A solution of acetic acid is 1.0% ionized. Determine the molar concentration of acid $\left({\mathrm{K}}_{\mathrm{a}} = 1.8 \times {10}^{-5}\right)$ and also the $\left[{\mathrm{H}}^{+}\right]$

$\underset{\mathrm{C} - \mathrm{C\alpha }}{{\mathrm{CH}}_3\mathrm{COOH}} \rightleftharpoons \underset{\mathrm{C\alpha }}{{\mathrm{CH}}_3{\mathrm{COO}}^{-}} + \underset{\mathrm{C\alpha }}{{\mathrm{H}}^{+}}$     where ‘C’ is the concentration of the acid and $\alpha$ is the degree of dissociation.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{C\alpha }\times \mathrm{C\alpha }}{\mathrm{C}(1-\mathrm{\alpha })} \\ \mathrm{As} \mathrm{\alpha } \mathrm{is} \mathrm{very} \mathrm{small} \mathrm{can} \mathrm{be} \mathrm{neglected} \mathrm{in} \mathrm{the} \mathrm{denominator} \\ {\mathrm{K}}_{\mathrm{a}}={\mathrm{C\alpha }}^2 \end{gathered}$$   

$K_a = 1.8 \times {10}^{-5}= C{\alpha }^2$

$\therefore 1.8 \times {10}^{-5}= C.{\left(0.01\right)}^2$

or        $\mathrm{C}=1.8 \times {10}^{-5} \times {10}^4$

               =   $1.8 \times {10}^{-1}M$

$\left[H^{+}\right] = C\alpha or \left[H^{+}\right] =1.8 \times {10}^{-1} \times 0.01$

         $=0.18 \times 0.01=0.0018 M or 1.8 \times {10}^{-3} M$