Question

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 $HCl$ ?

Answer 1

$p{K}_a=-log{K}_a=4.74$
$K_a={10}^{-pKa}={10}^{-4.74}=1.8\times {10}^{-5}$
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
$\left[H^{+}\right]=0.01+x$
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.
$\left[C{H}_{3}CO{O}^{-}\right]=x$
$\left[C{H}_{3}COOH\right]=0.05-x$
$K_a=\frac{\left[H^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$
$1.8\times {10}^{-5}=\frac{(0.01+x)x}{(0.05-x)}$.....(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 $\frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}=\frac{x}{c}=\frac{9.0\times {10}^{-5}}{0.05}=1.8\times {10}^{-3}=0.0018$.
(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
$\left[H^{+}\right]=0.1+x$
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.
$\left[C{H}_{3}CO{O}^{-}\right]=x$
$\left[C{H}_{3}COOH\right]=0.05-x$
$K_a=\frac{\left[H^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$
$1.8\times {10}^{-5}=\frac{(0.1+x)x}{(0.05-x)}$.....(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 $\frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}=\frac{x}{c}=\frac{9.0\times {10}^{-6}}{0.05}=1.8\times {10}^{-4}=0.00018$.