Dung dịch T phản ứng với $A{l_2}{O_3}$, suy ra T có HCl dư hoặc NaOH dư
${n_{A{l_2}{O_3}}} = 0.51/102 = 0.005\,mol$
TH1: HCl dư
$\begin{gathered} NaOH\, + \,HCl\, \to \,NaCl\, + \,{H_2}O \hfill \\ \,\,\,0.2\,\,\,\, \to \,\,0.2\,\,\,\,\,\,\,\,(mol) \hfill \\ 6HC{l_{du}} + A{l_2}{O_3} \to 2AlC{l_3} + 3{H_2}O \hfill \\ (0.5a - 0.2) \to (0.5a - 0.2)/6 \hfill \\ (0.5a - 0.2)/6 = 0.005\, \to a = 0.46M \hfill \\ \end{gathered}$
TH2:NaOH dư
$\begin{gathered} NaOH\, + \,HCl\, \to \,NaCl\, + \,{H_2}O \hfill \\ 0.5a\,\,\, \leftarrow \,\,0.5a\,\,\,(mol) \hfill \\ 2NaO{H_{du}} + A{l_2}{O_3} \to NaAl{O_2} + {H_2}O \hfill \\ (0.2 - 0.5a) \to (0.1 - 0.25a)\,\,\,\,(mol) \hfill \\ 0.1 - 0.25a = 0.005\, \to a = 0.38M \hfill \\ \end{gathered} $$