Question

What is the percentage dissociation of 0.1M solution of Acetic acid? (\[{K_a} = {10^{ - 5}}\])

(A) 10%

(B) 100%

(C) 1%

(D) 0.01%

Answer 1

Hint: Any acid dissociates into ions according to the dissociation constant of that acid. Not all the acid molecules are dissociated in the solution. We can calculate the % dissociation by knowing the concentration of both acetic acid and its ions in the solution.

Complete step by step answer:

Let’s see what happens when acetic acid dissociates.

\[C{H_3}COOH \rightleftharpoons C{H_3}CO{O^ - }_{(aq)} + {H^ + }_{(aq)}\]

Acetic acid molecules dissociate into two ions as shown in the above given reaction. As it is an acid, it gives one proton and its conjugate base acetate ion. This reaction generally occurs in aqueous media. The reverse arrow suggests that the reaction is reversible means it can give the reverse reaction also.

Now, the extent of dissociation of acid can be expressed by a formula. 

We know the formula of Dissociation constant of acetic acid which is

\[{K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }_{(aq)}} \right]\left[ {{H^ + }_{(aq)}} \right]}}{{\left[ {C{H_3}COOH} \right]}}\]………………..(1)

It is given in the question that \[{K_a} = {10^{ - 5}}\] and concentration of Acetic acid is 0.1M. Now putting all these value in equation (1), we get

\[{10^{ - 5}} = \dfrac{{\left[ {C{H_3}CO{O^ - }_{(aq)}} \right]\left[ {{H^ + }_{(aq)}} \right]}}{{0.1}}\]

So we can simplify this as

\[\left[ {C{H_3}CO{O^ - }_{(aq)}} \right]\left[ {{H^ + }_{(aq)}} \right] = {10^{ - 6}}\]

Now we can see in the reaction that concentration of acetate ions and protons will be the same.

\[{\left[ {C{H_3}CO{O^ - }_{(aq)}} \right]^2} = {\left[ {{H^ + }_{(aq)}} \right]^2} = {10^{ - 6}}\] so,

\[\left[ {C{H_3}CO{O^ - }_{(aq)}} \right] = \left[ {{H^ + }_{(aq)}} \right] = {10^{ - 3}}\]

As we know the concentration of acid and ions in the solution, we can calculate the % dissociation as well.

\[\mathop {C{H_3}COOH}\limits_{0.1M}  \rightleftarrows \mathop {C{H_3}CO{O^ - }_{(aq)}}\limits_{{{10}^{ - 3}}M}  + \mathop {{H^ + }_{(aq)}}\limits_{{{10}^{ - 3}}M} \]

$\% Dissociation = \dfrac{{{\text{Concentration of dissociated ions}} \times {\text{100}}}}{{{\text{Concentration of acid in solution}}}}$ ………..(2)

Hence, putting the values in the equation (2), we get

% Dissociation = $\dfrac{{{{10}^{ - 3}} \times 100}}{{0.1}}$ 

% Dissociation = $\dfrac{0.1}{0.1}$ 

% Dissociation = $1\% $ 

So, the correct answer is “Option C”.

- In the formula of Dissociation constant of acid, concentration of all the species is described in Molarity.

- As shown in the reaction, the dissociated ions and the undissociated acid are always in equilibrium.

- With the value of Dissociation constant of acid, we can measure how much acid will dissociate when it is dissolved in the solution. So, basically we can measure the strength of any acid by its dissociation constant. 

Note: - Acetic acid is a monoprotic acid as it has only one acidic proton. Do not consider it something else as it has more than one hydrogen atom in its structure. 

- Acetic acid dissociates into two ions only, acetate ion and hydrogen ion. Any other kind of dissociation is not possible.