0 . 01 moles of a weak acid H A( K _ a = 2 . 0 × 10 ^ - 6 ) is...

$\begin{cases}HA ⇌ {H}^{+} + & {A}^{-} & \\ \text{Initial conc.} & 0 . 1M & 0M & \\ \text{Equ. conc.} & ( 0 . 01 - x ) ( 0 . 1 + x ) & xM & \\ \approx 0 . 01M \approx 0 . 1M & \end{cases}$

Now, ${K}_{a} = \dfrac{\left[{H}^{+}\right]\left[{A}^{-}\right]}{[ HA ]} ⇒ 2 \times {10}^{- 6} = \dfrac{0 . 1 \times x}{0 . 01}$

$∴ x = 2 \times {10}^{- 7}$

Now, $\alpha = \dfrac{x}{0 . 01} = \dfrac{2 \times {10}^{- 7}}{0 . 01} = 2 \times {10}^{- 5}$