Method 1

$1. {\mathrm{CH}}_3\mathrm{COOH}\leftrightarrow {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}^{+} {\mathrm{K}}_{\mathrm{a}}=1.74\times {10}^{-5}$
$2. {\mathrm{H}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}\leftrightarrow {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{OH}}^{-} {\mathrm{K}}_{\mathrm{w}}=1.0\times {10}^{-14}$
$\mathrm{Since} {\mathrm{K}}_{\mathrm{a}}>>{\mathrm{K}}_{\mathrm{w}} :$
${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}\leftrightarrow {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$
${\mathrm{C}}_{\mathrm{i}}=0.05 0 0$
$0.05\mathrm{\alpha }-0.5\mathrm{\alpha } 0.05\mathrm{\alpha } 0.05\mathrm{\alpha }$
${\mathrm{K}}_{\mathrm{a}}=\frac{\left(.05\mathrm{\alpha }\right)\left(0.5\mathrm{\alpha }\right)}{\left(0.5-0.05\mathrm{\alpha }\right)}$
$=\frac{\left(.05\mathrm{\alpha }\right)\left(0.05\mathrm{\alpha }\right)}{0.5\left(1-\mathrm{\alpha }\right)}$
$=\frac{0.5{\mathrm{\alpha }}^2}{1-\mathrm{\alpha }}$
$\begin{array}{l}1.74\times {10}^{-5}=\frac{0.05{\alpha }^2}{1-\alpha }\\ 1.74\times {10}^{-5}-1.74\times {10}^{-5}\alpha =0.05{\alpha }^2\\ 0.05{\alpha }^2+1.74\times {10}^{-5}\alpha -1.74\times {10}^{-5}\\ D=b^2-4ac\\ ={(1.74\times {10}^{-5})}^2-4\left(.05\right)(1.74\times {10}^{-5})\\ =3.02\times {10}^{-25}+.348\times {10}^{-5}\\ \alpha =\sqrt{\frac{K_a}{c}}\\ \alpha =\sqrt{\frac{1.74\times {10}^{-5}}{.05}}\\ =\sqrt{\frac{34.8\times {10}^{-5}\times 10}{10}}\\ =\sqrt{3.48\times {10}^{-6}}\end{array}$
$={\mathrm{CH}}_3\mathrm{COOH}\leftrightarrow {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}^{{}_{+}}$

$$\begin{gathered} \mathrm{\alpha }1.86\times {10}^{-3} \\ \left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]=0.05\times 1.86\times {10}^{-3} \\ =\frac{0.93\times {10}^{-3}}{1000} \\ =.000093 \\ \mathrm{Method} 2 \\ \mathrm{Degree} \mathrm{of} \mathrm{dissociation}, \\ \mathrm{\alpha }=\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{c}}} \\ \mathrm{c}=0.05 \mathrm{M} \\ {\mathrm{K}}_{\mathrm{a}}=1.74\times {10}^{-5} \\ \mathrm{Then}, \begin{array}{l}\alpha =\sqrt{\frac{1.74\times {10}^{-5}}{.05}}\\ \alpha =\sqrt{34.8\times {10}^{-5}}\\ \alpha =\sqrt{3.48}\times {10}^{-4}\\ \alpha ={1.8610}^{-2}\end{array} \\ {\mathrm{CH}}_3\mathrm{COOH}\leftrightarrow {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}^{+} \\ \mathrm{Thus}, \mathrm{concentration} \mathrm{of} {\mathrm{CH}}_3{\mathrm{COO}}^{-}=\mathrm{c} \\ =.05\times 1.86\times {10}^{-2} \\ =.093\times {10}^{-2} \\ =.00093 \\ \mathrm{Since} \left[{\mathrm{OAc}}^{-}\right]=\left[{\mathrm{H}}^{+}\right], \\ \left[{\mathrm{H}}^{+}\right]=.00093=.093\times {10}^{-2}. \\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ = -\log \left(.093\times -{10}^{-2}\right) \\ \therefore \mathrm{pH}=3.03 \end{gathered}$$

Hence, the concentration of acetate ion in the solution is 0.00093 M and is PH is 3.03.