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

$p{K}_a=-log{K}_a=4.74$
$K_a=10^{-pKa}=10^{-4.74}=1.8\times 10^{-5}$
Let x be the degree of dissociation. The concentration of acetic acid solution, C = 0.05 M
The degree of dissociation, $x=\sqrt{\frac{K_a}{C}}=\sqrt{\frac{1.8\times 10^{-5}}{0.05}}=0.019$


(a) The solution is also 0.01 M in HCl.
Let x M be the hydrogen ion concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration
$[H^{+}]=0.01+x$
The acetate ion concentration is equal to the  hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.
$[C{H}_{3}CO{O}^{-}]=x$
$[C{H}_{3}COOH]=0.05-x$
$K_a=\frac{[H^{+}][C{H}_{3}CO{O}^{-}]}{[C{H}_{3}COOH]}$
$1.8\times 10^{-5}=\frac{(0.01+x)x}{(0.05-x)}$.....(1)
As x is very small,  $0.01+x\approx 0.01$
$0.05-x\approx 0.05$
Hence, the equation (i) becomes
$1.8\times 10^{-5}=\frac{0.01x}{0.05}$
$x=9.0\times 10^{-5}M$
The degree of ionization is $\frac{[C{H}_{3}CO{O}^{-}]}{[C{H}_{3}COOH]}=\frac{x}{c}=\frac{9.0\times 10^{-5}}{0.05}=1.8\times 10^{-3}=0.0018$.

(b) The solution is also 0.1 M in HCl.
Let x M be the hydrogen ion concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration
$[H^{+}]=0.1+x$
The acetate ion concentration is equal to the  hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.
$[C{H}_{3}CO{O}^{-}]=x$
$[C{H}_{3}COOH]=0.05-x$
$K_a=\frac{[H^{+}][C{H}_{3}CO{O}^{-}]}{[C{H}_{3}COOH]}$
$1.8\times 10^{-5}=\frac{(0.1+x)x}{(0.05-x)}$.....(1)
As x is very small,  $0.1+x\approx 0.1$
$0.05-x\approx 0.05$
Hence, the equation (i) becomes
$1.8\times 10^{-5}=\frac{0.1x}{0.05}$
$x=9.0\times 10^{-6}M$
The degree of ionization is $\frac{[C{H}_{3}CO{O}^{-}]}{[C{H}_{3}COOH]}=\frac{x}{c}=\frac{9.0\times 10^{-6}}{0.05}=1.8\times 10^{-4}=0.00018$.