\textbf{230 --} If the value of $\alpha$ for acid HA is equal to $10\%$, the pH of its solution is 3, and the value of $K_a$ is equal to $\text{mol.L}^{-1}$, approximately:
\begin{align*}
&(1)\quad 1/11\times10^{-6}\ ,\ 9\times10^{-3} \qquad\qquad (2)\quad 1/11\times10^{-6}\ ,\ 1\times10^{-2}\\
&(3)\quad 1/11\times10^{-4}\ ,\ 9\times10^{-3} \qquad\qquad (4)\quad 1/11\times10^{-4}\ ,\ 1\times10^{-2}
\end{align*}