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 $pH$.

(N/A) The dissociation of acetic acid is given by:
$CH_{3}COOH \longleftrightarrow CH_{3}COO^{-} + H^{+}$
Given $K_{a} = 1.74 \times 10^{-5}$ and $c = 0.05 \ M$.
Since $K_{a}$ is very small,we use the approximation $\alpha = \sqrt{\frac{K_{a}}{c}}$.
$\alpha = \sqrt{\frac{1.74 \times 10^{-5}}{0.05}} = \sqrt{34.8 \times 10^{-5}} = \sqrt{3.48 \times 10^{-4}} \approx 1.86 \times 10^{-2}$.
The concentration of acetate ion $[CH_{3}COO^{-}] = c \times \alpha = 0.05 \times 1.86 \times 10^{-2} = 9.3 \times 10^{-4} \ M$.
Since $[H^{+}] = [CH_{3}COO^{-}] = 9.3 \times 10^{-4} \ M$,
$pH = -\log[H^{+}] = -\log(9.3 \times 10^{-4}) = 4 - \log(9.3) \approx 4 - 0.968 = 3.032$.
Thus,the degree of dissociation is $0.0186$,the concentration of acetate ion is $9.3 \times 10^{-4} \ M$,and the $pH$ is $3.03$.