Đáp án đúng: C
0,45 lít.
${{n}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}}=0,125.0,5=0,0625\ mol$⇒ nAl3+ = 0,0625.2 = 0,125 mol$\begin{matrix}A{{l}^{3+}} & + & 3O{{H}^{-}} & \to & Al{{(OH)}_{3}}\downarrow \\0,125 & \to & 0,375 & {} & 0,125\\\end{matrix}$$\begin{array}{l}Al{{(OH)}_{3}}\,\,\,\,\,\,\,+\,\,\,\,\,\,\,O{{H}^{-}}\,\xrightarrow{{}}\,Al(OH)_{4}^{-}\\(0,125-0,05)\,\to \,0,075\end{array}$$\Rightarrow {{n}_{O{{H}^{-}}}}\,=\,0,45\,mol\,\Rightarrow \,{{V}_{dd}}\,=\,0,45\,lit$