Calculate the degree of ionization of $0.05 \,M$ acetic acid if its $p K_{ a }$ value is $4.74$ 

How is the degree of dissociation affected when its solution also contains $(a)$ $0.01 \,M$ $(b)$ $0.1 \,M$ in $HCl$ ?

$c=0.05 \,M$

$p K_{a}=4.74$

$p K_{a}=-\log \left(K_{a}\right)$

$K_{a}=1.82 \times 10^{-5}$

$K_{a}=c \alpha^{2}$ $\alpha=\sqrt{\frac{K_{a}}{c}}$

$\alpha=\sqrt{\frac{1.82 \times 10^{-5}}{5 \times 10^{-2}}}=1.908 \times 10^{-2}$

When $HCI$ is added to the solution, the concentration of $H ^{+}$ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

Case $I:$ When $0.01 \,M$ $HCl$ is taken.

Let $x$ be the amount of acetic acid dissociated after the addition of $HCl$.

                                    $C{H_3}COOH\quad  \leftrightarrow \quad {H^ + }\quad  + \quad C{H_3}CO{O^ - }$

Initial conc.                      $0.05\,M$                          $0$                       $0$

After dissociation           $0.05-x$                     $0.01+x$          $x$

As the dissociation of a very small amount of acetic acid will take place, the values i.e., $0.05-x$ and $0.01+x$ can be taken as $0.05$ and $0.01$ respectively.

$K_{a}=\frac{\left[ CH _{3} COO ^{-}\right]\left[ H ^{+}\right]}{\left[ CH _{3} COOH \right]}$

$\therefore K_{a}=\frac{(0.01) x}{0.05}$

$x=\frac{1.82 \times 10^{-5} \times 0.05}{0.01}$

$x=1.82 \times 10^{-3} \times 0.05 \,M$

Now, $\alpha=\frac{\text { Amount of acid dissociated }}{\text { Amount of acid taken }}$

$=\frac{1.82 \times 10^{-3} \times 0.05}{0.05}$

$=1.82 \times 10^{-3}$

Case $II:$ When $0.1 \,M$ $HCl$ is taken.

Let the amount of acetic acid dissociated in this case be $X$. As we have done in the first case, the concentrations of various species involved in the reaction are:

$\left[ CH _{3} COOH \right]=0.05-X ; 0.05\, M$

$\left[ CH _{3} COO ^{-}\right]=X$

$\left[ H ^{+}\right]=0.1+X ; 0.1 \,M$

$K_{a}=\frac{\left[ CH _{3} COO ^{-}\right]\left[ H ^{+}\right]}{\left[ CH _{3} COOH \right]}$

$\therefore K_{a}=\frac{(0.1) X}{0.05}$

$x=\frac{1.82 \times 10^{-5} \times 0.05}{0.1}$

$x=1.82 \times 10^{-4} \times 0.05 \,M$

Now,  $\alpha=\frac{\text { Amount of acid dissociated }}{\text { Amount of acid taken }}$

$=\frac{1.82 \times 10^{-4} \times 0.05}{0.05}$

$=1.82 \times 10^{-4}$