**Question:**

Calculate the degree of ionization of $0.05 \mathrm{M}$ acetic acid if its $\mathrm{pK}_{\mathrm{a}}$ value is $4.74$.

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

**Solution:**

$c=0.05 \mathrm{M}$

$p K_{a}=4.74$

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

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

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

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

When $\mathrm{HCl}$ is added to the solution, the concentration of $\mathrm{H}^{+}$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.

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

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

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

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

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

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:

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