Question

The ionization constant of acetic acid is $1.74\times {10}^{-5}$. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its $pH$.

Answer 1

The dissociation equilibrium is $C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$.

Let $\alpha$ be the degree of dissociation.

The equilibrium concentrations of $C{H}_{3}COOH,C{H}_{3}CO{O}^{-}$ and $H^{+}$ are $c(1-\alpha ),c\left(\alpha \right)$ and $c\left(\alpha \right)$ respectively.

The equilibrium constant expression is $K_c=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$.

$K_c=\frac{\left(c\alpha \right)\left(c\alpha \right)}{c(1-\alpha )}\approx c{\alpha }^2$
$\alpha =\sqrt{\frac{K_a}{c}}=\sqrt{\frac{1.74\times {10}^{-5}}{0.05}}=1.865\times {10}^{-2}$

$\left[C{H}_{3}C{O}^{-}\right]=\left[H^{+}\right]=c\alpha =0.05\times 1.865\times {10}^{-2}=9.33\times {10}^{-4}M$

$pH=-log\left[H^{+}\right]=-log(9.33\times {10}^{-4})=3.03$

The concentration of acetate ion and its pH are $9.33\times {10}^{-4}$ and $3.03$ respectively.