explanation

Solution

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 Ka{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 0.1M0.1M and that HCl is a strong acid which dissociation coefficient α=1\alpha = 1 , the dissociation of HCl can be shown as:
HCl→H++Cl−HCl \to {H^ + } + C{l^ - }
0.1 0.1 0.1
The concentration of acetic acid is given as 0.2M0.2M . Acetic acid is a weak acid and will dissociate partially. Consider the amount dissociated to be ‘x’. The dissociation can be shown as:
CH3COOH⇌CH3COO−+H+C{H_3}COOH \rightleftharpoons C{H_3}CO{O^ - } + {H^ + }

T=00.2--
T=equilibrium0.2−x0.2 - xxxxx

Here, H+{H^ + } is the common ion, and will exert a common ion effect. The total concentration of H+{H^ + } ions now will be =x+0.1= x + 0.1 (both from acetic acid and Hydrochloric acid). The dissociation Constant Ka{K_a} of acetic acid can be given as:
Ka=[CH3COO−][H+][CH3COOH]{K_a} = \dfrac{{[C{H_3}CO{O^ - }][{H^ + }]}}{{[C{H_3}COOH]}}
Substituting the values we get,
1.8×10−5=(x)(0.1+x)0.21.8 \times {10^{ - 5}} = \dfrac{{(x)(0.1 + x)}}{{0.2}}
0.36×10−5=(x)×(x+0.1)0.36 \times {10^{ - 5}} = (x) \times (x + 0.1)
x2+0.1x−0.36×10−5=0{x^2} + 0.1x - 0.36 \times {10^{ - 5}} = 0
x=0.35×10−4x = 0.35 \times {10^{ - 4}}
To find the percentage dissociation the formula would be: Dissociated AmountInitial Amount×100\dfrac{{Dissociated{\text{ }}Amount}}{{Initial{\text{ }}Amount}} \times 100
Therefore, the percentage dissociation of acetic acid will be =0.35×10−40.2×100=0.0175%= \dfrac{{0.35 \times {{10}^{ - 4}}}}{{0.2}} \times 100 = 0.0175\%
This value is approximately equal to 0.018%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+{H^ + } is the common ion, hence its overall concentration increases. Also, remember that if Ka{K_a} of any electrolyte is given, then that electrolyte is always a weak electrolyte. Strong electrolytes have unity as the dissociation constant.