Em tham khảo nha:
\(\begin{array}{l}
a)\\
HCl \to {H^ + } + C{l^ - }\\
[{H^ + }] = [C{l^ - }] = {C_M}HCl = 0,05M\\
b)\\
F{e_2}{(S{O_4})_3} \to 2F{e^{3 + }} + 3S{O_4}^{2 - }\\
[F{e^{3 + }}] = 2{C_M}F{e_2}{(S{O_4})_3} = 0,1M\\
[S{O_4}^{2 - }] = 3{C_M}F{e_2}{(S{O_4})_3} = 0,15M\\
c)\\
{n_{NaCl}} = \dfrac{{7,02}}{{58,5}} = 0,12\,mol\\
{C_M}NaCl = \dfrac{{0,12}}{{1,5}} = 0,08M\\
NaCl \to N{a^ + } + C{l^ - }\\
[N{a^ + }] = [C{l^ - }] = 0,08M
\end{array}\)