Question

Calculate the degree of ionization of 0.05 M acetic acid of its ${}_p{K}_a$ value is 4.74. How is the degree of dissociation affected when its solution is also (a) 0.01 M and (b) 0.1 M in hydrochloric acid?

Answer 1

$C=0.05M$

$p{K}_a=4.74$

$\Rightarrow p{K}_a=-\log K_a$

$K_a=1.82\times {10}^{-5}$

$K_a=C{\propto }^2$;$\propto =\sqrt{\frac{K_a}{C}}$

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

When $HCl$ is added to the $so{l}^n$, the $con{c}^n$ of $H^{+}$ ions will increase. Therefore, the equilibrium will shift in the backward direction, ie; dissociation of acetic acid will decrease.

Case (a) When $0.01M$ $HCl$ taken

$C{H}_{3}COOH⟷H^{+}+C{H}_{3}CO{O}^{-}$

initial $e{q}^n$ $0.05$ $O$ $O$

$0.05-x$ $0.01+x$ $x$

as the dissociation of a very small amount of $C{H}_{3}COOH$ will take place , the values $0.05-x$ and $0.01+x$ cn be taken as $0.05$ and $0.01$ respectively.

$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$

$\therefore$ , $K_a=\frac{\left(0.01x\right)}{0.05}$

$x=\frac{1.82\times {10}^{-5}\times 0.05}{0.01}=1.82\times {10}^{-3}\times 0.05M$

Now,

$\rho =\frac{amountofaciddissociated}{amountofacidtaken}$

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

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

Case (b): When $0.1M$ of $HCl$ taken

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

$\left[C{H}_{3}CO{O}^{-}\right]=X$

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

$K_a=\frac{0.1X}{0.05}$

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

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

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

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