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 $HCl$ ?
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, $\alpha=\frac{\text { Amount of acid dissociated }}{\text { Amount of acid taken }}$
$=\frac{1.82 \times 10^{-4} \times 0.05}{0.05}$
$=1.82 \times 10^{-4}$
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