If the ionization constant of acetic acid is 1.8 $\times {10}^{–5}$, at what concentration will it be dissociated to 2%?

 $C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$

$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$

$K_a=\frac{C\alpha \times C\alpha }{C(1-\alpha )}$

$K_a=\frac{C{\alpha }^2}{C(1-\alpha )}$

Since $\alpha$ is very small $1-\alpha \approx$ 1

Therefore, $K_a=C{\alpha }^2$

$1.8\times {10}^{–5}=C{(0.02)}^2$

C = $\frac{1.8\times {10}^{–5}}{0.02\times 0.02}$

C = 0.045 M

Hence, the correct option is option (D) 0.045 M.